[{"Name":"Introduction to Acids and Bases","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Acids and Bases - Definitions","Duration":"6m 24s","ChapterTopicVideoID":27346,"CourseChapterTopicPlaylistID":272178,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.944","Text":"In earlier videos, we learned about acids and bases."},{"Start":"00:03.944 ","End":"00:08.670","Text":"In this video, we review what we learned and expand on it."},{"Start":"00:08.670 ","End":"00:14.865","Text":"The oldest useful definition of an acid and base is that of Arrhenius,"},{"Start":"00:14.865 ","End":"00:17.040","Text":"and that\u0027s from 1884."},{"Start":"00:17.040 ","End":"00:20.325","Text":"Arrhenius acid is a compound,"},{"Start":"00:20.325 ","End":"00:22.242","Text":"not an ion,"},{"Start":"00:22.242 ","End":"00:27.750","Text":"that contains hydrogen and reacts with water to form H plus ions."},{"Start":"00:27.750 ","End":"00:36.370","Text":"An Arrhenius base is a compound that produces hydroxide ions in water, OH minus."},{"Start":"00:36.800 ","End":"00:40.210","Text":"Now we should recall that an H plus ion,"},{"Start":"00:40.210 ","End":"00:45.440","Text":"which is just a proton because it\u0027s lost its electron is always bonded to"},{"Start":"00:45.440 ","End":"00:52.200","Text":"a water molecule to produce the hydronium ion H_3O plus."},{"Start":"00:59.620 ","End":"01:06.110","Text":"That is often hydrogen bonded to even more water molecules."},{"Start":"01:06.110 ","End":"01:11.125","Text":"It has a hydration shell around it."},{"Start":"01:11.125 ","End":"01:14.855","Text":"Now this definition of Arrhenius only applies to"},{"Start":"01:14.855 ","End":"01:19.430","Text":"aqueous solutions and to compounds but not ions."},{"Start":"01:19.430 ","End":"01:22.180","Text":"Here are some examples."},{"Start":"01:22.180 ","End":"01:24.900","Text":"Let\u0027s start off with acids."},{"Start":"01:24.900 ","End":"01:30.610","Text":"HCl, hydrochloric acid is a stronger acid."},{"Start":"01:30.610 ","End":"01:38.710","Text":"It reacts with water to give H_3O plus and Cl minus completely dissociates."},{"Start":"01:38.710 ","End":"01:42.879","Text":"Whereas acetic acid is a weak acid."},{"Start":"01:42.879 ","End":"01:45.760","Text":"CH_3CO_2H is a weak acid,"},{"Start":"01:45.760 ","End":"01:55.320","Text":"reacts with water to give an equilibrium of H_3O plus and acetate anion."},{"Start":"01:55.320 ","End":"01:56.880","Text":"Here it\u0027s an equilibrium,"},{"Start":"01:56.880 ","End":"01:59.125","Text":"where C is a complete reaction."},{"Start":"01:59.125 ","End":"02:01.825","Text":"When it\u0027s complete, it\u0027s a strong acid,"},{"Start":"02:01.825 ","End":"02:05.825","Text":"when it\u0027s incomplete, it\u0027s a weak acid."},{"Start":"02:05.825 ","End":"02:08.520","Text":"Now here are 2 bases."},{"Start":"02:08.520 ","End":"02:15.180","Text":"NaOH reacts with water to give Na plus and OH minus."},{"Start":"02:15.180 ","End":"02:17.685","Text":"NH_3, which is a weak base,"},{"Start":"02:17.685 ","End":"02:23.505","Text":"it\u0027s ammonia, reacts with water to give NH_4 plus and OH minus."},{"Start":"02:23.505 ","End":"02:26.890","Text":"We have a strong base NaOH,"},{"Start":"02:26.890 ","End":"02:29.795","Text":"which completely dissociates to ions,"},{"Start":"02:29.795 ","End":"02:35.380","Text":"and NH_3, which reacts with water to give us an equilibrium."},{"Start":"02:35.380 ","End":"02:38.360","Text":"We have a strong base and a weak base."},{"Start":"02:38.360 ","End":"02:42.245","Text":"Sodium hydroxide is strong, ammonia is weak."},{"Start":"02:42.245 ","End":"02:44.910","Text":"We can see the definitions,"},{"Start":"02:44.910 ","End":"02:49.130","Text":"HCl contains H, that\u0027s an acidic hydrogen,"},{"Start":"02:49.130 ","End":"02:52.490","Text":"CH_3CO_2H contains this hydrogen,"},{"Start":"02:52.490 ","End":"02:54.290","Text":"which is an acidic hydrogen,"},{"Start":"02:54.290 ","End":"03:01.200","Text":"and NaOH gives us OH minus when it reacts with water."},{"Start":"03:01.200 ","End":"03:05.870","Text":"NH_3 also gives us OH minus when it reacts with water."},{"Start":"03:05.870 ","End":"03:09.390","Text":"These are both Arrhenius bases."},{"Start":"03:09.400 ","End":"03:12.855","Text":"Now, a more modern definition,"},{"Start":"03:12.855 ","End":"03:14.385","Text":"they\u0027re still very old,"},{"Start":"03:14.385 ","End":"03:17.990","Text":"is the Bronsted-Lowry definition of an acid and base."},{"Start":"03:17.990 ","End":"03:21.755","Text":"This is the work of Bronsted and Lowry separately."},{"Start":"03:21.755 ","End":"03:24.700","Text":"We usually just call it Bronsted definition."},{"Start":"03:24.700 ","End":"03:27.975","Text":"For short Bronsted acid and Bronsted base."},{"Start":"03:27.975 ","End":"03:31.440","Text":"This is dates from 1923."},{"Start":"03:31.440 ","End":"03:36.225","Text":"Bronsted acid is a proton donor,"},{"Start":"03:36.225 ","End":"03:40.385","Text":"whereas a Bronsted base is a proton acceptor."},{"Start":"03:40.385 ","End":"03:43.290","Text":"Proton, we just mean H plus."},{"Start":"03:43.900 ","End":"03:49.640","Text":"Now every Bronsted acid or base is also an Arrhenius acid or base,"},{"Start":"03:49.640 ","End":"03:51.965","Text":"but not vice versa."},{"Start":"03:51.965 ","End":"03:58.224","Text":"This definition is more general than the Arrhenius definition."},{"Start":"03:58.224 ","End":"04:00.870","Text":"Here\u0027s our first example."},{"Start":"04:00.870 ","End":"04:06.860","Text":"HCl, which we saw was certainly an Arrhenius acid,"},{"Start":"04:06.860 ","End":"04:13.350","Text":"reacts with water to give us H_3O plus and Cl minus."},{"Start":"04:13.350 ","End":"04:17.900","Text":"According to Bronsted, HCl is an acid as it donates"},{"Start":"04:17.900 ","End":"04:23.515","Text":"its hydrogen to water to form H_3O plus."},{"Start":"04:23.515 ","End":"04:26.375","Text":"Let\u0027s give them this hydrogen to water."},{"Start":"04:26.375 ","End":"04:33.335","Text":"In this case H_2O is a base as it accepts the proton."},{"Start":"04:33.335 ","End":"04:36.740","Text":"Going from H_20 to H_3O plus,"},{"Start":"04:36.740 ","End":"04:39.750","Text":"it\u0027s accepted a proton."},{"Start":"04:41.810 ","End":"04:46.300","Text":"Here\u0027s another example. NH_3,"},{"Start":"04:46.300 ","End":"04:48.800","Text":"which we saw in the previous video,"},{"Start":"04:48.800 ","End":"04:53.270","Text":"is certainly an Arrhenius base because it gives us OH minus."},{"Start":"04:53.270 ","End":"04:58.280","Text":"Here we\u0027ll see why it\u0027s also a Bronsted base."},{"Start":"04:58.280 ","End":"05:05.520","Text":"NH_3 reacts with water to give us NH_4 plus and OH minus."},{"Start":"05:05.520 ","End":"05:10.940","Text":"Now here, NH_3 is a base as it accepts a proton from water."},{"Start":"05:10.940 ","End":"05:15.270","Text":"It goes from NH_3 to NH_4 plus."},{"Start":"05:16.760 ","End":"05:26.145","Text":"In this case, the water is an acid because it donates its proton to NH_3."},{"Start":"05:26.145 ","End":"05:28.560","Text":"H itself becomes OH minus."},{"Start":"05:28.560 ","End":"05:32.740","Text":"It loses a proton to become OH minus."},{"Start":"05:36.460 ","End":"05:41.345","Text":"Now there\u0027s a third definition called the Lewis definition."},{"Start":"05:41.345 ","End":"05:45.455","Text":"We\u0027re going to talk about this extensively at a later video."},{"Start":"05:45.455 ","End":"05:48.920","Text":"Here it\u0027s sufficient to say that a Lewis acid is"},{"Start":"05:48.920 ","End":"05:56.255","Text":"an electron pair acceptor and a Lewis base is an electron pair donor."},{"Start":"05:56.255 ","End":"06:01.220","Text":"Now every Lewis acid or base is also Bronsted acid or base,"},{"Start":"06:01.220 ","End":"06:03.350","Text":"but not vice versa."},{"Start":"06:03.350 ","End":"06:07.610","Text":"We\u0027re going in order of generality."},{"Start":"06:07.610 ","End":"06:13.690","Text":"We went from Arrhenius to Bronsted to Lewis."},{"Start":"06:13.690 ","End":"06:18.620","Text":"Each time the definition becomes more general."},{"Start":"06:18.620 ","End":"06:24.480","Text":"In this video, we learned about the definitions of acids and bases."}],"ID":28461},{"Watched":false,"Name":"Conjugate Acids and Bases","Duration":"6m 25s","ChapterTopicVideoID":27347,"CourseChapterTopicPlaylistID":272178,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.175","Text":"In the previous video,"},{"Start":"00:02.175 ","End":"00:06.246","Text":"we reviewed the 3 common definitions of acids and bases,"},{"Start":"00:06.246 ","End":"00:11.070","Text":"and this video we\u0027ll expand on the Bronsted definition."},{"Start":"00:11.070 ","End":"00:14.550","Text":"We\u0027re going to talk about Bronsted acids and bases."},{"Start":"00:14.550 ","End":"00:18.135","Text":"This is the most common definition of acids and bases."},{"Start":"00:18.135 ","End":"00:21.495","Text":"Now Bronsted acid is a proton donor."},{"Start":"00:21.495 ","End":"00:24.610","Text":"Proton is just H plus."},{"Start":"00:25.190 ","End":"00:29.370","Text":"A Bronsted base is a proton acceptor."},{"Start":"00:29.370 ","End":"00:31.940","Text":"Generally when we say acid or base,"},{"Start":"00:31.940 ","End":"00:34.700","Text":"unless we specifically say otherwise,"},{"Start":"00:34.700 ","End":"00:38.005","Text":"we mean a Bronsted acid or base."},{"Start":"00:38.005 ","End":"00:44.420","Text":"Now an acid can only donate a proton if a base is present to accept it."},{"Start":"00:44.420 ","End":"00:47.855","Text":"We always have acid-base reactions."},{"Start":"00:47.855 ","End":"00:52.790","Text":"Now the acid contains at least 1 ionizable H atom,"},{"Start":"00:52.790 ","End":"00:55.595","Text":"which we often call an acidic H atom."},{"Start":"00:55.595 ","End":"01:02.050","Text":"That\u0027s the H atom which comes off when the acid reacts."},{"Start":"01:02.050 ","End":"01:09.635","Text":"A base contains an atom with a lone pair of electrons onto which a proton combined."},{"Start":"01:09.635 ","End":"01:14.155","Text":"Now we\u0027re going to talk about conjugate acids and bases."},{"Start":"01:14.155 ","End":"01:18.425","Text":"Now if we have an acid and a conjugate base,"},{"Start":"01:18.425 ","End":"01:22.745","Text":"they differ by a single H plus ion."},{"Start":"01:22.745 ","End":"01:29.900","Text":"We say that the acid and its conjugate base form a conjugate acid-base pair."},{"Start":"01:29.900 ","End":"01:32.850","Text":"Here\u0027s the first example."},{"Start":"01:33.380 ","End":"01:41.479","Text":"HCl plus H_2O to give H_3O plus and Cl minus."},{"Start":"01:41.479 ","End":"01:44.200","Text":"This is hydrochloric acid."},{"Start":"01:44.200 ","End":"01:47.175","Text":"Now HCl is an acid."},{"Start":"01:47.175 ","End":"01:50.015","Text":"Here is it\u0027s ionizable hydrogen."},{"Start":"01:50.015 ","End":"01:54.155","Text":"It donates a proton to water."},{"Start":"01:54.155 ","End":"01:57.110","Text":"Water becomes H_3O plus."},{"Start":"01:57.110 ","End":"02:00.620","Text":"The HCl becomes Cl minus,"},{"Start":"02:00.620 ","End":"02:02.575","Text":"it\u0027s lost the proton."},{"Start":"02:02.575 ","End":"02:06.785","Text":"HCl and Cl minus differ by a proton."},{"Start":"02:06.785 ","End":"02:09.200","Text":"We see that HCl,"},{"Start":"02:09.200 ","End":"02:11.143","Text":"that\u0027s an acid,"},{"Start":"02:11.143 ","End":"02:12.384","Text":"and Cl minus,"},{"Start":"02:12.384 ","End":"02:18.900","Text":"that\u0027s a base, form a conjugate acid-base pair."},{"Start":"02:25.510 ","End":"02:30.245","Text":"This is a pair, acid and base."},{"Start":"02:30.245 ","End":"02:39.540","Text":"Now the water here accepts a proton from the HCl becomes H_3O plus."},{"Start":"02:39.540 ","End":"02:45.605","Text":"These 2 H_2O and H_3O plus differ by a proton."},{"Start":"02:45.605 ","End":"02:53.835","Text":"The water here is a base and the H_3O plus is an acid,"},{"Start":"02:53.835 ","End":"02:56.920","Text":"and they form a pair."},{"Start":"02:59.390 ","End":"03:02.725","Text":"We have 2 pairs."},{"Start":"03:02.725 ","End":"03:11.405","Text":"Here\u0027s another example, ammonia reacting with water to give NH_4 plus and OH minus."},{"Start":"03:11.405 ","End":"03:15.305","Text":"Ammonia accepts a proton from water."},{"Start":"03:15.305 ","End":"03:21.450","Text":"So that\u0027s a base and it becomes NH_4 plus."},{"Start":"03:21.450 ","End":"03:27.255","Text":"You\u0027ll see the NH_3 and NH_4 plus differ by a proton."},{"Start":"03:27.255 ","End":"03:34.010","Text":"NH_4 plus is an acid and they form a conjugate pair,"},{"Start":"03:34.010 ","End":"03:37.710","Text":"NH_3 and NH_4 plus."},{"Start":"03:41.870 ","End":"03:46.485","Text":"Now water donates a proton in this case,"},{"Start":"03:46.485 ","End":"03:51.610","Text":"so it\u0027s an acid and it becomes OH minus,"},{"Start":"03:51.610 ","End":"03:53.690","Text":"which is a base."},{"Start":"03:53.690 ","End":"03:59.245","Text":"They differ by a proton H_2O and OH minus differ by proton."},{"Start":"03:59.245 ","End":"04:07.220","Text":"An H_2O and OH minus form a conjugate acid-base pair."},{"Start":"04:12.120 ","End":"04:14.890","Text":"Now here\u0027s a third example."},{"Start":"04:14.890 ","End":"04:20.080","Text":"This time we have an ion HCO_3 minus that\u0027s"},{"Start":"04:20.080 ","End":"04:24.590","Text":"bicarbonate and it\u0027s reacting with water to give"},{"Start":"04:24.590 ","End":"04:30.305","Text":"us H_3O plus and CO_3^2 minus That\u0027s carbonate."},{"Start":"04:30.305 ","End":"04:35.450","Text":"The first thing to note is that ions can be Bronsted acids or bases,"},{"Start":"04:35.450 ","End":"04:37.040","Text":"but not Arrhenius ones."},{"Start":"04:37.040 ","End":"04:40.640","Text":"In the definition for Arrhenius it said just compounds."},{"Start":"04:40.640 ","End":"04:45.120","Text":"Here, HCO_3 minus is an acid,"},{"Start":"04:45.190 ","End":"04:54.630","Text":"CO_3^2 minus is a base and these form a conjugate acid-base pair."},{"Start":"04:58.070 ","End":"05:08.410","Text":"Under this case water is a base and H_3O plus an acid and they form a pair."},{"Start":"05:09.240 ","End":"05:14.740","Text":"Now let\u0027s talk about amphiprotic compounds."},{"Start":"05:14.740 ","End":"05:23.070","Text":"Amphiprotic compound is a substance has both an acidic H atom and a lone pair."},{"Start":"05:23.070 ","End":"05:28.080","Text":"It can act as an acid or as"},{"Start":"05:28.080 ","End":"05:34.310","Text":"a base and it said to be amphiprotic."},{"Start":"05:34.310 ","End":"05:37.825","Text":"It can act as either an acid or a base."},{"Start":"05:37.825 ","End":"05:42.460","Text":"Sometimes people mix up this definition,"},{"Start":"05:42.460 ","End":"05:52.595","Text":"amphiprotic with amphoteric which means that it can react with an acid or a base."},{"Start":"05:52.595 ","End":"05:59.820","Text":"Often people say amphoteric with should\u0027ve said amphiprotic and so on."},{"Start":"05:59.830 ","End":"06:05.345","Text":"An example of an amphiprotic compound is water."},{"Start":"06:05.345 ","End":"06:14.010","Text":"We saw before that in Example 2 water acts as an acid whereas in Examples 1 and 3,"},{"Start":"06:14.010 ","End":"06:15.815","Text":"water acts as a base."},{"Start":"06:15.815 ","End":"06:20.165","Text":"Water is an amphiprotic compounds."},{"Start":"06:20.165 ","End":"06:25.620","Text":"In this video, we talked about conjugate acids and bases."}],"ID":28462},{"Watched":false,"Name":"Exercise 1","Duration":"3m 18s","ChapterTopicVideoID":29795,"CourseChapterTopicPlaylistID":272178,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31443},{"Watched":false,"Name":"Exercise 2","Duration":"3m 1s","ChapterTopicVideoID":29796,"CourseChapterTopicPlaylistID":272178,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31444},{"Watched":false,"Name":"Exercise 3","Duration":"3m 59s","ChapterTopicVideoID":29797,"CourseChapterTopicPlaylistID":272178,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31445},{"Watched":false,"Name":"Exercise 4","Duration":"5m 44s","ChapterTopicVideoID":29798,"CourseChapterTopicPlaylistID":272178,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31446}],"Thumbnail":null,"ID":272178},{"Name":"Aqueous Solutions of Acids and Bases","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Self - Ionization of Water","Duration":"6m 44s","ChapterTopicVideoID":27350,"CourseChapterTopicPlaylistID":272179,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.010","Text":"In the previous video,"},{"Start":"00:02.010 ","End":"00:04.955","Text":"we talked about conjugate acids and bases."},{"Start":"00:04.955 ","End":"00:09.900","Text":"In this video we\u0027ll talk about the self-ionization of water."},{"Start":"00:10.100 ","End":"00:14.760","Text":"We\u0027re going to talk about the self-ionization of water."},{"Start":"00:14.760 ","End":"00:18.480","Text":"Now we said that water\u0027s amphiprotic."},{"Start":"00:18.480 ","End":"00:22.170","Text":"It can act as an acid or as a base."},{"Start":"00:22.170 ","End":"00:25.245","Text":"If 2 molecules of water react,"},{"Start":"00:25.245 ","End":"00:29.640","Text":"1 can act as an acid and 1 as a base."},{"Start":"00:29.640 ","End":"00:38.375","Text":"These 2 water molecules in equilibrium with H_3O plus and OH minus,"},{"Start":"00:38.375 ","End":"00:46.335","Text":"so H_2O has turned into H_3O plus an acid and a base."},{"Start":"00:46.335 ","End":"00:50.960","Text":"H_30 plus is an acid and OH minus is a base."},{"Start":"00:50.960 ","End":"00:58.230","Text":"That means that 1 molecule of water acts as a base and 1 acted as an acid."},{"Start":"00:58.520 ","End":"01:02.810","Text":"Here\u0027s 1 water molecule and another 1."},{"Start":"01:02.810 ","End":"01:09.460","Text":"Say that this hydrogen here is going to be"},{"Start":"01:09.460 ","End":"01:15.731","Text":"donated to this water molecule to give us H_3O plus,"},{"Start":"01:15.731 ","End":"01:18.670","Text":"this is H_3O plus."},{"Start":"01:18.680 ","End":"01:23.580","Text":"Here, the hydrogen is being donated,"},{"Start":"01:23.580 ","End":"01:26.200","Text":"so that\u0027s an acid."},{"Start":"01:26.730 ","End":"01:30.520","Text":"This water molecule is accepting the hydrogen,"},{"Start":"01:30.520 ","End":"01:32.630","Text":"so it\u0027s a base."},{"Start":"01:33.270 ","End":"01:36.400","Text":"Then this is an acid,"},{"Start":"01:36.400 ","End":"01:38.875","Text":"and this is a base."},{"Start":"01:38.875 ","End":"01:40.849","Text":"We have 2 pairs,"},{"Start":"01:40.849 ","End":"01:49.520","Text":"1 pair here and another pair here."},{"Start":"01:52.670 ","End":"01:56.965","Text":"Now since the 2 ions are produced just from water,"},{"Start":"01:56.965 ","End":"02:00.440","Text":"this is called self-ionization of water."},{"Start":"02:00.440 ","End":"02:03.330","Text":"Pure water contains ions,"},{"Start":"02:03.330 ","End":"02:07.250","Text":"but in very very low concentrations."},{"Start":"02:07.250 ","End":"02:11.870","Text":"Let\u0027s talk about the ion product of water."},{"Start":"02:11.990 ","End":"02:15.700","Text":"The equilibrium constants for self-ionization of"},{"Start":"02:15.700 ","End":"02:19.230","Text":"water is called the ion product of water."},{"Start":"02:19.230 ","End":"02:21.680","Text":"We write it K_w,"},{"Start":"02:21.680 ","End":"02:24.050","Text":"w for water, of course,"},{"Start":"02:24.050 ","End":"02:30.700","Text":"and it\u0027s equal to the activity of H_3O plus times the activity of OH minus."},{"Start":"02:30.700 ","End":"02:37.085","Text":"These are the products divided by the activity of water squared."},{"Start":"02:37.085 ","End":"02:39.260","Text":"That\u0027s the reactants."},{"Start":"02:39.260 ","End":"02:49.410","Text":"These are equal to the concentration of H_3O plus in molar concentrations divide by C_0,"},{"Start":"02:49.410 ","End":"02:53.645","Text":"which remember, is 1 mole per liter,"},{"Start":"02:53.645 ","End":"03:00.920","Text":"times OH minus concentration divide by C_0 and as water is a liquid,"},{"Start":"03:00.920 ","End":"03:03.425","Text":"the activity is just 1."},{"Start":"03:03.425 ","End":"03:09.355","Text":"If we just write the concentrations without units,"},{"Start":"03:09.355 ","End":"03:13.250","Text":"then we can write them H_3O plus times OH minus"},{"Start":"03:13.250 ","End":"03:18.110","Text":"and these are dimensionless, K_w is dimensionless."},{"Start":"03:18.110 ","End":"03:25.054","Text":"K_w is equal to H_3O plus times OH minus."},{"Start":"03:25.054 ","End":"03:27.545","Text":"Now, at 25-degrees Celsius,"},{"Start":"03:27.545 ","End":"03:32.495","Text":"K_w is equal to 1 times 10^minus 14."},{"Start":"03:32.495 ","End":"03:35.045","Text":"The changes with temperature,"},{"Start":"03:35.045 ","End":"03:38.615","Text":"but this is its value, 25-degrees Celsius."},{"Start":"03:38.615 ","End":"03:45.875","Text":"Now since the concentration of H_3O plus is equal to concentration of OH minus,"},{"Start":"03:45.875 ","End":"03:54.420","Text":"we always get 1 molecule of H_3O plus and 1 molecule of OH minus,"},{"Start":"03:54.420 ","End":"03:56.310","Text":"the 1 to 1 ratios,"},{"Start":"03:56.310 ","End":"03:58.155","Text":"we can call them x."