Common-ion Effect in Acid-Base Equilibria
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Buffer Solutions
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Acid-Base Indicators
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Titration Curves
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- Titrations Basic Definitions
- Titration of a Strong Acid with a Strong Base
- Titration of a Weak Polyprotic Acid with a Strong Base 1
- Titration of a Weak Polyprotic Acid with a Strong Base 2
- Titration of a Weak Acid with a Strong Base 1
- Titration of a Weak Acid with a Strong Base 2
- Exercise 1
- Exercise 2 part a
- Exercise 2 part b
- Exercise 3
- Exercise 4 part a
- Exercise 4 part b

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[{"Name":"Common-ion Effect in Acid-Base Equilibria","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Weak and Strong Acid","Duration":"4m 46s","ChapterTopicVideoID":28384,"CourseChapterTopicPlaylistID":282559,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/28384.jpeg","UploadDate":"2022-01-16T08:46:40.7970000","DurationForVideoObject":"PT4M46S","Description":null,"MetaTitle":"Weak and Strong Acid: Video + Workbook | Proprep","MetaDescription":"Buffers and Titrations - Common-ion Effect in Acid-Base Equilibria. 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/buffers-and-titrations/common_ion-effect-in-acid_base-equilibria/vid31519","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.115","Text":"In previous videos, we learned about weak and strong acids."},{"Start":"00:05.115 ","End":"00:10.920","Text":"In this video, we\u0027ll talk about a solution containing both a strong and a weak acid."},{"Start":"00:10.920 ","End":"00:17.940","Text":"Let\u0027s recall what we calculated before in a previous video about a weak acid."},{"Start":"00:17.940 ","End":"00:20.250","Text":"The question we solved there was,"},{"Start":"00:20.250 ","End":"00:27.495","Text":"calculate the ion concentration and pH of a 0.02 molar solution of acetic acid."},{"Start":"00:27.495 ","End":"00:29.415","Text":"Here\u0027s our equilibrium."},{"Start":"00:29.415 ","End":"00:35.910","Text":"Acetic acid in water gives us H_3O plus and acetate anion."},{"Start":"00:35.910 ","End":"00:40.500","Text":"The Ka for this is 1.8 times 10 to the power minus 5."},{"Start":"00:40.500 ","End":"00:44.405","Text":"The answer we got was the concentration of H_3O plus,"},{"Start":"00:44.405 ","End":"00:48.125","Text":"which is equal to the concentration of CH_3CO_2 minus,"},{"Start":"00:48.125 ","End":"00:51.875","Text":"is equal to 6 times 10 to the power minus 4."},{"Start":"00:51.875 ","End":"00:55.520","Text":"That gives us a pH of 3.2."},{"Start":"00:55.520 ","End":"00:59.480","Text":"Let\u0027s go on and consider the same weak acid,"},{"Start":"00:59.480 ","End":"01:03.155","Text":"acetic acid, and a strong acid together,"},{"Start":"01:03.155 ","End":"01:05.930","Text":"that\u0027s called a mixed solution."},{"Start":"01:05.930 ","End":"01:12.380","Text":"Now the question is to calculate the ion concentration and pH of a solution that contains"},{"Start":"01:12.380 ","End":"01:19.590","Text":"0.02 molar acetic acid and also 0.02 molar HCl."},{"Start":"01:19.590 ","End":"01:22.355","Text":"We have a weak acid and a strong acid."},{"Start":"01:22.355 ","End":"01:26.315","Text":"Now the hydrochloric acid dissociates completely."},{"Start":"01:26.315 ","End":"01:31.055","Text":"We only need to consider the ionization of acetic acid equilibrium."},{"Start":"01:31.055 ","End":"01:33.020","Text":"That\u0027s the only equilibrium we have."},{"Start":"01:33.020 ","End":"01:34.760","Text":"That of acetic acid."},{"Start":"01:34.760 ","End":"01:39.725","Text":"Now Le Châtelier\u0027s principle tells us that when we add the HCl,"},{"Start":"01:39.725 ","End":"01:42.320","Text":"here we\u0027re adding a strong acid,"},{"Start":"01:42.320 ","End":"01:46.714","Text":"this will push the equilibrium to the left."},{"Start":"01:46.714 ","End":"01:49.070","Text":"Here\u0027s our ICE table."},{"Start":"01:49.070 ","End":"01:50.765","Text":"We have citric acid,"},{"Start":"01:50.765 ","End":"01:53.855","Text":"H_3O plus, and acetate."},{"Start":"01:53.855 ","End":"01:59.645","Text":"We have 0.02 molar of the acetic acid."},{"Start":"01:59.645 ","End":"02:08.150","Text":"Now we have 0.02 molar of the H_3O plus that comes from the dissociation of the HCl."},{"Start":"02:08.150 ","End":"02:10.955","Text":"At the beginning, we have no acetate."},{"Start":"02:10.955 ","End":"02:15.650","Text":"Now the changes in the way to equilibrium are minus x,"},{"Start":"02:15.650 ","End":"02:17.840","Text":"plus x, and plus x."},{"Start":"02:17.840 ","End":"02:21.290","Text":"At equilibrium, for acetic acid,"},{"Start":"02:21.290 ","End":"02:23.580","Text":"we have 0.02 minus x."},{"Start":"02:24.010 ","End":"02:29.360","Text":"For H_3O plus we have 0.02 plus x,"},{"Start":"02:29.360 ","End":"02:32.495","Text":"and for the acetate, we have just plus x."},{"Start":"02:32.495 ","End":"02:34.715","Text":"Now we can write the Ka,"},{"Start":"02:34.715 ","End":"02:39.320","Text":"which we know to be equal to 1.8 times 10 to the power minus 5,"},{"Start":"02:39.320 ","End":"02:42.575","Text":"is equal to 0.02 plus x,"},{"Start":"02:42.575 ","End":"02:46.085","Text":"times x, divided by 0.02 minus x."},{"Start":"02:46.085 ","End":"02:50.360","Text":"Here we have 0.02 plus x, times x,"},{"Start":"02:50.360 ","End":"02:52.640","Text":"divided by 0.02 minus x."},{"Start":"02:52.640 ","End":"02:55.805","Text":"Now if x is small,"},{"Start":"02:55.805 ","End":"02:59.030","Text":"we can ignore it in the numerator,"},{"Start":"02:59.030 ","End":"03:00.890","Text":"and in the denominator."},{"Start":"03:00.890 ","End":"03:04.040","Text":"0.02 cancels with 0.02."},{"Start":"03:04.040 ","End":"03:10.060","Text":"We\u0027re left just with x. Ka is approximately equal to x."},{"Start":"03:10.060 ","End":"03:14.040","Text":"We know Ka is 1.8 times 10 to the power minus 5,"},{"Start":"03:14.040 ","End":"03:17.830","Text":"so x is 1.8 times 10 to the power minus 5."},{"Start":"03:17.830 ","End":"03:24.895","Text":"We can write that the concentration of H_3O plus is 0.02 plus x,"},{"Start":"03:24.895 ","End":"03:28.860","Text":"and that\u0027s approximately 0.02 molar."},{"Start":"03:28.860 ","End":"03:32.370","Text":"That gives us a pH of 1.7,"},{"Start":"03:32.370 ","End":"03:38.545","Text":"and the concentration of the acetate is 1.8 times 10 to the power minus 5 molar."},{"Start":"03:38.545 ","End":"03:43.805","Text":"Now, we saw the H_3O plus is formed in both the ionization processes,"},{"Start":"03:43.805 ","End":"03:48.200","Text":"comes from HCl and from acetic acid."},{"Start":"03:48.200 ","End":"03:51.020","Text":"We see it\u0027s a common ion."},{"Start":"03:51.020 ","End":"03:58.870","Text":"We can see that the strong acid suppresses the ionization of the weak acid."},{"Start":"03:58.870 ","End":"04:03.600","Text":"The concentration of H_3OCO2 minus is low now."},{"Start":"04:03.600 ","End":"04:07.185","Text":"It\u0027s 1.8 times 10 to the power minus 5."},{"Start":"04:07.185 ","End":"04:10.010","Text":"Originally, with acetic acid on its own,"},{"Start":"04:10.010 ","End":"04:13.660","Text":"it was 6 times 10 to the power minus 4."},{"Start":"04:13.660 ","End":"04:18.335","Text":"The strong acid has suppressed the ionization of the weak acid."},{"Start":"04:18.335 ","End":"04:22.970","Text":"This is an example of the common ion effect."},{"Start":"04:22.970 ","End":"04:25.970","Text":"Now if we instead of a strong acid,"},{"Start":"04:25.970 ","End":"04:27.090","Text":"we have a strong base,"},{"Start":"04:27.090 ","End":"04:30.350","Text":"it will also suppress the ionization of a weak base."},{"Start":"04:30.350 ","End":"04:34.655","Text":"For example, if we had ammonia and sodium hydroxide,"},{"Start":"04:34.655 ","End":"04:39.650","Text":"the hydroxide would suppress the ionization of the ammonia."},{"Start":"04:39.650 ","End":"04:46.620","Text":"In this video, we talked about the common ion in a solution of weak and strong acids."}],"ID":31519},{"Watched":false,"Name":"Weak Acid and its Salt","Duration":"4m 41s","ChapterTopicVideoID":28383,"CourseChapterTopicPlaylistID":282559,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.965","Text":"In the previous video,"},{"Start":"00:01.965 ","End":"00:07.035","Text":"we learned about the common ion effect in a solution of weak and strong acids."},{"Start":"00:07.035 ","End":"00:13.110","Text":"In this video, we\u0027ll consider a solution of a weak acid and its salt."},{"Start":"00:13.110 ","End":"00:16.890","Text":"We\u0027re going to consider a weak acid and its salt."},{"Start":"00:16.890 ","End":"00:20.235","Text":"This is another example of a mixed solution."},{"Start":"00:20.235 ","End":"00:22.785","Text":"Here is our salt."},{"Start":"00:22.785 ","End":"00:29.835","Text":"Sodium acetate reacts with water to give us sodium plus and acetate."},{"Start":"00:29.835 ","End":"00:36.995","Text":"It completely decomposes into sodium plus and acetate."},{"Start":"00:36.995 ","End":"00:41.598","Text":"Now here\u0027s our weak acid, acetic acid."},{"Start":"00:41.598 ","End":"00:46.900","Text":"It reacts with water to give us H_3O plus and acetate."},{"Start":"00:46.900 ","End":"00:50.300","Text":"We can see that we have a common ion."},{"Start":"00:50.300 ","End":"00:54.505","Text":"The acetate is common to both the salt and to the weak acid."},{"Start":"00:54.505 ","End":"00:56.180","Text":"I\u0027ve written it here,"},{"Start":"00:56.180 ","End":"01:01.640","Text":"the salt ionizes completely and the common ion is the acetate anion."},{"Start":"01:01.640 ","End":"01:09.620","Text":"Now the acetate anion from the salt will push the acetic acid equilibrium to the left."},{"Start":"01:09.620 ","End":"01:15.795","Text":"It\u0027s going to increase the amount of acetate pushing the equilibrium to the left."},{"Start":"01:15.795 ","End":"01:21.125","Text":"That means the ionization of acetic acid will be suppressed."},{"Start":"01:21.125 ","End":"01:25.550","Text":"Less acetic acid will ionize than before."},{"Start":"01:25.550 ","End":"01:27.170","Text":"Here\u0027s an example."},{"Start":"01:27.170 ","End":"01:36.995","Text":"Calculate the ion concentrations of 0.02 molar acetic acid and 0.02 molar sodium acetate."},{"Start":"01:36.995 ","End":"01:41.045","Text":"You can see that the both have the same concentrations."},{"Start":"01:41.045 ","End":"01:44.015","Text":"Here\u0027s our ICE table."},{"Start":"01:44.015 ","End":"01:46.690","Text":"We have acetic acid,"},{"Start":"01:46.690 ","End":"01:50.770","Text":"H_3O plus, and acetate."},{"Start":"01:50.770 ","End":"01:57.020","Text":"Now at the beginning we have 0.02 molar acetic acid. That\u0027s here."},{"Start":"01:57.020 ","End":"02:06.515","Text":"We have no H_3O plus and we have 0.02 molar of acetate."},{"Start":"02:06.515 ","End":"02:10.745","Text":"Now what\u0027s the change when we approach equilibrium?"},{"Start":"02:10.745 ","End":"02:13.505","Text":"We have minus x for acetic acid,"},{"Start":"02:13.505 ","End":"02:16.265","Text":"plus x for the H_3O plus,"},{"Start":"02:16.265 ","End":"02:19.895","Text":"and plus x to the CH_3CO_2 minus."},{"Start":"02:19.895 ","End":"02:21.815","Text":"Here\u0027s our equation here."},{"Start":"02:21.815 ","End":"02:30.885","Text":"We can see that the acetic acid is changing into H_3O plus and acetate."},{"Start":"02:30.885 ","End":"02:34.670","Text":"At equilibrium, we have for acetic acid,"},{"Start":"02:34.670 ","End":"02:39.660","Text":"0.02 minus x minus x."},{"Start":"02:39.660 ","End":"02:42.351","Text":"For each H_3O plus, we have plus x,"},{"Start":"02:42.351 ","End":"02:48.640","Text":"and for acetate, we have 0.02 plus x."},{"Start":"02:48.640 ","End":"02:52.660","Text":"So now we can write the K_ a which has the value of 1.8 times"},{"Start":"02:52.660 ","End":"02:56.680","Text":"10 to the power minus 5 for acetic acid is equal"},{"Start":"02:56.680 ","End":"03:05.890","Text":"to 0.02 plus x times x divided by 0.02 minus x."},{"Start":"03:05.890 ","End":"03:08.665","Text":"Now if x is small,"},{"Start":"03:08.665 ","End":"03:14.530","Text":"0.02 plus x will be almost 0.02."},{"Start":"03:14.530 ","End":"03:16.525","Text":"If x is small,"},{"Start":"03:16.525 ","End":"03:22.345","Text":"0.02 minus x will also be similar to 0.02."},{"Start":"03:22.345 ","End":"03:24.860","Text":"The two of these can cancel."},{"Start":"03:24.860 ","End":"03:27.035","Text":"We\u0027re left with just x,"},{"Start":"03:27.035 ","End":"03:37.305","Text":"so x is equal to 1.8 times 10 to the power minus 5 because x is equal to K_ a."},{"Start":"03:37.305 ","End":"03:41.310","Text":"We see that the concentration of H_3O plus is"},{"Start":"03:41.310 ","End":"03:47.810","Text":"now 1.8 times 10 to the power minus 5 in the presence of the salt."},{"Start":"03:47.810 ","End":"03:49.954","Text":"Whereas we calculated before,"},{"Start":"03:49.954 ","End":"03:55.025","Text":"it was 6 times 10 to the power minus 4 in the absence of the salt."},{"Start":"03:55.025 ","End":"04:01.335","Text":"That means the ionization of the weak acid is suppressed by adding the salt."},{"Start":"04:01.335 ","End":"04:07.385","Text":"Very important result is that when the concentrations of weak acid and salt are equal,"},{"Start":"04:07.385 ","End":"04:12.860","Text":"0.02 for acetic acid and 0.02 for the sodium acetate,"},{"Start":"04:12.860 ","End":"04:18.575","Text":"then we got the result that the pH is exactly equal to pK_a."},{"Start":"04:18.575 ","End":"04:22.930","Text":"Now if we were to do a similar calculation for a weak base and its salt,"},{"Start":"04:22.930 ","End":"04:26.171","Text":"for example, ammonia and ammonium chloride,"},{"Start":"04:26.171 ","End":"04:28.250","Text":"we\u0027d get the same idea."},{"Start":"04:28.250 ","End":"04:34.355","Text":"The ammonium chloride would suppress the ionization of the ammonia."},{"Start":"04:34.355 ","End":"04:36.290","Text":"In this video, we talked about"},{"Start":"04:36.290 ","End":"04:42.030","Text":"the common ion effect in a solution of a weak acid and its salt."}],"ID":31520},{"Watched":false,"Name":"Exercise 1","Duration":"6m 41s","ChapterTopicVideoID":29815,"CourseChapterTopicPlaylistID":282559,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31521},{"Watched":false,"Name":"Exercise 2","Duration":"4m 43s","ChapterTopicVideoID":29810,"CourseChapterTopicPlaylistID":282559,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31522},{"Watched":false,"Name":"Exercise 3","Duration":"4m 56s","ChapterTopicVideoID":29811,"CourseChapterTopicPlaylistID":282559,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31523},{"Watched":false,"Name":"Exercise 4 Part a","Duration":"3m 49s","ChapterTopicVideoID":29812,"CourseChapterTopicPlaylistID":282559,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31524},{"Watched":false,"Name":"Exercise 4 Part b","Duration":"3m 41s","ChapterTopicVideoID":29813,"CourseChapterTopicPlaylistID":282559,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31525},{"Watched":false,"Name":"Exercise 5","Duration":"5m 12s","ChapterTopicVideoID":29814,"CourseChapterTopicPlaylistID":282559,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31526}],"Thumbnail":null,"ID":282559},{"Name":"Buffer Solutions","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Buffer Solutions","Duration":"4m 12s","ChapterTopicVideoID":28391,"CourseChapterTopicPlaylistID":282560,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:02.370","Text":"In the previous video,"},{"Start":"00:02.370 ","End":"00:06.270","Text":"we talked about a solution containing a weak acid and salt."},{"Start":"00:06.270 ","End":"00:12.195","Text":"In this video, we\u0027ll show that such a solution can act as a buffer solution."},{"Start":"00:12.195 ","End":"00:15.270","Text":"We\u0027re going to talk about buffer solutions."},{"Start":"00:15.270 ","End":"00:19.350","Text":"A buffer solution is a solution whose pH changes"},{"Start":"00:19.350 ","End":"00:24.315","Text":"only slightly when a small amount of acid or base is added."},{"Start":"00:24.315 ","End":"00:26.975","Text":"Now there are many buffer solutions in nature."},{"Start":"00:26.975 ","End":"00:31.400","Text":"For example, human blood plasma is buffered to pH equal to"},{"Start":"00:31.400 ","End":"00:35.810","Text":"7.4 and seawater is buffered to approximately"},{"Start":"00:35.810 ","End":"00:45.230","Text":"pH 8.4 and saliva in your mouth is buffered to pH between 6.2 and 7.6."},{"Start":"00:45.230 ","End":"00:52.429","Text":"Now, a mixed solution we met in the last two videos containing a weak acid and its salt,"},{"Start":"00:52.429 ","End":"00:54.920","Text":"or a weak base and its salt,"},{"Start":"00:54.920 ","End":"00:57.425","Text":"can act as a buffer solution."},{"Start":"00:57.425 ","End":"01:01.985","Text":"A mixed solution containing a weak acid and its salt,"},{"Start":"01:01.985 ","End":"01:06.485","Text":"or a weak base and its salt can act as a buffer solution."},{"Start":"01:06.485 ","End":"01:08.675","Text":"Here\u0027s an example of a buffer solution,"},{"Start":"01:08.675 ","End":"01:11.660","Text":"acetic acid and sodium acetate,"},{"Start":"01:11.660 ","End":"01:14.059","Text":"a weak acid and its salt."},{"Start":"01:14.059 ","End":"01:17.445","Text":"That\u0027s the acid and that\u0027s the salt."},{"Start":"01:17.445 ","End":"01:20.465","Text":"Here\u0027s the equilibrium for acetic acid."},{"Start":"01:20.465 ","End":"01:26.190","Text":"Acetic acid in water gives us H_3O plus and acetate."},{"Start":"01:26.190 ","End":"01:28.355","Text":"We\u0027ve met this many times before."},{"Start":"01:28.355 ","End":"01:32.510","Text":"Now, a buffer solution is formed when the concentrations"},{"Start":"01:32.510 ","End":"01:36.740","Text":"of acetic acid and it\u0027s acetate are similar,"},{"Start":"01:36.740 ","End":"01:41.280","Text":"approximately not less than 10 percent, 1 of the other."},{"Start":"01:41.280 ","End":"01:44.925","Text":"This can be 10% of that or that 10% of that."},{"Start":"01:44.925 ","End":"01:48.320","Text":"Now supposing we add a few drops of HCl to"},{"Start":"01:48.320 ","End":"01:53.155","Text":"the buffer solution and then the H_3O plus we are adding,"},{"Start":"01:53.155 ","End":"01:59.165","Text":"that comes from the HCl will react with the acetate because the acetate is a base."},{"Start":"01:59.165 ","End":"02:03.200","Text":"Here we have the HCl plus that came from HCl"},{"Start":"02:03.200 ","End":"02:07.940","Text":"reacting with the acetate to give us acetic acid and water."},{"Start":"02:07.940 ","End":"02:13.625","Text":"If this happens, the concentration of the acetate will go down,"},{"Start":"02:13.625 ","End":"02:21.140","Text":"the concentration of the acetic acid will go up and now we have more acid than salt,"},{"Start":"02:21.140 ","End":"02:23.510","Text":"so the pH will decrease,"},{"Start":"02:23.510 ","End":"02:25.205","Text":"it will become more acidic."},{"Start":"02:25.205 ","End":"02:29.630","Text":"Now supposing we add a few drops of NaOH to the buffer solution,"},{"Start":"02:29.630 ","End":"02:34.790","Text":"then the OH minus that comes from the NaOH will react with the acetic acid."},{"Start":"02:34.790 ","End":"02:40.385","Text":"Here\u0027s the hydroxide will react with acetic acid to give us acetate and water."},{"Start":"02:40.385 ","End":"02:44.420","Text":"The acetic acid will go down,"},{"Start":"02:44.420 ","End":"02:47.980","Text":"it\u0027s reacted, so its concentration will decrease."},{"Start":"02:47.980 ","End":"02:50.850","Text":"The acetate has been produced,"},{"Start":"02:50.850 ","End":"02:53.870","Text":"so its concentration will increase."},{"Start":"02:53.870 ","End":"02:56.990","Text":"Now we have less acid than we had before,"},{"Start":"02:56.990 ","End":"03:00.930","Text":"so the pH will go up. Here\u0027s a drawing."},{"Start":"03:00.930 ","End":"03:05.555","Text":"Supposing we have the same amount of salt and acid, same concentrations,"},{"Start":"03:05.555 ","End":"03:08.810","Text":"and we add a little amount of acid,"},{"Start":"03:08.810 ","End":"03:13.070","Text":"then we\u0027ll have less salt than acid and the pH will decrease."},{"Start":"03:13.070 ","End":"03:18.095","Text":"In fact, we\u0027ll see that the pH will now be less than the pK_a of acetic acid."},{"Start":"03:18.095 ","End":"03:19.805","Text":"If on the other hand,"},{"Start":"03:19.805 ","End":"03:21.680","Text":"we add a base,"},{"Start":"03:21.680 ","End":"03:24.350","Text":"then we\u0027ll have more salt than acid."},{"Start":"03:24.350 ","End":"03:25.550","Text":"The amount of salt will go up,"},{"Start":"03:25.550 ","End":"03:31.725","Text":"the amount of acid will go down and the pH will now be greater than pK_a."},{"Start":"03:31.725 ","End":"03:34.695","Text":"Thus the amount of salt and acid are equal."},{"Start":"03:34.695 ","End":"03:36.815","Text":"As we saw in our previous video,"},{"Start":"03:36.815 ","End":"03:40.135","Text":"the pH will be equal to the pK_a."},{"Start":"03:40.135 ","End":"03:41.955","Text":"Add a little acid,"},{"Start":"03:41.955 ","End":"03:45.480","Text":"the pH will go below the pK_a."},{"Start":"03:45.480 ","End":"03:50.360","Text":"Add a little base and the pH will go above pK_a."},{"Start":"03:50.360 ","End":"03:53.870","Text":"But these are small amounts, small differences."},{"Start":"03:53.870 ","End":"03:57.890","Text":"Now it\u0027s often said that the acetic acid is a source of"},{"Start":"03:57.890 ","End":"04:02.330","Text":"protons and it\u0027s acetate is a sink for protons."},{"Start":"04:02.330 ","End":"04:07.280","Text":"These are useful terms that we\u0027ll use in the following videos."},{"Start":"04:07.280 ","End":"04:12.510","Text":"In this video, we began to talk about buffer solutions."}],"ID":31527},{"Watched":false,"Name":"Henderson-Hasselbach Equation","Duration":"8m 7s","ChapterTopicVideoID":28388,"CourseChapterTopicPlaylistID":282560,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.620 ","End":"00:02.790","Text":"In the previous video,"},{"Start":"00:02.790 ","End":"00:04.800","Text":"we talked about buffer solutions."},{"Start":"00:04.800 ","End":"00:12.040","Text":"In this video, we\u0027ll derive an approximate equation for the pH of buffer solutions."},{"Start":"00:12.200 ","End":"00:16.875","Text":"Let\u0027s first consider an acid buffer solution."},{"Start":"00:16.875 ","End":"00:22.455","Text":"We\u0027re going to consider a solution with similar concentrations of the acid,"},{"Start":"00:22.455 ","End":"00:28.000","Text":"acetic acid and the conjugate base acetate."},{"Start":"00:28.850 ","End":"00:33.255","Text":"We\u0027re going to write K_a for acetic acid."},{"Start":"00:33.255 ","End":"00:36.750","Text":"That\u0027s the concentration of H_3O+ times the concentration of"},{"Start":"00:36.750 ","End":"00:41.520","Text":"the acetate divided by the concentration of the acetic acid."},{"Start":"00:41.520 ","End":"00:48.390","Text":"We\u0027re going to write the H_3O+ concentration of H_3O+ is equal to"},{"Start":"00:48.390 ","End":"00:57.005","Text":"K_a times the concentration of acetic acid divided by the concentration of the acetate."},{"Start":"00:57.005 ","End":"01:00.605","Text":"Here, we have an equation for each field plot."},{"Start":"01:00.605 ","End":"01:02.705","Text":"Here\u0027s the equation written again."},{"Start":"01:02.705 ","End":"01:05.540","Text":"Now, if we take the minus log of"},{"Start":"01:05.540 ","End":"01:09.125","Text":"a left-hand side and the minus log of the right-hand side,"},{"Start":"01:09.125 ","End":"01:14.000","Text":"we get minus log of H_3O+ is equal to minus log of"},{"Start":"01:14.000 ","End":"01:21.320","Text":"K_a times minus log of the acetic acid divide by the acetate."},{"Start":"01:21.320 ","End":"01:23.465","Text":"Since there\u0027s a minus,"},{"Start":"01:23.465 ","End":"01:30.065","Text":"I can interchange the numerator and denominator and write a plus sign."},{"Start":"01:30.065 ","End":"01:33.980","Text":"Here we have minus log of H_3O+ is equal to minus log of"},{"Start":"01:33.980 ","End":"01:39.400","Text":"K_a plus the log of the acetate divided by the acetic acid."},{"Start":"01:39.400 ","End":"01:44.610","Text":"Now, we know that minus a log of H_3O+ is the pH,"},{"Start":"01:44.610 ","End":"01:47.830","Text":"minus log of K_a is a pK_a."},{"Start":"01:47.830 ","End":"01:53.495","Text":"Then we still have this plus log of acetate divided by acetic acid."},{"Start":"01:53.495 ","End":"01:55.605","Text":"Here\u0027s a very useful equation,"},{"Start":"01:55.605 ","End":"02:00.240","Text":"pH is equal to pK_a plus log of"},{"Start":"02:00.240 ","End":"02:06.545","Text":"the ratio of acetate and acetic acid."},{"Start":"02:06.545 ","End":"02:09.920","Text":"Now, suppose if we call this ratio"},{"Start":"02:09.920 ","End":"02:17.725","Text":"R. That\u0027s the conjugate base divided by the acid if we use some other acid."},{"Start":"02:17.725 ","End":"02:20.625","Text":"Now if R is equal to 1,"},{"Start":"02:20.625 ","End":"02:26.700","Text":"then the log of R will be 0 and we get pH equal to pK_a."},{"Start":"02:26.700 ","End":"02:31.100","Text":"Now if we added a little acid, what would happen?"},{"Start":"02:31.100 ","End":"02:36.530","Text":"The acid would react with the base so that will decrease and the amount of"},{"Start":"02:36.530 ","End":"02:42.580","Text":"acid would increase so R would now be less than 1."