},{"Start":"03:58.155 ","End":"03:59.640","Text":"Then K_w=x^2."},{"Start":"03:59.640 ","End":"04:09.850","Text":"X is equal to the square root of K_w and the square root of 10^minus 14 is 10^minus 7."},{"Start":"04:09.850 ","End":"04:14.360","Text":"We can say that the concentration of H_3O plus is equal to"},{"Start":"04:14.360 ","End":"04:20.815","Text":"the concentration of OH minus and that\u0027s equal to 10^minus 7."},{"Start":"04:20.815 ","End":"04:25.330","Text":"All this is in pure water."},{"Start":"04:25.330 ","End":"04:33.950","Text":"Now we can draw very important conclusions from these values of 10^minus 7."},{"Start":"04:33.950 ","End":"04:36.635","Text":"In all aqueous solutions,"},{"Start":"04:36.635 ","End":"04:42.935","Text":"this product is always 1 times 10^minus 14,"},{"Start":"04:42.935 ","End":"04:47.000","Text":"of course, at 25-degrees Celsius."},{"Start":"04:47.000 ","End":"04:49.895","Text":"If we have an acidic solution,"},{"Start":"04:49.895 ","End":"04:55.850","Text":"then we have more H_3O plus than in the neutral solution,"},{"Start":"04:55.850 ","End":"05:01.760","Text":"so H_3O plus concentration will be greater than 10^minus 7,"},{"Start":"05:01.760 ","End":"05:06.650","Text":"which means that the concentration of OH minus must be less than"},{"Start":"05:06.650 ","End":"05:11.810","Text":"10^minus 7 because the product is always 10^minus14,"},{"Start":"05:11.810 ","End":"05:13.745","Text":"so if one goes up,"},{"Start":"05:13.745 ","End":"05:16.075","Text":"the other must come down."},{"Start":"05:16.075 ","End":"05:18.395","Text":"This is an acid."},{"Start":"05:18.395 ","End":"05:20.420","Text":"Now if we have a base,"},{"Start":"05:20.420 ","End":"05:24.860","Text":"then we have a higher concentration of OH minus."},{"Start":"05:24.860 ","End":"05:29.435","Text":"The concentration of OH minus will be greater than 10^minus 7."},{"Start":"05:29.435 ","End":"05:36.190","Text":"That means the concentration of H_3O plus must be less than 10^minus 7."},{"Start":"05:36.190 ","End":"05:38.730","Text":"It\u0027s a bit like a seesaw."},{"Start":"05:38.730 ","End":"05:40.500","Text":"If one goes up,"},{"Start":"05:40.500 ","End":"05:42.075","Text":"the other must come down."},{"Start":"05:42.075 ","End":"05:45.350","Text":"Now we\u0027re going to talk a moment about the suppression of"},{"Start":"05:45.350 ","End":"05:48.850","Text":"self-ionization in acids or bases."},{"Start":"05:48.850 ","End":"05:50.715","Text":"Here\u0027s our equation."},{"Start":"05:50.715 ","End":"05:57.440","Text":"2 water molecules in equilibrium with H_3O plus and OH minus."},{"Start":"05:57.440 ","End":"06:00.125","Text":"Now if we add an acid or base,"},{"Start":"06:00.125 ","End":"06:02.509","Text":"if we increase H_3O plus,"},{"Start":"06:02.509 ","End":"06:04.640","Text":"then according to Le Chatelier\u0027s,"},{"Start":"06:04.640 ","End":"06:07.220","Text":"the reaction will go to the left."},{"Start":"06:07.220 ","End":"06:09.980","Text":"If we increase OH minus,"},{"Start":"06:09.980 ","End":"06:12.355","Text":"it will also go to the left."},{"Start":"06:12.355 ","End":"06:16.640","Text":"If it goes to left that means there\u0027s less ions,"},{"Start":"06:16.640 ","End":"06:20.930","Text":"so self-ionization is partially suppressed."},{"Start":"06:20.930 ","End":"06:24.150","Text":"There will be less ions."},{"Start":"06:24.260 ","End":"06:32.135","Text":"This means self-ionization is negligible in aqueous acidic or basic solutions,"},{"Start":"06:32.135 ","End":"06:35.870","Text":"apart from very very weak ones."},{"Start":"06:35.870 ","End":"06:43.740","Text":"In this video, we talked about self-ionization of water and its consequences."}],"ID":28465},{"Watched":false,"Name":"pH and pOH","Duration":"7m 15s","ChapterTopicVideoID":27349,"CourseChapterTopicPlaylistID":272179,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.830","Text":"In the previous video,"},{"Start":"00:01.830 ","End":"00:05.355","Text":"we learned about the self ionization of water."},{"Start":"00:05.355 ","End":"00:11.595","Text":"In this video, we will define pH and pOH."},{"Start":"00:11.595 ","End":"00:15.315","Text":"Water pH and pOH."},{"Start":"00:15.315 ","End":"00:20.830","Text":"These terms were introduced by Soren Sorensen in 1909."},{"Start":"00:21.340 ","End":"00:29.420","Text":"We need to find pH as minus the log to the base 10 of the concentration of H_3O"},{"Start":"00:29.420 ","End":"00:38.810","Text":"plus and pOH is minus the log to the base 10 of the concentration of OH minus."},{"Start":"00:38.810 ","End":"00:43.640","Text":"Now to find the relationship between pH and pOH,"},{"Start":"00:43.640 ","End":"00:48.740","Text":"we need to remember that the concentration of H_3O plus times"},{"Start":"00:48.740 ","End":"00:55.205","Text":"the concentration of OH minus is equal to 1 times 10^minus 14."},{"Start":"00:55.205 ","End":"01:00.670","Text":"We learned this when we learned about self ionization of water."},{"Start":"01:00.800 ","End":"01:05.765","Text":"Now if we take the minus the log of this expression,"},{"Start":"01:05.765 ","End":"01:11.765","Text":"minus the log of the concentration of H_3O plus times the concentration of OH minus,"},{"Start":"01:11.765 ","End":"01:17.905","Text":"that\u0027s equal to minus the log of 1 times 10^minus 14."},{"Start":"01:17.905 ","End":"01:20.640","Text":"We have minus outside,"},{"Start":"01:20.640 ","End":"01:26.480","Text":"and the log of 10^minus 14 is just minus 14;"},{"Start":"01:26.480 ","End":"01:29.345","Text":"so we have minus times minus 14,"},{"Start":"01:29.345 ","End":"01:32.135","Text":"and that\u0027s equal to 14."},{"Start":"01:32.135 ","End":"01:35.000","Text":"Now if we take this line here,"},{"Start":"01:35.000 ","End":"01:46.080","Text":"and we recall that the log of a product a times b is equal to the log of a plus log of b."},{"Start":"01:46.450 ","End":"01:53.540","Text":"Then we can write minus the log of the product is equal to minus log of H_3O plus,"},{"Start":"01:53.540 ","End":"01:57.454","Text":"plus, minus log of OH minus."},{"Start":"01:57.454 ","End":"02:00.034","Text":"Recall that\u0027s 14."},{"Start":"02:00.034 ","End":"02:01.850","Text":"Now if we look at this slide,"},{"Start":"02:01.850 ","End":"02:08.165","Text":"we see that minus log of H_3O plus is just the definition of the pH."},{"Start":"02:08.165 ","End":"02:12.905","Text":"Minus log of OH minus is just the definition of pOH."},{"Start":"02:12.905 ","End":"02:19.075","Text":"We have finally, the sum of the pH and the pOH is 14."},{"Start":"02:19.075 ","End":"02:23.310","Text":"This is a very important relationship."},{"Start":"02:23.350 ","End":"02:26.944","Text":"Let\u0027s solve an example."},{"Start":"02:26.944 ","End":"02:29.705","Text":"The first part of the question is,"},{"Start":"02:29.705 ","End":"02:31.684","Text":"in an acidic solution,"},{"Start":"02:31.684 ","End":"02:37.390","Text":"the concentration of H_3O plus is 2.5 times 10^minus 3."},{"Start":"02:37.390 ","End":"02:40.360","Text":"Calculate the pH and pOH."},{"Start":"02:40.360 ","End":"02:45.330","Text":"The second part is if the pOH is 3.2,"},{"Start":"02:45.330 ","End":"02:52.925","Text":"calculate H_3O plus concentration of H_3O plus the question continues,"},{"Start":"02:52.925 ","End":"02:56.125","Text":"is the solution acidic or basic?"},{"Start":"02:56.125 ","End":"02:59.120","Text":"Let\u0027s look at the first part of the problem."},{"Start":"02:59.120 ","End":"03:06.485","Text":"We\u0027re given the H_3O plus concentration is 2.5 times 10^minus 3."},{"Start":"03:06.485 ","End":"03:13.850","Text":"We can write that the pH is minus the log of 2.5 times 10^minus 3."},{"Start":"03:13.850 ","End":"03:15.845","Text":"If we use our calculator,"},{"Start":"03:15.845 ","End":"03:19.565","Text":"we find out that that\u0027s 2.60."},{"Start":"03:19.565 ","End":"03:22.225","Text":"The pH is 2.60."},{"Start":"03:22.225 ","End":"03:25.490","Text":"Now we want to find the pOH."},{"Start":"03:25.490 ","End":"03:30.185","Text":"Recall that the sum of the pH and pOH is 14,"},{"Start":"03:30.185 ","End":"03:34.055","Text":"so the pOH is 14 minus the pH,"},{"Start":"03:34.055 ","End":"03:36.335","Text":"and the pH is 2.60,"},{"Start":"03:36.335 ","End":"03:40.055","Text":"so the result is 11.40."},{"Start":"03:40.055 ","End":"03:43.510","Text":"Now we have found the pOH."},{"Start":"03:43.790 ","End":"03:47.335","Text":"Let\u0027s look at the second part of the problem."},{"Start":"03:47.335 ","End":"03:49.510","Text":"We\u0027re given the pOH,"},{"Start":"03:49.510 ","End":"03:51.655","Text":"we can calculate the pH."},{"Start":"03:51.655 ","End":"03:58.045","Text":"The pH is 14 minus 3.2 and that\u0027s 10.8."},{"Start":"03:58.045 ","End":"04:02.709","Text":"We have the pH. Once we have the pH,"},{"Start":"04:02.709 ","End":"04:06.910","Text":"we can discover the concentration of H_3O plus."},{"Start":"04:06.910 ","End":"04:13.855","Text":"We need the relationship between pH and H_3O plus concentration."},{"Start":"04:13.855 ","End":"04:15.789","Text":"We\u0027d use this expression."},{"Start":"04:15.789 ","End":"04:23.720","Text":"The concentration of H_3O plus is equal to 10 to the power minus pH."},{"Start":"04:25.430 ","End":"04:33.400","Text":"The concentration of H_3O plus is equal to 10^minus 10.8."},{"Start":"04:33.400 ","End":"04:41.130","Text":"If we look up our calculator that\u0027s 1.58 times 10^minus 11."},{"Start":"04:41.530 ","End":"04:46.070","Text":"If this is less than 10^minus 7,"},{"Start":"04:46.070 ","End":"04:48.875","Text":"the solution is basic."},{"Start":"04:48.875 ","End":"04:52.260","Text":"It\u0027s less than 10^minus 7."},{"Start":"04:53.530 ","End":"04:56.780","Text":"Let\u0027s look at the pH scale."},{"Start":"04:56.780 ","End":"05:01.205","Text":"We know whether a solution is acidic, basic, or neutral."},{"Start":"05:01.205 ","End":"05:05.020","Text":"The left-hand side we have pH."},{"Start":"05:05.020 ","End":"05:08.625","Text":"The right-hand side, pOH."},{"Start":"05:08.625 ","End":"05:13.305","Text":"The central black line relates to 7,"},{"Start":"05:13.305 ","End":"05:15.660","Text":"so pH is 7 here in the middle."},{"Start":"05:15.660 ","End":"05:18.780","Text":"Here, pOH is 7 in the middle."},{"Start":"05:18.780 ","End":"05:27.790","Text":"Now, red is acid and blue is base."},{"Start":"05:29.610 ","End":"05:32.050","Text":"Now in pure water,"},{"Start":"05:32.050 ","End":"05:36.130","Text":"the concentration of H_3O plus is 10^minus 7."},{"Start":"05:36.130 ","End":"05:39.830","Text":"We can work out the pH is 7."},{"Start":"05:39.890 ","End":"05:44.675","Text":"Log of 10^minus 7 is minus 7."},{"Start":"05:44.675 ","End":"05:46.570","Text":"We get rid of the minus sign,"},{"Start":"05:46.570 ","End":"05:48.295","Text":"so we just have 7."},{"Start":"05:48.295 ","End":"05:51.550","Text":"Neutrality is this black line,"},{"Start":"05:51.550 ","End":"05:54.080","Text":"pH equal to 7."},{"Start":"05:54.080 ","End":"05:56.865","Text":"Of course, if the pH is 7,"},{"Start":"05:56.865 ","End":"06:00.370","Text":"the pOH is also 7."},{"Start":"06:00.460 ","End":"06:03.425","Text":"Now if we have an acidic solution,"},{"Start":"06:03.425 ","End":"06:08.560","Text":"the concentration of H_3O plus is greater than 10^minus 7."},{"Start":"06:08.560 ","End":"06:16.360","Text":"That means the pH is less than 7 and the pOH will be greater than 7."},{"Start":"06:17.320 ","End":"06:19.970","Text":"If you look at our diagram,"},{"Start":"06:19.970 ","End":"06:22.350","Text":"this will be acid."},{"Start":"06:23.360 ","End":"06:26.475","Text":"For the pH the pH will be less than 7,"},{"Start":"06:26.475 ","End":"06:32.350","Text":"and the pOH will be greater than 7 for acids, and this is acid."},{"Start":"06:32.870 ","End":"06:35.090","Text":"In a basic solution,"},{"Start":"06:35.090 ","End":"06:39.515","Text":"the concentration of H_3O plus is less than 10^minus 7."},{"Start":"06:39.515 ","End":"06:47.640","Text":"So the pH will be greater than 7 and the pOH will be less than 7."},{"Start":"06:47.710 ","End":"06:54.220","Text":"Here\u0027s our base and here\u0027s our base."},{"Start":"06:54.220 ","End":"07:02.320","Text":"Here we have that the acid has a pH less than 7 and a base greater than 7,"},{"Start":"07:02.320 ","End":"07:06.140","Text":"and pOH of an acid is greater than 7,"},{"Start":"07:06.140 ","End":"07:10.475","Text":"and pOH of a base is less than 7."},{"Start":"07:10.475 ","End":"07:15.990","Text":"In this video, we discuss pH and pOH."}],"ID":28464},{"Watched":false,"Name":"Ionization of Acids and Bases in Water","Duration":"6m 54s","ChapterTopicVideoID":27348,"CourseChapterTopicPlaylistID":272179,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.785","Text":"In the previous video,"},{"Start":"00:01.785 ","End":"00:05.040","Text":"we learnt about pH and pOH."},{"Start":"00:05.040 ","End":"00:11.020","Text":"In this video, we\u0027ll learn about the ionization of acids and bases in water."},{"Start":"00:11.030 ","End":"00:16.335","Text":"We\u0027re going to start off with the ionization of acids."},{"Start":"00:16.335 ","End":"00:19.950","Text":"We\u0027re going to denote a monoprotic acid,"},{"Start":"00:19.950 ","End":"00:24.515","Text":"that\u0027s 1 acidic hydrogen as HA."},{"Start":"00:24.515 ","End":"00:30.838","Text":"H is the acidic hydrogen and A is the rest of the molecule."},{"Start":"00:30.838 ","End":"00:34.355","Text":"Its ionization in water is written as"},{"Start":"00:34.355 ","End":"00:40.220","Text":"HA(aq) plus H2O(l) in equilibrium with H3O plus and Aminus."},{"Start":"00:40.220 ","End":"00:44.945","Text":"We can define an equilibrium constant called K_a,"},{"Start":"00:44.945 ","End":"00:49.835","Text":"which is equal to the activity of H_3O plus times the activity of"},{"Start":"00:49.835 ","End":"00:56.195","Text":"A minus divided by the activity of HA multiplied by the activity of H_2O."},{"Start":"00:56.195 ","End":"01:02.075","Text":"Once again, we can write this in terms of concentrations in dimensionless units."},{"Start":"01:02.075 ","End":"01:06.080","Text":"It\u0027s the concentration of H3O plus times"},{"Start":"01:06.080 ","End":"01:11.225","Text":"the concentration of A minus divided by the concentration of HA."},{"Start":"01:11.225 ","End":"01:16.640","Text":"Of course, the activity of water being a pure liquid is just 1."},{"Start":"01:16.640 ","End":"01:21.950","Text":"K_a is called the acid ionization constant."},{"Start":"01:21.950 ","End":"01:25.835","Text":"It ranges from very small values to very high values,"},{"Start":"01:25.835 ","End":"01:30.265","Text":"and gives us information about the strength of the acid."},{"Start":"01:30.265 ","End":"01:34.380","Text":"It\u0027s a very important constant."},{"Start":"01:34.380 ","End":"01:40.515","Text":"We can define pK_a in analogy with pH and pOH."},{"Start":"01:40.515 ","End":"01:43.830","Text":"pK_a is equal to minus logK_a."},{"Start":"01:43.830 ","End":"01:49.500","Text":"If we want to go in the opposite direction, K_a=10^minus pK_a."},{"Start":"01:52.660 ","End":"01:57.515","Text":"We can also talk about the ionization of bases."},{"Start":"01:57.515 ","End":"02:02.200","Text":"We\u0027re going to denote the base as B."},{"Start":"02:02.200 ","End":"02:06.710","Text":"Its ionization in water can be written B plus"},{"Start":"02:06.710 ","End":"02:12.414","Text":"H_2O in equilibrium with BH plus and OH minus."},{"Start":"02:12.414 ","End":"02:15.850","Text":"Now we define K_b,"},{"Start":"02:15.850 ","End":"02:18.730","Text":"just like we defined K_a before."},{"Start":"02:18.730 ","End":"02:21.565","Text":"We have the equilibrium constant for the base,"},{"Start":"02:21.565 ","End":"02:26.418","Text":"and that\u0027s the activity of OH minus times the activity of BH"},{"Start":"02:26.418 ","End":"02:32.125","Text":"plus divided by the activity of B times the activity of water."},{"Start":"02:32.125 ","End":"02:39.025","Text":"That can be written in terms of concentrations without noting the units as"},{"Start":"02:39.025 ","End":"02:42.242","Text":"the concentration of BH plus times the concentration of OH"},{"Start":"02:42.242 ","End":"02:46.670","Text":"minus divided by the concentration of B."},{"Start":"02:46.670 ","End":"02:51.250","Text":"It\u0027s called the base ionization constant."},{"Start":"02:51.250 ","End":"02:54.340","Text":"Once again, it ranges from very small values to"},{"Start":"02:54.340 ","End":"03:00.450","Text":"very high values and gives us information about the strength of the base."},{"Start":"03:01.850 ","End":"03:08.250","Text":"Just as we defined pK_a and pH and pOH,"},{"Start":"03:08.250 ","End":"03:15.360","Text":"we can define pK_b and pK_b= minus lgK_b."},{"Start":"03:15.360 ","End":"03:20.220","Text":"If we want to go in the opposite direction, K_b=10^minus pK_b."},{"Start":"03:21.610 ","End":"03:26.230","Text":"Let\u0027s talk about strong acids and bases."},{"Start":"03:26.230 ","End":"03:31.920","Text":"Strong acids or bases have K_a or K_b much greater than unity,"},{"Start":"03:31.920 ","End":"03:37.870","Text":"very high values, and they undergo almost complete ionization of water."},{"Start":"03:37.870 ","End":"03:40.925","Text":"We can just assume it\u0027s complete ionization."},{"Start":"03:40.925 ","End":"03:44.983","Text":"The actual values of K_a or K_b are irrelevant,"},{"Start":"03:44.983 ","End":"03:47.720","Text":"because we don\u0027t have an equilibrium,"},{"Start":"03:47.720 ","End":"03:51.365","Text":"just a reaction going to completion."},{"Start":"03:51.365 ","End":"03:56.600","Text":"Examples of strong acids are HCl, HBr, HI,"},{"Start":"03:56.600 ","End":"03:58.670","Text":"HNO_3, nitric acid,"},{"Start":"03:58.670 ","End":"04:01.955","Text":"and H_2SO_4, sulfuric acid."},{"Start":"04:01.955 ","End":"04:04.295","Text":"Examples of strong bases,"},{"Start":"04:04.295 ","End":"04:07.255","Text":"you see they all have OH in them: LiOH,"},{"Start":"04:07.255 ","End":"04:15.919","Text":"NAOH, KOH, magnesium hydroxide and calcium hydroxide."},{"Start":"04:15.919 ","End":"04:20.414","Text":"These are all hydroxides."},{"Start":"04:20.414 ","End":"04:24.205","Text":"Now let\u0027s look at weak acids and bases."},{"Start":"04:24.205 ","End":"04:30.180","Text":"Now they have K_a or K_b much smaller than 1,"},{"Start":"04:30.180 ","End":"04:34.840","Text":"much smaller than unity, and they undergo only partial ionization in water."},{"Start":"04:34.840 ","End":"04:38.664","Text":"They\u0027re quite different from the strong acids and bases."},{"Start":"04:38.664 ","End":"04:42.865","Text":"Let\u0027s look at several examples of weak acids and bases."},{"Start":"04:42.865 ","End":"04:45.340","Text":"Let\u0027s start off with acids."},{"Start":"04:45.340 ","End":"04:48.796","Text":"Here are few examples: Iodic acid,"},{"Start":"04:48.796 ","End":"04:55.615","Text":"with a K_a of 1.6 times 10^minus 1 and a pK_a of 0.80."},{"Start":"04:55.615 ","End":"04:58.780","Text":"This is a weak acid,"},{"Start":"04:58.780 ","End":"05:00.790","Text":"but fairly strong one."},{"Start":"05:00.790 ","End":"05:02.705","Text":"Then nitrous acid,"},{"Start":"05:02.705 ","End":"05:06.465","Text":"and that\u0027s 7.2 times 10^minus 4,"},{"Start":"05:06.465 ","End":"05:08.970","Text":"so it\u0027s weaker than iodic acid,"},{"Start":"05:08.970 ","End":"05:12.780","Text":"and look, that it\u0027s pK_a is higher."},{"Start":"05:12.780 ","End":"05:21.300","Text":"It\u0027s 3.14. That means that if we have pK_a goes"},{"Start":"05:21.300 ","End":"05:30.720","Text":"up as the acid strength decreases."},{"Start":"05:30.720 ","End":"05:33.330","Text":"Another example is acetic acid."},{"Start":"05:33.330 ","End":"05:35.760","Text":"We\u0027ll meet this many times."},{"Start":"05:35.760 ","End":"05:43.725","Text":"It\u0027s K_a is 1.8 times 10^minus 5 and a pK_a of 4.74."},{"Start":"05:43.725 ","End":"05:47.469","Text":"We\u0027ll use these values many times."},{"Start":"05:47.469 ","End":"05:53.975","Text":"The final example I have here is hydrocyanic acid, HCN."},{"Start":"05:53.975 ","End":"05:58.850","Text":"Its K_a is 6.2 times 10^minus 10."},{"Start":"05:58.850 ","End":"06:05.860","Text":"It\u0027s a pretty weak acid and its pK_a is 9.21."},{"Start":"06:05.860 ","End":"06:08.070","Text":"Now here are 2 bases,"},{"Start":"06:08.070 ","End":"06:11.250","Text":"ammonia that will meet many times."},{"Start":"06:11.250 ","End":"06:15.600","Text":"It\u0027s K_b is 1.8 times 10^minus 5,"},{"Start":"06:15.600 ","End":"06:19.140","Text":"and its pK_b, 4.74."},{"Start":"06:19.140 ","End":"06:21.710","Text":"Now just by chance,"},{"Start":"06:21.710 ","End":"06:28.105","Text":"it has the same values as K_a and pK_a of acetic acid."},{"Start":"06:28.105 ","End":"06:30.665","Text":"We\u0027ll meet these too many times,"},{"Start":"06:30.665 ","End":"06:34.490","Text":"as examples of weak acids and weak bases."},{"Start":"06:34.490 ","End":"06:36.680","Text":"Then we have pyridine,"},{"Start":"06:36.680 ","End":"06:39.410","Text":"for example, which is much weaker."},{"Start":"06:39.410 ","End":"06:43.820","Text":"It has a K_b of 1.5 times 10^minus 9,"},{"Start":"06:43.820 ","End":"06:47.915","Text":"and a pK_b of 8.82."},{"Start":"06:47.915 ","End":"06:54.720","Text":"In this video, we talked about the ionization of acids and bases in water."}],"ID":28463},{"Watched":false,"Name":"Degree of Ionization","Duration":"5m 3s","ChapterTopicVideoID":27351,"CourseChapterTopicPlaylistID":272179,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.815","Text":"In the previous video,"},{"Start":"00:01.815 ","End":"00:05.895","Text":"we talked about the ionization of acids and bases in water."},{"Start":"00:05.895 ","End":"00:10.500","Text":"This video, we\u0027ll talk about the degree of ionization."},{"Start":"00:10.500 ","End":"00:14.595","Text":"We\u0027re going to talk about the degree of ionization."},{"Start":"00:14.595 ","End":"00:19.110","Text":"For an acid HA that ionizes according to"},{"Start":"00:19.110 ","End":"00:23.895","Text":"HA plus water in equilibrium with H_3O plus and A minus,"},{"Start":"00:23.895 ","End":"00:29.460","Text":"the degree of ionization is defined as Alpha"},{"Start":"00:29.460 ","End":"00:36.960","Text":"equal to the molarity of A minus divided by the initial molarity of HA."},{"Start":"00:36.960 ","End":"00:43.295","Text":"Now if we have a base that ionizes according to B plus water in equilibrium,"},{"Start":"00:43.295 ","End":"00:45.590","Text":"BH plus and OH minus,"},{"Start":"00:45.590 ","End":"00:49.660","Text":"the degree of ionization is defined in a similar way,"},{"Start":"00:49.660 ","End":"00:53.870","Text":"and it\u0027s written as Alpha is equal to the molarity of"},{"Start":"00:53.870 ","End":"00:58.310","Text":"BH plus divided by the initial molarity of"},{"Start":"00:58.310 ","End":"01:07.610","Text":"B. Alpha gives us the fraction of an acid or base in its ionized form."},{"Start":"01:07.610 ","End":"01:13.210","Text":"As a fraction, it lies between 0 and 1,"},{"Start":"01:13.210 ","End":"01:15.560","Text":"it\u0027s never exactly 0,"},{"Start":"01:15.560 ","End":"01:17.660","Text":"and it\u0027s never exactly 1."},{"Start":"01:17.660 ","End":"01:22.230","Text":"So I\u0027ve just written as lying between 0 and 1."},{"Start":"01:23.090 ","End":"01:28.340","Text":"Now, this is very useful for weak acids and bases,"},{"Start":"01:28.340 ","End":"01:33.365","Text":"but it\u0027s almost 1 for strong acids and bases so there it\u0027s not really relevant."},{"Start":"01:33.365 ","End":"01:36.200","Text":"Just as we have a fraction above,"},{"Start":"01:36.200 ","End":"01:40.355","Text":"we can also talk about our percent ionization."},{"Start":"01:40.355 ","End":"01:46.120","Text":"The percent ionization is Alpha multiplied by 100%."},{"Start":"01:46.120 ","End":"01:48.355","Text":"Here\u0027s an example."},{"Start":"01:48.355 ","End":"01:55.292","Text":"Calculate the percent ionization of HA in 1 M HA, of course,"},{"Start":"01:55.292 ","End":"02:03.985","Text":"in an aqueous solution and we\u0027re given K_a is equal to 1 times 10^minus 6."},{"Start":"02:03.985 ","End":"02:07.515","Text":"We can draw an ICE table."},{"Start":"02:07.515 ","End":"02:12.110","Text":"The initial concentration of HA is just 1"},{"Start":"02:12.110 ","End":"02:22.255","Text":"M. The concentration initially of H_3O plus is 0 and A minus is also 0."},{"Start":"02:22.255 ","End":"02:30.710","Text":"Now the change when the acid ionizes is minus X for the HA,"},{"Start":"02:30.710 ","End":"02:37.100","Text":"and plus X for H_3O plus and plus X for A minus."},{"Start":"02:37.100 ","End":"02:45.270","Text":"At equilibrium, the concentration of HA will be 1 minus X."},{"Start":"02:45.430 ","End":"02:50.670","Text":"The concentration of H_3O plus will be X."},{"Start":"02:51.080 ","End":"02:56.290","Text":"The concentration of A minus will also be X."},{"Start":"02:56.420 ","End":"02:59.145","Text":"Now we can write K_a,"},{"Start":"02:59.145 ","End":"03:02.625","Text":"which we know is 1 times 10^minus 6,"},{"Start":"03:02.625 ","End":"03:05.510","Text":"and that\u0027s the concentration of A minus times"},{"Start":"03:05.510 ","End":"03:10.370","Text":"the concentration of H_3O plus divided by the concentration of HA."},{"Start":"03:10.370 ","End":"03:14.650","Text":"Now we can substitute the values we have in the table."},{"Start":"03:14.650 ","End":"03:17.490","Text":"Instead of A minus we have X."},{"Start":"03:17.490 ","End":"03:20.220","Text":"Instead of H_3O plus we also have X,"},{"Start":"03:20.220 ","End":"03:24.840","Text":"that\u0027s X^2 divided by 1 minus X."},{"Start":"03:24.840 ","End":"03:27.510","Text":"Now, again, assuming that X is very,"},{"Start":"03:27.510 ","End":"03:30.225","Text":"very much smaller than 1,"},{"Start":"03:30.225 ","End":"03:39.120","Text":"then we can write the X^2 is equal to 1 times 10^minus 6 times 1."},{"Start":"03:39.120 ","End":"03:42.510","Text":"So that\u0027s just 1 times 10^minus 6."},{"Start":"03:42.510 ","End":"03:46.410","Text":"X^2 is 1 times 10^minus 6,"},{"Start":"03:46.410 ","End":"03:51.375","Text":"so X will be 1 times 10^minus 3."},{"Start":"03:51.375 ","End":"03:54.000","Text":"Now we can define Alpha."},{"Start":"03:54.000 ","End":"03:56.745","Text":"Alpha is X,"},{"Start":"03:56.745 ","End":"04:06.784","Text":"that\u0027s the concentration of A minus divided by the concentration of HA."},{"Start":"04:06.784 ","End":"04:12.005","Text":"It was initially 1 and then some of it ionizes 1 minus X."},{"Start":"04:12.005 ","End":"04:17.430","Text":"If we substitute the values we found for X,"},{"Start":"04:17.430 ","End":"04:19.080","Text":"10 to power minus 3,"},{"Start":"04:19.080 ","End":"04:25.930","Text":"we get 10^minus 3 divided by 1 minus 10^minus 3."},{"Start":"04:25.930 ","End":"04:30.930","Text":"This is just approximately 10^minus 3."},{"Start":"04:30.930 ","End":"04:36.080","Text":"So the percent ionization is 0.1%."},{"Start":"04:36.080 ","End":"04:38.180","Text":"We multiply this by 100,"},{"Start":"04:38.180 ","End":"04:47.805","Text":"that\u0027s 10^2,10^minus 3 times 10^2 is equal to 10^minus 1."},{"Start":"04:47.805 ","End":"04:50.200","Text":"That\u0027s 0.1%."},{"Start":"04:50.200 ","End":"04:57.545","Text":"The percentage ionization of this acid HA is 0.1%."},{"Start":"04:57.545 ","End":"05:03.030","Text":"In this video, we talked about the degree of ionization."}],"ID":28466},{"Watched":false,"Name":"Percent Ionization vs Initial Concentration","Duration":"5m 15s","ChapterTopicVideoID":27366,"CourseChapterTopicPlaylistID":272179,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.230 ","End":"00:02.445","Text":"In the previous video,"},{"Start":"00:02.445 ","End":"00:05.220","Text":"we learned about percent ionization."},{"Start":"00:05.220 ","End":"00:06.735","Text":"In this video, we\u0027ll study"},{"Start":"00:06.735 ","End":"00:12.280","Text":"the percent ionization as a function of the initial concentration."},{"Start":"00:12.530 ","End":"00:19.575","Text":"We\u0027re going to look at the percent ionization as a function of the initial concentration."},{"Start":"00:19.575 ","End":"00:23.685","Text":"Let\u0027s go back to the example in the previous video."},{"Start":"00:23.685 ","End":"00:32.985","Text":"We were asked to calculate the percent ionization of HA in 1M HA solution."},{"Start":"00:32.985 ","End":"00:39.770","Text":"We were given that K_a=1 times 10^minus 6."},{"Start":"00:39.770 ","End":"00:44.225","Text":"The answer we worked out was 0.1%."},{"Start":"00:44.225 ","End":"00:48.395","Text":"Now let\u0027s go back to how we solved the problem."},{"Start":"00:48.395 ","End":"00:52.900","Text":"K_a was equal 1 times 10^minus 6."},{"Start":"00:52.900 ","End":"00:56.960","Text":"That was equal to the concentration of A^minus times"},{"Start":"00:56.960 ","End":"01:03.260","Text":"the concentration of H3O^plus divided by the concentration of HA."},{"Start":"01:03.260 ","End":"01:08.570","Text":"Then we substituted the values in our table."},{"Start":"01:08.570 ","End":"01:19.610","Text":"A^minus was x and H3O^plus was x and HA was the initial concentration of HA minus x."},{"Start":"01:19.610 ","End":"01:25.660","Text":"Now, if we assume the x is much smaller than HA initial,"},{"Start":"01:25.660 ","End":"01:33.515","Text":"then we can write this as approximately x^2 divided by the initial concentration of HA."},{"Start":"01:33.515 ","End":"01:36.650","Text":"Now, it\u0027s obviously not valid for"},{"Start":"01:36.650 ","End":"01:41.390","Text":"very dilute solutions where we can\u0027t make this approximation."},{"Start":"01:41.390 ","End":"01:45.960","Text":"It\u0027s only valid where we can make the approximation."},{"Start":"01:48.520 ","End":"01:58.590","Text":"Now we have that x^2=K_a times the initial concentration of HA."},{"Start":"01:59.300 ","End":"02:01.580","Text":"We can work out from that,"},{"Start":"02:01.580 ","End":"02:09.600","Text":"that x itself is equal to the square root of K_a times the initial concentration of HA."},{"Start":"02:11.470 ","End":"02:18.769","Text":"Then Alpha, which is the concentration of A^minus, that\u0027s x,"},{"Start":"02:18.769 ","End":"02:26.640","Text":"divided by the concentration of the acid is HA_initial minus x."},{"Start":"02:26.640 ","End":"02:29.165","Text":"Again, if we make this approximation,"},{"Start":"02:29.165 ","End":"02:32.805","Text":"that x is much smaller than HA_initial,"},{"Start":"02:32.805 ","End":"02:40.310","Text":"this approximation, then we get that Alpha=x over HA_initial."},{"Start":"02:40.310 ","End":"02:46.815","Text":"Now we can substitute into this value of x."},{"Start":"02:46.815 ","End":"02:54.110","Text":"X we saw was the square root of K_a times the concentration of HA_initial."},{"Start":"02:54.110 ","End":"03:03.720","Text":"Instead of x, we have the square root of K_a times the square root of HA_initial,"},{"Start":"03:03.720 ","End":"03:08.700","Text":"and that\u0027s divided by the initial concentration of HA."},{"Start":"03:08.700 ","End":"03:14.840","Text":"Instead of HA to the square root divided by HA,"},{"Start":"03:14.840 ","End":"03:20.460","Text":"we can write that as 1 over the square root of HA_initial."},{"Start":"03:20.460 ","End":"03:24.030","Text":"We still have the K_a square root."},{"Start":"03:24.030 ","End":"03:28.860","Text":"We end up with an expression for Alpha,"},{"Start":"03:28.860 ","End":"03:39.700","Text":"that Alpha=square root of K_a divided by the square root of HA_initial."},{"Start":"03:39.700 ","End":"03:45.030","Text":"Now we can calculate Alpha for different concentrations."},{"Start":"03:45.030 ","End":"03:51.130","Text":"For example, if we take HA_initial to be 1 M,"},{"Start":"03:51.130 ","End":"03:56.380","Text":"just as we had in the previous example, Alpha was 0.10%."},{"Start":"03:56.380 ","End":"04:01.170","Text":"If it\u0027s 0.1, Alpha\u0027s 0.32%."},{"Start":"04:01.170 ","End":"04:05.730","Text":"If it\u0027s 0.01, Alpha\u0027s 1%."},{"Start":"04:05.730 ","End":"04:09.875","Text":"If the initial concentration\u0027s 0.001,"},{"Start":"04:09.875 ","End":"04:13.510","Text":"then Alpha will be 3.2%."},{"Start":"04:13.510 ","End":"04:21.665","Text":"We can see that as a solution becomes more dilute, Alpha increases."},{"Start":"04:21.665 ","End":"04:24.170","Text":"As the solution becomes more dilute,"},{"Start":"04:24.170 ","End":"04:28.504","Text":"there\u0027s greater ionization of the acid."},{"Start":"04:28.504 ","End":"04:31.792","Text":"We can plot these numbers;"},{"Start":"04:31.792 ","End":"04:36.770","Text":"percentage ionization versus initial concentration."},{"Start":"04:36.770 ","End":"04:41.090","Text":"Remember, we mustn\u0027t go to too low initial concentrations,"},{"Start":"04:41.090 ","End":"04:43.995","Text":"otherwise, our approximations are invalid."},{"Start":"04:43.995 ","End":"04:45.885","Text":"Here\u0027s our graph."},{"Start":"04:45.885 ","End":"04:51.275","Text":"We see here quite clearly that as the concentration gets lower,"},{"Start":"04:51.275 ","End":"04:55.800","Text":"the percentage ionization increases."},{"Start":"04:56.380 ","End":"05:00.170","Text":"This is as if we\u0027re going from right to left,"},{"Start":"05:00.170 ","End":"05:03.035","Text":"lower and lower concentrations,"},{"Start":"05:03.035 ","End":"05:06.770","Text":"the percentage of ionization increases."},{"Start":"05:06.770 ","End":"05:10.220","Text":"In this video, we looked at how the percent ionization"},{"Start":"05:10.220 ","End":"05:14.610","Text":"changes with the initial concentration of the acid."}],"ID":28467},{"Watched":false,"Name":"Exercise 1","Duration":"2m ","ChapterTopicVideoID":31998,"CourseChapterTopicPlaylistID":272179,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34272},{"Watched":false,"Name":"Exercise 2","Duration":"1m 51s","ChapterTopicVideoID":31999,"CourseChapterTopicPlaylistID":272179,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34273},{"Watched":false,"Name":"Exercise 3","Duration":"3m 12s","ChapterTopicVideoID":31996,"CourseChapterTopicPlaylistID":272179,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34274},{"Watched":false,"Name":"Exercise 4","Duration":"2m 57s","ChapterTopicVideoID":31997,"CourseChapterTopicPlaylistID":272179,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34275}],"Thumbnail":null,"ID":272179},{"Name":"Strong Acids and Bases","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Calculating Ion Concentrations in Strong Acids and Bases","Duration":"5m 22s","ChapterTopicVideoID":27352,"CourseChapterTopicPlaylistID":272180,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.875","Text":"In our previous video,"},{"Start":"00:01.875 ","End":"00:07.215","Text":"we talked about aqueous solutions of strong acids and bases and in this video,"},{"Start":"00:07.215 ","End":"00:13.425","Text":"we\u0027ll calculate the ion concentrations and pH\u0027s of such solutions."},{"Start":"00:13.425 ","End":"00:17.190","Text":"Let\u0027s consider strong acids first."},{"Start":"00:17.190 ","End":"00:23.685","Text":"We learnt already that a strong acid is almost completely ionized in an aqueous solution."},{"Start":"00:23.685 ","End":"00:26.565","Text":"In a solution of strong acid in water,"},{"Start":"00:26.565 ","End":"00:29.040","Text":"there are 2 sources of H_3O+,"},{"Start":"00:29.040 ","End":"00:32.550","Text":"the acid and the self ionization of the water."},{"Start":"00:32.550 ","End":"00:35.355","Text":"Now when we add a strong acid to water,"},{"Start":"00:35.355 ","End":"00:38.720","Text":"we saw before in the previous video that this suppresses"},{"Start":"00:38.720 ","End":"00:44.130","Text":"self ionization of the water so there is less H_3O+."},{"Start":"00:50.060 ","End":"00:55.085","Text":"We can learn from this that the H_3O+ from self ionization"},{"Start":"00:55.085 ","End":"01:00.610","Text":"can be ignored except for extremely dilute solutions."},{"Start":"01:00.610 ","End":"01:02.970","Text":"Here\u0027s an example,"},{"Start":"01:02.970 ","End":"01:07.680","Text":"calculate the concentration of H_3O+, Cl-,"},{"Start":"01:07.680 ","End":"01:14.100","Text":"OH- and the pH in 0.02 M HCl."},{"Start":"01:14.100 ","End":"01:18.120","Text":"Here\u0027s the equation, HCl plus H_2O,"},{"Start":"01:18.120 ","End":"01:23.445","Text":"going completely to completion, H_30+ plus Cl-."},{"Start":"01:23.445 ","End":"01:27.235","Text":"So we have almost complete ionization."},{"Start":"01:27.235 ","End":"01:37.460","Text":"That means that there will be no HCl left at all and we will have H_3O+ and Cl- instead."},{"Start":"01:37.460 ","End":"01:41.815","Text":"The concentration of H_3O+ will be equal to the concentration of Cl-,"},{"Start":"01:41.815 ","End":"01:48.775","Text":"and both of them will be equal to 0.02 M. From this we can calculate the pH."},{"Start":"01:48.775 ","End":"01:53.925","Text":"The pH is minus the log of the concentration of H_30+,"},{"Start":"01:53.925 ","End":"02:00.950","Text":"and that\u0027s a minus the log(0.02) and if we look at our calculator, that\u0027s 1.7."},{"Start":"02:02.270 ","End":"02:07.805","Text":"Now the only thing we have left to calculate is the concentration of OH-."},{"Start":"02:07.805 ","End":"02:12.605","Text":"Now we know that the multiplication of the product of"},{"Start":"02:12.605 ","End":"02:19.295","Text":"H_3O+ concentration and OH- concentration is 10^-14."},{"Start":"02:19.295 ","End":"02:23.570","Text":"So we can work out the concentration of OH-"},{"Start":"02:23.570 ","End":"02:28.145","Text":"is 10^-14 divided by the concentration of H_3O+."},{"Start":"02:28.145 ","End":"02:35.900","Text":"The concentration of H_3O+ was 0.02 so that\u0027s 10^-14 divided by"},{"Start":"02:35.900 ","End":"02:43.890","Text":"2 times 10^-2 and that\u0027s 5 times 10^-13."},{"Start":"02:43.890 ","End":"02:49.845","Text":"We have the concentration of OH- is 5 times 10^-13."},{"Start":"02:49.845 ","End":"02:52.830","Text":"Let\u0027s go on to strong bases."},{"Start":"02:52.830 ","End":"02:54.420","Text":"Like a strong acid,"},{"Start":"02:54.420 ","End":"02:58.520","Text":"a strong base is almost completely ionized in an aqueous solution."},{"Start":"02:58.520 ","End":"03:01.580","Text":"Again, when we have a solution of strong base in water,"},{"Start":"03:01.580 ","End":"03:03.575","Text":"there are 2 sources of OH-,"},{"Start":"03:03.575 ","End":"03:06.695","Text":"the base and self ionization of water."},{"Start":"03:06.695 ","End":"03:09.770","Text":"We saw in the previous video that adding a strong base to"},{"Start":"03:09.770 ","End":"03:13.055","Text":"water suppresses the self ionization of the water,"},{"Start":"03:13.055 ","End":"03:15.810","Text":"so there\u0027s less OH-."},{"Start":"03:17.870 ","End":"03:22.790","Text":"The conclusion is the OH- from self ionization can be"},{"Start":"03:22.790 ","End":"03:27.215","Text":"ignored except for extremely dilute solutions."},{"Start":"03:27.215 ","End":"03:29.235","Text":"Here\u0027s an example,"},{"Start":"03:29.235 ","End":"03:32.835","Text":"calculate the concentrations of OH-, Na+,"},{"Start":"03:32.835 ","End":"03:41.580","Text":"H_3O+ and the pH in 0.02 M NaOH."},{"Start":"03:41.580 ","End":"03:47.040","Text":"NaOH completely decomposes to Na+ and OH-."},{"Start":"03:47.480 ","End":"03:51.510","Text":"There is no NaOH left,"},{"Start":"03:51.510 ","End":"03:59.655","Text":"that means the concentration of Na+ and OH- are equal and equal to 0.02."},{"Start":"03:59.655 ","End":"04:02.600","Text":"From that we can calculate the pOH,"},{"Start":"04:02.600 ","End":"04:07.250","Text":"that\u0027s minus the log of the concentration of OH- which is"},{"Start":"04:07.250 ","End":"04:13.660","Text":"equal to minus log(0.02) which we saw before was 1.7."},{"Start":"04:13.660 ","End":"04:17.980","Text":"The pOH is 1.7."},{"Start":"04:17.980 ","End":"04:23.615","Text":"Now we need to calculate the concentration of H_3O+."},{"Start":"04:23.615 ","End":"04:29.345","Text":"We know that this product H_3O+ and OH- is 10^-14,"},{"Start":"04:29.345 ","End":"04:33.980","Text":"that means the concentration of H_3O+ is 10^-14 divided by"},{"Start":"04:33.980 ","End":"04:41.910","Text":"the concentration of OH- and that\u0027s 2 times 10^-2."},{"Start":"04:41.910 ","End":"04:45.585","Text":"When we divide them, we get 5 times 10^-13,"},{"Start":"04:45.585 ","End":"04:49.480","Text":"and the pH is therefore 12.3."},{"Start":"04:49.480 ","End":"04:59.375","Text":"Now we could have calculated the pH directly from the pOH because pH is pK_w minus pOH."},{"Start":"04:59.375 ","End":"05:08.930","Text":"pK_w is 14 and pOH we found was 1.7 so 14 minus 1.7 is 12.3."},{"Start":"05:08.930 ","End":"05:12.919","Text":"So that\u0027s another way we could calculate the pH."},{"Start":"05:12.919 ","End":"05:16.880","Text":"In this video, we calculate the ion concentrations of pH"},{"Start":"05:16.880 ","End":"05:21.720","Text":"in aqueous solutions of strong acids and bases."}],"ID":28468},{"Watched":false,"Name":"Exercise 1 Part a","Duration":"4m ","ChapterTopicVideoID":29799,"CourseChapterTopicPlaylistID":272180,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31447},{"Watched":false,"Name":"Exercise 1 Part b","Duration":"3m 22s","ChapterTopicVideoID":29800,"CourseChapterTopicPlaylistID":272180,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31448},{"Watched":false,"Name":"Exercise 1 Part c","Duration":"2m 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45s","ChapterTopicVideoID":29803,"CourseChapterTopicPlaylistID":272180,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31451},{"Watched":false,"Name":"Exercise 3 Part a","Duration":"4m 1s","ChapterTopicVideoID":29804,"CourseChapterTopicPlaylistID":272180,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31452},{"Watched":false,"Name":"Exercise 3 Part b","Duration":"3m 37s","ChapterTopicVideoID":29805,"CourseChapterTopicPlaylistID":272180,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31453},{"Watched":false,"Name":"Exercise 4","Duration":"5m 59s","ChapterTopicVideoID":29806,"CourseChapterTopicPlaylistID":272180,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31454},{"Watched":false,"Name":"Exercise 5","Duration":"4m 56s","ChapterTopicVideoID":29807,"CourseChapterTopicPlaylistID":272180,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31455}],"Thumbnail":null,"ID":272180},{"Name":"Weak Acids and Bases","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Calculating pH of a Weak Acid","Duration":"5m 12s","ChapterTopicVideoID":27354,"CourseChapterTopicPlaylistID":272181,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.020 ","End":"00:02.325","Text":"In the previous video,"},{"Start":"00:02.325 ","End":"00:05.520","Text":"we calculated the pH of a strong acid."