},{"Start":"02:42.580 ","End":"02:46.260","Text":"Log of R would be negative."},{"Start":"02:46.260 ","End":"02:50.745","Text":"Then the pH would be less than the pK_a."},{"Start":"02:50.745 ","End":"02:52.905","Text":"Here this would be negative,"},{"Start":"02:52.905 ","End":"02:57.000","Text":"so pH would be less than"},{"Start":"02:57.000 ","End":"03:02.525","Text":"the pK_a because it would be a pH equal to pK_a minus a small amount."},{"Start":"03:02.525 ","End":"03:05.269","Text":"The other hand, if we added a little base,"},{"Start":"03:05.269 ","End":"03:09.260","Text":"the ratio of R would now increase be greater than"},{"Start":"03:09.260 ","End":"03:13.500","Text":"1 because base would react with the acid."},{"Start":"03:13.500 ","End":"03:15.750","Text":"The amount of acid would decrease,"},{"Start":"03:15.750 ","End":"03:20.190","Text":"the amount of base would increase so R would be greater than 1 and"},{"Start":"03:20.190 ","End":"03:26.110","Text":"log R would be positive and then pH would be greater than pK_a."},{"Start":"03:26.110 ","End":"03:27.950","Text":"We\u0027re going to write what\u0027s called"},{"Start":"03:27.950 ","End":"03:33.215","Text":"the Henderson-Hasselbach equation for an acid buffer solution."},{"Start":"03:33.215 ","End":"03:37.400","Text":"Now, if we assume that adding a little acid or base has no effect on"},{"Start":"03:37.400 ","End":"03:41.860","Text":"the concentrations of the acetic acid and its conjugate base,"},{"Start":"03:41.860 ","End":"03:44.795","Text":"that\u0027s the acetate anion."},{"Start":"03:44.795 ","End":"03:49.240","Text":"We can rewrite the equation we had before."},{"Start":"03:49.240 ","End":"03:53.120","Text":"This is the equation we had before and we can rewrite it as"},{"Start":"03:53.120 ","End":"03:56.840","Text":"pH is equal to pK_a plus the log of"},{"Start":"03:56.840 ","End":"03:59.630","Text":"the initial concentration of"},{"Start":"03:59.630 ","End":"04:04.735","Text":"the conjugate base divide by the initial concentration of the acid."},{"Start":"04:04.735 ","End":"04:08.675","Text":"This is called the Henderson-Hasselbach equation."},{"Start":"04:08.675 ","End":"04:13.865","Text":"Suppose we want to write a similar equation for a base buffer solution."},{"Start":"04:13.865 ","End":"04:19.640","Text":"We\u0027re going to consider a solution with similar concentrations of the base ammonia and"},{"Start":"04:19.640 ","End":"04:25.715","Text":"the conjugate acid NH_4+ ammonium."},{"Start":"04:25.715 ","End":"04:29.085","Text":"K_b, very similar to what we did before,"},{"Start":"04:29.085 ","End":"04:32.270","Text":"K_b is the concentration of OH- times"},{"Start":"04:32.270 ","End":"04:36.620","Text":"the concentration NH_4+ divide by concentration of ammonia."},{"Start":"04:36.620 ","End":"04:39.655","Text":"We isolate the OH- we get"},{"Start":"04:39.655 ","End":"04:47.735","Text":"OH- concentrations K_b times concentration of the ammonia divided by the ammonium."},{"Start":"04:47.735 ","End":"04:50.905","Text":"Here\u0027s the equation again."},{"Start":"04:50.905 ","End":"04:55.555","Text":"Once again, we could write minus the log of both sides of the equation."},{"Start":"04:55.555 ","End":"05:00.220","Text":"There we get minus log of OH- is equal to minus log of K_b plus"},{"Start":"05:00.220 ","End":"05:05.555","Text":"the log of the ratio of the ammonium and the ammonia."},{"Start":"05:05.555 ","End":"05:15.370","Text":"We know that minus the log of OH- is pOH and minus the log of K_b is pK_b."},{"Start":"05:15.370 ","End":"05:19.375","Text":"Then we still have this log of the ratio."},{"Start":"05:19.375 ","End":"05:27.460","Text":"We have pOH is equal to pK_b plus the log of the conjugate acid divided by the base."},{"Start":"05:27.460 ","End":"05:30.060","Text":"This was a specific example."},{"Start":"05:30.060 ","End":"05:35.110","Text":"Here, NH_3 was a base and NH_4+ conjugate acid."},{"Start":"05:35.110 ","End":"05:38.915","Text":"Once again, if we assume that adding a little acid or base has"},{"Start":"05:38.915 ","End":"05:44.000","Text":"no effect on the concentrations of the base and its conjugate acid,"},{"Start":"05:44.000 ","End":"05:46.460","Text":"we can write this in terms of"},{"Start":"05:46.460 ","End":"05:48.950","Text":"the initial concentrations of"},{"Start":"05:48.950 ","End":"05:53.000","Text":"the conjugate acid divided by the initial concentration of the base."},{"Start":"05:53.000 ","End":"05:54.470","Text":"We have 2 equations,"},{"Start":"05:54.470 ","End":"05:56.915","Text":"1 for the acid and 1 for the base."},{"Start":"05:56.915 ","End":"06:03.245","Text":"Now it might be a good idea to have 1 equation for both the acid and the base buffers."},{"Start":"06:03.245 ","End":"06:09.825","Text":"If we take the equation we have for the base and we rewrite it."},{"Start":"06:09.825 ","End":"06:13.710","Text":"pOH we rewrite as pK_w minus"},{"Start":"06:13.710 ","End":"06:20.910","Text":"pH and pK_b we write as pK_w minus pK_a."},{"Start":"06:20.910 ","End":"06:28.055","Text":"Then we still have plus log of acid divide by conjugate base."},{"Start":"06:28.055 ","End":"06:33.290","Text":"I want the equation to look like the equation we had for the acid."},{"Start":"06:33.290 ","End":"06:35.990","Text":"I\u0027ve just written acid divide by conjugate base."},{"Start":"06:35.990 ","End":"06:41.665","Text":"It really doesn\u0027t matter which one is conjugate to which, they\u0027re interchanging."},{"Start":"06:41.665 ","End":"06:47.800","Text":"Now we can get rid of the pK_w and we can change the sign."},{"Start":"06:47.800 ","End":"06:51.080","Text":"Get rid of this negative sign."},{"Start":"06:51.080 ","End":"06:56.355","Text":"We have pH equal to pK_a."},{"Start":"06:56.355 ","End":"07:00.535","Text":"We got rid of the negative sign and we have a negative sign here."},{"Start":"07:00.535 ","End":"07:06.740","Text":"We can just interchange the numerator and denominator get a positive sign."},{"Start":"07:06.740 ","End":"07:16.490","Text":"Now we have pH is equal to pK_a plus the log of the conjugate base divided by the acid."},{"Start":"07:16.490 ","End":"07:20.075","Text":"Because I don\u0027t want this negative sign so I\u0027ve"},{"Start":"07:20.075 ","End":"07:24.815","Text":"changed it to positive and interchanged the new region denominator."},{"Start":"07:24.815 ","End":"07:29.254","Text":"This is now the same equation as we had for the acid."},{"Start":"07:29.254 ","End":"07:33.215","Text":"Here\u0027s some examples of how we can use this equation."},{"Start":"07:33.215 ","End":"07:37.880","Text":"If we have again, acetic acid and sodium acetate as a buffer solution,"},{"Start":"07:37.880 ","End":"07:40.745","Text":"we use the pK_a of acetic acid."},{"Start":"07:40.745 ","End":"07:46.590","Text":"The acid would be acetic acid and the conjugate base will be acetate ion."},{"Start":"07:46.750 ","End":"07:51.980","Text":"If we have ammonia and ammonium chloride buffer solution,"},{"Start":"07:51.980 ","End":"07:55.370","Text":"the pK_a will be of ammonium ion,"},{"Start":"07:55.370 ","End":"08:00.210","Text":"the acid will be ammonium ion and the base ammonia."},{"Start":"08:00.550 ","End":"08:07.080","Text":"In this video, we derived the Henderson-Hasselbach equation."}],"ID":31528},{"Watched":false,"Name":"Calculating pH of Buffer Solution","Duration":"4m 6s","ChapterTopicVideoID":28387,"CourseChapterTopicPlaylistID":282560,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.650","Text":"In the previous videos,"},{"Start":"00:01.650 ","End":"00:03.765","Text":"we talked about buffer solutions."},{"Start":"00:03.765 ","End":"00:07.965","Text":"In this video, we\u0027ll calculate the pH of a buffer solution."},{"Start":"00:07.965 ","End":"00:11.190","Text":"We\u0027re going to calculate the pH of a buffer solution."},{"Start":"00:11.190 ","End":"00:12.675","Text":"Here\u0027s an example."},{"Start":"00:12.675 ","End":"00:17.640","Text":"What\u0027s the pH of a solution containing 0.05 molar sodium"},{"Start":"00:17.640 ","End":"00:23.385","Text":"acetate and also 0.10 molar acetic acid."},{"Start":"00:23.385 ","End":"00:26.940","Text":"Here\u0027s our usual equilibrium for acetic acid."},{"Start":"00:26.940 ","End":"00:32.070","Text":"Acetic acid in water gives us H_3O plus and acetate."},{"Start":"00:32.070 ","End":"00:35.200","Text":"Here\u0027s our ICE table."},{"Start":"00:35.200 ","End":"00:41.645","Text":"We have 0.10 molar for acetic acid."},{"Start":"00:41.645 ","End":"00:44.675","Text":"At the beginning there\u0027s no H_3O plus."},{"Start":"00:44.675 ","End":"00:49.475","Text":"For acetate, we have 0.05 molar."},{"Start":"00:49.475 ","End":"00:54.620","Text":"The change is minus x for the acetic acid,"},{"Start":"00:54.620 ","End":"00:56.915","Text":"plus x for H_3O plus,"},{"Start":"00:56.915 ","End":"00:59.545","Text":"and plus x for the acetate."},{"Start":"00:59.545 ","End":"01:03.270","Text":"At equilibrium, for acetic acid,"},{"Start":"01:03.270 ","End":"01:06.585","Text":"we have 0.10 minus x,"},{"Start":"01:06.585 ","End":"01:09.810","Text":"for H_3O plus we have plus x,"},{"Start":"01:09.810 ","End":"01:14.195","Text":"and for acetate, we have 0.05 plus x."},{"Start":"01:14.195 ","End":"01:15.620","Text":"We can write that K_a,"},{"Start":"01:15.620 ","End":"01:19.775","Text":"which is 1.8 times 10 to the power minus 5 for acetic acid,"},{"Start":"01:19.775 ","End":"01:25.455","Text":"is equal to 0.05 plus x times x,"},{"Start":"01:25.455 ","End":"01:26.740","Text":"that\u0027s the numerator,"},{"Start":"01:26.740 ","End":"01:31.160","Text":"divide by 0.10 minus x, that\u0027s the denominator."},{"Start":"01:31.160 ","End":"01:33.515","Text":"If x is small,"},{"Start":"01:33.515 ","End":"01:36.605","Text":"we can ignore it in the numerator and the denominator,"},{"Start":"01:36.605 ","End":"01:43.685","Text":"and we get 0.05 divide by 0.10, which is 1/2."},{"Start":"01:43.685 ","End":"01:47.690","Text":"The x remains, so we get x divided by 2."},{"Start":"01:47.690 ","End":"01:52.545","Text":"That means that x=K_a times 2,"},{"Start":"01:52.545 ","End":"01:56.350","Text":"3.6 times 10 to the power minus 5."},{"Start":"01:56.350 ","End":"02:04.375","Text":"We see that it was correct to ignore it in comparison to 0.05 and 0.10,"},{"Start":"02:04.375 ","End":"02:06.215","Text":"is a good approximation."},{"Start":"02:06.215 ","End":"02:10.450","Text":"The concentration of H_3O plus is x. I\u0027ve written here,"},{"Start":"02:10.450 ","End":"02:12.890","Text":"concentration of H_3O plus is x,"},{"Start":"02:12.890 ","End":"02:17.320","Text":"and that\u0027s 3.6 times 10 to the power minus 5 molar."},{"Start":"02:17.320 ","End":"02:22.530","Text":"That means that the pH is 4.44."},{"Start":"02:22.530 ","End":"02:27.135","Text":"4.44 is less than pK_a,"},{"Start":"02:27.135 ","End":"02:31.710","Text":"pK_a acetic acid is 4.74."},{"Start":"02:31.710 ","End":"02:41.925","Text":"We have a pH that\u0027s smaller than pK_a."},{"Start":"02:41.925 ","End":"02:49.219","Text":"When we neglected this x is the same as assuming concentrations haven\u0027t changed."},{"Start":"02:49.219 ","End":"02:56.830","Text":"The concentration stayed 0.05 for the acetate,"},{"Start":"02:56.830 ","End":"03:00.675","Text":"and 0.10 for the acetic acid."},{"Start":"03:00.675 ","End":"03:06.120","Text":"Tentatively, we could\u0027ve used the Henderson-Hasselbach equation."},{"Start":"03:06.120 ","End":"03:13.670","Text":"That says that the pH=pK_a plus the log of the acetate,"},{"Start":"03:13.670 ","End":"03:16.040","Text":"the initial concentration of the acetate"},{"Start":"03:16.040 ","End":"03:19.250","Text":"divide by the initial concentration of acetic acid,"},{"Start":"03:19.250 ","End":"03:26.235","Text":"and that\u0027s 4.74 for the pK_a plus the log,"},{"Start":"03:26.235 ","End":"03:34.460","Text":"and then we have 0.05 for the acetate divided by 0.10 for the acetic acid."},{"Start":"03:34.460 ","End":"03:37.480","Text":"That\u0027s the log of 1/2."},{"Start":"03:37.480 ","End":"03:42.930","Text":"We work that out, it\u0027s 4.74 minus 0.30."},{"Start":"03:42.930 ","End":"03:46.070","Text":"That gives us 4.44,"},{"Start":"03:46.070 ","End":"03:49.265","Text":"which is identical to what we got before."},{"Start":"03:49.265 ","End":"03:53.460","Text":"We got the same answers before and again,"},{"Start":"03:53.460 ","End":"03:56.285","Text":"in writing the Henderson-Hasselbach equation,"},{"Start":"03:56.285 ","End":"04:00.995","Text":"we\u0027ve assumed that the initial concentrations haven\u0027t changed."},{"Start":"04:00.995 ","End":"04:05.850","Text":"In this video, we calculated pH of a buffer solution."