},{"Start":"00:05.520 ","End":"00:10.455","Text":"In this video, we\u0027ll calculate the pH of a weak acid."},{"Start":"00:10.455 ","End":"00:13.620","Text":"We\u0027re going to talk about weak acids."},{"Start":"00:13.620 ","End":"00:20.310","Text":"A few points that we\u0027ve learned already just to go over about weak acids."},{"Start":"00:20.310 ","End":"00:25.180","Text":"Weak acids undergo partial ionization of water."},{"Start":"00:25.400 ","End":"00:29.940","Text":"The H_3O plus from self ionization of water"},{"Start":"00:29.940 ","End":"00:34.530","Text":"can be ignored except for extremely dilute solutions."},{"Start":"00:34.530 ","End":"00:40.410","Text":"We can often make approximations to simplify the solution of the problem."},{"Start":"00:40.410 ","End":"00:43.005","Text":"Here\u0027s an example."},{"Start":"00:43.005 ","End":"00:50.915","Text":"Calculate the pH of a 0.02 molar solution of acetic acid."},{"Start":"00:50.915 ","End":"00:53.000","Text":"Here\u0027s the equation."},{"Start":"00:53.000 ","End":"00:59.515","Text":"Acetic acid reacts with water to give us H_3O plus, and acetate anion."},{"Start":"00:59.515 ","End":"01:04.380","Text":"The K_a for this is 1.8 times 10^minus 5."},{"Start":"01:04.380 ","End":"01:07.095","Text":"This is partial ionization."},{"Start":"01:07.095 ","End":"01:10.360","Text":"You can see the equilibrium sign here."},{"Start":"01:10.450 ","End":"01:13.910","Text":"Let\u0027s draw our ICE table."},{"Start":"01:13.910 ","End":"01:20.055","Text":"We have acetic acid with 0.02 molar at the beginning,"},{"Start":"01:20.055 ","End":"01:26.310","Text":"and there\u0027s no H_3O plus and no acetate anion, zero and zero."},{"Start":"01:26.310 ","End":"01:30.645","Text":"Now, there\u0027s a change to reach equilibrium,"},{"Start":"01:30.645 ","End":"01:33.355","Text":"minus x for acetic acid,"},{"Start":"01:33.355 ","End":"01:34.905","Text":"plus x for the H_3O plus,"},{"Start":"01:34.905 ","End":"01:37.740","Text":"and plus x for the acetate."},{"Start":"01:37.740 ","End":"01:40.145","Text":"Finally, at equilibrium,"},{"Start":"01:40.145 ","End":"01:45.475","Text":"the concentration of acetic acid is 0.02 minus x,"},{"Start":"01:45.475 ","End":"01:47.930","Text":"and H_3O plus,"},{"Start":"01:47.930 ","End":"01:51.240","Text":"plus x, and acetate plus x."},{"Start":"01:51.980 ","End":"01:54.405","Text":"We can write the K_a,"},{"Start":"01:54.405 ","End":"02:00.000","Text":"the equilibrium constant for the acid is 1.8 times 10^minus 5,"},{"Start":"02:00.000 ","End":"02:05.465","Text":"and that\u0027s equal to the concentration of H_3O plus x,"},{"Start":"02:05.465 ","End":"02:09.605","Text":"times the concentration of CH_3CO_2 minus, that\u0027s also x."},{"Start":"02:09.605 ","End":"02:11.700","Text":"That gives us x^2,"},{"Start":"02:11.700 ","End":"02:14.975","Text":"divided by the concentration of acetic acid,"},{"Start":"02:14.975 ","End":"02:19.115","Text":"which is 0.02 minus x. Here we have it."},{"Start":"02:19.115 ","End":"02:23.810","Text":"K_a is equal to x^2 divided by 0.02 minus x."},{"Start":"02:23.810 ","End":"02:29.310","Text":"Now, if we assume that x is much smaller than 0.02,"},{"Start":"02:32.000 ","End":"02:35.145","Text":"then we can ignore this x."},{"Start":"02:35.145 ","End":"02:39.300","Text":"Then we have x^2 divided by 0.02."},{"Start":"02:39.300 ","End":"02:47.220","Text":"X^2 is equal to 0.02 times 1.8 times 10^minus 5."},{"Start":"02:47.220 ","End":"02:53.040","Text":"That\u0027s 3.6 times 10^minus 7."},{"Start":"02:53.040 ","End":"03:00.960","Text":"X^2 is 3.6 times 10^minus 7."},{"Start":"03:00.960 ","End":"03:04.828","Text":"Remember x is the concentration of H_3O plus,"},{"Start":"03:04.828 ","End":"03:07.590","Text":"is equal to 6 times 10^minus 4,"},{"Start":"03:07.590 ","End":"03:12.240","Text":"the square root of 3.6 times 10^minus 7."},{"Start":"03:12.240 ","End":"03:15.630","Text":"Now, we can calculate the pH."},{"Start":"03:15.630 ","End":"03:20.730","Text":"The pH is minus the log of x,"},{"Start":"03:20.730 ","End":"03:23.190","Text":"times 10^minus 4,"},{"Start":"03:23.190 ","End":"03:26.440","Text":"and that\u0027s equal to 3.2."},{"Start":"03:26.780 ","End":"03:30.595","Text":"The pH is 3.2."},{"Start":"03:30.595 ","End":"03:35.230","Text":"Now, just a few words about the approximation."},{"Start":"03:35.230 ","End":"03:45.045","Text":"We see the x, 6 times 10^minus 4 is much smaller than 2 times 10^minus 2."},{"Start":"03:45.045 ","End":"03:48.425","Text":"This approximation is justified."},{"Start":"03:48.425 ","End":"03:51.935","Text":"Now, when we neglect x in the denominator,"},{"Start":"03:51.935 ","End":"03:56.720","Text":"it means that we\u0027re assuming that the acid is essentially non-ionized."},{"Start":"03:56.720 ","End":"04:03.410","Text":"We\u0027re taking 0.02 as the concentration of the acid as if nothing had happened,"},{"Start":"04:03.410 ","End":"04:06.100","Text":"as if there was no ionization."},{"Start":"04:06.100 ","End":"04:10.150","Text":"Now, this approximation works in general,"},{"Start":"04:10.150 ","End":"04:19.210","Text":"is the concentration of the acid divided by K_a is greater than about 100."},{"Start":"04:19.210 ","End":"04:23.449","Text":"In general, it works for these conditions;"},{"Start":"04:23.449 ","End":"04:29.705","Text":"concentration of HA divided by K_a greater than 100."},{"Start":"04:29.705 ","End":"04:35.660","Text":"Now, supposing we were given the pH and asked to calculate K_a."},{"Start":"04:35.810 ","End":"04:45.310","Text":"We can use the pH to get the value of x. X is the concentration of H_3O plus."},{"Start":"04:51.370 ","End":"04:54.290","Text":"Once we have the volume of x,"},{"Start":"04:54.290 ","End":"05:02.525","Text":"we can calculate K_a from the equation K_a is equal to x squared divided by 0.02 minus x."},{"Start":"05:02.525 ","End":"05:06.140","Text":"In this video, we learned how to calculate the pH from"},{"Start":"05:06.140 ","End":"05:11.160","Text":"the acid ionization constant and vice versa."}],"ID":28470},{"Watched":false,"Name":"Calculating pH of a Weak Base","Duration":"4m 46s","ChapterTopicVideoID":27353,"CourseChapterTopicPlaylistID":272181,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.860","Text":"In the previous video,"},{"Start":"00:01.860 ","End":"00:04.485","Text":"we calculated the pH of a weak acid."},{"Start":"00:04.485 ","End":"00:08.910","Text":"In this video, we\u0027ll calculate the pH for a weak base."},{"Start":"00:08.910 ","End":"00:12.134","Text":"We\u0027re going to talk about a weak base."},{"Start":"00:12.134 ","End":"00:14.985","Text":"A few points for weak bases."},{"Start":"00:14.985 ","End":"00:19.050","Text":"Weak bases undergo partial ionization of water."},{"Start":"00:19.050 ","End":"00:24.975","Text":"We can ignore self ionization except for extremely dilute solutions."},{"Start":"00:24.975 ","End":"00:30.420","Text":"We can often make approximations to simplify the solution of the problem."},{"Start":"00:30.420 ","End":"00:32.985","Text":"Here\u0027s an example,"},{"Start":"00:32.985 ","End":"00:39.304","Text":"calculate the pH of a 0.02 molar solution of ammonia."},{"Start":"00:39.304 ","End":"00:41.465","Text":"Here\u0027s the equation."},{"Start":"00:41.465 ","End":"00:47.675","Text":"Ammonia reacts with water to give us NH_4 plus ions and OH minus,"},{"Start":"00:47.675 ","End":"00:52.970","Text":"and the K_b for this process is 1.8 times 10 to the power minus 5,"},{"Start":"00:52.970 ","End":"00:55.480","Text":"ad it\u0027s partial ionization,"},{"Start":"00:55.480 ","End":"00:59.430","Text":"we can see that from the equilibrium sign."},{"Start":"00:59.900 ","End":"01:03.360","Text":"Here\u0027s ICE table."},{"Start":"01:03.360 ","End":"01:09.920","Text":"We start off with NH_3 concentration of 0.02 molar,"},{"Start":"01:09.920 ","End":"01:14.260","Text":"NH4 plus 0 and OH minus 0."},{"Start":"01:14.260 ","End":"01:20.110","Text":"The change to reach equilibrium is minus x plus x and plus x,"},{"Start":"01:20.110 ","End":"01:21.755","Text":"and so at equilibrium,"},{"Start":"01:21.755 ","End":"01:31.770","Text":"the concentration of NH_3 is 0.02 minus x of NH_4 plus x and the OH minus plus x."},{"Start":"01:32.510 ","End":"01:34.890","Text":"We can write the K_b,"},{"Start":"01:34.890 ","End":"01:38.335","Text":"which is equal to 1.8 times 10 to the power minus 5,"},{"Start":"01:38.335 ","End":"01:40.130","Text":"is equal to x squared."},{"Start":"01:40.130 ","End":"01:45.050","Text":"That\u0027s the concentration of NH_4 plus times the concentration of OH minus,"},{"Start":"01:45.050 ","End":"01:48.350","Text":"divided by the concentration of NH_3,"},{"Start":"01:48.350 ","End":"01:51.170","Text":"which is 0.02 minus x."},{"Start":"01:51.170 ","End":"01:57.020","Text":"If we make the usual approximation that x is negligible compared to 0.02,"},{"Start":"01:57.020 ","End":"02:00.530","Text":"we get x squared divided by 0.02."},{"Start":"02:00.530 ","End":"02:07.190","Text":"That means the x squared is equal to 0.02 times 1.8 times 10 to the power minus 5,"},{"Start":"02:07.190 ","End":"02:11.700","Text":"and that\u0027s 3.6 times 10 to the power of minus 7."},{"Start":"02:13.430 ","End":"02:17.225","Text":"Now that we\u0027ve got x squared, which we worked out,"},{"Start":"02:17.225 ","End":"02:23.205","Text":"and x is equal to square root of 3.6 times 10 to power minus 7,"},{"Start":"02:23.205 ","End":"02:27.310","Text":"and that\u0027s 6 times 10 to the power of minus 4."},{"Start":"02:28.100 ","End":"02:34.990","Text":"Now x in this problem is the concentration of OH minus."},{"Start":"02:34.990 ","End":"02:39.300","Text":"From that we can work out the pOH."},{"Start":"02:39.300 ","End":"02:47.590","Text":"The pOH is minus the log of 6 times 10 to the power minus 4, and that\u0027s 3.2."},{"Start":"02:49.690 ","End":"02:54.770","Text":"We have the pOH and now we can calculate the pH."},{"Start":"02:54.770 ","End":"03:02.520","Text":"The pH is 14, that\u0027s pKw minus 3.2,"},{"Start":"03:02.520 ","End":"03:05.565","Text":"and that gives us 11.8."},{"Start":"03:05.565 ","End":"03:09.270","Text":"The pH is 11.8."},{"Start":"03:09.270 ","End":"03:11.655","Text":"As we would expect,"},{"Start":"03:11.655 ","End":"03:13.865","Text":"since we\u0027re talking about a base,"},{"Start":"03:13.865 ","End":"03:17.090","Text":"the pH is greater than 7."},{"Start":"03:17.090 ","End":"03:21.385","Text":"Now just a note about the approximation we made."},{"Start":"03:21.385 ","End":"03:27.360","Text":"We can see that x equal to 6 times 10 to the power of minus 4 is indeed good,"},{"Start":"03:27.360 ","End":"03:30.870","Text":"but smaller than 2 times 10 to the power of minus 2."},{"Start":"03:30.870 ","End":"03:33.595","Text":"The approximation is valid."},{"Start":"03:33.595 ","End":"03:37.340","Text":"Now, we neglected x in the denominator,"},{"Start":"03:37.340 ","End":"03:42.020","Text":"and that means that we\u0027re assuming that the base is essentially non-ionized."},{"Start":"03:42.020 ","End":"03:48.185","Text":"We took the concentration of the base of 0.02 instead of 0.02 minus x,"},{"Start":"03:48.185 ","End":"03:52.880","Text":"so we\u0027re ignoring the fact that it\u0027s ionized."},{"Start":"03:52.880 ","End":"03:58.160","Text":"Now in general, this approximation works if the concentration of"},{"Start":"03:58.160 ","End":"04:03.365","Text":"the base divided by K_b is greater than a 100,"},{"Start":"04:03.365 ","End":"04:06.450","Text":"that\u0027s the general rule."},{"Start":"04:06.940 ","End":"04:13.525","Text":"Now supposing we were given the pH and asked to calculate K_b."},{"Start":"04:13.525 ","End":"04:19.535","Text":"We\u0027re given the pH, we can calculate pOH from it."},{"Start":"04:19.535 ","End":"04:23.540","Text":"The pOH gives us the volume x."},{"Start":"04:23.540 ","End":"04:26.210","Text":"We can calculate K_b from the equation,"},{"Start":"04:26.210 ","End":"04:31.885","Text":"K_b is equal to x squared divided by 0.02 minus x."},{"Start":"04:31.885 ","End":"04:37.895","Text":"This is how we can calculate the K_b if we\u0027re given pH."},{"Start":"04:37.895 ","End":"04:41.180","Text":"In this video, we learned how to calculate the pH of"},{"Start":"04:41.180 ","End":"04:46.559","Text":"the base ionization constant and vice versa."}],"ID":28469},{"Watched":false,"Name":"Exercise 1","Duration":"4m 8s","ChapterTopicVideoID":32000,"CourseChapterTopicPlaylistID":272181,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34276},{"Watched":false,"Name":"Exercise 2","Duration":"2m 29s","ChapterTopicVideoID":32001,"CourseChapterTopicPlaylistID":272181,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34277}],"Thumbnail":null,"ID":272181},{"Name":"Polyprotic Acids","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Polyprotic Acid - Phosphoric Acid","Duration":"7m 45s","ChapterTopicVideoID":27356,"CourseChapterTopicPlaylistID":272182,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.130","Text":"In previous videos,"},{"Start":"00:02.130 ","End":"00:04.859","Text":"we talked about monoprotic acids."},{"Start":"00:04.859 ","End":"00:09.269","Text":"At this video, we\u0027ll discuss polyprotic acids."},{"Start":"00:09.269 ","End":"00:12.480","Text":"What\u0027s a monoprotic acid?"},{"Start":"00:12.480 ","End":"00:18.720","Text":"Monoprotic acids have a single ionizable H atom,"},{"Start":"00:18.720 ","End":"00:22.470","Text":"things like HCl and so on,"},{"Start":"00:22.470 ","End":"00:27.690","Text":"and polyprotic acids have more than 1 ionizable H atom."},{"Start":"00:27.690 ","End":"00:31.335","Text":"Examples of phosphoric acid and sulfuric acid."},{"Start":"00:31.335 ","End":"00:34.215","Text":"We\u0027ll talk about phosphoric acid in this video,"},{"Start":"00:34.215 ","End":"00:37.215","Text":"and sulfuric acid and the next one."},{"Start":"00:37.215 ","End":"00:41.175","Text":"Phosphoric acid is H_3PO_4,"},{"Start":"00:41.175 ","End":"00:45.510","Text":"it has 3 ionizable H atoms."},{"Start":"00:45.510 ","End":"00:49.120","Text":"We have 3 equilibria,"},{"Start":"00:49.730 ","End":"00:57.875","Text":"H_3PO_4 plus water to give us H3_O plus and H_2PO_4 minus."},{"Start":"00:57.875 ","End":"01:06.410","Text":"H_3PO_4 has lost 1 hydrogen and produces H_2PO_4 minus."},{"Start":"01:06.410 ","End":"01:11.205","Text":"The K for that which we\u0027re going to call K_a_1,"},{"Start":"01:11.205 ","End":"01:15.105","Text":"is 7.1 times 10^-3."},{"Start":"01:15.105 ","End":"01:19.905","Text":"It\u0027s pretty strong acid in its first step."},{"Start":"01:19.905 ","End":"01:23.665","Text":"In the second step,"},{"Start":"01:23.665 ","End":"01:27.310","Text":"H_2PO_4 minus reacts with water."},{"Start":"01:27.310 ","End":"01:35.780","Text":"It loses 1 hydrogen to become HPO_4 2 minus and H_3O plus is produced."},{"Start":"01:35.780 ","End":"01:37.755","Text":"That\u0027s the K_a_2."},{"Start":"01:37.755 ","End":"01:40.845","Text":"We see it\u0027s much smaller than the K_a_1,"},{"Start":"01:40.845 ","End":"01:45.255","Text":"it\u0027s only 6.3 times 10^-8."},{"Start":"01:45.255 ","End":"01:50.900","Text":"So it\u0027s 5 orders of magnitude smaller than K_a_1."},{"Start":"01:51.050 ","End":"01:57.160","Text":"Third step, HPO_4 2 minus reacts with"},{"Start":"01:57.160 ","End":"02:02.500","Text":"water and loses the final hydrogen to become PO_4 3 minus,"},{"Start":"02:02.500 ","End":"02:08.835","Text":"phosphate, and producers of course, H_3O_."},{"Start":"02:08.835 ","End":"02:16.390","Text":"Here the K_a for the third step is extremely small."},{"Start":"02:16.390 ","End":"02:19.189","Text":"It\u0027s 5 orders of magnitude,"},{"Start":"02:19.189 ","End":"02:22.660","Text":"again, smaller than K_a_2."},{"Start":"02:22.660 ","End":"02:29.190","Text":"We see that K_a_1 is much greater than K_a_2 which is much greater than K_a_3."},{"Start":"02:29.190 ","End":"02:35.540","Text":"We can assume that almost all the H_3O plus comes from the very first ionization step,"},{"Start":"02:35.540 ","End":"02:37.800","Text":"from this step here."},{"Start":"02:38.060 ","End":"02:41.750","Text":"Since the K_a_2 is so small,"},{"Start":"02:41.750 ","End":"02:46.445","Text":"very little of H_2PO_4 minus ionizes."},{"Start":"02:46.445 ","End":"02:48.020","Text":"In the second step,"},{"Start":"02:48.020 ","End":"02:56.030","Text":"we can assume that H_2PO_4 minus is just equal"},{"Start":"02:56.030 ","End":"03:01.265","Text":"to H_3O plus"},{"Start":"03:01.265 ","End":"03:08.015","Text":"but the same molarity as H_2PO_4 minus."},{"Start":"03:08.015 ","End":"03:13.130","Text":"The H_2PO_4 minus almost doesn\u0027t ionize,"},{"Start":"03:13.130 ","End":"03:19.380","Text":"so H_3O plus is equal to H_2PO_4 minus. Here we have it here."},{"Start":"03:20.180 ","End":"03:23.220","Text":"We can write K_a_2,"},{"Start":"03:23.220 ","End":"03:27.470","Text":"which is H_3O times HPO_4 2 minus divided by"},{"Start":"03:27.470 ","End":"03:33.170","Text":"H_2PO_4 minus as just equal to HPO_4 2 minus,"},{"Start":"03:33.170 ","End":"03:38.960","Text":"because H3_O plus is equal to the concentration of H_2PO_4 minus,"},{"Start":"03:38.960 ","End":"03:44.585","Text":"these 2 are equal so we\u0027re just left with the HPO_4 2 minus."},{"Start":"03:44.585 ","End":"03:48.390","Text":"So we immediately know the concentration of"},{"Start":"03:48.390 ","End":"03:53.700","Text":"HPO_4 2 minus because we know the volume of K_a_2."},{"Start":"03:53.710 ","End":"03:57.695","Text":"That means we know the volume of the concentration of"},{"Start":"03:57.695 ","End":"04:03.205","Text":"HPO_4 2 minus regardless of the molarity of the acid."},{"Start":"04:03.205 ","End":"04:06.200","Text":"Here\u0027s an example."},{"Start":"04:06.200 ","End":"04:09.640","Text":"Calculate the ion concentrations,"},{"Start":"04:09.640 ","End":"04:12.295","Text":"that\u0027s all the concentrations of the ions,"},{"Start":"04:12.295 ","End":"04:17.440","Text":"for a 4.0 molar solution of phosphoric acid."},{"Start":"04:17.440 ","End":"04:20.845","Text":"Here\u0027s our table. In the question,"},{"Start":"04:20.845 ","End":"04:25.180","Text":"we\u0027ve taken a rather high concentration of phosphoric acid,"},{"Start":"04:25.180 ","End":"04:29.005","Text":"because otherwise our approximations won\u0027t be valid."},{"Start":"04:29.005 ","End":"04:37.405","Text":"The initial concentration of H_3PO_4 is 4 and that of H_3O plus and H_2PO_4 minus 0."},{"Start":"04:37.405 ","End":"04:46.225","Text":"Again, we have the change necessary to reach equilibrium, -x+x+x."},{"Start":"04:46.225 ","End":"04:54.610","Text":"Finally, take equilibrium, we get 4-x+x+x."},{"Start":"04:54.610 ","End":"05:01.520","Text":"We can write K_a as x^2 divided by 4.0-x,"},{"Start":"05:01.520 ","End":"05:04.625","Text":"and we\u0027re going to assume the x is smaller than 4."},{"Start":"05:04.625 ","End":"05:08.855","Text":"Because the K_a_1 is rather large,"},{"Start":"05:08.855 ","End":"05:12.460","Text":"this is just about a good approximation."},{"Start":"05:12.460 ","End":"05:18.685","Text":"It\u0027s x^2 divided by 4 equal to 7.1 times 10^-3."},{"Start":"05:18.685 ","End":"05:26.675","Text":"X squared is 2.84 times 10^-2 is 4 times 7.1 times 10^3,"},{"Start":"05:26.675 ","End":"05:29.480","Text":"and x is 0.169."},{"Start":"05:29.480 ","End":"05:33.840","Text":"That\u0027s approximately 4 percent of the initial concentration."},{"Start":"05:33.840 ","End":"05:40.020","Text":"It\u0027s tolerable approximation, but not a wonderful one."},{"Start":"05:41.680 ","End":"05:47.930","Text":"We know that the concentration of H_3O plus is 0.169 molar."},{"Start":"05:47.930 ","End":"05:51.140","Text":"Assuming all of it comes to the first step,"},{"Start":"05:51.140 ","End":"05:54.210","Text":"that\u0027s the assumption we\u0027re going to make."},{"Start":"05:54.710 ","End":"05:57.980","Text":"The next assumption is that concentration of"},{"Start":"05:57.980 ","End":"06:01.620","Text":"H_2PO_4 minus is equal to the concentration of H_3O"},{"Start":"06:01.620 ","End":"06:08.035","Text":"plus which we\u0027ve just calculated as 0.169 molar."},{"Start":"06:08.035 ","End":"06:17.270","Text":"We can work out that the concentration of HPO_4 2 minus is approximately like K_a_2,"},{"Start":"06:17.270 ","End":"06:21.260","Text":"and that\u0027s 6.3 times 10^-8."},{"Start":"06:21.260 ","End":"06:26.720","Text":"Now we have the concentration of the H_2PO_4 minus and we have"},{"Start":"06:26.720 ","End":"06:32.505","Text":"the concentration of the HPO_4 2 minus,"},{"Start":"06:32.505 ","End":"06:36.825","Text":"and that\u0027s 6.3 times 10^-8."},{"Start":"06:36.825 ","End":"06:43.210","Text":"Now, we have to calculate the concentration of PO_4 3 minus."},{"Start":"06:43.210 ","End":"06:45.450","Text":"We\u0027ll look at the last step."},{"Start":"06:45.450 ","End":"06:50.540","Text":"The K_a_3 is 4.2 times 10^-13."},{"Start":"06:50.540 ","End":"06:54.080","Text":"We\u0027re going to assume that all the concentration of H_3O plus comes from"},{"Start":"06:54.080 ","End":"06:57.785","Text":"the first step, so 0.169,"},{"Start":"06:57.785 ","End":"07:01.100","Text":"and the concentration of HPO_4 2 minus,"},{"Start":"07:01.100 ","End":"07:06.335","Text":"we\u0027ve just calculated as 6.3 times 10^-8."},{"Start":"07:06.335 ","End":"07:10.850","Text":"We\u0027re left with the concentration of PO_4 3 minus."},{"Start":"07:10.850 ","End":"07:19.265","Text":"We can work out the concentration PO_4 3 minus is 4.2 times 10^-13 from here,"},{"Start":"07:19.265 ","End":"07:22.865","Text":"times 6.3 times 10^-8 from here,"},{"Start":"07:22.865 ","End":"07:26.330","Text":"divided by 0.169 from up here."},{"Start":"07:26.330 ","End":"07:34.860","Text":"When we do that, we get 1.5 times 10^-19 is extremely small."},{"Start":"07:36.820 ","End":"07:44.940","Text":"In this video, we discussed phosphoric acid which is a polyprotic acid."}],"ID":28472},{"Watched":false,"Name":"Polyprotic Acid - Sulfuric Acid","Duration":"4m 9s","ChapterTopicVideoID":27355,"CourseChapterTopicPlaylistID":272182,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.590","Text":"In the previous video,"},{"Start":"00:01.590 ","End":"00:03.650","Text":"we talked about phosphoric acid."},{"Start":"00:03.650 ","End":"00:08.880","Text":"In this video, we\u0027ll discuss another polyprotic acid, sulfuric acid."},{"Start":"00:08.880 ","End":"00:13.620","Text":"We\u0027re continuing our discussion of polyprotic acids."},{"Start":"00:13.620 ","End":"00:18.810","Text":"These are acids that have more than one ionizable each atom."},{"Start":"00:18.810 ","End":"00:22.500","Text":"In this video, we\u0027ll talk about sulfuric acid."},{"Start":"00:22.500 ","End":"00:25.343","Text":"Sulfuric acid, H_2SO_4,"},{"Start":"00:25.343 ","End":"00:28.455","Text":"has two ionizable each atoms."},{"Start":"00:28.455 ","End":"00:29.955","Text":"Here\u0027s the 1st step;"},{"Start":"00:29.955 ","End":"00:36.060","Text":"H_2SO_4 plus water to give H_3O plus and HSO_4 minus."},{"Start":"00:36.060 ","End":"00:40.580","Text":"This is going to be arrow because goes almost to completion."},{"Start":"00:40.580 ","End":"00:44.090","Text":"K_a1 is very large."},{"Start":"00:44.090 ","End":"00:45.665","Text":"The 2nd step;"},{"Start":"00:45.665 ","End":"00:49.040","Text":"HSO_4 minus reacts with water and there\u0027s"},{"Start":"00:49.040 ","End":"00:54.520","Text":"an equilibrium giving H_3O plus and SO_4^2 minus."},{"Start":"00:54.520 ","End":"00:58.520","Text":"K_a2 is pretty large for weak acid,"},{"Start":"00:58.520 ","End":"01:00.080","Text":"but still it\u0027s a weak acid,"},{"Start":"01:00.080 ","End":"01:03.245","Text":"1.1 times 10_minus 2."},{"Start":"01:03.245 ","End":"01:08.315","Text":"We can say that sulfuric acid is different from most polyprotic acids."},{"Start":"01:08.315 ","End":"01:14.465","Text":"It acts as a strong acid in its 1st ionization and a weak acid in its second."},{"Start":"01:14.465 ","End":"01:16.220","Text":"Pretty strong, weak acid,"},{"Start":"01:16.220 ","End":"01:18.690","Text":"but nevertheless a weak acid."},{"Start":"01:19.210 ","End":"01:22.550","Text":"H_2SO_4 completely ionizes in"},{"Start":"01:22.550 ","End":"01:27.139","Text":"the 1st ionization so that none of the original acid remains,"},{"Start":"01:27.139 ","End":"01:29.510","Text":"it all turns into ions."},{"Start":"01:29.510 ","End":"01:31.340","Text":"Here\u0027s an example."},{"Start":"01:31.340 ","End":"01:37.860","Text":"Calculate the ion concentrations for 1 molar solution of sulfuric acid."},{"Start":"01:37.990 ","End":"01:40.190","Text":"In the 1st step,"},{"Start":"01:40.190 ","End":"01:46.265","Text":"all the sulfuric acid is converted into H_3O plus and HSO_4 minus."},{"Start":"01:46.265 ","End":"01:48.515","Text":"At the end of this process,"},{"Start":"01:48.515 ","End":"01:52.940","Text":"the concentration of H_2SO_4 is 0 and the concentration of"},{"Start":"01:52.940 ","End":"01:58.255","Text":"H_3O plus and HSO_4 minus are equal and they\u0027re both 1 molar."},{"Start":"01:58.255 ","End":"02:00.630","Text":"Now in the 2nd step,"},{"Start":"02:00.630 ","End":"02:07.955","Text":"HSO_4 minus reacts with water to give us H_3O plus and SO_4^2 minus."},{"Start":"02:07.955 ","End":"02:13.020","Text":"K_a2 is 1.1 times 10_ minus 2."},{"Start":"02:13.220 ","End":"02:17.355","Text":"Let\u0027s draw our ICE table."},{"Start":"02:17.355 ","End":"02:25.980","Text":"At the beginning, the concentration of HSO_4 minus and H_3O plus are both 1 molar."},{"Start":"02:25.980 ","End":"02:28.565","Text":"There\u0027s no SO_4^2 minus."},{"Start":"02:28.565 ","End":"02:34.520","Text":"The change necessary to reach equilibrium is minus x plus x and plus x."},{"Start":"02:34.520 ","End":"02:39.220","Text":"At equilibrium, we have 1 minus x,"},{"Start":"02:39.220 ","End":"02:46.830","Text":"1 plus x, and x."},{"Start":"02:46.830 ","End":"02:51.530","Text":"We can write K_a2 is equal to the concentration of H_3O plus,"},{"Start":"02:51.530 ","End":"02:53.075","Text":"that\u0027s 1 plus x,"},{"Start":"02:53.075 ","End":"02:57.350","Text":"times the concentration SO_4^2 minus, that\u0027s x,"},{"Start":"02:57.350 ","End":"03:00.110","Text":"divided by concentration HSO_4 minus,"},{"Start":"03:00.110 ","End":"03:02.245","Text":"that\u0027s 1 minus x."},{"Start":"03:02.245 ","End":"03:06.680","Text":"We have 1 plus x times x divided by 1 minus x."},{"Start":"03:06.680 ","End":"03:09.245","Text":"Now if we assume that x is very small,"},{"Start":"03:09.245 ","End":"03:12.420","Text":"x is less than 1,"},{"Start":"03:12.420 ","End":"03:19.110","Text":"we can neglect these two x\u0027s and we\u0027re left with 1 over 1,"},{"Start":"03:19.110 ","End":"03:21.435","Text":"which is 1, times x."},{"Start":"03:21.435 ","End":"03:23.855","Text":"All this is just equal to x."},{"Start":"03:23.855 ","End":"03:31.685","Text":"We know that x is equal to K_a2 and K_a2 is 1.1 times 10 to the power minus 2."},{"Start":"03:31.685 ","End":"03:36.170","Text":"We have x equal to 1.1 times 10_minus 2."},{"Start":"03:36.170 ","End":"03:44.515","Text":"We can write that the concentration of H_3O plus is 1 plus x, that\u0027s 1.01,"},{"Start":"03:44.515 ","End":"03:49.090","Text":"the concentration HSO_4 minus is 1 minus x,"},{"Start":"03:49.090 ","End":"03:59.175","Text":"that\u0027s 0.99, and the concentration of SO_4^2 minus is 0.011."},{"Start":"03:59.175 ","End":"04:02.750","Text":"Here we have all the concentrations."},{"Start":"04:02.750 ","End":"04:06.515","Text":"In this video, we discussed sulfuric acid,"},{"Start":"04:06.515 ","End":"04:09.750","Text":"which is a polyprotic acid."}],"ID":28471},{"Watched":false,"Name":"Exercise 1","Duration":"3m 53s","ChapterTopicVideoID":32002,"CourseChapterTopicPlaylistID":272182,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34278},{"Watched":false,"Name":"Exercise 2","Duration":"4m ","ChapterTopicVideoID":32003,"CourseChapterTopicPlaylistID":272182,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34279}],"Thumbnail":null,"ID":272182},{"Name":"Ions as Acids and Bases","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Ions as Acids and Bases","Duration":"7m 33s","ChapterTopicVideoID":27358,"CourseChapterTopicPlaylistID":272183,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.800","Text":"In a previous video,"},{"Start":"00:01.800 ","End":"00:05.235","Text":"we saw that ions could act as Bronsted acids or bases."},{"Start":"00:05.235 ","End":"00:08.340","Text":"In this video, we\u0027ll expand on this theme."},{"Start":"00:08.340 ","End":"00:12.720","Text":"We\u0027re going to talk about ions as acids and bases."},{"Start":"00:12.720 ","End":"00:16.035","Text":"An ion can behave as a Bronsted acid or base,"},{"Start":"00:16.035 ","End":"00:18.405","Text":"but not an Arrhenius acid or base."},{"Start":"00:18.405 ","End":"00:21.990","Text":"There they talked only about compounds."},{"Start":"00:21.990 ","End":"00:24.715","Text":"Here\u0027s the first example."},{"Start":"00:24.715 ","End":"00:35.500","Text":"Here the ammonium cation is going to act as an acid and ammonia is its conjugate base."},{"Start":"00:38.180 ","End":"00:48.210","Text":"In this case, water is going to act as a base and H_3O plus as its conjugate acid."},{"Start":"00:51.830 ","End":"00:56.415","Text":"Here it\u0027s written down ammonium is an acid,"},{"Start":"00:56.415 ","End":"00:59.510","Text":"ammonia is its conjugate base."},{"Start":"00:59.510 ","End":"01:01.745","Text":"Water acts as a base,"},{"Start":"01:01.745 ","End":"01:06.155","Text":"and the hydronium ion is it\u0027s conjugate acid."},{"Start":"01:06.155 ","End":"01:09.125","Text":"Here\u0027s our second example."},{"Start":"01:09.125 ","End":"01:17.630","Text":"Acetate reacts with water to give us hydroxide ion and acetic acid."},{"Start":"01:17.630 ","End":"01:22.370","Text":"Here, the acetate is acting as a base,"},{"Start":"01:22.370 ","End":"01:27.510","Text":"and acetic acid is it\u0027s conjugate acid."},{"Start":"01:31.610 ","End":"01:40.230","Text":"In this case, water is acting as an acid and OH minus is it\u0027s conjugate base."},{"Start":"01:40.360 ","End":"01:44.795","Text":"Here it\u0027s written acetate is a base,"},{"Start":"01:44.795 ","End":"01:49.315","Text":"and acetic acid is its conjugate acid."},{"Start":"01:49.315 ","End":"01:52.430","Text":"Water here acts as an acid,"},{"Start":"01:52.430 ","End":"01:56.245","Text":"and hydroxide is its conjugate base."},{"Start":"01:56.245 ","End":"01:59.360","Text":"Now let\u0027s try to find the relation between"},{"Start":"01:59.360 ","End":"02:05.730","Text":"the acid and base ionization constants between K_a and K_b."},{"Start":"02:08.570 ","End":"02:11.760","Text":"Let\u0027s go back to Example 1."},{"Start":"02:11.760 ","End":"02:15.595","Text":"Let\u0027s write K_a for this reaction."},{"Start":"02:15.595 ","End":"02:21.965","Text":"K_a is equal to the concentration of ammonia times the concentration of H_3O plus,"},{"Start":"02:21.965 ","End":"02:25.585","Text":"divided by the concentration of NH_4 plus."},{"Start":"02:25.585 ","End":"02:30.125","Text":"We\u0027re going to multiply the top and bottom by OH minus."},{"Start":"02:30.125 ","End":"02:36.600","Text":"Then we can see the H_3O plus times OH minus is K_w."},{"Start":"02:39.650 ","End":"02:46.605","Text":"Then we have NH_3 divided by NH_4 plus times OH minus."},{"Start":"02:46.605 ","End":"02:49.695","Text":"That\u0027s 1 over K_b."},{"Start":"02:49.695 ","End":"02:55.960","Text":"Here\u0027s K_b for ammonia and each 4 plus times OH minus divided by NH_3."},{"Start":"02:55.960 ","End":"02:58.840","Text":"We can see here we have NH_3,"},{"Start":"02:58.840 ","End":"03:01.465","Text":"the numerator instead of denominator,"},{"Start":"03:01.465 ","End":"03:06.610","Text":"NH_4 plus and OH minus in the denominator instead of the numerator."},{"Start":"03:06.610 ","End":"03:10.190","Text":"We have K_w divided by K_b."},{"Start":"03:10.190 ","End":"03:18.420","Text":"We know that K_b for ammonia is 1.8 times 10^-5."},{"Start":"03:18.420 ","End":"03:27.215","Text":"That means that K_a is equal to 1 times 10^-14 divided by K_b."},{"Start":"03:27.215 ","End":"03:31.040","Text":"K_b is 1.8 times 10^-5."},{"Start":"03:31.040 ","End":"03:37.260","Text":"That means that K_a is 5.6 times 10^-10."},{"Start":"03:38.780 ","End":"03:45.780","Text":"Now we can see that K_a is equal to K_w divided by K_b."},{"Start":"03:45.780 ","End":"03:52.140","Text":"That means that K_w is equal to K_a times K_b."},{"Start":"03:52.140 ","End":"03:57.815","Text":"So K_a for a particular acid times K_b for its conjugate base,"},{"Start":"03:57.815 ","End":"03:59.540","Text":"is equal to K_w."},{"Start":"03:59.540 ","End":"04:02.970","Text":"Very important relationship."},{"Start":"04:03.980 ","End":"04:10.714","Text":"We can see the conjugate of a weak acid is a weak base."},{"Start":"04:10.714 ","End":"04:14.915","Text":"This NH_4 plus is a weak acid,"},{"Start":"04:14.915 ","End":"04:19.580","Text":"and ammonia is a weak base."},{"Start":"04:19.580 ","End":"04:22.250","Text":"Now ammonia is a moderately strong weak base,"},{"Start":"04:22.250 ","End":"04:26.150","Text":"is 1.8 times 10^-5 for K_b."},{"Start":"04:26.150 ","End":"04:34.185","Text":"K_a is much smaller because the product of K_a and K_b has to be K_w."},{"Start":"04:34.185 ","End":"04:39.725","Text":"K_a we saw was 5.6 times 10^-10."},{"Start":"04:39.725 ","End":"04:45.235","Text":"Which is much smaller than K_b for ammonia."},{"Start":"04:45.235 ","End":"04:49.055","Text":"So ammonia is a moderately strong, weak base,"},{"Start":"04:49.055 ","End":"04:52.820","Text":"and the ammonium ion is a very weak acid."},{"Start":"04:52.820 ","End":"04:56.180","Text":"Let\u0027s look at the second example."},{"Start":"04:56.180 ","End":"05:01.190","Text":"This time we can write K_b for the acetate."},{"Start":"05:01.190 ","End":"05:07.220","Text":"K_b is equal to the concentration of acetic acid times"},{"Start":"05:07.220 ","End":"05:13.348","Text":"the concentration of OH minus divided by the concentration of acetate."},{"Start":"05:13.348 ","End":"05:18.450","Text":"Now, we can multiply the top and bottom by H_3O plus,"},{"Start":"05:20.780 ","End":"05:25.580","Text":"and then we can see that OH minus times H_3O plus,"},{"Start":"05:25.580 ","End":"05:29.805","Text":"this times this, that\u0027s equal to K_w."},{"Start":"05:29.805 ","End":"05:32.930","Text":"If we look at the rest, what remains,"},{"Start":"05:32.930 ","End":"05:38.405","Text":"we see that we have CH_3_CO_2H on the numerator."},{"Start":"05:38.405 ","End":"05:42.370","Text":"Whereas in K_a for acetic acid it\u0027s in the denominator."},{"Start":"05:42.370 ","End":"05:45.180","Text":"In the denominator for K_b,"},{"Start":"05:45.180 ","End":"05:48.720","Text":"we have CH_3_CO_2 minus times H_3O plus,"},{"Start":"05:48.720 ","End":"05:52.095","Text":"and these are in the numerator for K_a."},{"Start":"05:52.095 ","End":"05:58.440","Text":"Now we can write the K_b is equal to K_w divided by K_a."},{"Start":"05:58.440 ","End":"06:06.745","Text":"We can work out K_b is 1 times 10^-14 divided by 1.8 times 10^-5."},{"Start":"06:06.745 ","End":"06:08.570","Text":"It\u0027s the same numbers as before,"},{"Start":"06:08.570 ","End":"06:16.740","Text":"but that\u0027s just because acetic acid and ammonia have the same values for the K_a, K_b."},{"Start":"06:16.740 ","End":"06:22.680","Text":"Now we have K_b is equal to 5.6 times 10^-10."},{"Start":"06:22.680 ","End":"06:27.660","Text":"In this case, we can see the K_b times K_a is equal to K_w."},{"Start":"06:27.660 ","End":"06:33.910","Text":"So K_b for a base times K_a for conjugate acid is equal to K_w."},{"Start":"06:33.910 ","End":"06:36.555","Text":"We can see that in general,"},{"Start":"06:36.555 ","End":"06:38.345","Text":"K_a times K_b,"},{"Start":"06:38.345 ","End":"06:41.510","Text":"these are conjugate pair is equal to K_w."},{"Start":"06:41.510 ","End":"06:43.760","Text":"K_b times K_a,"},{"Start":"06:43.760 ","End":"06:46.825","Text":"where this are conjugate pair, is equal to K_w."},{"Start":"06:46.825 ","End":"06:51.760","Text":"We have K_a times K_b equal to K_w."},{"Start":"06:51.920 ","End":"06:58.084","Text":"We see here that the conjugate of a weak base is a weak acid."},{"Start":"06:58.084 ","End":"07:02.330","Text":"We saw that acetic acid is a moderately strong weak acid."},{"Start":"07:02.330 ","End":"07:05.719","Text":"The acetate ion is a very weak base."},{"Start":"07:05.719 ","End":"07:10.385","Text":"If we have a moderately strong weak acid or base,"},{"Start":"07:10.385 ","End":"07:18.930","Text":"the conjugate will be very weak because the multiplication of the two has to equal K_w."},{"Start":"07:18.930 ","End":"07:22.920","Text":"K_a is relatively large,"},{"Start":"07:22.920 ","End":"07:27.155","Text":"K_b will be relatively small, and vice versa."},{"Start":"07:27.155 ","End":"07:33.300","Text":"In this video, we saw that ions could act as Bronsted acids or bases."}],"ID":28474},{"Watched":false,"Name":"Hydrolysis of Ions and pH of Salt Solutions","Duration":"5m 28s","ChapterTopicVideoID":27357,"CourseChapterTopicPlaylistID":272183,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.740","Text":"In the previous video,"},{"Start":"00:01.740 ","End":"00:04.725","Text":"we showed that ions can act as acids and bases."},{"Start":"00:04.725 ","End":"00:08.280","Text":"In this video, we\u0027ll talk about the hydrolysis of ions,"},{"Start":"00:08.280 ","End":"00:10.605","Text":"and the pH of their solution."},{"Start":"00:10.605 ","End":"00:14.550","Text":"We\u0027re going to talk about the hydrolysis of ions."},{"Start":"00:14.550 ","End":"00:20.310","Text":"Now all salts that dissolve in water completely dissociate into ions,"},{"Start":"00:20.310 ","End":"00:23.460","Text":"and some of these ions react with water."},{"Start":"00:23.460 ","End":"00:28.545","Text":"Now the reaction of an ion with water is called hydrolysis."},{"Start":"00:28.545 ","End":"00:32.445","Text":"The ion is said to hydrolyze."},{"Start":"00:32.445 ","End":"00:36.990","Text":"Let\u0027s look at some examples of hydrolysis of ions,"},{"Start":"00:36.990 ","End":"00:40.130","Text":"and see if we can find out what the pH of the solutions of"},{"Start":"00:40.130 ","End":"00:43.675","Text":"the salts containing these ions is likely to be."},{"Start":"00:43.675 ","End":"00:46.605","Text":"Let\u0027s start with NaCl."},{"Start":"00:46.605 ","End":"00:50.000","Text":"Now when NaCl goes into water,"},{"Start":"00:50.000 ","End":"00:54.690","Text":"it completely dissolves, giving us Na+ and Cl-."},{"Start":"00:54.690 ","End":"00:57.495","Text":"But neither of these react with water,"},{"Start":"00:57.495 ","End":"01:01.230","Text":"now the Na+ or Cl- react with water,"},{"Start":"01:01.230 ","End":"01:03.840","Text":"so the pH stays 7,"},{"Start":"01:03.840 ","End":"01:06.700","Text":"just like the pH of water."},{"Start":"01:06.980 ","End":"01:11.540","Text":"We can say, that salts are derived from a strong acid,"},{"Start":"01:11.540 ","End":"01:13.375","Text":"in this case HCl,"},{"Start":"01:13.375 ","End":"01:15.610","Text":"and a strong base NaOH,"},{"Start":"01:15.610 ","End":"01:18.190","Text":"when these 2 react we get NaCl,"},{"Start":"01:18.190 ","End":"01:21.140","Text":"do not hydrolyze in water."},{"Start":"01:21.140 ","End":"01:25.370","Text":"Here\u0027s another example, NH_4Cl."},{"Start":"01:25.370 ","End":"01:29.120","Text":"Now when NH_4Cl dissolves in water,"},{"Start":"01:29.120 ","End":"01:32.795","Text":"we get NH_4+ and Cl-."},{"Start":"01:32.795 ","End":"01:35.560","Text":"Now Cl-, as we saw before,"},{"Start":"01:35.560 ","End":"01:37.775","Text":"does not react with water,"},{"Start":"01:37.775 ","End":"01:42.275","Text":"but NH_4 does, as we saw in the previous video."},{"Start":"01:42.275 ","End":"01:48.110","Text":"NH_4+ reacts with water to give us NH_3 and H_3O+."},{"Start":"01:48.110 ","End":"01:52.575","Text":"In this case NH_4+ is an acid,"},{"Start":"01:52.575 ","End":"01:55.995","Text":"and it hydrolyzes forming H_3O+,"},{"Start":"01:55.995 ","End":"01:58.485","Text":"so the solution is acidic,"},{"Start":"01:58.485 ","End":"02:02.025","Text":"with pH less than 7."},{"Start":"02:02.025 ","End":"02:04.095","Text":"We get ammonia,"},{"Start":"02:04.095 ","End":"02:06.820","Text":"which is a weak base,"},{"Start":"02:07.160 ","End":"02:12.580","Text":"and H_3O+ which is a strong acid,"},{"Start":"02:13.430 ","End":"02:16.310","Text":"so the solution is acidic."},{"Start":"02:16.310 ","End":"02:18.720","Text":"It goes with according to the strong acid,"},{"Start":"02:18.720 ","End":"02:21.860","Text":"so the pH will be less than 7."},{"Start":"02:21.860 ","End":"02:25.030","Text":"Again, we can form a rule,"},{"Start":"02:25.030 ","End":"02:27.830","Text":"that salts are derived from a strong acid."},{"Start":"02:27.830 ","End":"02:30.145","Text":"In this case it\u0027s HCl,"},{"Start":"02:30.145 ","End":"02:32.860","Text":"and a weak base NH_3,"},{"Start":"02:32.860 ","End":"02:38.210","Text":"form acidic solutions that goes as a strong acid."},{"Start":"02:38.210 ","End":"02:43.410","Text":"Of course the pH will be less than 7."},{"Start":"02:43.990 ","End":"02:48.475","Text":"Here\u0027s another example, sodium acetate."},{"Start":"02:48.475 ","End":"02:50.930","Text":"Now when this dissolves in water,"},{"Start":"02:50.930 ","End":"02:54.880","Text":"we get Na+ and CH_3CO_2-."},{"Start":"02:54.880 ","End":"02:57.930","Text":"Now we saw from Example 1,"},{"Start":"02:57.930 ","End":"03:00.935","Text":"that Na+ does not hydrolyze in water,"},{"Start":"03:00.935 ","End":"03:05.515","Text":"but CH_3CO_2- does react with water."},{"Start":"03:05.515 ","End":"03:08.535","Text":"We talked about this in the last video."