}],"ID":31529},{"Watched":false,"Name":"Preparing Buffer Solution with Specific pH","Duration":"4m 57s","ChapterTopicVideoID":28389,"CourseChapterTopicPlaylistID":282560,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.145","Text":"In previous videos,"},{"Start":"00:02.145 ","End":"00:04.305","Text":"we talked about buffer solutions."},{"Start":"00:04.305 ","End":"00:10.215","Text":"In this video, we\u0027ll show how to prepare a buffer solution with a specific pH."},{"Start":"00:10.215 ","End":"00:13.665","Text":"We\u0027re going to talk about how to prepare a buffer solution"},{"Start":"00:13.665 ","End":"00:18.630","Text":"of a specific pH and this will be an acidic pH."},{"Start":"00:18.630 ","End":"00:26.865","Text":"What we need to do is to choose an acid with a pK_a close to the desired pH."},{"Start":"00:26.865 ","End":"00:30.465","Text":"Then we go to the Henderson-Hasselbach equation,"},{"Start":"00:30.465 ","End":"00:35.720","Text":"which says that pH is equal to pK_a plus the log of"},{"Start":"00:35.720 ","End":"00:39.061","Text":"the initial volume of the concentration of"},{"Start":"00:39.061 ","End":"00:43.595","Text":"the conjugate base divided by the initial value of the concentration of the acid,"},{"Start":"00:43.595 ","End":"00:50.180","Text":"and we calculate the ratio of base to acid that we need to prepare."},{"Start":"00:50.180 ","End":"00:55.250","Text":"Then we prepare a buffer solution with this ratio."},{"Start":"00:55.250 ","End":"00:56.865","Text":"Here\u0027s an example."},{"Start":"00:56.865 ","End":"00:59.085","Text":"What mass of sodium acetate,"},{"Start":"00:59.085 ","End":"01:03.275","Text":"we\u0027re given that the molar mass is 82 grams per mole,"},{"Start":"01:03.275 ","End":"01:09.050","Text":"must be dissolved in 250 milliliters of 0.25"},{"Start":"01:09.050 ","End":"01:15.100","Text":"molar acetic acid to produce a solution with pH equal to 5."},{"Start":"01:15.100 ","End":"01:20.845","Text":"We\u0027re going to assume that the volume of the buffer solution stays at 250 milliliters."},{"Start":"01:20.845 ","End":"01:26.140","Text":"With adding the sodium acetate doesn\u0027t make any difference to the volume."},{"Start":"01:26.140 ","End":"01:32.410","Text":"We are also given the information that the pK_a of acetic acid is 4.74."},{"Start":"01:32.410 ","End":"01:36.265","Text":"The pH we\u0027re aiming for is 5."},{"Start":"01:36.265 ","End":"01:43.735","Text":"We can calculate from the Henderson-Hasselbach equation that the log of this ratio,"},{"Start":"01:43.735 ","End":"01:48.340","Text":"which is equal to the pH minus pK_a is 5."},{"Start":"01:48.340 ","End":"01:50.080","Text":"That\u0027s the pH we\u0027re looking for,"},{"Start":"01:50.080 ","End":"01:54.325","Text":"minus the value of the pK_a, which is 4.74."},{"Start":"01:54.325 ","End":"02:00.270","Text":"So 5 minus 4.74 is 0.26."},{"Start":"02:00.270 ","End":"02:06.755","Text":"We need this log of this ratio to be 0.26."},{"Start":"02:06.755 ","End":"02:15.970","Text":"That means that the ratio itself is 10^0.26, and that\u0027s 1.82."},{"Start":"02:15.970 ","End":"02:21.355","Text":"We need the ratio itself to be 1.82."},{"Start":"02:21.355 ","End":"02:28.900","Text":"Now, we\u0027re told that the initial concentration of acetic acid is 0.25 molar."},{"Start":"02:28.900 ","End":"02:32.880","Text":"That means that the concentration of"},{"Start":"02:32.880 ","End":"02:38.975","Text":"the sodium acetate has to be"},{"Start":"02:38.975 ","End":"02:45.410","Text":"0.250 times 1.82 from this ratio here."},{"Start":"02:45.410 ","End":"02:52.925","Text":"Here we have the acetate is equal to the concentration of the acetic acid times 1.82,"},{"Start":"02:52.925 ","End":"02:57.855","Text":"and that\u0027s 0.455 molar."},{"Start":"02:57.855 ","End":"03:04.010","Text":"We know that we need the acetate to have this concentration."},{"Start":"03:04.010 ","End":"03:08.075","Text":"Now, we need to know the number of moles."},{"Start":"03:08.075 ","End":"03:16.070","Text":"Now, the number of moles of the acetate is equal to the molarity times the volume."},{"Start":"03:16.070 ","End":"03:19.100","Text":"Remember, the volume has to be in liters."},{"Start":"03:19.100 ","End":"03:26.160","Text":"That\u0027s 0.455 times 0.250 liters."},{"Start":"03:26.160 ","End":"03:31.055","Text":"If we work that out, we get 0.114 moles."},{"Start":"03:31.055 ","End":"03:36.500","Text":"The number of moles we need to prepare is 0.114."},{"Start":"03:36.500 ","End":"03:39.350","Text":"We know that the mass of"},{"Start":"03:39.350 ","End":"03:45.370","Text":"the acetate is equal to the number of moles times the molar mass."},{"Start":"03:45.370 ","End":"03:49.845","Text":"The number of moles is 0.114,"},{"Start":"03:49.845 ","End":"03:52.190","Text":"and the molar mass,"},{"Start":"03:52.190 ","End":"03:57.065","Text":"we were given that that\u0027s 82 grams per mole."},{"Start":"03:57.065 ","End":"04:03.005","Text":"Now, multiply these 2 and we get 9.35 grams."},{"Start":"04:03.005 ","End":"04:07.840","Text":"That\u0027s the mass of sodium acetate we need to prepare."},{"Start":"04:07.840 ","End":"04:09.900","Text":"Now a note."},{"Start":"04:09.900 ","End":"04:14.955","Text":"In practice as the Henderson-Hasselbach equation is only approximate,"},{"Start":"04:14.955 ","End":"04:21.370","Text":"it is used to make a first estimate of the acid and conjugate base concentrations."},{"Start":"04:21.370 ","End":"04:24.965","Text":"This is the first estimate of how much we need."},{"Start":"04:24.965 ","End":"04:31.565","Text":"Then we can add more acid or base while monitoring the pH with a pH meter."},{"Start":"04:31.565 ","End":"04:36.575","Text":"pH meter is a very accurate way of finding the pH."},{"Start":"04:36.575 ","End":"04:38.795","Text":"If we have a pH meter,"},{"Start":"04:38.795 ","End":"04:45.965","Text":"then we can adjust the amount of acid or acetate so that we get the correct pH,"},{"Start":"04:45.965 ","End":"04:48.845","Text":"the pH we desire."},{"Start":"04:48.845 ","End":"04:56.520","Text":"In this video, we learned how to prepare a buffer solution with a specific pH."}],"ID":31530},{"Watched":false,"Name":"Calculating Change in pH of Buffer Solution","Duration":"5m 54s","ChapterTopicVideoID":28386,"CourseChapterTopicPlaylistID":282560,"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 a previous video,"},{"Start":"00:02.010 ","End":"00:05.445","Text":"we calculated the pH of a buffer solution."},{"Start":"00:05.445 ","End":"00:08.070","Text":"We\u0027re going to calculate the change in the pH of"},{"Start":"00:08.070 ","End":"00:12.180","Text":"a buffer solution when we add a small amount of acid."},{"Start":"00:12.180 ","End":"00:13.785","Text":"Here\u0027s an example,"},{"Start":"00:13.785 ","End":"00:21.890","Text":"what is the pH when 0.005 moles of HCL is added to 500 milliliters of"},{"Start":"00:21.890 ","End":"00:26.630","Text":"a buffer solution and the buffer solution consists of 0.05"},{"Start":"00:26.630 ","End":"00:32.555","Text":"molar sodium acetate and 0.10 molar acetic acid?"},{"Start":"00:32.555 ","End":"00:34.295","Text":"Now in a previous video,"},{"Start":"00:34.295 ","End":"00:37.970","Text":"we calculated the pH of a buffer solution like this."},{"Start":"00:37.970 ","End":"00:42.890","Text":"Ph is equal to pK_a plus the log of the ratio of"},{"Start":"00:42.890 ","End":"00:48.330","Text":"the concentration of the acetate and the concentration of the acetic acid,"},{"Start":"00:48.330 ","End":"00:50.690","Text":"that\u0027s the initial concentrations."},{"Start":"00:50.690 ","End":"00:57.225","Text":"That\u0027s equal to 4.74 for the pK_a plus the log of this ratio,"},{"Start":"00:57.225 ","End":"01:06.220","Text":"so the acetate, it\u0027s 0.05 and for the acetic acid is 0.10."},{"Start":"01:06.560 ","End":"01:15.695","Text":"The ratio is 0.05 divided by 0.10 and that of course, is a half."},{"Start":"01:15.695 ","End":"01:20.925","Text":"The pH has a pK_a of 4.74 plus the log of this ratio,"},{"Start":"01:20.925 ","End":"01:22.760","Text":"that means the log of a 1/2,"},{"Start":"01:22.760 ","End":"01:26.960","Text":"and that\u0027s 4.74 minus 0.30,"},{"Start":"01:26.960 ","End":"01:30.755","Text":"and the answer is 4.44."},{"Start":"01:30.755 ","End":"01:33.875","Text":"That\u0027s the pH of our buffer solution."},{"Start":"01:33.875 ","End":"01:38.780","Text":"Now, let\u0027s calculate how many moles of acetate"},{"Start":"01:38.780 ","End":"01:44.495","Text":"we have and how many moles of acetic acid we have in this buffer solution."},{"Start":"01:44.495 ","End":"01:55.130","Text":"For the acetate, M is 0.05 and the volume in liters is 0.500."},{"Start":"01:55.130 ","End":"02:00.145","Text":"The multiplication gives us 0.025 moles."},{"Start":"02:00.145 ","End":"02:02.225","Text":"Now, for the acetic acid,"},{"Start":"02:02.225 ","End":"02:04.385","Text":"again, it\u0027s M times V,"},{"Start":"02:04.385 ","End":"02:08.565","Text":"that\u0027s 0.10 M times"},{"Start":"02:08.565 ","End":"02:16.329","Text":"0.500 liters and that product is 0.050 moles."},{"Start":"02:16.329 ","End":"02:25.960","Text":"Now we\u0027re going to add the HCL and the number of moles of HCL is 0.005,"},{"Start":"02:25.960 ","End":"02:32.775","Text":"so we\u0027ll have 0.005 moles of H_3O plus."},{"Start":"02:32.775 ","End":"02:39.520","Text":"H_3O plus will react with the acetate to give us acetic acid and water."},{"Start":"02:39.520 ","End":"02:42.360","Text":"Here\u0027s our table, initially,"},{"Start":"02:42.360 ","End":"02:47.760","Text":"we have 0.005 moles of H_3O plus."},{"Start":"02:47.760 ","End":"02:52.470","Text":"We also have 0.025 moles of"},{"Start":"02:52.470 ","End":"02:58.820","Text":"acetate and 0.050 moles of acetic acid."},{"Start":"02:58.820 ","End":"03:05.270","Text":"Now we can see here that the acetate is in excess of the H_3O plus,"},{"Start":"03:05.270 ","End":"03:12.405","Text":"so only part of the acetate will react and all of the H_3O plus will react."},{"Start":"03:12.405 ","End":"03:21.420","Text":"Now, the amount of H_3O plus will decrease to 0 by 0.005,"},{"Start":"03:21.420 ","End":"03:27.630","Text":"all of it will react and 0.005 of the acetate will react,"},{"Start":"03:27.630 ","End":"03:30.870","Text":"so that\u0027s again a minus sign and"},{"Start":"03:30.870 ","End":"03:39.850","Text":"0.05 moles of acetic acid will be produced."},{"Start":"03:39.850 ","End":"03:43.040","Text":"In the final reckoning,"},{"Start":"03:43.040 ","End":"03:45.470","Text":"we have no H_3O plus,"},{"Start":"03:45.470 ","End":"03:47.630","Text":"F is for final."},{"Start":"03:47.630 ","End":"03:57.520","Text":"We have 0.025 minus 0.005 acetate, that\u0027s 0.020."},{"Start":"03:57.520 ","End":"04:04.490","Text":"Acetic acid we have 0.050 plus 0.005."},{"Start":"04:04.490 ","End":"04:08.360","Text":"Now, this is the amount of the number of moles of"},{"Start":"04:08.360 ","End":"04:14.170","Text":"acetate versus number of moles of acetic acid that we have."},{"Start":"04:14.170 ","End":"04:20.725","Text":"We can use Henderson-Hasselbach equation to calculate the pH."},{"Start":"04:20.725 ","End":"04:30.920","Text":"The pH is the pK_a 4.74 plus the log of the acetate divided by the acetic acid."},{"Start":"04:30.920 ","End":"04:33.200","Text":"Now, we can calculate"},{"Start":"04:33.200 ","End":"04:37.820","Text":"the concentrations but there\u0027s no necessity because the volume is the"},{"Start":"04:37.820 ","End":"04:45.305","Text":"same in the initial value of the acetate and the acetic acid,"},{"Start":"04:45.305 ","End":"04:46.955","Text":"it\u0027s the same volumes."},{"Start":"04:46.955 ","End":"04:50.105","Text":"So we can just divide the number of moles."},{"Start":"04:50.105 ","End":"04:55.790","Text":"It\u0027s 4.74 plus the log of the number of moles of acetate"},{"Start":"04:55.790 ","End":"05:01.800","Text":"0.020 divide by the number of moles of acetic acid,"},{"Start":"05:01.800 ","End":"05:11.180","Text":"0.055, here it is and the result is 4.74 plus this log of this ratio,"},{"Start":"05:11.180 ","End":"05:16.835","Text":"that\u0027s 4.74 minus 0.44."},{"Start":"05:16.835 ","End":"05:20.090","Text":"That gives us 4.30."},{"Start":"05:20.090 ","End":"05:26.955","Text":"We can see that initially the pH of the buffer solution was 4.44,"},{"Start":"05:26.955 ","End":"05:30.710","Text":"and now it\u0027s decreased to 4.30."},{"Start":"05:30.710 ","End":"05:35.105","Text":"When we added the acid, the pH decreased,"},{"Start":"05:35.105 ","End":"05:36.800","Text":"but not by enormous amount,"},{"Start":"05:36.800 ","End":"05:38.330","Text":"by a small amount,"},{"Start":"05:38.330 ","End":"05:40.805","Text":"so our buffer solution is very good."