},{"Start":"03:08.535 ","End":"03:10.815","Text":"Acetate reacts with water,"},{"Start":"03:10.815 ","End":"03:18.960","Text":"to give us OH- which is strongly basic and CH_3CO_2H,"},{"Start":"03:18.960 ","End":"03:22.559","Text":"which is weakly acidic."},{"Start":"03:22.559 ","End":"03:26.885","Text":"That means when the acetate hydrolyzes,"},{"Start":"03:26.885 ","End":"03:28.865","Text":"we get the OH-,"},{"Start":"03:28.865 ","End":"03:35.250","Text":"so the solution is basic with pH greater than 7."},{"Start":"03:36.520 ","End":"03:41.820","Text":"We can say that salts that derive from strong bases,"},{"Start":"03:41.820 ","End":"03:44.115","Text":"in this case it\u0027s NaOH,"},{"Start":"03:44.115 ","End":"03:45.735","Text":"and weak acids,"},{"Start":"03:45.735 ","End":"03:49.020","Text":"citric acid, form basic solutions."},{"Start":"03:49.020 ","End":"03:51.920","Text":"It goes according to the stronger of the 2,"},{"Start":"03:51.920 ","End":"03:55.290","Text":"so strong base and a weak acid,"},{"Start":"03:56.840 ","End":"04:01.780","Text":"so the pH will be greater than 7."},{"Start":"04:02.600 ","End":"04:10.000","Text":"Now what happens when we have salts derived from weak acids and weak bases?"},{"Start":"04:10.000 ","End":"04:13.870","Text":"Now these can be either acidic or basic,"},{"Start":"04:13.870 ","End":"04:18.830","Text":"depending on the relative strengths of the weak acid and weak base."},{"Start":"04:18.830 ","End":"04:21.695","Text":"If K_a is larger than K_b,"},{"Start":"04:21.695 ","End":"04:22.989","Text":"it will be acidic."},{"Start":"04:22.989 ","End":"04:25.240","Text":"If K_b is larger than K_a,"},{"Start":"04:25.240 ","End":"04:26.845","Text":"it will be basic."},{"Start":"04:26.845 ","End":"04:32.753","Text":"We need to look at the relative values of K_a and K_b."},{"Start":"04:32.753 ","End":"04:39.470","Text":"Now sometimes we have salts containing small highly charged metal ions."},{"Start":"04:39.470 ","End":"04:42.620","Text":"Now the metal ions in Groups 1 and 2,"},{"Start":"04:42.620 ","End":"04:45.355","Text":"usually do not affect the pH of a solution,"},{"Start":"04:45.355 ","End":"04:50.540","Text":"so Na+ we saw it before and say we have Ca^2+,"},{"Start":"04:50.540 ","End":"04:54.080","Text":"usually these do not affect the pH of the solution."},{"Start":"04:54.080 ","End":"04:58.580","Text":"However, sometimes we have metal ions with higher charges like aluminum,"},{"Start":"04:58.580 ","End":"04:59.870","Text":"chromium, iron,"},{"Start":"04:59.870 ","End":"05:07.125","Text":"and these can affect the pH of a solution via interaction with hydration shell."},{"Start":"05:07.125 ","End":"05:11.285","Text":"They interact with the hydration shell and that produces H_3O+,"},{"Start":"05:11.285 ","End":"05:14.510","Text":"so the solution is acidic."},{"Start":"05:14.510 ","End":"05:21.275","Text":"We\u0027ll talk more about that when we talk about Lewis acids and bases."},{"Start":"05:21.275 ","End":"05:28.260","Text":"In this video we talked about hydrolysis of ions and the pH of salt solutions."}],"ID":28473},{"Watched":false,"Name":"Calculating pH of a Salt Solution","Duration":"8m 5s","ChapterTopicVideoID":27359,"CourseChapterTopicPlaylistID":272183,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.175","Text":"In the previous video,"},{"Start":"00:02.175 ","End":"00:06.555","Text":"we talked about hydrolysis of ions and the pH of salt solutions."},{"Start":"00:06.555 ","End":"00:09.600","Text":"In this video, we\u0027ll calculate the pH of"},{"Start":"00:09.600 ","End":"00:14.715","Text":"an acidic salt solution and then of a basic salt solution."},{"Start":"00:14.715 ","End":"00:18.465","Text":"The first example is an acidic cation."},{"Start":"00:18.465 ","End":"00:26.295","Text":"The question is, what is the pH of 0.2 molar ammonium chloride?"},{"Start":"00:26.295 ","End":"00:30.090","Text":"Now, we know that the ammonium ion is acidic,"},{"Start":"00:30.090 ","End":"00:33.855","Text":"whereas the chloride ion does not react with water."},{"Start":"00:33.855 ","End":"00:37.800","Text":"We think that the solution should be acidic"},{"Start":"00:37.800 ","End":"00:47.305","Text":"because NH_4Cl dissolves to give us NH_4 plus and Cl minus,"},{"Start":"00:47.305 ","End":"00:50.940","Text":"and Cl minus doesn\u0027t react with water."},{"Start":"00:50.940 ","End":"00:53.990","Text":"Here\u0027s our equation, NH_4 plus,"},{"Start":"00:53.990 ","End":"00:58.670","Text":"plus water in equilibrium with NH_3 and H_3O plus."},{"Start":"00:58.670 ","End":"01:03.215","Text":"We can work out that the K_a for the ammonium ion,"},{"Start":"01:03.215 ","End":"01:10.970","Text":"NH_4 plus is equal to K_w divided by K_b of its conjugate base."},{"Start":"01:10.970 ","End":"01:17.260","Text":"NH_4 plus and NH_3 are conjugate to each other."},{"Start":"01:18.980 ","End":"01:24.825","Text":"K_a of ammonium ion is K_w divided by K_b."},{"Start":"01:24.825 ","End":"01:30.650","Text":"That\u0027s 1 times 10^-14 divided by 1.8 times 10^-5,"},{"Start":"01:30.650 ","End":"01:36.030","Text":"that gives us 5.6 times 10^-10."},{"Start":"01:36.640 ","End":"01:40.940","Text":"Now we can go on with the second part,"},{"Start":"01:40.940 ","End":"01:44.315","Text":"which is NH_4 plus reacting with water."},{"Start":"01:44.315 ","End":"01:49.610","Text":"The first part is the formation of NH_4 plus and Cl minus and water."},{"Start":"01:49.610 ","End":"01:56.240","Text":"Now the NH_4 plus reacts with water to give us NH_3 and H_3O plus."},{"Start":"01:56.240 ","End":"02:01.790","Text":"We can solve this as an equilibrium problem."},{"Start":"02:01.790 ","End":"02:06.005","Text":"We have NH_4 plus is 0.2 molar,"},{"Start":"02:06.005 ","End":"02:09.935","Text":"because all the ammonium chloride dissolved,"},{"Start":"02:09.935 ","End":"02:16.684","Text":"and NH_3, 0 at the beginning and H_3O plus 0."},{"Start":"02:16.684 ","End":"02:23.690","Text":"Now, we can write the change that leads to equilibrium as minus x,"},{"Start":"02:23.690 ","End":"02:25.340","Text":"where NH_4 plus,"},{"Start":"02:25.340 ","End":"02:30.935","Text":"plus x for NH_3 and plus x for H_3O plus."},{"Start":"02:30.935 ","End":"02:41.950","Text":"At equilibrium, we have 0.2 minus x plus x for ammonia and plus x for H_3O plus."},{"Start":"02:41.950 ","End":"02:48.820","Text":"K_a, which we discovered is 5.6 times 10^-10,"},{"Start":"02:48.820 ","End":"02:52.610","Text":"and we can write that as the concentration of ammonia times"},{"Start":"02:52.610 ","End":"02:57.275","Text":"the concentration ratio plus divided by the concentration of NH_4 plus,"},{"Start":"02:57.275 ","End":"03:07.534","Text":"that\u0027s x^2 x times x divided by 0.2 minus x for the ammonia."},{"Start":"03:07.534 ","End":"03:10.760","Text":"We have x^2 divided by 0.2 minus x."},{"Start":"03:10.760 ","End":"03:16.640","Text":"If we make an approximation that x is much smaller than 0.2,"},{"Start":"03:16.640 ","End":"03:20.870","Text":"we\u0027ll get x^2 divided by 0.2."},{"Start":"03:20.870 ","End":"03:27.515","Text":"That x^2 is 0.2 times 5.6 times 10^-10."},{"Start":"03:27.515 ","End":"03:32.855","Text":"That\u0027s 1.12 times 10^-10."},{"Start":"03:32.855 ","End":"03:39.560","Text":"We can work out that x is the square root of 1.12 times 10^-10,"},{"Start":"03:39.560 ","End":"03:45.005","Text":"and that\u0027s 1.06 times 10^-5."},{"Start":"03:45.005 ","End":"03:55.223","Text":"Since x is the concentration of H_3O plus,"},{"Start":"03:55.223 ","End":"04:01.455","Text":"we can find out the pH and that\u0027s 4.97."},{"Start":"04:01.455 ","End":"04:09.435","Text":"The pH of our solution of NH_4Cl is 4.97,"},{"Start":"04:09.435 ","End":"04:12.850","Text":"which is clearly less than 7."},{"Start":"04:14.790 ","End":"04:17.965","Text":"As we thought at the beginning,"},{"Start":"04:17.965 ","End":"04:21.050","Text":"we have an acidic solution."},{"Start":"04:22.770 ","End":"04:25.675","Text":"Now let us take another example."},{"Start":"04:25.675 ","End":"04:29.025","Text":"This time of a basic anion."},{"Start":"04:29.025 ","End":"04:39.581","Text":"The question is, what is the pH of 0.2 molar NaCH_3CO_2, sodium acetate?"},{"Start":"04:39.581 ","End":"04:43.730","Text":"Now, we can expect that as the acetate ion"},{"Start":"04:43.730 ","End":"04:48.215","Text":"is basic and the sodium ion doesn\u0027t react with water,"},{"Start":"04:48.215 ","End":"04:50.450","Text":"the solution should be basic."},{"Start":"04:50.450 ","End":"04:52.340","Text":"Let\u0027s see if that\u0027s correct."},{"Start":"04:52.340 ","End":"04:54.635","Text":"Here\u0027s our equation,"},{"Start":"04:54.635 ","End":"04:59.683","Text":"acetate reacting with water to give us OH minus and acetic acid."},{"Start":"04:59.683 ","End":"05:10.210","Text":"The K_b for the acidic acetate ion is just K_w divided by K_a of its conjugate acid."},{"Start":"05:10.210 ","End":"05:16.200","Text":"Here we have a base and its conjugate acid."},{"Start":"05:18.620 ","End":"05:24.995","Text":"K_b is equal to K_w divided by K_a is the same numbers as before,"},{"Start":"05:24.995 ","End":"05:32.115","Text":"1 times 10^-14 for K_w divided by 1.8 times 10^-5 for K_a."},{"Start":"05:32.115 ","End":"05:38.410","Text":"That gives us 5.6 times 10^-10."},{"Start":"05:39.030 ","End":"05:47.950","Text":"The first step, the sodium acetate completely dissociated to Na plus and acetate ion."},{"Start":"05:47.950 ","End":"05:50.409","Text":"Now, since we have an equilibrium,"},{"Start":"05:50.409 ","End":"05:53.095","Text":"we can do an equilibrium calculation."},{"Start":"05:53.095 ","End":"05:57.460","Text":"Initially, the concentration of the acetate is 0.2"},{"Start":"05:57.460 ","End":"06:02.645","Text":"molar and OH minus 0 and acetic acid is 0."},{"Start":"06:02.645 ","End":"06:07.820","Text":"That the change towards the equilibrium is minus x plus x plus x."},{"Start":"06:07.820 ","End":"06:13.885","Text":"At equilibrium we have 0.2 minus x for the acetate,"},{"Start":"06:13.885 ","End":"06:16.380","Text":"plus x for OH minus,"},{"Start":"06:16.380 ","End":"06:19.365","Text":"and plus x for acetic acid."},{"Start":"06:19.365 ","End":"06:23.605","Text":"We can write the K_b for this reaction."},{"Start":"06:23.605 ","End":"06:27.295","Text":"That\u0027s equal to 5.6 times 10^-10."},{"Start":"06:27.295 ","End":"06:30.545","Text":"That\u0027s the concentration of acetic acid times that of"},{"Start":"06:30.545 ","End":"06:35.285","Text":"the OH minus divided by concentration of acetate."},{"Start":"06:35.285 ","End":"06:41.415","Text":"That\u0027s x times x x^2 divided by 0.2 minus x."},{"Start":"06:41.415 ","End":"06:46.670","Text":"Again, neglecting the x compared to 0.2,"},{"Start":"06:46.670 ","End":"06:50.410","Text":"we get x^2 divided by 0.2."},{"Start":"06:50.410 ","End":"06:56.295","Text":"So x^2 is 0.2 times 5.6 times 10^-10."},{"Start":"06:56.295 ","End":"07:00.295","Text":"That\u0027s 1.12 times 10^-10,"},{"Start":"07:00.295 ","End":"07:02.630","Text":"and x is the square root of that,"},{"Start":"07:02.630 ","End":"07:06.710","Text":"that\u0027s 1.06 times 10^-5."},{"Start":"07:06.710 ","End":"07:13.200","Text":"Now, this time x is the concentration of OH minus."},{"Start":"07:16.850 ","End":"07:23.775","Text":"What we calculate from x is pOH not pH as before."},{"Start":"07:23.775 ","End":"07:29.905","Text":"We saw that when we calculate it, it\u0027s 4.97."},{"Start":"07:29.905 ","End":"07:31.794","Text":"This time it\u0027s a pOH,"},{"Start":"07:31.794 ","End":"07:35.985","Text":"that\u0027s 4.97, not the pH."},{"Start":"07:35.985 ","End":"07:38.595","Text":"Now we can calculate the pH."},{"Start":"07:38.595 ","End":"07:45.510","Text":"The pH is 14 minus the pOH."},{"Start":"07:45.510 ","End":"07:49.660","Text":"That gives us 9.03."},{"Start":"07:49.660 ","End":"07:54.930","Text":"As we expected, this solution is basic."},{"Start":"07:57.310 ","End":"08:04.980","Text":"In this video, we\u0027ll calculated the pH of acidic and basic salt solutions."}],"ID":28475},{"Watched":false,"Name":"Exercise 1","Duration":"7m 21s","ChapterTopicVideoID":32007,"CourseChapterTopicPlaylistID":272183,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34281},{"Watched":false,"Name":"Exercise 2","Duration":"4m 15s","ChapterTopicVideoID":32005,"CourseChapterTopicPlaylistID":272183,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34282},{"Watched":false,"Name":"Exercise 3","Duration":"4m 22s","ChapterTopicVideoID":32006,"CourseChapterTopicPlaylistID":272183,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34283}],"Thumbnail":null,"ID":272183},{"Name":"General Approach to Acid Base Calculations","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Acid-Base Calculations Without Approximations","Duration":"6m 26s","ChapterTopicVideoID":27360,"CourseChapterTopicPlaylistID":272184,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/27360.jpeg","UploadDate":"2022-02-08T09:05:19.8970000","DurationForVideoObject":"PT6M26S","Description":null,"MetaTitle":"Acid-Base Calculations Without Approximations: Video + Workbook | Proprep","MetaDescription":"Acids and Bases - General Approach to Acid Base Calculations. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/general-chemistry/acids-and-bases/general-approach-to-acid-base-calculations/vid28476","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.520","Text":"In previous videos, we use the approximations to solve acids and base problems."},{"Start":"00:05.520 ","End":"00:08.790","Text":"In this video, we\u0027ll outline a strategy for"},{"Start":"00:08.790 ","End":"00:12.810","Text":"solving such problems without making approximations."},{"Start":"00:12.810 ","End":"00:17.985","Text":"We\u0027re going to outline a strategy for solving acid-base problems."},{"Start":"00:17.985 ","End":"00:20.190","Text":"Here\u0027s the first step."},{"Start":"00:20.190 ","End":"00:25.455","Text":"Write down all the species involved apart from water."},{"Start":"00:25.455 ","End":"00:31.215","Text":"Then write down the balanced equations for all the reactions involved."},{"Start":"00:31.215 ","End":"00:38.000","Text":"Step 2, we\u0027re going to assume that the concentrations are our unknowns."},{"Start":"00:38.000 ","End":"00:41.409","Text":"These are the quantities we want to determine."},{"Start":"00:41.409 ","End":"00:46.710","Text":"We\u0027re going to write equations for the concentrations of all species,"},{"Start":"00:46.710 ","End":"00:52.105","Text":"and we know that the number of equations required is equal to the number of unknowns,"},{"Start":"00:52.105 ","End":"00:57.500","Text":"if we want to find all the absolute values of the unknowns."},{"Start":"00:57.500 ","End":"01:01.730","Text":"Now, what types of equations can we use?"},{"Start":"01:01.730 ","End":"01:06.570","Text":"We can use the expressions for equilibrium constants,"},{"Start":"01:06.830 ","End":"01:10.992","Text":"we can use the equation for the material balance,"},{"Start":"01:10.992 ","End":"01:15.665","Text":"and we can use an equation for charge balance."},{"Start":"01:15.665 ","End":"01:17.810","Text":"Let\u0027s consider an example."},{"Start":"01:17.810 ","End":"01:19.790","Text":"The example is phosphoric acid,"},{"Start":"01:19.790 ","End":"01:24.890","Text":"which we talked about before, using many approximations."},{"Start":"01:24.890 ","End":"01:29.870","Text":"The question is to determine all the ion concentrations for"},{"Start":"01:29.870 ","End":"01:35.269","Text":"0.1 molar aqueous solution of phosphoric acid."},{"Start":"01:35.269 ","End":"01:37.595","Text":"Now there are 6 unknowns."},{"Start":"01:37.595 ","End":"01:39.115","Text":"What are these?"},{"Start":"01:39.115 ","End":"01:46.115","Text":"H_3O plus concentration, H_3PO_4 concentration, that\u0027s phosphoric acid."},{"Start":"01:46.115 ","End":"01:52.400","Text":"Then we have the various ions that derive from phosphoric acid,"},{"Start":"01:52.400 ","End":"01:55.169","Text":"when we remove 1 hydrogen,"},{"Start":"01:55.169 ","End":"01:57.585","Text":"2 hydrogens, and 3 hydrogens."},{"Start":"01:57.585 ","End":"01:59.310","Text":"Here we\u0027ve removed 1,"},{"Start":"01:59.310 ","End":"02:01.095","Text":"here we\u0027ve removed 2,"},{"Start":"02:01.095 ","End":"02:03.790","Text":"and here we\u0027ve removed all 3."},{"Start":"02:03.790 ","End":"02:10.145","Text":"We see that the charge goes from minus to 2 minus to 3 minus."},{"Start":"02:10.145 ","End":"02:17.045","Text":"Of course, just as we have H_3O plus we always have OH minus."},{"Start":"02:17.045 ","End":"02:20.465","Text":"Here are our 6 equations."},{"Start":"02:20.465 ","End":"02:24.575","Text":"Start with the expressions for the equilibrium constants."},{"Start":"02:24.575 ","End":"02:26.465","Text":"Here\u0027s the first one,"},{"Start":"02:26.465 ","End":"02:33.245","Text":"for phosphoric acid reacting with water to give us H_2PO_4 minus."},{"Start":"02:33.245 ","End":"02:35.950","Text":"We can work out K_a1,"},{"Start":"02:35.950 ","End":"02:39.420","Text":"whose dealt with all this previously."},{"Start":"02:39.420 ","End":"02:45.105","Text":"That\u0027s H_3O plus times H_2PO_4 minus divide base H_3O_4,"},{"Start":"02:45.105 ","End":"02:49.610","Text":"and the volume is 7.1 times 10^-3."},{"Start":"02:49.610 ","End":"02:56.775","Text":"Then the H_2PO_4 minus reacts with water to lose one of its hydrogens,"},{"Start":"02:56.775 ","End":"03:01.500","Text":"so then we have H_3O plus and HPO_4 2 minus,"},{"Start":"03:01.500 ","End":"03:03.315","Text":"and the K_a_2 for this,"},{"Start":"03:03.315 ","End":"03:07.260","Text":"is 6.3 times 10^-8."},{"Start":"03:07.260 ","End":"03:11.395","Text":"Then the HPO_4 2 minus reacts with water,"},{"Start":"03:11.395 ","End":"03:13.765","Text":"again loses a proton."},{"Start":"03:13.765 ","End":"03:15.690","Text":"The only remaining proton,"},{"Start":"03:15.690 ","End":"03:19.035","Text":"to give us H_3O plus and PO_4 3 minus,"},{"Start":"03:19.035 ","End":"03:25.605","Text":"and K_a_3 for this is 4.2 times 10^-13."},{"Start":"03:25.605 ","End":"03:33.330","Text":"Each time we go down by a factor of 10^-5,"},{"Start":"03:33.330 ","End":"03:36.690","Text":"K_a_2 is much smaller than K_a_1."},{"Start":"03:36.690 ","End":"03:40.270","Text":"K_a_3 is much smaller than K_a_2."},{"Start":"03:40.270 ","End":"03:46.310","Text":"The fourth equation can be the self ionization of water,"},{"Start":"03:46.310 ","End":"03:51.595","Text":"and that is K_w equal to 10^-14."},{"Start":"03:51.595 ","End":"03:55.725","Text":"We have 4 equations so far for."},{"Start":"03:55.725 ","End":"03:58.230","Text":"Now, we can have more equations,"},{"Start":"03:58.230 ","End":"04:01.860","Text":"one from the equation for material balance,"},{"Start":"04:01.860 ","End":"04:03.890","Text":"there\u0027s the fifth equation."},{"Start":"04:03.890 ","End":"04:10.670","Text":"Now the initial concentration of H_3PO_4 phosphoric acid was 0.1."},{"Start":"04:10.670 ","End":"04:17.510","Text":"After ionization, the phosphoric acid gives us H_3PO_4,"},{"Start":"04:17.510 ","End":"04:21.965","Text":"whatever is remaining plus each of the ions"},{"Start":"04:21.965 ","End":"04:28.265","Text":"H_2PO_4 minus HPO_4 2 minus and PO_4 3 minus."},{"Start":"04:28.265 ","End":"04:31.880","Text":"The concentration of all these ions together must be"},{"Start":"04:31.880 ","End":"04:36.005","Text":"equal to the initial concentration of H_3PO_4,"},{"Start":"04:36.005 ","End":"04:38.855","Text":"which was 0.1 molar."},{"Start":"04:38.855 ","End":"04:41.290","Text":"So that\u0027s our fifth equation."},{"Start":"04:41.290 ","End":"04:46.385","Text":"Sixth equation comes from the equation for charge balance."},{"Start":"04:46.385 ","End":"04:49.445","Text":"Now our solution is neutral."},{"Start":"04:49.445 ","End":"04:55.745","Text":"The number of positive charges must be equal to the number of negative charges."},{"Start":"04:55.745 ","End":"05:00.875","Text":"The only positively charged ion is H_3O plus,"},{"Start":"05:00.875 ","End":"05:03.715","Text":"and that has a charge of 1."},{"Start":"05:03.715 ","End":"05:11.010","Text":"Then the anions we have H_2PO_4 minus has a charge of 1."},{"Start":"05:11.090 ","End":"05:16.035","Text":"HPO_4 2 minus a charge of 2."},{"Start":"05:16.035 ","End":"05:19.200","Text":"Of course, 1 minus and 2 minus,"},{"Start":"05:19.200 ","End":"05:26.575","Text":"we multiply it by 2 times the concentration to get the number of charges."},{"Start":"05:26.575 ","End":"05:29.260","Text":"We\u0027re looking for the concentration of charges,"},{"Start":"05:29.260 ","End":"05:33.415","Text":"and PO_4 3 minus has a charge of 3 minus,"},{"Start":"05:33.415 ","End":"05:36.950","Text":"so that\u0027s 3 times its concentration."},{"Start":"05:36.950 ","End":"05:41.285","Text":"OH minuses charge of just 1."},{"Start":"05:41.285 ","End":"05:45.940","Text":"We add all the negatively charged ions"},{"Start":"05:45.940 ","End":"05:50.170","Text":"that has to be equal to the positively charged ones."},{"Start":"05:50.170 ","End":"05:54.710","Text":"So this gives us our sixth equation."},{"Start":"05:54.710 ","End":"05:58.460","Text":"Now how can we solve such a problem?"},{"Start":"05:58.460 ","End":"06:04.700","Text":"Well, we can go to our computer and try to get a numerical solution."