},{"Start":"05:40.805 ","End":"05:45.320","Text":"This is just to say that since the volume is the same for both the acid and base,"},{"Start":"05:45.320 ","End":"05:48.065","Text":"we can use moles rather molarity."},{"Start":"05:48.065 ","End":"05:54.060","Text":"In this video, we calculated the change in pH of a buffer solution."}],"ID":31531},{"Watched":false,"Name":"Buffer Capacity and Buffer Range","Duration":"4m 13s","ChapterTopicVideoID":28390,"CourseChapterTopicPlaylistID":282560,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.260 ","End":"00:02.610","Text":"In the previous video,"},{"Start":"00:02.610 ","End":"00:07.155","Text":"we talked about preparing a buffer solution with a particular pH,"},{"Start":"00:07.155 ","End":"00:12.419","Text":"and in this video, we\u0027ll talk about buffer capacity and buffer range."},{"Start":"00:12.419 ","End":"00:15.510","Text":"Let\u0027s begin with buffer capacity."},{"Start":"00:15.510 ","End":"00:19.410","Text":"There are limits to how much strong acid or base can be added"},{"Start":"00:19.410 ","End":"00:23.640","Text":"to a buffer solution without changing its pH drastically."},{"Start":"00:23.640 ","End":"00:27.840","Text":"This is because the buffer\u0027s proton sources and sinks"},{"Start":"00:27.840 ","End":"00:32.220","Text":"become exhausted if too much strong acid or base is added."},{"Start":"00:32.220 ","End":"00:36.510","Text":"Now the buffer capacity is the maximum amount of acid or"},{"Start":"00:36.510 ","End":"00:42.000","Text":"base that can be added without changing the pH drastically,"},{"Start":"00:42.000 ","End":"00:49.200","Text":"and of course, high buffer capacity is better than low buffer capacity."},{"Start":"00:49.390 ","End":"00:53.480","Text":"Now the buffer action ceases when most of"},{"Start":"00:53.480 ","End":"00:57.005","Text":"the weak acid has been converted to its conjugate base,"},{"Start":"00:57.005 ","End":"01:01.310","Text":"or a weak base has been converted to its conjugate acid."},{"Start":"01:01.310 ","End":"01:06.110","Text":"The final comment here is that the concentrated buffer solution has"},{"Start":"01:06.110 ","End":"01:11.930","Text":"a greater capacity than the same volume of a dilute solution of the same buffer."},{"Start":"01:11.930 ","End":"01:14.915","Text":"Now we\u0027re going to talk about the buffer range."},{"Start":"01:14.915 ","End":"01:18.680","Text":"Now it can be shown experimentally that the buffer capacity is"},{"Start":"01:18.680 ","End":"01:23.840","Text":"high as long as the ratio of the concentration of"},{"Start":"01:23.840 ","End":"01:33.490","Text":"the conjugate base to the concentration of the acid lies between 1 over 10 and 10."},{"Start":"01:33.490 ","End":"01:36.710","Text":"We\u0027re talking about the initial values."},{"Start":"01:36.710 ","End":"01:40.160","Text":"Now if we take the log of the left hand side,"},{"Start":"01:40.160 ","End":"01:44.300","Text":"the log of 1 over 10 is minus 1."},{"Start":"01:44.300 ","End":"01:46.235","Text":"Then we have in the middle,"},{"Start":"01:46.235 ","End":"01:48.790","Text":"the log of the ratio."},{"Start":"01:48.790 ","End":"01:51.865","Text":"The log of 10 is 1."},{"Start":"01:51.865 ","End":"01:57.320","Text":"So we see that the buffer capacity is high as long"},{"Start":"01:57.320 ","End":"02:02.770","Text":"as the log of this ratio lies between minus 1 and 1."},{"Start":"02:02.770 ","End":"02:06.800","Text":"Now if we use the Henderson-Hasselbach equation that says that pH"},{"Start":"02:06.800 ","End":"02:11.105","Text":"is equal to pK_a plus the log of the ratio,"},{"Start":"02:11.105 ","End":"02:19.010","Text":"we can see that the buffer range is pH equal to pK_a plus or minus 1."},{"Start":"02:19.010 ","End":"02:20.585","Text":"Instead of the log,"},{"Start":"02:20.585 ","End":"02:22.955","Text":"I\u0027ve written plus or minus 1."},{"Start":"02:22.955 ","End":"02:26.270","Text":"So this is the buffer range."},{"Start":"02:26.270 ","End":"02:29.045","Text":"Here\u0027s some examples."},{"Start":"02:29.045 ","End":"02:31.160","Text":"Now, for the acetic acid,"},{"Start":"02:31.160 ","End":"02:33.200","Text":"sodium acetate buffer solution,"},{"Start":"02:33.200 ","End":"02:37.525","Text":"the pH is 4.74 plus or minus 1."},{"Start":"02:37.525 ","End":"02:39.150","Text":"That\u0027s the range."},{"Start":"02:39.150 ","End":"02:47.605","Text":"This is 4.74 is the pK_a of acetic acid and this is the range."},{"Start":"02:47.605 ","End":"02:53.825","Text":"Now for a basic buffer such as ammonia and ammonium chloride,"},{"Start":"02:53.825 ","End":"02:58.810","Text":"the pH is 9.26 plus or minus 1."},{"Start":"02:58.810 ","End":"03:01.170","Text":"Now how did we get to that?"},{"Start":"03:01.170 ","End":"03:10.160","Text":"The pK_a of ammonium is equal to pK_w minus pK_b of ammonia."},{"Start":"03:10.160 ","End":"03:15.035","Text":"That\u0027s 14 minus 4.74,"},{"Start":"03:15.035 ","End":"03:17.675","Text":"and it gives us 9.26."},{"Start":"03:17.675 ","End":"03:20.585","Text":"So that\u0027s how we got the 9.26."},{"Start":"03:20.585 ","End":"03:25.330","Text":"Now you might wonder how I got this relationship,"},{"Start":"03:25.330 ","End":"03:27.560","Text":"and in order to get it,"},{"Start":"03:27.560 ","End":"03:34.880","Text":"we have to recall that K_a times K_b is equal to K_w."},{"Start":"03:34.880 ","End":"03:38.405","Text":"So minus log of K_a,"},{"Start":"03:38.405 ","End":"03:46.234","Text":"minus log of K_b is equal to minus log of K_w."},{"Start":"03:46.234 ","End":"03:55.410","Text":"Minus log of K_a is pK_a minus log of K_b is pK_b,"},{"Start":"03:55.410 ","End":"04:00.300","Text":"and minus log of K_w is pK_w."},{"Start":"04:00.300 ","End":"04:02.610","Text":"That\u0027s how we got this expression,"},{"Start":"04:02.610 ","End":"04:07.070","Text":"pK_a plus pK_b is equal to pK_w."},{"Start":"04:07.070 ","End":"04:08.570","Text":"So in this video,"},{"Start":"04:08.570 ","End":"04:13.560","Text":"we talked about buffer capacity and buffer range."}],"ID":31532},{"Watched":false,"Name":"Exercise 1","Duration":"5m 40s","ChapterTopicVideoID":29816,"CourseChapterTopicPlaylistID":282560,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31533},{"Watched":false,"Name":"Exercise 2","Duration":"6m 53s","ChapterTopicVideoID":29817,"CourseChapterTopicPlaylistID":282560,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31534},{"Watched":false,"Name":"Exercise 2 Alternative","Duration":"2m 50s","ChapterTopicVideoID":29818,"CourseChapterTopicPlaylistID":282560,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31535},{"Watched":false,"Name":"Exercise 3","Duration":"6m 18s","ChapterTopicVideoID":29819,"CourseChapterTopicPlaylistID":282560,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31536},{"Watched":false,"Name":"Exercise 4","Duration":"5m 26s","ChapterTopicVideoID":29820,"CourseChapterTopicPlaylistID":282560,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31537}],"Thumbnail":null,"ID":282560},{"Name":"Acid-Base Indicators","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Indicators","Duration":"7m 12s","ChapterTopicVideoID":28385,"CourseChapterTopicPlaylistID":282561,"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":"00:05.940","Text":"we talked about buffer capacity and buffer range."},{"Start":"00:05.940 ","End":"00:10.740","Text":"In this video, we\u0027ll learn about acid-base indicators."},{"Start":"00:10.740 ","End":"00:14.790","Text":"We\u0027re going to talk about acid-base indicators."},{"Start":"00:14.790 ","End":"00:19.410","Text":"Now, these are indicators that change color according"},{"Start":"00:19.410 ","End":"00:24.630","Text":"to the pH of the solution to which they are added."},{"Start":"00:24.630 ","End":"00:28.200","Text":"Acid-base indicators change color."},{"Start":"00:28.200 ","End":"00:30.330","Text":"We\u0027ll see how they change color."},{"Start":"00:30.330 ","End":"00:32.490","Text":"Now, they have 2 forms,"},{"Start":"00:32.490 ","End":"00:34.190","Text":"a weak acid form,"},{"Start":"00:34.190 ","End":"00:35.630","Text":"which we\u0027re going to call HIn,"},{"Start":"00:35.630 ","End":"00:38.460","Text":"and that\u0027s one color."},{"Start":"00:38.460 ","End":"00:40.565","Text":"Then there\u0027s another form,"},{"Start":"00:40.565 ","End":"00:44.540","Text":"and that\u0027s its conjugate base, In minus."},{"Start":"00:44.540 ","End":"00:46.580","Text":"We\u0027re just going to call it In minus,"},{"Start":"00:46.580 ","End":"00:48.230","Text":"which has a different color."},{"Start":"00:48.230 ","End":"00:54.845","Text":"Two colors; an HIn color and its conjugate base In minus color,"},{"Start":"00:54.845 ","End":"00:59.645","Text":"and sometimes there\u0027s an intermediate color when it changes from one to the other."},{"Start":"00:59.645 ","End":"01:04.340","Text":"Now, we only use a very small amount of indicator so that"},{"Start":"01:04.340 ","End":"01:09.020","Text":"it doesn\u0027t change the pH of the solution we\u0027re trying to investigate."},{"Start":"01:09.020 ","End":"01:14.485","Text":"The indicator changes color according to the pH of the solution."},{"Start":"01:14.485 ","End":"01:21.665","Text":"Let\u0027s look at the ionization equilibrium of this HIn, of this indicator."},{"Start":"01:21.665 ","End":"01:25.370","Text":"Once again, an indicator is a weak acid which has"},{"Start":"01:25.370 ","End":"01:29.365","Text":"a different color in its acid and conjugate base forms."},{"Start":"01:29.365 ","End":"01:35.110","Text":"We can use Henderson-Hasselbach equation to describe it."},{"Start":"01:35.110 ","End":"01:42.380","Text":"Here\u0027s HIn reacts with water to give us H_3O plus, plus In minus."},{"Start":"01:42.380 ","End":"01:44.405","Text":"Since it\u0027s a very small amount,"},{"Start":"01:44.405 ","End":"01:48.815","Text":"it\u0027s not going to change the H_3O plus of our solution."},{"Start":"01:48.815 ","End":"01:52.400","Text":"This is the acid form and this is the base form,"},{"Start":"01:52.400 ","End":"01:56.705","Text":"and here\u0027s our Henderson-Hasselbach equation for the indicator."},{"Start":"01:56.705 ","End":"02:01.195","Text":"The pH is equal to the pK of the HIn,"},{"Start":"02:01.195 ","End":"02:05.317","Text":"the pK of the indicator plus the log of the In"},{"Start":"02:05.317 ","End":"02:11.030","Text":"minus concentration divided by the HIn concentration."},{"Start":"02:11.030 ","End":"02:15.870","Text":"Now, experimentally it can be shown that if this ratio In"},{"Start":"02:15.870 ","End":"02:21.800","Text":"minus divided by HIn is less than 0.1,"},{"Start":"02:21.800 ","End":"02:26.560","Text":"that means the log is less than minus 1,"},{"Start":"02:26.560 ","End":"02:32.980","Text":"so that the pH is less than the pKa minus 1,"},{"Start":"02:32.980 ","End":"02:35.230","Text":"we\u0027ll get an acid color."},{"Start":"02:35.230 ","End":"02:37.270","Text":"Now if on the other hand,"},{"Start":"02:37.270 ","End":"02:39.940","Text":"the ratio is greater than 10,"},{"Start":"02:39.940 ","End":"02:46.145","Text":"that means the log of the ratio will be greater than plus 1."},{"Start":"02:46.145 ","End":"02:53.985","Text":"Then the pH will be greater than the pK of the indicator plus 1."},{"Start":"02:53.985 ","End":"02:59.455","Text":"Then when that pH is greater than pK of the indicator plus 1,"},{"Start":"02:59.455 ","End":"03:01.640","Text":"we get the base color."},{"Start":"03:01.640 ","End":"03:04.930","Text":"If the ratio is approximately 1,"},{"Start":"03:04.930 ","End":"03:07.705","Text":"that means the log is approximately 0."},{"Start":"03:07.705 ","End":"03:13.985","Text":"The pH will be approximately equal to the pK of the indicator,"},{"Start":"03:13.985 ","End":"03:16.880","Text":"and then we\u0027ll have an intermediate color."},{"Start":"03:16.880 ","End":"03:20.000","Text":"What\u0027s the range of the indicator,"},{"Start":"03:20.000 ","End":"03:23.650","Text":"within which range will it change color?"},{"Start":"03:23.650 ","End":"03:28.400","Text":"We can see from what we got before that the indicator range is"},{"Start":"03:28.400 ","End":"03:34.820","Text":"pH equal to pK of the indicator plus or minus 1."},{"Start":"03:34.820 ","End":"03:36.860","Text":"This is approximate,"},{"Start":"03:36.860 ","End":"03:40.580","Text":"it\u0027s slightly different for different indicators."},{"Start":"03:40.580 ","End":"03:44.825","Text":"The exact range varies from indicator to indicator,"},{"Start":"03:44.825 ","End":"03:50.500","Text":"and it can also change according to the solvent indicators dissolved in."},{"Start":"03:50.500 ","End":"03:52.685","Text":"Now, what\u0027s the endpoint?"