},{"Start":"06:04.700 ","End":"06:08.390","Text":"That\u0027s without making any approximations."},{"Start":"06:08.390 ","End":"06:13.370","Text":"Or if we want to solve the problem by hand,"},{"Start":"06:13.370 ","End":"06:17.000","Text":"then we have to introduce approximations."},{"Start":"06:17.000 ","End":"06:18.830","Text":"In this video,"},{"Start":"06:18.830 ","End":"06:25.860","Text":"we discussed how to solve an acid-base problem without making any approximations."}],"ID":28476},{"Watched":false,"Name":"Exercise 1","Duration":"4m 28s","ChapterTopicVideoID":32004,"CourseChapterTopicPlaylistID":272184,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34280}],"Thumbnail":null,"ID":272184},{"Name":"Acid Base Reactions","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Qualitative Aspects of Acid-Base Reactions","Duration":"7m 39s","ChapterTopicVideoID":27361,"CourseChapterTopicPlaylistID":272185,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.835","Text":"In previous videos we talked about aqueous solutions of acids and bases."},{"Start":"00:04.835 ","End":"00:10.245","Text":"In this video we\u0027ll talk about qualitative aspects of acid-base reactions."},{"Start":"00:10.245 ","End":"00:14.130","Text":"We\u0027re going to talk about acid-base reactions."},{"Start":"00:14.130 ","End":"00:18.360","Text":"We\u0027re going to indicate an acid-base reaction by HA,"},{"Start":"00:18.360 ","End":"00:21.345","Text":"which is a typical acid,"},{"Start":"00:21.345 ","End":"00:28.215","Text":"reacting with B which is a base to give us BH plus and A minus."},{"Start":"00:28.215 ","End":"00:36.080","Text":"HA and A minus are conjugate acid-base pair."},{"Start":"00:36.080 ","End":"00:39.710","Text":"B and BH plus are also"},{"Start":"00:39.710 ","End":"00:44.840","Text":"a conjugate acid-base pair or base-acid pair because it starts with the base,"},{"Start":"00:44.840 ","End":"00:46.750","Text":"but it\u0027s the same thing."},{"Start":"00:46.750 ","End":"00:51.260","Text":"Here we\u0027ve written it HA and A minus a conjugate acid-base pair,"},{"Start":"00:51.260 ","End":"00:54.800","Text":"B and BH plus a conjugate base-acid pair."},{"Start":"00:54.800 ","End":"00:57.560","Text":"Now we\u0027re going to write the equilibrium constant for"},{"Start":"00:57.560 ","End":"01:03.530","Text":"an acid-base reaction in terms of three separate processes."},{"Start":"01:03.530 ","End":"01:11.135","Text":"The first is the reaction of the acid with water to give us H_3_O plus and A minus."},{"Start":"01:11.135 ","End":"01:15.260","Text":"So here\u0027s our acid HA,"},{"Start":"01:15.260 ","End":"01:19.800","Text":"and here\u0027s our conjugate base A minus."},{"Start":"01:19.800 ","End":"01:27.920","Text":"We can write that K_a(HA) is equal to K_w divided by K_b(A minus)."},{"Start":"01:27.920 ","End":"01:35.170","Text":"This is from the expression we learned before that K_a times K_b=K_w."},{"Start":"01:35.860 ","End":"01:45.740","Text":"Second process involves the base reacting with water to give us BH plus and OH minus."},{"Start":"01:45.740 ","End":"01:52.770","Text":"So B is the base and BH plus is a conjugate acid."},{"Start":"01:53.960 ","End":"02:02.495","Text":"We can write K_b for the base is equal to K_w divided by K_a for the conjugate acid."},{"Start":"02:02.495 ","End":"02:07.355","Text":"So we have base BH plus is the conjugate acid."},{"Start":"02:07.355 ","End":"02:12.860","Text":"The third expression is just the self ionization of water."},{"Start":"02:12.860 ","End":"02:17.450","Text":"H_3_O plus plus OH minus gives us 2 H_2_O."},{"Start":"02:17.450 ","End":"02:20.000","Text":"Since we\u0027ve written this backwards,"},{"Start":"02:20.000 ","End":"02:22.520","Text":"not the self ionization of water,"},{"Start":"02:22.520 ","End":"02:24.783","Text":"but the inverse,"},{"Start":"02:24.783 ","End":"02:28.300","Text":"we can write this as 1 divided by K_w."},{"Start":"02:28.300 ","End":"02:32.815","Text":"So if we sum these 3 reactions,"},{"Start":"02:32.815 ","End":"02:35.180","Text":"we have water,"},{"Start":"02:35.180 ","End":"02:39.305","Text":"twice H_2_O and H_2_O and here\u0027s 2 H_2_O on the other side."},{"Start":"02:39.305 ","End":"02:42.479","Text":"These cancel."},{"Start":"02:42.479 ","End":"02:47.090","Text":"Now let\u0027s look at H_3_O plus cancels with H_3_O"},{"Start":"02:47.090 ","End":"02:52.550","Text":"plus here and O minus cancels with OH minus here."},{"Start":"02:52.550 ","End":"02:57.140","Text":"We\u0027re left with HA plus B,"},{"Start":"02:57.140 ","End":"03:00.980","Text":"giving BH plus and A minus."},{"Start":"03:00.980 ","End":"03:04.715","Text":"Now, how do we write the equilibrium constant for this?"},{"Start":"03:04.715 ","End":"03:08.165","Text":"So it\u0027s the product of the 3 reactions."},{"Start":"03:08.165 ","End":"03:15.674","Text":"It\u0027s K_a(HA) times K_b(B) divided by K_w."},{"Start":"03:15.674 ","End":"03:17.990","Text":"Now, let\u0027s rewrite this expression."},{"Start":"03:17.990 ","End":"03:24.560","Text":"K is equal to K_a(HA) times K_b(B) divided by K_w."},{"Start":"03:24.560 ","End":"03:28.784","Text":"We\u0027re going to rewrite K_a(HA) by this expression,"},{"Start":"03:28.784 ","End":"03:32.110","Text":"K_w divided by K_b."},{"Start":"03:32.450 ","End":"03:39.060","Text":"Now we have K_w divided by K_b times K_b divided by K_w."},{"Start":"03:39.060 ","End":"03:40.875","Text":"K_w cancels."},{"Start":"03:40.875 ","End":"03:45.185","Text":"We get K_b(B) divided by K_b(A minus)."},{"Start":"03:45.185 ","End":"03:48.930","Text":"So here we\u0027re comparing 2 bases,"},{"Start":"03:48.930 ","End":"03:51.858","Text":"B and A minus."},{"Start":"03:51.858 ","End":"03:56.280","Text":"Now, if instead of rewriting K_a(HA),"},{"Start":"03:56.280 ","End":"03:58.680","Text":"we write K_b(B),"},{"Start":"03:58.680 ","End":"04:00.870","Text":"then instead of writing K_b(B),"},{"Start":"04:00.870 ","End":"04:08.140","Text":"we can write K_b(B) is equal to K_w divided by K_a(BH plus)."},{"Start":"04:08.210 ","End":"04:10.890","Text":"When we do that,"},{"Start":"04:10.890 ","End":"04:15.765","Text":"we get K_w divided by K_a(BH plus)."},{"Start":"04:15.765 ","End":"04:17.820","Text":"The K_ws cancel."},{"Start":"04:17.820 ","End":"04:24.220","Text":"So we get K_a(HA) divided K_a(BH plus)."},{"Start":"04:25.400 ","End":"04:28.850","Text":"We have 2 equivalent expressions for"},{"Start":"04:28.850 ","End":"04:33.770","Text":"K. The equilibrium constant of the acid-base reaction,"},{"Start":"04:33.770 ","End":"04:40.510","Text":"1 in terms of bases and 1 in terms of acids."},{"Start":"04:40.510 ","End":"04:43.744","Text":"Let\u0027s first look at the acids."},{"Start":"04:43.744 ","End":"04:50.510","Text":"Supposing K_a(HA) is greater than K_a(BH plus),"},{"Start":"04:50.510 ","End":"04:54.550","Text":"then K will be greater than 1."},{"Start":"04:54.550 ","End":"04:56.940","Text":"If the reverse is true,"},{"Start":"04:56.940 ","End":"05:00.330","Text":"if K_a(HA) is less than K_a(BH plus),"},{"Start":"05:00.330 ","End":"05:02.715","Text":"then K will be less than 1."},{"Start":"05:02.715 ","End":"05:05.170","Text":"If K is greater than 1,"},{"Start":"05:05.170 ","End":"05:11.025","Text":"then the equilibrium will be towards the products here."},{"Start":"05:11.025 ","End":"05:13.850","Text":"Whereas if K is less than 1,"},{"Start":"05:13.850 ","End":"05:18.150","Text":"the equilibrium will be to the side of the reactants."},{"Start":"05:18.310 ","End":"05:24.499","Text":"We can say that the reaction favors the formation of the weaker acid."},{"Start":"05:24.499 ","End":"05:28.895","Text":"Because in the first case where it\u0027s towards the products,"},{"Start":"05:28.895 ","End":"05:33.020","Text":"then it\u0027s towards BH plus,"},{"Start":"05:33.020 ","End":"05:35.510","Text":"which is weaker than HA."},{"Start":"05:35.510 ","End":"05:37.145","Text":"In the second case,"},{"Start":"05:37.145 ","End":"05:40.025","Text":"where it\u0027s towards the reactants,"},{"Start":"05:40.025 ","End":"05:44.990","Text":"then the equilibrium will be more than the direction of HA,"},{"Start":"05:44.990 ","End":"05:48.365","Text":"and this case HA is a weaker acid."},{"Start":"05:48.365 ","End":"05:53.244","Text":"The reaction favors the formation of the weaker acid."},{"Start":"05:53.244 ","End":"05:56.255","Text":"Now if HA is a very strong acid,"},{"Start":"05:56.255 ","End":"06:00.275","Text":"and instead of K greater than 1 or K much greater than 1,"},{"Start":"06:00.275 ","End":"06:03.685","Text":"then the reaction will go to completion."},{"Start":"06:03.685 ","End":"06:06.180","Text":"Let\u0027s talk about the bases."},{"Start":"06:06.180 ","End":"06:12.135","Text":"If K_b(B) is greater than K_b(A minus),"},{"Start":"06:12.135 ","End":"06:14.065","Text":"then K will be greater than 1."},{"Start":"06:14.065 ","End":"06:18.245","Text":"The reaction will go towards the products."},{"Start":"06:18.245 ","End":"06:26.630","Text":"If K_b(B) is less than K_b(A minus) the reaction will go towards the reactants."},{"Start":"06:26.630 ","End":"06:29.990","Text":"Because K will be less than 1."},{"Start":"06:29.990 ","End":"06:35.467","Text":"We can say that the reaction favors the formation of the weaker base."},{"Start":"06:35.467 ","End":"06:41.090","Text":"In this case, A minus reaction will go towards A minus,"},{"Start":"06:41.090 ","End":"06:44.274","Text":"and A minus is the weaker base."},{"Start":"06:44.274 ","End":"06:49.430","Text":"In this case it will go towards B and B is the weaker base."},{"Start":"06:49.430 ","End":"06:54.145","Text":"We can say the reaction favors the formation of the weaker base."},{"Start":"06:54.145 ","End":"06:59.420","Text":"If B is a very strong base instead of K greater than 1,"},{"Start":"06:59.420 ","End":"07:02.725","Text":"reaction will go to completion."},{"Start":"07:02.725 ","End":"07:06.249","Text":"We can formulate 2 rules."},{"Start":"07:06.249 ","End":"07:14.060","Text":"We can say the reaction goes into the direction that leads to weak acid and weak base."},{"Start":"07:14.060 ","End":"07:19.910","Text":"We can say that if a strong acid or base is involved in the reaction,"},{"Start":"07:19.910 ","End":"07:22.415","Text":"the reaction will go to completion."},{"Start":"07:22.415 ","End":"07:27.110","Text":"This is something we\u0027ll use a great deal in the next chapter."},{"Start":"07:27.110 ","End":"07:30.555","Text":"Strong base or acid is involved,"},{"Start":"07:30.555 ","End":"07:32.450","Text":"reaction will go to completion."},{"Start":"07:32.450 ","End":"07:39.300","Text":"In this video, we talked about qualitative aspects of acid-base reactions."}],"ID":28477},{"Watched":false,"Name":"Exercise 1","Duration":"6m 5s","ChapterTopicVideoID":32012,"CourseChapterTopicPlaylistID":272185,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34286}],"Thumbnail":null,"ID":272185},{"Name":"Lewis Acids and Bases","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Lewis Acids and Bases","Duration":"6m 26s","ChapterTopicVideoID":27362,"CourseChapterTopicPlaylistID":272186,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.980","Text":"In previous videos,"},{"Start":"00:01.980 ","End":"00:05.805","Text":"we defined Arrhenius and Bronsted-Lowry acids and bases."},{"Start":"00:05.805 ","End":"00:10.590","Text":"In this video, we\u0027ll talk about Lewis acids and bases."},{"Start":"00:10.590 ","End":"00:14.384","Text":"We\u0027re going to talk about Lewis acids and bases."},{"Start":"00:14.384 ","End":"00:15.990","Text":"As we\u0027ve said before,"},{"Start":"00:15.990 ","End":"00:19.725","Text":"a Lewis acid is an electron pair acceptor,"},{"Start":"00:19.725 ","End":"00:23.195","Text":"and a Lewis base is an electron pair donor."},{"Start":"00:23.195 ","End":"00:26.580","Text":"There are a pair, an acceptor, and a donor."},{"Start":"00:26.580 ","End":"00:30.975","Text":"Now the definition is more general than that of Bronsted and Lowry."},{"Start":"00:30.975 ","End":"00:35.355","Text":"Every Lewis acid or base is also Bronsted acid or base,"},{"Start":"00:35.355 ","End":"00:37.590","Text":"but not vice versa."},{"Start":"00:37.590 ","End":"00:40.515","Text":"Let\u0027s give some examples."},{"Start":"00:40.515 ","End":"00:42.945","Text":"An oxide ion,"},{"Start":"00:42.945 ","End":"00:44.240","Text":"here\u0027s an oxide ion,"},{"Start":"00:44.240 ","End":"00:47.570","Text":"is a Lewis base, that\u0027s a base."},{"Start":"00:47.570 ","End":"00:53.015","Text":"It supplies a pair of electrons to a proton which has a Lewis acid."},{"Start":"00:53.015 ","End":"00:54.955","Text":"Here\u0027s our Lewis acid."},{"Start":"00:54.955 ","End":"00:59.420","Text":"Here the pair of electrons go into the Lewis acid,"},{"Start":"00:59.420 ","End":"01:03.660","Text":"and then afterward this bond breaks."},{"Start":"01:03.820 ","End":"01:08.840","Text":"We\u0027re left with OH minus"},{"Start":"01:08.840 ","End":"01:15.135","Text":"from the oxygen 2 minus and OH minus from the water."},{"Start":"01:15.135 ","End":"01:20.054","Text":"Here we have 2 OH minuses."},{"Start":"01:20.054 ","End":"01:23.600","Text":"Between the O and the H is a coordinate covalent bond."},{"Start":"01:23.600 ","End":"01:26.870","Text":"That\u0027s a bond in which 1 of the atoms"},{"Start":"01:26.870 ","End":"01:31.435","Text":"donates both of the electrons, donates electron pair."},{"Start":"01:31.435 ","End":"01:35.925","Text":"The oxygen has donated an electron pair to the hydrogen."},{"Start":"01:35.925 ","End":"01:38.500","Text":"Here\u0027s an electron pair."},{"Start":"01:38.830 ","End":"01:44.548","Text":"Now ammonia is a Lewis base,"},{"Start":"01:44.548 ","End":"01:50.049","Text":"and it supplies a pair of electrons to a proton which has a Lewis acid."},{"Start":"01:50.049 ","End":"01:54.570","Text":"Here\u0027s the electrons going to the hydrogen."},{"Start":"01:54.570 ","End":"02:00.060","Text":"Afterwards, again as before, this bond breaks."},{"Start":"02:00.060 ","End":"02:06.190","Text":"Now we have NH_4 plus and OH minus."},{"Start":"02:06.730 ","End":"02:09.575","Text":"The ammonia is the base,"},{"Start":"02:09.575 ","End":"02:14.730","Text":"and the hydrogen is the acid,"},{"Start":"02:15.200 ","End":"02:18.970","Text":"or rather the proton is the acid."},{"Start":"02:18.970 ","End":"02:23.180","Text":"Now we should note that the proton itself is a Lewis acid."},{"Start":"02:23.180 ","End":"02:26.675","Text":"This is slightly different from the Bronsted acids."},{"Start":"02:26.675 ","End":"02:32.630","Text":"The Bronsted acid is a compound or an ion that possesses an acidic hydrogen,"},{"Start":"02:32.630 ","End":"02:36.710","Text":"and the whole molecule ion is called a Bronsted acid."},{"Start":"02:36.710 ","End":"02:39.880","Text":"Just a small difference."},{"Start":"02:40.640 ","End":"02:46.755","Text":"Now ammonia as we\u0027ve said before is a Lewis base,"},{"Start":"02:46.755 ","End":"02:51.580","Text":"and boron trifluoride is a Lewis acid."},{"Start":"02:51.620 ","End":"02:57.685","Text":"The ammonia donates the pair of electrons to the boron."},{"Start":"02:57.685 ","End":"03:01.885","Text":"Remember boron is an electron-deficient atom."},{"Start":"03:01.885 ","End":"03:09.800","Text":"They react forming a coordinate covalent bond between N and B."},{"Start":"03:09.800 ","End":"03:12.755","Text":"Now we have this molecule."},{"Start":"03:12.755 ","End":"03:17.930","Text":"Now, also in addition to these molecules we\u0027ve talked about,"},{"Start":"03:17.930 ","End":"03:23.970","Text":"complex ions can also act as Lewis acids."},{"Start":"03:23.970 ","End":"03:25.530","Text":"What are complex ions?"},{"Start":"03:25.530 ","End":"03:29.330","Text":"Complex ions are polyatomic ions containing"},{"Start":"03:29.330 ","End":"03:36.365","Text":"a central metal ion to which other ions or molecules called ligands are attached."},{"Start":"03:36.365 ","End":"03:38.660","Text":"The whole field of chemistry,"},{"Start":"03:38.660 ","End":"03:41.345","Text":"and here we\u0027re just touching on it."},{"Start":"03:41.345 ","End":"03:46.250","Text":"These metal ions can act as Lewis acids and"},{"Start":"03:46.250 ","End":"03:51.280","Text":"the water as Lewis bases in hydrated metal ions."},{"Start":"03:51.280 ","End":"03:55.040","Text":"We\u0027re first going to talk about hydrated metal ions."},{"Start":"03:55.040 ","End":"03:56.900","Text":"Here\u0027s an example."},{"Start":"03:56.900 ","End":"04:02.720","Text":"Aluminum 3 plus with 6 water molecules around it in"},{"Start":"04:02.720 ","End":"04:08.960","Text":"an octahedral positions reacts with water in its shell,"},{"Start":"04:08.960 ","End":"04:11.083","Text":"in the hydration shell,"},{"Start":"04:11.083 ","End":"04:15.335","Text":"and then we get aluminum OH,"},{"Start":"04:15.335 ","End":"04:18.660","Text":"OH minus here and"},{"Start":"04:18.660 ","End":"04:25.045","Text":"5 water molecules because we have the minus instead of 3 plus is now 2 plus."},{"Start":"04:25.045 ","End":"04:33.185","Text":"We\u0027re left with H_3O plus from the water that was in the hydration shell."},{"Start":"04:33.185 ","End":"04:38.585","Text":"Here\u0027s our aluminum 3 pus with the 6 water molecules."},{"Start":"04:38.585 ","End":"04:45.245","Text":"I\u0027ve just drawn the arrows to where they are without indicating these 5 water molecules."},{"Start":"04:45.245 ","End":"04:48.335","Text":"I\u0027ve just indicated at the top water molecule."},{"Start":"04:48.335 ","End":"04:51.010","Text":"Here I\u0027ve OH_2."},{"Start":"04:51.850 ","End":"04:58.820","Text":"The lone pair on the water goes to capture this hydrogen."},{"Start":"04:58.820 ","End":"05:03.090","Text":"Then the hydrogen bond breaks."},{"Start":"05:03.370 ","End":"05:12.945","Text":"Then we have AL with the OH minus giving a charge of 2 plus instead of 3 plus."},{"Start":"05:12.945 ","End":"05:19.400","Text":"Of course, we have all the same aluminum ligands,"},{"Start":"05:19.400 ","End":"05:24.300","Text":"all the water ligands surrounding the aluminum."},{"Start":"05:29.540 ","End":"05:32.330","Text":"The water has taken the hydrogen,"},{"Start":"05:32.330 ","End":"05:36.160","Text":"so now it becomes H_3O plus."},{"Start":"05:36.160 ","End":"05:43.040","Text":"We see the aluminum with its 6 water molecules has reacted with the water in"},{"Start":"05:43.040 ","End":"05:50.000","Text":"the hydration shell to give us this molecule with Al and OH at the top,"},{"Start":"05:50.000 ","End":"05:53.940","Text":"and of course, H_3O plus."},{"Start":"05:53.940 ","End":"05:58.090","Text":"This is a Lewis acid."},{"Start":"05:59.120 ","End":"06:04.625","Text":"Metal ions can also act in Lewis acids in complex ions"},{"Start":"06:04.625 ","End":"06:09.680","Text":"such as zinc with 4 ammonia molecules as ligands around it."},{"Start":"06:09.680 ","End":"06:14.675","Text":"This is zinc 2 plus with 4 ammonia molecules as ligands."},{"Start":"06:14.675 ","End":"06:21.274","Text":"Zn 2 plus is a Lewis acid and ammonia is a Lewis base."},{"Start":"06:21.274 ","End":"06:26.790","Text":"In this video we talked about Lewis acids and bases."}],"ID":28478},{"Watched":false,"Name":"Exercise 1","Duration":"2m 8s","ChapterTopicVideoID":32009,"CourseChapterTopicPlaylistID":272186,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34285}],"Thumbnail":null,"ID":272186},{"Name":"Molecular Structure and Acid Strength","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Strength of Binary Acids 1","Duration":"5m 3s","ChapterTopicVideoID":27364,"CourseChapterTopicPlaylistID":272187,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.280","Text":"In this video and the next one,"},{"Start":"00:02.280 ","End":"00:07.124","Text":"we\u0027ll try to rationalize the relative strengths of binary acids."},{"Start":"00:07.124 ","End":"00:10.980","Text":"Let\u0027s talk about the strength of acids."},{"Start":"00:10.980 ","End":"00:14.910","Text":"Now K_a the equilibrium constant for acids,"},{"Start":"00:14.910 ","End":"00:18.360","Text":"is related to the Gibbs free energy of proton transfer."},{"Start":"00:18.360 ","End":"00:21.870","Text":"So the entropy should be taken into account."},{"Start":"00:21.870 ","End":"00:24.450","Text":"This is rather difficult to do."},{"Start":"00:24.450 ","End":"00:28.390","Text":"We usually just talk about the enthalpy."},{"Start":"00:28.610 ","End":"00:32.880","Text":"In addition the solvent plays a role in proton transfer,"},{"Start":"00:32.880 ","End":"00:35.594","Text":"so it should also be considered."},{"Start":"00:35.594 ","End":"00:38.070","Text":"Because of these two difficulties,"},{"Start":"00:38.070 ","End":"00:41.780","Text":"we\u0027re just going to look for trends in acid strength rather"},{"Start":"00:41.780 ","End":"00:46.115","Text":"than absolute values which are rather difficult to predict."},{"Start":"00:46.115 ","End":"00:49.805","Text":"We\u0027re going to talk about binary acids."},{"Start":"00:49.805 ","End":"00:56.405","Text":"We\u0027re going to compare binary acids consisting of hydrogen and 1 other atom,"},{"Start":"00:56.405 ","End":"01:02.210","Text":"such binary acids, as HCl, HBr, H_2O etc."