},{"Start":"03:52.685 ","End":"03:56.573","Text":"The endpoint of an indicator is a point at which In"},{"Start":"03:56.573 ","End":"04:01.090","Text":"minus concentration is equal to the concentration of HIn."},{"Start":"04:01.090 ","End":"04:06.215","Text":"At that point, the pH is equal to pK of the indicator."},{"Start":"04:06.215 ","End":"04:09.680","Text":"This is the point at which it\u0027s changing its color,"},{"Start":"04:09.680 ","End":"04:13.820","Text":"it\u0027s very, very important when you\u0027re doing titrations."},{"Start":"04:13.820 ","End":"04:16.910","Text":"When we use an indicator in a titration,"},{"Start":"04:16.910 ","End":"04:21.305","Text":"we\u0027re going to talk a lot in the next few videos about titrations."},{"Start":"04:21.305 ","End":"04:25.490","Text":"We should choose one whose endpoint is close to"},{"Start":"04:25.490 ","End":"04:29.735","Text":"what\u0027s called the stoichiometric or equivalence point of the titration."},{"Start":"04:29.735 ","End":"04:36.515","Text":"The point at which the acids and base have the same number of moles,"},{"Start":"04:36.515 ","End":"04:39.260","Text":"in which they react with each other."},{"Start":"04:39.260 ","End":"04:43.460","Text":"The endpoint belongs to the indicator and"},{"Start":"04:43.460 ","End":"04:48.215","Text":"the stoichiometric or equivalence point belongs to the acid and base."},{"Start":"04:48.215 ","End":"04:50.200","Text":"Here\u0027s some examples."},{"Start":"04:50.200 ","End":"04:54.620","Text":"Bromothymol blue has a pK of 7.1."},{"Start":"04:54.620 ","End":"04:58.395","Text":"At 6.1, less than 6.1, it\u0027s yellow."},{"Start":"04:58.395 ","End":"05:00.830","Text":"At 7.1 is green,"},{"Start":"05:00.830 ","End":"05:05.990","Text":"and then at greater than 8.1 it changes to blue."},{"Start":"05:05.990 ","End":"05:08.960","Text":"We\u0027ll see when we do the titration,"},{"Start":"05:08.960 ","End":"05:12.235","Text":"it rapidly goes from yellow to green to blue."},{"Start":"05:12.235 ","End":"05:14.520","Text":"At the endpoint it\u0027s green."},{"Start":"05:14.520 ","End":"05:19.324","Text":"Methyl orange has a pK of 3.5 in water,"},{"Start":"05:19.324 ","End":"05:23.570","Text":"and then it goes from red when the pH is less than"},{"Start":"05:23.570 ","End":"05:28.820","Text":"2.5 to orange when the pH is 3.5 approximately."},{"Start":"05:28.820 ","End":"05:34.100","Text":"Then to yellow when the pH is greater than 4.4."},{"Start":"05:34.100 ","End":"05:41.150","Text":"The endpoint is going very rapidly from red to orange to yellow."},{"Start":"05:41.150 ","End":"05:43.100","Text":"By very rapidly, I mean,"},{"Start":"05:43.100 ","End":"05:47.645","Text":"as we add a tiny bit more of the base or acid."},{"Start":"05:47.645 ","End":"05:50.215","Text":"Now here\u0027s some applications."},{"Start":"05:50.215 ","End":"05:55.370","Text":"These indicators are very useful when only the approximate pH is acquired."},{"Start":"05:55.370 ","End":"05:58.130","Text":"This could be in field studies."},{"Start":"05:58.130 ","End":"06:00.380","Text":"For example, when we\u0027re outside the lab,"},{"Start":"06:00.380 ","End":"06:05.030","Text":"we dissolve the indicator in a solvent and add a few drops to"},{"Start":"06:05.030 ","End":"06:10.370","Text":"the solution that we want to titrate or whose pH needs to be determined."},{"Start":"06:10.370 ","End":"06:16.820","Text":"Another thing we could do is to take porous paper and drip some of the indicator onto it,"},{"Start":"06:16.820 ","End":"06:21.020","Text":"impregnate the porous paper with the indicator and then dry it."},{"Start":"06:21.020 ","End":"06:28.460","Text":"You\u0027ve often seen these sorts of indicators used by doctors or outside the lab."},{"Start":"06:28.460 ","End":"06:33.515","Text":"An example, the oldest example perhaps, is litmus paper."},{"Start":"06:33.515 ","End":"06:35.960","Text":"Here\u0027s some examples of the use,"},{"Start":"06:35.960 ","End":"06:40.415","Text":"the applications of the indicators."},{"Start":"06:40.415 ","End":"06:42.860","Text":"We can test the pH of a swimming pool."},{"Start":"06:42.860 ","End":"06:45.800","Text":"For example, the pH could be 7.4."},{"Start":"06:45.800 ","End":"06:50.015","Text":"Just put the indicator on paper"},{"Start":"06:50.015 ","End":"06:55.535","Text":"and dip it in and we should see that the pH is approximately 7.4."},{"Start":"06:55.535 ","End":"06:57.830","Text":"Or if we want to grow plants,"},{"Start":"06:57.830 ","End":"07:02.090","Text":"we can test whether the pH of the soil is suitable for a particular crop."},{"Start":"07:02.090 ","End":"07:05.690","Text":"Every crop has a different pH."},{"Start":"07:05.690 ","End":"07:11.100","Text":"In this video, we talked about acid-base indicators."}],"ID":29775},{"Watched":false,"Name":"Exercise 1","Duration":"3m 23s","ChapterTopicVideoID":31906,"CourseChapterTopicPlaylistID":282561,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34180}],"Thumbnail":null,"ID":282561},{"Name":"Titration Curves","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Titrations Basic Definitions","Duration":"2m 35s","ChapterTopicVideoID":28396,"CourseChapterTopicPlaylistID":282562,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.920","Text":"In the previous videos,"},{"Start":"00:01.920 ","End":"00:03.840","Text":"we talked about buffer solutions."},{"Start":"00:03.840 ","End":"00:05.580","Text":"In the next few videos,"},{"Start":"00:05.580 ","End":"00:07.785","Text":"we\u0027ll talk about titrations."},{"Start":"00:07.785 ","End":"00:13.995","Text":"In this video, we\u0027ll talk about some basic definitions related to titrations."},{"Start":"00:13.995 ","End":"00:16.560","Text":"Now, in an acid-base titration,"},{"Start":"00:16.560 ","End":"00:21.720","Text":"a base or acid of known concentration in a buret."},{"Start":"00:21.720 ","End":"00:28.920","Text":"Now, buret is a graduated glass tube with a stopper at the end."},{"Start":"00:28.920 ","End":"00:31.245","Text":"This is a graduated,"},{"Start":"00:31.245 ","End":"00:37.245","Text":"we know exactly how much has dripped out into the flask."},{"Start":"00:37.245 ","End":"00:39.590","Text":"In an acid-base titration,"},{"Start":"00:39.590 ","End":"00:44.840","Text":"a base or acid of known concentration in a buret is added to"},{"Start":"00:44.840 ","End":"00:51.080","Text":"a flask or a beaker containing a known volume of an acid or base."},{"Start":"00:51.080 ","End":"00:54.425","Text":"We want to find out its concentration."},{"Start":"00:54.425 ","End":"00:59.690","Text":"Here, we know the concentration and we know the volume."},{"Start":"00:59.690 ","End":"01:02.465","Text":"Here, we know the volume,"},{"Start":"01:02.465 ","End":"01:04.535","Text":"but we don\u0027t know the concentration."},{"Start":"01:04.535 ","End":"01:06.590","Text":"That\u0027s what we want to find out."},{"Start":"01:06.590 ","End":"01:08.840","Text":"The solution in the flask is called"},{"Start":"01:08.840 ","End":"01:14.370","Text":"the analyte and the solution in the buret is called the titrant."},{"Start":"01:14.370 ","End":"01:21.810","Text":"The point at which the hydroxide and the titrant exactly balances the acid,"},{"Start":"01:21.810 ","End":"01:25.775","Text":"the H3O plus in the analyte, or vice versa,"},{"Start":"01:25.775 ","End":"01:30.800","Text":"is called the stoichiometric or equivalence point."},{"Start":"01:30.800 ","End":"01:33.395","Text":"When we get to neutrality,"},{"Start":"01:33.395 ","End":"01:37.115","Text":"strong acid versus a strong base,"},{"Start":"01:37.115 ","End":"01:42.155","Text":"that\u0027s called the stoichiometric or equivalence point."},{"Start":"01:42.155 ","End":"01:46.325","Text":"It\u0027s very useful in titrations to use the millimole."},{"Start":"01:46.325 ","End":"01:52.325","Text":"That\u0027s because only small quantities of acids and bases are involved in the titration."},{"Start":"01:52.325 ","End":"01:55.430","Text":"It\u0027s useful to define the millimole,"},{"Start":"01:55.430 ","End":"01:57.755","Text":"which we\u0027ll call mmol."},{"Start":"01:57.755 ","End":"02:01.160","Text":"1 millimole is 10^ -3 of a mole,"},{"Start":"02:01.160 ","End":"02:02.870","Text":"1/1000 of a mole."},{"Start":"02:02.870 ","End":"02:08.290","Text":"Now, we can say that the molarity is equal to moles divided by liters."},{"Start":"02:08.290 ","End":"02:12.890","Text":"If we divide top and bottom numerator and denominator by 1000,"},{"Start":"02:12.890 ","End":"02:16.115","Text":"we get moles divided by 1000 and that\u0027s millimoles,"},{"Start":"02:16.115 ","End":"02:19.070","Text":"liter divided by 1000, that\u0027s milliliters."},{"Start":"02:19.070 ","End":"02:25.820","Text":"You can either say that the molarity is a mole per liter or millimoles per milliliter."},{"Start":"02:25.820 ","End":"02:31.835","Text":"We will often use this when we talk about titrations."},{"Start":"02:31.835 ","End":"02:36.330","Text":"This video, we\u0027ll start to learn about titrations."}],"ID":29782},{"Watched":false,"Name":"Titration of a Strong Acid with a Strong Base","Duration":"7m 4s","ChapterTopicVideoID":28397,"CourseChapterTopicPlaylistID":282562,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.115","Text":"In the previous video,"},{"Start":"00:02.115 ","End":"00:04.440","Text":"we began to study titrations."},{"Start":"00:04.440 ","End":"00:09.930","Text":"In this video, we\u0027ll study the titration of a strong acid and a strong base."},{"Start":"00:09.930 ","End":"00:16.380","Text":"We\u0027re going to study the titration of a strong acid with a strong base using the example."},{"Start":"00:16.380 ","End":"00:21.390","Text":"Calculate the pH at each of the following points in the titration of"},{"Start":"00:21.390 ","End":"00:28.410","Text":"50 milliliters of 0.100 M HCl with 0.100 M NaOH."},{"Start":"00:28.410 ","End":"00:32.895","Text":"Then plot the pH versus the volume of NaOH added."},{"Start":"00:32.895 ","End":"00:37.865","Text":"The first part is to calculate the initial pH of HCl."},{"Start":"00:37.865 ","End":"00:42.760","Text":"After the addition of 49 milliliters of NaOH,"},{"Start":"00:42.760 ","End":"00:47.085","Text":"then after the additional 50 milliliters of NaOH,"},{"Start":"00:47.085 ","End":"00:52.250","Text":"and finally, after the additional 51 milliliters of NaOH."},{"Start":"00:52.250 ","End":"00:55.955","Text":"Let\u0027s look at the chemical equation for the titration."},{"Start":"00:55.955 ","End":"00:59.180","Text":"First thing to note is the reaction goes to"},{"Start":"00:59.180 ","End":"01:04.279","Text":"completion as it involves a strong acid and a strong base."},{"Start":"01:04.279 ","End":"01:06.445","Text":"Here\u0027s the full equation,"},{"Start":"01:06.445 ","End":"01:11.785","Text":"HCl plus NaOH giving us NaCl plus water,"},{"Start":"01:11.785 ","End":"01:13.730","Text":"and that\u0027s the full equation."},{"Start":"01:13.730 ","End":"01:20.840","Text":"We can write it in terms of ions H_3O+ plus Cl- plus Na+ plus"},{"Start":"01:20.840 ","End":"01:28.330","Text":"OH- and that\u0027s giving us Na+ plus Cl- and 2 water molecules."},{"Start":"01:28.330 ","End":"01:34.025","Text":"That\u0027s the full ionic equation."},{"Start":"01:34.025 ","End":"01:37.700","Text":"Then we can cancel the spectators,"},{"Start":"01:37.700 ","End":"01:41.510","Text":"that\u0027s the ions that appear both sides of the equation."},{"Start":"01:41.510 ","End":"01:45.770","Text":"We have Cl-, Cl-, Na+, Na+."},{"Start":"01:45.770 ","End":"01:49.785","Text":"We\u0027re left with H_3O+ plus OH-,"},{"Start":"01:49.785 ","End":"01:52.630","Text":"giving us 2 water molecules."},{"Start":"01:52.630 ","End":"01:57.890","Text":"The acid is going to neutralize the base or vice-versa,"},{"Start":"01:57.890 ","End":"02:03.405","Text":"and we\u0027ll get 2 water molecules and the ratio is 1 mole to 1 mole."},{"Start":"02:03.405 ","End":"02:09.085","Text":"The number of millimoles of H3O+ will be equal to the millimoles of OH-."},{"Start":"02:09.085 ","End":"02:14.105","Text":"Let\u0027s start calculating the points in the titration curve."},{"Start":"02:14.105 ","End":"02:18.185","Text":"The very first point is before we add any base."},{"Start":"02:18.185 ","End":"02:24.135","Text":"The concentration of H_3O+ is 10^minus 1 molar,"},{"Start":"02:24.135 ","End":"02:25.960","Text":"so the pH is just 1."},{"Start":"02:25.960 ","End":"02:32.510","Text":"Now, the second point is after the addition of 49 milliliters of NaOH."},{"Start":"02:32.510 ","End":"02:35.720","Text":"Let\u0027s look at the acid."},{"Start":"02:35.720 ","End":"02:45.040","Text":"The millimoles of the acid of H_3O+ is 0.100 molar times 50 milliliters."},{"Start":"02:45.040 ","End":"02:47.945","Text":"That\u0027s equal to 5 millimoles."},{"Start":"02:47.945 ","End":"02:53.420","Text":"Millimoles of OH- is 0.100 molar"},{"Start":"02:53.420 ","End":"02:59.430","Text":"times 49 milliliters, that\u0027s 4.90 millimoles."},{"Start":"02:59.430 ","End":"03:07.865","Text":"We have 5 millimoles of the acid and 4.9 millimoles of the base."},{"Start":"03:07.865 ","End":"03:10.025","Text":"If all the base reacts,"},{"Start":"03:10.025 ","End":"03:16.490","Text":"we\u0027ll be left with 5 minus 4.9 millimoles of acid."},{"Start":"03:16.490 ","End":"03:19.451","Text":"That\u0027s equal to 0.1."},{"Start":"03:19.451 ","End":"03:28.150","Text":"The millimoles of acid that are left is 0.10 millimoles."},{"Start":"03:28.150 ","End":"03:36.625","Text":"The concentration of H_3O+ is 0.10 millimoles divided by 99 milliliters,"},{"Start":"03:36.625 ","End":"03:42.900","Text":"because we have 50 milliliters of acid and 49 of base, total 99."},{"Start":"03:42.900 ","End":"03:48.420","Text":"We will divide these 2 we get 1.01 times 10^minus 3 molar."},{"Start":"03:48.420 ","End":"03:56.070","Text":"If we calculate the pH, it\u0027s 3."},{"Start":"03:56.070 ","End":"04:02.525","Text":"We\u0027ve gone from a pH of 1 to a pH of 3."},{"Start":"04:02.525 ","End":"04:05.905","Text":"Now, the stoichiometric or equivalence point,"},{"Start":"04:05.905 ","End":"04:08.720","Text":"the base completely neutralizes the acid."},{"Start":"04:08.720 ","End":"04:12.140","Text":"We have 50 milliliters of each."},{"Start":"04:12.140 ","End":"04:14.175","Text":"This will now be 50 milliliters."},{"Start":"04:14.175 ","End":"04:17.170","Text":"They completely neutralize each other,"},{"Start":"04:17.170 ","End":"04:21.095","Text":"and therefore, we get water which has a pH of 7."},{"Start":"04:21.095 ","End":"04:26.330","Text":"Here\u0027s equation H_3O+ plus OH- giving us water,"},{"Start":"04:26.330 ","End":"04:28.955","Text":"which of course is a pH 7."},{"Start":"04:28.955 ","End":"04:31.420","Text":"Now, we\u0027ve gone to pH 7."},{"Start":"04:31.420 ","End":"04:34.115","Text":"We\u0027ve done this extremely rapidly."},{"Start":"04:34.115 ","End":"04:36.320","Text":"Just adding 1 more milliliter,"},{"Start":"04:36.320 ","End":"04:38.360","Text":"we went from 3 to 7."},{"Start":"04:38.360 ","End":"04:42.355","Text":"Now, let\u0027s look at the fourth point, that\u0027s 51 milliliters."},{"Start":"04:42.355 ","End":"04:47.475","Text":"Again, we have millimoles of acid, 5 millimoles,"},{"Start":"04:47.475 ","End":"04:53.735","Text":"and millimoles of base is 0.100 times 51 milliliters,"},{"Start":"04:53.735 ","End":"04:57.275","Text":"and that\u0027s 5.1 millimoles."},{"Start":"04:57.275 ","End":"05:00.275","Text":"Now the base is in excess,"},{"Start":"05:00.275 ","End":"05:07.985","Text":"all the acid will react and we\u0027ll be left with 0.10 millimoles of base."},{"Start":"05:07.985 ","End":"05:14.215","Text":"Finally, we have 0.10 millimoles of base."},{"Start":"05:14.215 ","End":"05:21.995","Text":"Now we can calculate the concentration that 0.10 millimoles divided by 101 milliliters,"},{"Start":"05:21.995 ","End":"05:27.090","Text":"because we have 50 of acid and 51 of base,"},{"Start":"05:27.090 ","End":"05:31.170","Text":"so 50 plus 51, 101."},{"Start":"05:31.170 ","End":"05:36.360","Text":"Divide these 2, we get 0.99 times 10^minus 3."},{"Start":"05:36.360 ","End":"05:43.470","Text":"This gives us the pOH which is 3 and the pH is 11."},{"Start":"05:43.470 ","End":"05:52.830","Text":"We\u0027ve gone within adding just 1 milliliter from 3 to 7 to 11,"},{"Start":"05:52.830 ","End":"05:55.650","Text":"we added 1, we got from 3 to 7."},{"Start":"05:55.650 ","End":"05:57.240","Text":"We added another milliliter,"},{"Start":"05:57.240 ","End":"05:59.430","Text":"we went from 7 to 11."},{"Start":"05:59.430 ","End":"06:02.430","Text":"Now we can draw the pH curve."},{"Start":"06:02.430 ","End":"06:04.864","Text":"I\u0027ve added a few additional points."},{"Start":"06:04.864 ","End":"06:11.705","Text":"Here\u0027s the pH as a function of the milliliters of sodium hydroxide was added."},{"Start":"06:11.705 ","End":"06:16.145","Text":"At first, it went very slowly,"},{"Start":"06:16.145 ","End":"06:20.780","Text":"and then suddenly a very steep slope,"},{"Start":"06:20.780 ","End":"06:22.940","Text":"and then again, slowly."},{"Start":"06:22.940 ","End":"06:28.245","Text":"This is a stoichiometric point with pH equal to 7."},{"Start":"06:28.245 ","End":"06:31.775","Text":"We see that the pH rises slowly at first,"},{"Start":"06:31.775 ","End":"06:37.190","Text":"then steeply from just before the stoichiometric point until just after it,"},{"Start":"06:37.190 ","End":"06:39.800","Text":"and then it rises slowly again."},{"Start":"06:39.800 ","End":"06:42.355","Text":"Now a few words about the indicator,"},{"Start":"06:42.355 ","End":"06:47.360","Text":"1 possibility is to use bromothymol blue that has an endpoint at"},{"Start":"06:47.360 ","End":"06:53.510","Text":"7.1 and that\u0027s very near the stoichiometric point of pH 7."},{"Start":"06:53.510 ","End":"06:56.615","Text":"That\u0027s a useful indicator."},{"Start":"06:56.615 ","End":"06:58.730","Text":"There are other indicators as well."},{"Start":"06:58.730 ","End":"07:04.620","Text":"In this video, we learned about the titration of a strong acid and a strong base."}],"ID":29783},{"Watched":false,"Name":"Titration of a Weak Polyprotic Acid with a Strong Base 1","Duration":"6m 36s","ChapterTopicVideoID":28394,"CourseChapterTopicPlaylistID":282562,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.540","Text":"In previous videos we talked about the titration of"},{"Start":"00:03.540 ","End":"00:06.510","Text":"a weak monoprotic acid with a strong base,"},{"Start":"00:06.510 ","End":"00:10.635","Text":"and in this video, we\u0027ll learn about the titration of a weak polyprotic acid."},{"Start":"00:10.635 ","End":"00:14.790","Text":"Then we\u0027re going to talk about the titration of weak polyprotic acids."},{"Start":"00:14.790 ","End":"00:20.235","Text":"Now, the titration of a monoprotic acid has 1 stoichiometric or equivalence point,"},{"Start":"00:20.235 ","End":"00:23.490","Text":"whereas the titration of a polyprotic acid has"},{"Start":"00:23.490 ","End":"00:28.065","Text":"1 stoichiometric or equivalence point for each acidic hydrogen."},{"Start":"00:28.065 ","End":"00:30.935","Text":"Let\u0027s take the example of phosphoric acid,"},{"Start":"00:30.935 ","End":"00:35.345","Text":"H_3PO_4, which has 3 acidic hydrogens."},{"Start":"00:35.345 ","End":"00:39.860","Text":"Now, to neutralize each acidic hydrogen in 1 mole of acid,"},{"Start":"00:39.860 ","End":"00:41.660","Text":"we need 1 mole of base."},{"Start":"00:41.660 ","End":"00:47.125","Text":"We need 3 moles to neutralize all 3 hydrogens."},{"Start":"00:47.125 ","End":"00:51.050","Text":"Let\u0027s look at the first stoichiometric or equivalence point."},{"Start":"00:51.050 ","End":"00:57.605","Text":"The full equation is H_3PO_4 plus NaOH to give NaH_2PO_4 and water,"},{"Start":"00:57.605 ","End":"00:59.635","Text":"that\u0027s the full equation."},{"Start":"00:59.635 ","End":"01:07.325","Text":"The net equation is H_3PO_4 plus OH- to give H_2PO_4- plus water,"},{"Start":"01:07.325 ","End":"01:09.215","Text":"so that\u0027s our net equation."},{"Start":"01:09.215 ","End":"01:13.655","Text":"Now, once the H_2PO_4- is formed,"},{"Start":"01:13.655 ","End":"01:19.820","Text":"then it can undergo another hydrolysis."},{"Start":"01:19.820 ","End":"01:21.275","Text":"It reacts with water,"},{"Start":"01:21.275 ","End":"01:25.970","Text":"H_2PO_4- and can lose another hydrogen."},{"Start":"01:25.970 ","End":"01:30.443","Text":"This is second K_a2 that\u0027s relevant."},{"Start":"01:30.443 ","End":"01:38.320","Text":"But H_2PO_4- reacts with water to give us H_3O plus, plus HPO_4^2-."},{"Start":"01:38.320 ","End":"01:41.594","Text":"That\u0027s the second ionization constant,"},{"Start":"01:41.594 ","End":"01:44.985","Text":"and it\u0027s 6.3 times 10^minus 8."},{"Start":"01:44.985 ","End":"01:47.495","Text":"This is an acid hydrolysis."},{"Start":"01:47.495 ","End":"01:48.875","Text":"On the other hand,"},{"Start":"01:48.875 ","End":"01:56.825","Text":"H_2PO_4- can also react with water to give us H_3PO_4 plus OH-."},{"Start":"01:56.825 ","End":"01:59.125","Text":"Here, it\u0027s reacting like a base,"},{"Start":"01:59.125 ","End":"02:02.625","Text":"so that\u0027s base hydrolysis, and K_b1,"},{"Start":"02:02.625 ","End":"02:07.905","Text":"the constant for this base is K_w divided by K_a1,"},{"Start":"02:07.905 ","End":"02:11.645","Text":"and that\u0027s 1.4 times 10^minus 12,"},{"Start":"02:11.645 ","End":"02:13.525","Text":"a very small number."},{"Start":"02:13.525 ","End":"02:15.705","Text":"We have 2 possibilities,"},{"Start":"02:15.705 ","End":"02:18.485","Text":"reacting like an acid or reacting like a base."},{"Start":"02:18.485 ","End":"02:24.010","Text":"Now, we can see that the K_a2 is much greater than the K_b1."},{"Start":"02:24.010 ","End":"02:25.980","Text":"This is 10^minus 8,"},{"Start":"02:25.980 ","End":"02:27.995","Text":"this is 10^minus 12."},{"Start":"02:27.995 ","End":"02:29.660","Text":"Since this is the case,"},{"Start":"02:29.660 ","End":"02:32.690","Text":"K_a2 is much greater than K_b1,"},{"Start":"02:32.690 ","End":"02:37.085","Text":"the first stoichiometric point will have an acidic pH."},{"Start":"02:37.085 ","End":"02:40.400","Text":"Here, methyl orange can be used as an indicator."},{"Start":"02:40.400 ","End":"02:42.680","Text":"It will change from red to orange."},{"Start":"02:42.680 ","End":"02:45.185","Text":"That\u0027s the first stoichiometric point."},{"Start":"02:45.185 ","End":"02:48.610","Text":"Let us look at the second stoichiometric point."},{"Start":"02:48.610 ","End":"02:56.520","Text":"Now, H_2PO_4- is reacting with a hydroxide to give us HPO_4^2- plus water."},{"Start":"02:56.520 ","End":"03:03.170","Text":"Of course, this reaction goes to completion because we have a strong base."},{"Start":"03:03.170 ","End":"03:07.100","Text":"This is our net equation for the second reaction,"},{"Start":"03:07.100 ","End":"03:12.300","Text":"losing the second hydrogen H_2 to H. Now,"},{"Start":"03:12.300 ","End":"03:16.825","Text":"let\u0027s look at the reaction of the HPO_4^2-."},{"Start":"03:16.825 ","End":"03:23.525","Text":"It can react with water to give us H_2PO_4- plus hydroxide."},{"Start":"03:23.525 ","End":"03:26.380","Text":"Here, it\u0027s behaving like a base,"},{"Start":"03:26.380 ","End":"03:28.965","Text":"and this is base hydrolysis."},{"Start":"03:28.965 ","End":"03:34.670","Text":"We need K_b2 and that\u0027s equal to K_w divided by K_a2,"},{"Start":"03:34.670 ","End":"03:38.575","Text":"and that\u0027s 1.6 times 10^-7."},{"Start":"03:38.575 ","End":"03:40.100","Text":"On the other hand,"},{"Start":"03:40.100 ","End":"03:44.360","Text":"it can react with water and lose another hydrogen,"},{"Start":"03:44.360 ","End":"03:50.585","Text":"lose a third hydrogen to give us PO_4^3-, plus H_3O plus."},{"Start":"03:50.585 ","End":"03:56.220","Text":"Now this is the third ionization constant of the acid,"},{"Start":"03:56.220 ","End":"04:01.710","Text":"K_a3, and that\u0027s 4.2 times 10^minus 13."},{"Start":"04:01.710 ","End":"04:08.030","Text":"We have HPO_4^2- reacting as a base or reacting as an acid."},{"Start":"04:08.030 ","End":"04:12.775","Text":"We see the K_b2 is much greater than K_a3."},{"Start":"04:12.775 ","End":"04:16.785","Text":"Now, since K_b2 is much greater than K_a3,"},{"Start":"04:16.785 ","End":"04:21.860","Text":"the second stoichiometric point will have a basic pH."},{"Start":"04:21.860 ","End":"04:24.955","Text":"It\u0027s reacting as a base this time."},{"Start":"04:24.955 ","End":"04:28.170","Text":"The first point it reacted like an acid,"},{"Start":"04:28.170 ","End":"04:30.955","Text":"the second point like a base."},{"Start":"04:30.955 ","End":"04:35.200","Text":"First point will be acidic and the second point will be basic."},{"Start":"04:35.200 ","End":"04:39.290","Text":"Here, phenolphthalein can be used as an indicator and it"},{"Start":"04:39.290 ","End":"04:43.660","Text":"changes from colorless to pink at the second point."},{"Start":"04:43.660 ","End":"04:47.705","Text":"Now, let\u0027s look at the third equivalence point."},{"Start":"04:47.705 ","End":"04:53.275","Text":"Here, HPO_4^2- reacts with hydroxide to give us PO_4^3-,"},{"Start":"04:53.275 ","End":"04:55.930","Text":"it\u0027s lost its final hydrogen, plus water."},{"Start":"04:55.930 ","End":"04:57.680","Text":"That\u0027s the net equation."},{"Start":"04:57.680 ","End":"05:01.305","Text":"The PO_4^3- can react with water,"},{"Start":"05:01.305 ","End":"05:03.675","Text":"it\u0027s a very strong base,"},{"Start":"05:03.675 ","End":"05:08.130","Text":"to give us HPO_4^2- plus OH-."