},{"Start":"01:02.210 ","End":"01:07.160","Text":"We\u0027re going to start by determining the enthalpy for proton transfer"},{"Start":"01:07.160 ","End":"01:12.905","Text":"from HA to water in terms of a series of steps."},{"Start":"01:12.905 ","End":"01:16.180","Text":"So here\u0027s our overall process."},{"Start":"01:16.180 ","End":"01:21.995","Text":"An acid and water reacts with water to give us H_3O plus and A minus,"},{"Start":"01:21.995 ","End":"01:26.555","Text":"the acid is dissociating in the water."},{"Start":"01:26.555 ","End":"01:32.825","Text":"The total enthalpy for this process is Delta H total."},{"Start":"01:32.825 ","End":"01:37.235","Text":"Now we\u0027re going to look at the contributions to this enthalpy."},{"Start":"01:37.235 ","End":"01:41.450","Text":"Contributions to the enthalpy of proton transfer."},{"Start":"01:41.450 ","End":"01:44.450","Text":"These are listed in this table."},{"Start":"01:44.450 ","End":"01:47.375","Text":"If we add up all these processes,"},{"Start":"01:47.375 ","End":"01:50.045","Text":"we should get the total process."},{"Start":"01:50.045 ","End":"01:53.135","Text":"If we add up all the enthalpy changes,"},{"Start":"01:53.135 ","End":"01:55.385","Text":"we will get Delta H total."},{"Start":"01:55.385 ","End":"01:58.600","Text":"Let us look at the separate processes."},{"Start":"01:58.600 ","End":"02:03.680","Text":"The first one is HA in the aqueous phase,"},{"Start":"02:03.680 ","End":"02:07.055","Text":"going to HA in the gas phase."},{"Start":"02:07.055 ","End":"02:12.140","Text":"This is a negative of the Delta H of hydration of HA."},{"Start":"02:12.140 ","End":"02:14.960","Text":"The sign will be positive,"},{"Start":"02:14.960 ","End":"02:19.510","Text":"the sign of the enthalpy for this process will be positive."},{"Start":"02:19.510 ","End":"02:25.710","Text":"The second reaction is HA dissociating into H and A,"},{"Start":"02:25.710 ","End":"02:31.444","Text":"we\u0027ll call this Delta H of dissociation of HA, that\u0027s also positive."},{"Start":"02:31.444 ","End":"02:34.640","Text":"Then we have H going to H plus,"},{"Start":"02:34.640 ","End":"02:37.835","Text":"plus an electron that\u0027s ionization."},{"Start":"02:37.835 ","End":"02:42.230","Text":"So we have I, the ionization potential of H,"},{"Start":"02:42.230 ","End":"02:46.940","Text":"or ionization energy of H and that\u0027s also positive."},{"Start":"02:46.940 ","End":"02:52.225","Text":"Then A capture your electron to turn into"},{"Start":"02:52.225 ","End":"02:59.165","Text":"A minus and the enthalpy for this is Delta H_ea electron affinity,"},{"Start":"02:59.165 ","End":"03:03.627","Text":"which is sometimes written as minus E_ea(A)."},{"Start":"03:03.627 ","End":"03:10.220","Text":"Electron affinity is sometimes written as Delta H_ea and sometimes E_ea."},{"Start":"03:10.220 ","End":"03:17.600","Text":"In any case, this is negative energy is gained when the electron is attached to A."},{"Start":"03:17.600 ","End":"03:21.510","Text":"Then we have the hydration of H plus,"},{"Start":"03:21.510 ","End":"03:25.485","Text":"that\u0027s Delta H hydration of H plus,"},{"Start":"03:25.485 ","End":"03:27.870","Text":"or the hydration of A minus,"},{"Start":"03:27.870 ","End":"03:32.060","Text":"that\u0027s A minus gas phase A minus aqueous phase."},{"Start":"03:32.060 ","End":"03:35.240","Text":"Both of these two are negative."},{"Start":"03:35.240 ","End":"03:39.755","Text":"So the total enthalpy is the sum of all these quantities"},{"Start":"03:39.755 ","End":"03:45.425","Text":"minus Delta H of hydration of HA plus Delta H of dissociation of HA"},{"Start":"03:45.425 ","End":"03:51.660","Text":"plus the ionization potential of H minus Delta H_ea of A that\u0027s"},{"Start":"03:51.660 ","End":"03:59.425","Text":"electron affinity that Delta H hydration of H plus and Delta H hydration of A minus."},{"Start":"03:59.425 ","End":"04:01.655","Text":"Now in this table,"},{"Start":"04:01.655 ","End":"04:05.450","Text":"there are some quantities highlighted in yellow here."},{"Start":"04:05.450 ","End":"04:10.685","Text":"The ionization potential H and the hydration of H plus."},{"Start":"04:10.685 ","End":"04:13.040","Text":"These are the same for all HA acids."},{"Start":"04:13.040 ","End":"04:15.365","Text":"So they\u0027re not going to be important."},{"Start":"04:15.365 ","End":"04:19.535","Text":"Then there\u0027s the quantity in blue that\u0027s"},{"Start":"04:19.535 ","End":"04:24.590","Text":"hydration of HA and that doesn\u0027t change very much."},{"Start":"04:24.590 ","End":"04:29.149","Text":"So neither the ones in yellow are going to be important to distinguish"},{"Start":"04:29.149 ","End":"04:35.817","Text":"between various acids or the ones in blue."},{"Start":"04:35.817 ","End":"04:36.950","Text":"We hope to find,"},{"Start":"04:36.950 ","End":"04:44.325","Text":"we expect to find that this Delta H total kilojoules per mole becomes more negative,"},{"Start":"04:44.325 ","End":"04:47.248","Text":"the acid becomes stronger."},{"Start":"04:47.248 ","End":"04:51.455","Text":"We\u0027re going to test this on 1 or 2 examples."},{"Start":"04:51.455 ","End":"04:56.630","Text":"In this video, we described the contributions to the enthalpy of proton transfer to"},{"Start":"04:56.630 ","End":"05:03.600","Text":"water and in then the next video we\u0027ll consider some examples of binary acids."}],"ID":30088},{"Watched":false,"Name":"Strength of Binary Acids 2","Duration":"5m 8s","ChapterTopicVideoID":27365,"CourseChapterTopicPlaylistID":272187,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.630","Text":"In the previous video,"},{"Start":"00:01.630 ","End":"00:07.090","Text":"we described the contributions to the entropy proton transfer to water and this video,"},{"Start":"00:07.090 ","End":"00:11.545","Text":"we\u0027ll apply this theory to 2 series of binary acids."},{"Start":"00:11.545 ","End":"00:13.090","Text":"In the previous video,"},{"Start":"00:13.090 ","End":"00:19.165","Text":"we said we expect to find that is Delta H total becomes more negative,"},{"Start":"00:19.165 ","End":"00:21.820","Text":"the acid will become stronger."},{"Start":"00:21.820 ","End":"00:26.574","Text":"Let\u0027s first start with group 17 binary acids."},{"Start":"00:26.574 ","End":"00:32.740","Text":"Now the order of acid strengths in group 17 is HF,"},{"Start":"00:32.740 ","End":"00:35.335","Text":"which is weaker than HCl,"},{"Start":"00:35.335 ","End":"00:37.405","Text":"which is weaker than HBr,"},{"Start":"00:37.405 ","End":"00:39.655","Text":"which is weaker than HI,"},{"Start":"00:39.655 ","End":"00:44.205","Text":"so the strongest acid is HI."},{"Start":"00:44.205 ","End":"00:47.685","Text":"Let\u0027s look at Delta H total."},{"Start":"00:47.685 ","End":"00:52.530","Text":"We see that for HF is minus 15 for HCI minus 41,"},{"Start":"00:52.530 ","End":"00:57.585","Text":"HBr minus 51 and for HI minus 43."},{"Start":"00:57.585 ","End":"01:01.145","Text":"Indeed it becomes more and more negative except for"},{"Start":"01:01.145 ","End":"01:06.725","Text":"HI with other considerations obviously have to be taken into account."},{"Start":"01:06.725 ","End":"01:12.365","Text":"However, if we look at Delta H of dissociation of HA,"},{"Start":"01:12.365 ","End":"01:16.730","Text":"which is one of the components contributing to Delta H total."},{"Start":"01:16.730 ","End":"01:19.445","Text":"We see that it works very well."},{"Start":"01:19.445 ","End":"01:26.610","Text":"HF delta H of dissociation is 565 kilojoules per mole."},{"Start":"01:26.610 ","End":"01:28.980","Text":"Then for HCl 431,"},{"Start":"01:28.980 ","End":"01:33.460","Text":"for HBr 366 and for HI 299."},{"Start":"01:33.460 ","End":"01:42.785","Text":"In other words, the acid strength increases as it becomes easier to separate the 2 atoms."},{"Start":"01:42.785 ","End":"01:48.575","Text":"As HA decreases becomes easier to separate,"},{"Start":"01:48.575 ","End":"01:53.555","Text":"then we see that the acid strength increases."},{"Start":"01:53.555 ","End":"02:02.290","Text":"The acid strength increases as Delta H of dissociation decreases."},{"Start":"02:02.290 ","End":"02:05.810","Text":"We solved that the more negative Delta H total,"},{"Start":"02:05.810 ","End":"02:11.030","Text":"the stronger the acid and the HI was an except to this."},{"Start":"02:11.030 ","End":"02:16.955","Text":"However, the acid strength correlates very well with Delta H of dissociation."},{"Start":"02:16.955 ","End":"02:21.499","Text":"This is called the homolytic bond dissociation energy."},{"Start":"02:21.499 ","End":"02:24.680","Text":"The weaker the HA bond,"},{"Start":"02:24.680 ","End":"02:26.795","Text":"and indeed the longer it is,"},{"Start":"02:26.795 ","End":"02:31.869","Text":"the stronger the acid and this seems perfectly rational."},{"Start":"02:31.869 ","End":"02:35.600","Text":"Now let\u0027s look at period 2 binary acids."},{"Start":"02:35.600 ","End":"02:39.485","Text":"We\u0027re going along the period rather than down the group."},{"Start":"02:39.485 ","End":"02:42.605","Text":"We\u0027re talking about the 3 acids,"},{"Start":"02:42.605 ","End":"02:46.030","Text":"NH_3, H_2_O, and HF."},{"Start":"02:46.030 ","End":"02:48.530","Text":"NH_3 is usually a base,"},{"Start":"02:48.530 ","End":"02:49.700","Text":"it\u0027s hardly an acid."},{"Start":"02:49.700 ","End":"02:52.745","Text":"H_2_O can be either an acid or a base,"},{"Start":"02:52.745 ","End":"02:56.930","Text":"and HF in water is definitely an acid."},{"Start":"02:56.930 ","End":"03:04.710","Text":"They become stronger as we go along the rule."},{"Start":"03:05.740 ","End":"03:08.975","Text":"Let\u0027s look at Delta H total again."},{"Start":"03:08.975 ","End":"03:13.670","Text":"We see it starts off positive 125 becomes less"},{"Start":"03:13.670 ","End":"03:18.770","Text":"positive 44 until it becomes negative for HF so we"},{"Start":"03:18.770 ","End":"03:23.420","Text":"can see that this agrees with our assumption that"},{"Start":"03:23.420 ","End":"03:29.795","Text":"Delta H total would become more negative as the acid strength increases."},{"Start":"03:29.795 ","End":"03:33.920","Text":"A quantity that really agrees well"},{"Start":"03:33.920 ","End":"03:38.014","Text":"with the order of the acid strength is the electron affinity."},{"Start":"03:38.014 ","End":"03:41.390","Text":"Electron affinity E_ea(A) is 71,"},{"Start":"03:41.390 ","End":"03:45.680","Text":"then a 178 and then 328."},{"Start":"03:45.680 ","End":"03:49.370","Text":"We saw here that the more negative Delta H total,"},{"Start":"03:49.370 ","End":"03:51.755","Text":"the stronger the acid,"},{"Start":"03:51.755 ","End":"03:57.500","Text":"the acid strength correlates well with the electron affinity of A."},{"Start":"03:57.500 ","End":"03:59.870","Text":"The higher the electron affinity,"},{"Start":"03:59.870 ","End":"04:01.810","Text":"the stronger the acid."},{"Start":"04:01.810 ","End":"04:04.850","Text":"Now the electron affinity correlates with"},{"Start":"04:04.850 ","End":"04:10.055","Text":"electronegativity of A and the polarity of the HA bond."},{"Start":"04:10.055 ","End":"04:14.840","Text":"The more polar the bond is easier to remove the acidic hydrogen."},{"Start":"04:14.840 ","End":"04:19.940","Text":"Here we have electronegativity of A minus electronegativity"},{"Start":"04:19.940 ","End":"04:25.085","Text":"of the hydrogen and it\u0027s 0.8 increasing to 1.4 for water,"},{"Start":"04:25.085 ","End":"04:31.970","Text":"increasing to 1.8 for HF so the bond is becoming more and more polar."},{"Start":"04:31.970 ","End":"04:34.010","Text":"If it\u0027s more and more polar,"},{"Start":"04:34.010 ","End":"04:37.975","Text":"it\u0027s easier to take off the H plus,"},{"Start":"04:37.975 ","End":"04:40.550","Text":"it becomes more and more polar"},{"Start":"04:40.550 ","End":"04:47.169","Text":"Delta plus Delta minus and then it\u0027s easier to remove the H plus."},{"Start":"04:47.169 ","End":"04:50.960","Text":"This also seems to be rationale."},{"Start":"04:50.960 ","End":"04:57.405","Text":"Suppose what we know experimentally the HF is an acid in water."},{"Start":"04:57.405 ","End":"04:59.915","Text":"H_2_O can act as an acid or a base,"},{"Start":"04:59.915 ","End":"05:03.430","Text":"and NH_3 is usually a base."},{"Start":"05:03.430 ","End":"05:09.000","Text":"In this video, we talked about examples of series of binary acid."}],"ID":30089},{"Watched":false,"Name":"Strength of Oxoacids and Carboxylic Acids","Duration":"6m 35s","ChapterTopicVideoID":27363,"CourseChapterTopicPlaylistID":272187,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.250","Text":"In the previous videos,"},{"Start":"00:02.250 ","End":"00:05.130","Text":"we talked about the strengths of binary acids."},{"Start":"00:05.130 ","End":"00:11.910","Text":"In this video, we\u0027ll talk about the strength of oxoacids, and carboxylic acids."},{"Start":"00:11.910 ","End":"00:15.540","Text":"Let\u0027s start with oxoacids."},{"Start":"00:15.540 ","End":"00:18.690","Text":"Oxoacids usually contain hydrogen,"},{"Start":"00:18.690 ","End":"00:21.255","Text":"oxygen, and 1 other element."},{"Start":"00:21.255 ","End":"00:25.240","Text":"We\u0027ll first talk about hypohalous acids."},{"Start":"00:25.240 ","End":"00:27.855","Text":"We\u0027ll consider the hypohalous acids,"},{"Start":"00:27.855 ","End":"00:31.800","Text":"HClO, HBrO, and HIO."},{"Start":"00:31.800 ","End":"00:34.905","Text":"Here they are Cl with an OH group,"},{"Start":"00:34.905 ","End":"00:36.450","Text":"Br with an OH group,"},{"Start":"00:36.450 ","End":"00:38.850","Text":"and iodine with an OH group."},{"Start":"00:38.850 ","End":"00:40.940","Text":"These are all weak acids,"},{"Start":"00:40.940 ","End":"00:49.540","Text":"and the order of the acid strength is HClO stronger than HBrO which is stronger than HIO."},{"Start":"00:49.540 ","End":"00:52.160","Text":"This correlates with the order of"},{"Start":"00:52.160 ","End":"00:55.950","Text":"electronegativities. Cl has electronegative of 3, Br 2.8 and iodine 2.5."},{"Start":"01:00.800 ","End":"01:04.475","Text":"The most electronegative is chlorine,"},{"Start":"01:04.475 ","End":"01:09.640","Text":"and it is the most strong acid of the 3."},{"Start":"01:09.640 ","End":"01:14.930","Text":"What\u0027s happening is its electrons are pulled away from the OH bond,"},{"Start":"01:14.930 ","End":"01:18.810","Text":"weakening the bond so that H is more acidic,"},{"Start":"01:18.810 ","End":"01:24.300","Text":"and it\u0027s easier to take it off as H^+."},{"Start":"01:24.300 ","End":"01:28.870","Text":"Electrons are pulled away, to this direction."},{"Start":"01:33.050 ","End":"01:38.430","Text":"We\u0027re going to talk about the Chlorine oxoacids and we are going to"},{"Start":"01:38.430 ","End":"01:43.880","Text":"consider the series of Chlorine oxoacids with increasing number of oxygen atoms."},{"Start":"01:43.880 ","End":"01:46.910","Text":"HClO, HClO_2,"},{"Start":"01:46.910 ","End":"01:49.215","Text":"HClO_3, HClO_4,"},{"Start":"01:49.215 ","End":"01:53.660","Text":"and the names we talked about when we talked about naming inorganic compounds,"},{"Start":"01:53.660 ","End":"02:00.295","Text":"hypochlorous, chlorous, chloric, and perchloric acid."},{"Start":"02:00.295 ","End":"02:03.185","Text":"Here\u0027s how they look,"},{"Start":"02:03.185 ","End":"02:06.625","Text":"Cl with an OH group, Cl with an OH group,"},{"Start":"02:06.625 ","End":"02:11.920","Text":"and also a double bond to another oxygen and here we have"},{"Start":"02:11.920 ","End":"02:18.435","Text":"Cl-O-H with 2 oxygens doubly bonded to Cl,"},{"Start":"02:18.435 ","End":"02:22.050","Text":"and here with 3 extra oxygens."},{"Start":"02:22.050 ","End":"02:27.195","Text":"Their oxidation numbers are +1, +3, +5, and +7."},{"Start":"02:27.195 ","End":"02:30.370","Text":"That\u0027s of the Cl of course."},{"Start":"02:30.530 ","End":"02:36.420","Text":"Now, HClO and HClO_2 are weak acids,"},{"Start":"02:36.420 ","End":"02:45.780","Text":"and HClO_3 and HClO_4 are strong acids and the order of acid strength is like this."},{"Start":"02:45.850 ","End":"02:49.385","Text":"HClO_2, is stronger than HClO,"},{"Start":"02:49.385 ","End":"02:56.125","Text":"and HClO_3 is stronger than HClO_2,"},{"Start":"02:56.125 ","End":"03:02.080","Text":"and this third, this one is the strongest HClO_4."},{"Start":"03:02.080 ","End":"03:05.030","Text":"Here\u0027s the order of the acid strength."},{"Start":"03:05.030 ","End":"03:08.030","Text":"As the number of oxygen atoms increases,"},{"Start":"03:08.030 ","End":"03:10.639","Text":"the acid becomes stronger."},{"Start":"03:10.639 ","End":"03:16.579","Text":"The acid strength increases with increasing number of oxygen atoms."},{"Start":"03:16.579 ","End":"03:24.170","Text":"What\u0027s happening is that the all atoms shift electron density away from the OH bond,"},{"Start":"03:24.170 ","End":"03:30.960","Text":"it\u0027s called electron-withdrawal effect so the hydrogen becomes more acidic."},{"Start":"03:31.330 ","End":"03:36.660","Text":"Electron density is shifting away from the OH bond."},{"Start":"03:43.460 ","End":"03:48.830","Text":"Of course, as the effect increases as the number of oxygen atoms increases."},{"Start":"03:48.830 ","End":"03:51.170","Text":"That\u0027s why they acids becomes stronger,"},{"Start":"03:51.170 ","End":"03:54.580","Text":"as they have greater number of oxygen atoms."},{"Start":"03:54.580 ","End":"03:57.090","Text":"The oxidation number,"},{"Start":"03:57.090 ","End":"04:03.500","Text":"ON [inaudible] of Chlorine also increases as the number of oxygen atoms increases."},{"Start":"04:03.500 ","End":"04:07.520","Text":"The acid strand we can see increases with"},{"Start":"04:07.520 ","End":"04:13.260","Text":"increasing oxidation number of chlorine; +1,+3,+5,+7."},{"Start":"04:14.140 ","End":"04:18.965","Text":"Let\u0027s say a few words about carboxylic acids."},{"Start":"04:18.965 ","End":"04:21.410","Text":"It\u0027s a similar effect."},{"Start":"04:21.410 ","End":"04:24.110","Text":"This taking away electron density,"},{"Start":"04:24.110 ","End":"04:28.060","Text":"removing electron density from the O-H bond."},{"Start":"04:28.060 ","End":"04:35.010","Text":"Carboxylic acids usually or probably know have the general formula RCO_2H."},{"Start":"04:35.010 ","End":"04:38.730","Text":"For example, for acetic acid R=CH3,"},{"Start":"04:38.730 ","End":"04:41.555","Text":"methyl group, and here\u0027s the structure."},{"Start":"04:41.555 ","End":"04:47.615","Text":"We have an OH group and a C double O carbonyl group."},{"Start":"04:47.615 ","End":"04:53.880","Text":"The C-O group shifts electron density away from the OH bond,"},{"Start":"04:54.880 ","End":"04:59.540","Text":"so that the hydrogen becomes more acidic."},{"Start":"04:59.540 ","End":"05:05.040","Text":"In addition, the acetate ion is formed once we take off the H+,"},{"Start":"05:05.470 ","End":"05:12.965","Text":"is also stabilized due to delocalization of charge over the CO_2- group."},{"Start":"05:12.965 ","End":"05:15.275","Text":"We have a single bond here,"},{"Start":"05:15.275 ","End":"05:17.135","Text":"another single bond here,"},{"Start":"05:17.135 ","End":"05:21.410","Text":"and there\u0027s"},{"Start":"05:21.410 ","End":"05:28.310","Text":"delocalization over the CO_2- group."},{"Start":"05:28.310 ","End":"05:33.050","Text":"You may remember, we talked about this previously when we talked about"},{"Start":"05:33.050 ","End":"05:39.985","Text":"molecular orbitals and the bond order of each of these CO bonds is 1.5."},{"Start":"05:39.985 ","End":"05:46.805","Text":"They\u0027re equivalent and intermediate between single and double bonds."},{"Start":"05:46.805 ","End":"05:52.794","Text":"That means that CO_2- is a very weak conjugate base."},{"Start":"05:52.794 ","End":"05:58.095","Text":"Now, if we replace the CH_3 by CCl_3,"},{"Start":"05:58.095 ","End":"06:01.930","Text":"that\u0027s Cl here instead of hydrogen,"},{"Start":"06:02.110 ","End":"06:06.740","Text":"then we get an additional electron withdrawal effect,"},{"Start":"06:06.740 ","End":"06:11.770","Text":"because chlorine is more electronegative than hydrogen."},{"Start":"06:11.770 ","End":"06:18.365","Text":"We expect that if more electron density will be taken away from this OH bond."},{"Start":"06:18.365 ","End":"06:25.010","Text":"That means that trichloroacetic acid is a stronger acid than acetic acid,"},{"Start":"06:25.010 ","End":"06:27.275","Text":"because the OH bond is weaker,"},{"Start":"06:27.275 ","End":"06:29.795","Text":"it\u0027s easier to remove the H+."},{"Start":"06:29.795 ","End":"06:36.720","Text":"In this video, we learned about the strength of Oxoacids, and Carboxylic acids."}],"ID":30090},{"Watched":false,"Name":"Exercise 1","Duration":"2m 1s","ChapterTopicVideoID":32010,"CourseChapterTopicPlaylistID":272187,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34287},{"Watched":false,"Name":"Exercise 2","Duration":"2m 40s","ChapterTopicVideoID":32011,"CourseChapterTopicPlaylistID":272187,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34288}],"Thumbnail":null,"ID":272187}]

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