},{"Start":"05:08.130 ","End":"05:12.705","Text":"The K_b3 for this is K_w divided by K_a3,"},{"Start":"05:12.705 ","End":"05:16.680","Text":"and that\u0027s 2.4 times 10^minus 2."},{"Start":"05:16.680 ","End":"05:18.990","Text":"It\u0027s a strong base,"},{"Start":"05:18.990 ","End":"05:22.660","Text":"PO_4^3- is a strong base."},{"Start":"05:22.660 ","End":"05:25.145","Text":"Here we have base hydrolysis,"},{"Start":"05:25.145 ","End":"05:26.896","Text":"and that\u0027s the only possibility,"},{"Start":"05:26.896 ","End":"05:29.195","Text":"we don\u0027t have another hydrogen to lose,"},{"Start":"05:29.195 ","End":"05:31.985","Text":"so it can only react as a base."},{"Start":"05:31.985 ","End":"05:36.980","Text":"PO_4^3- is a strong base so that the hydroxide using"},{"Start":"05:36.980 ","End":"05:39.110","Text":"the titration needs to be very strong for"},{"Start":"05:39.110 ","End":"05:42.080","Text":"the third point to show up at the titration curve."},{"Start":"05:42.080 ","End":"05:43.368","Text":"We\u0027ll see this generally,"},{"Start":"05:43.368 ","End":"05:46.790","Text":"you don\u0027t see the third in the titration curve."},{"Start":"05:46.790 ","End":"05:51.095","Text":"Now, let\u0027s look at the half stoichiometric or equivalence points."},{"Start":"05:51.095 ","End":"05:55.835","Text":"The half stoichiometric equivalence points occur in the buffer regions."},{"Start":"05:55.835 ","End":"05:57.320","Text":"They occur, remember,"},{"Start":"05:57.320 ","End":"06:01.225","Text":"just like we had in the monoprotic acid,"},{"Start":"06:01.225 ","End":"06:03.810","Text":"at pH equals pK_a1,"},{"Start":"06:03.810 ","End":"06:07.905","Text":"that\u0027s 2.15, pH equals pK_a2,"},{"Start":"06:07.905 ","End":"06:13.425","Text":"7.2 and pH equals pK_a, 12.38."},{"Start":"06:13.425 ","End":"06:18.890","Text":"This is acidic, basic, and very basic."},{"Start":"06:18.890 ","End":"06:22.385","Text":"The third half equivalence point is very basic,"},{"Start":"06:22.385 ","End":"06:26.045","Text":"and so we don\u0027t really see it in the titration curve."},{"Start":"06:26.045 ","End":"06:30.380","Text":"In this video, we learned about the titration of a weak polyprotic acid."},{"Start":"06:30.380 ","End":"06:32.045","Text":"In the next video,"},{"Start":"06:32.045 ","End":"06:36.450","Text":"we will learn about how to draw the titration curve."}],"ID":29784},{"Watched":false,"Name":"Titration of a Weak Polyprotic Acid with a Strong Base 2","Duration":"8m 3s","ChapterTopicVideoID":28395,"CourseChapterTopicPlaylistID":282562,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.905","Text":"In the previous video,"},{"Start":"00:01.905 ","End":"00:06.000","Text":"we talked about the titration of phosphoric acid with a strong base."},{"Start":"00:06.000 ","End":"00:10.064","Text":"In this video, we\u0027ll talk about the titration curve."},{"Start":"00:10.064 ","End":"00:16.500","Text":"Let\u0027s start by talking about the pH of the first and second stoichiometric points."},{"Start":"00:16.500 ","End":"00:20.580","Text":"Now, in both the first and second neutralization reactions,"},{"Start":"00:20.580 ","End":"00:27.030","Text":"the conjugate base that\u0027s formed can react with water as bases or as acids."},{"Start":"00:27.030 ","End":"00:33.165","Text":"Now, for solutions with concentrations greater than approximately 0.10 molar,"},{"Start":"00:33.165 ","End":"00:37.785","Text":"the pH values can be shown to be independent of concentration."},{"Start":"00:37.785 ","End":"00:46.610","Text":"The pH of the first stoichiometric point is equal to the average of the pK_a1 and pK_a2."},{"Start":"00:46.610 ","End":"00:53.365","Text":"The pH of the second point is the average of pK_a2 and pK_a3."},{"Start":"00:53.365 ","End":"00:59.690","Text":"Now, the salts formed at the first second points are NaH_2PO_4,"},{"Start":"00:59.690 ","End":"01:05.505","Text":"and Na_2HPO_4, and that gives us the ions,"},{"Start":"01:05.505 ","End":"01:12.835","Text":"H_2PO_4^-, and HPO_4^2-."},{"Start":"01:12.835 ","End":"01:16.280","Text":"Let\u0027s look at the first pH,"},{"Start":"01:16.280 ","End":"01:18.305","Text":"the first equivalence point."},{"Start":"01:18.305 ","End":"01:27.840","Text":"Now, the pH of H_2PO_4^- is 1/2 the pK_a1 and the pK_a2,"},{"Start":"01:27.840 ","End":"01:32.495","Text":"and that\u0027s 1/2 times 2.15 plus 7.20,"},{"Start":"01:32.495 ","End":"01:35.495","Text":"and that is equal to 4.68."},{"Start":"01:35.495 ","End":"01:38.270","Text":"We\u0027re talking about an acidic pH."},{"Start":"01:38.270 ","End":"01:43.755","Text":"The first equivalence point appears at acidic pH."},{"Start":"01:43.755 ","End":"01:53.670","Text":"The second point, that of HPO_4^2- appears at 1/2 times pK_a2 plus pK_a3,"},{"Start":"01:53.670 ","End":"01:57.855","Text":"and that\u0027s 1/2 7.2 plus 12.38,"},{"Start":"01:57.855 ","End":"02:02.295","Text":"and that gives a total of 9.79."},{"Start":"02:02.295 ","End":"02:05.220","Text":"So we\u0027re talking about a basic pH."},{"Start":"02:05.220 ","End":"02:08.435","Text":"I should point out for the rest of this video,"},{"Start":"02:08.435 ","End":"02:12.560","Text":"that this 12.38 is not very accurate."},{"Start":"02:12.560 ","End":"02:14.480","Text":"In all different sources,"},{"Start":"02:14.480 ","End":"02:16.745","Text":"you get slightly different values."},{"Start":"02:16.745 ","End":"02:18.905","Text":"We\u0027ll take that with a pinch of salt."},{"Start":"02:18.905 ","End":"02:23.450","Text":"Now let\u0027s look at the pH of the 3rd stoichiometric or equivalence point."},{"Start":"02:23.450 ","End":"02:30.185","Text":"Now, the salt formed in the 3rd neutralization reaction can only react as a base."},{"Start":"02:30.185 ","End":"02:34.055","Text":"We can calculate its pH in the usual way."},{"Start":"02:34.055 ","End":"02:36.815","Text":"Let\u0027s illustrate this by an example."},{"Start":"02:36.815 ","End":"02:43.665","Text":"What is the pH of the solution formed when 10 milliliters of 0.10 molar,"},{"Start":"02:43.665 ","End":"02:50.550","Text":"H_3PO_4 reacts with 30 milliliters of 0.10 molar NaOH?"},{"Start":"02:50.550 ","End":"02:56.115","Text":"What is the pH when 50 milliliters NaOH is added?"},{"Start":"02:56.115 ","End":"02:58.185","Text":"Here\u0027s the reaction."},{"Start":"02:58.185 ","End":"03:06.300","Text":"Phosphoric acid plus 3 moles of NaOH gives us Na_3PO_4,"},{"Start":"03:06.300 ","End":"03:08.760","Text":"and 3 water molecules."},{"Start":"03:08.760 ","End":"03:13.940","Text":"Before we start, we have 1 millimole of the phosphoric acid,"},{"Start":"03:13.940 ","End":"03:18.490","Text":"3 millimoles of sodium hydroxide."},{"Start":"03:18.490 ","End":"03:21.660","Text":"We can calculate that from here,"},{"Start":"03:21.660 ","End":"03:27.215","Text":"10 milliliters times 0.10 gives us 1 millimole,"},{"Start":"03:27.215 ","End":"03:33.360","Text":"and 30 milliliters times 0.10 molar of NaOH,"},{"Start":"03:33.360 ","End":"03:36.195","Text":"and that gives us 3 millimoles."},{"Start":"03:36.195 ","End":"03:38.460","Text":"These react completely,"},{"Start":"03:38.460 ","End":"03:43.545","Text":"and then afterwards, all the H_3PO_4 has reacted,"},{"Start":"03:43.545 ","End":"03:45.629","Text":"all the NaOH has reacted,"},{"Start":"03:45.629 ","End":"03:54.410","Text":"and we get 1 millimole of Na_3PO_4, sodium phosphate."},{"Start":"03:54.410 ","End":"04:02.990","Text":"Now, the concentration of the sodium phosphate is 1 millimole divided by 40 milliliters,"},{"Start":"04:02.990 ","End":"04:06.965","Text":"because we had 10 milliliters of the phosphoric acid,"},{"Start":"04:06.965 ","End":"04:10.105","Text":"and 30 milliliters of the NaOH."},{"Start":"04:10.105 ","End":"04:12.480","Text":"We had 40 milliliters in total,"},{"Start":"04:12.480 ","End":"04:15.555","Text":"and that gives us 0.025 molar."},{"Start":"04:15.555 ","End":"04:18.180","Text":"Now, the PO_4^3- is a base,"},{"Start":"04:18.180 ","End":"04:21.165","Text":"a very strong base."},{"Start":"04:21.165 ","End":"04:26.625","Text":"Reacts with water to give us HPO_4^2- plus OH^-,"},{"Start":"04:26.625 ","End":"04:28.560","Text":"and the K_b3,"},{"Start":"04:28.560 ","End":"04:32.210","Text":"the ionization constant for the strong base,"},{"Start":"04:32.210 ","End":"04:35.175","Text":"is K_w divided by K_a3,"},{"Start":"04:35.175 ","End":"04:40.530","Text":"and that\u0027s 1 times 10^- 14 divided by 4.2 times 10^-13,"},{"Start":"04:40.530 ","End":"04:45.160","Text":"and that gives us 2.4 times 10^- 2."},{"Start":"04:45.160 ","End":"04:47.900","Text":"You can see that it\u0027s a pretty strong base."},{"Start":"04:47.900 ","End":"04:51.575","Text":"Now we could do our ICE calculation."},{"Start":"04:51.575 ","End":"04:58.010","Text":"We have 0.025 molar of the phosphate ion,"},{"Start":"04:58.010 ","End":"05:01.955","Text":"and we have 0 at the beginning of this ion,"},{"Start":"05:01.955 ","End":"05:05.315","Text":"HPO_4^2-, and 0 of OH^-."},{"Start":"05:05.315 ","End":"05:07.160","Text":"The change is minus x,"},{"Start":"05:07.160 ","End":"05:08.945","Text":"plus x, and plus x."},{"Start":"05:08.945 ","End":"05:13.445","Text":"The equilibrium, 0.025 minus x for the PO_4^3-,"},{"Start":"05:13.445 ","End":"05:17.885","Text":"HPO_4^2 minus has plus x,"},{"Start":"05:17.885 ","End":"05:20.080","Text":"and OH^- has plus x."},{"Start":"05:20.080 ","End":"05:21.720","Text":"That means the K_b3,"},{"Start":"05:21.720 ","End":"05:26.100","Text":"which we found was 2.4 times 10^- 2 is equal to x^2,"},{"Start":"05:26.100 ","End":"05:29.985","Text":"x times x, divided by 0.025 minus x."},{"Start":"05:29.985 ","End":"05:31.835","Text":"Now, in this calculation,"},{"Start":"05:31.835 ","End":"05:35.210","Text":"because we have a very high K_b3,"},{"Start":"05:35.210 ","End":"05:40.340","Text":"you can\u0027t make the approximation that x is much smaller than 0.025."},{"Start":"05:40.340 ","End":"05:42.650","Text":"If you work it out by quadratic equation,"},{"Start":"05:42.650 ","End":"05:46.070","Text":"you get x is equal to 0.015."},{"Start":"05:46.070 ","End":"05:48.200","Text":"There are 2 roots, 1 positive, 1 negative."},{"Start":"05:48.200 ","End":"05:49.970","Text":"This is the positive root."},{"Start":"05:49.970 ","End":"05:57.335","Text":"You see that it\u0027s not negligible compared to the concentration 0.025."},{"Start":"05:57.335 ","End":"06:01.835","Text":"Now, the x relates to the concentration of OH^-."},{"Start":"06:01.835 ","End":"06:06.290","Text":"The concentration of OH^- is 0.015 molar,"},{"Start":"06:06.290 ","End":"06:09.475","Text":"and the pOH is 1.82,"},{"Start":"06:09.475 ","End":"06:12.510","Text":"giving a pH of 12.2."},{"Start":"06:12.510 ","End":"06:16.640","Text":"Now, that\u0027s strong base, that\u0027s very basic."},{"Start":"06:16.640 ","End":"06:19.880","Text":"Now, we can recall from the previous video that"},{"Start":"06:19.880 ","End":"06:24.545","Text":"the half equivalence point is approximately pH equal to 12.4."},{"Start":"06:24.545 ","End":"06:26.750","Text":"They\u0027re pretty similar, almost the same,"},{"Start":"06:26.750 ","End":"06:29.015","Text":"because as we said before,"},{"Start":"06:29.015 ","End":"06:32.380","Text":"this K_b3 is not too accurate."},{"Start":"06:32.380 ","End":"06:34.400","Text":"Now, we have the half equivalence,"},{"Start":"06:34.400 ","End":"06:38.015","Text":"point 12.4, the full equivalence, point 12.2."},{"Start":"06:38.015 ","End":"06:40.925","Text":"Now, what happens if we add 50 milliliters?"},{"Start":"06:40.925 ","End":"06:43.450","Text":"If 50 milliliters of NaOH is added,"},{"Start":"06:43.450 ","End":"06:47.910","Text":"then we\u0027ll just have 2 moles of OH^- remaining."},{"Start":"06:47.910 ","End":"06:51.190","Text":"We have a total volume of 60 milliliters."},{"Start":"06:51.190 ","End":"06:54.965","Text":"The concentration of OH^- is 2 divided by 60,"},{"Start":"06:54.965 ","End":"06:59.540","Text":"and that gives us 3.33 times 10^- 2,"},{"Start":"06:59.540 ","End":"07:02.825","Text":"and that gives a pOH of 1.48,"},{"Start":"07:02.825 ","End":"07:06.255","Text":"and a pH of approximately 12.5."},{"Start":"07:06.255 ","End":"07:12.429","Text":"You see all these 3 points are almost approximately equal,"},{"Start":"07:12.429 ","End":"07:17.885","Text":"so we won\u0027t see a clear 3rd equivalence point in the graph."},{"Start":"07:17.885 ","End":"07:22.820","Text":"Of course, there are other weak acids which do have a clear 3rd point."},{"Start":"07:22.820 ","End":"07:28.445","Text":"We see that the 3rd stoichiometric point is not seen clearly in the titration curve."},{"Start":"07:28.445 ","End":"07:30.740","Text":"Here, I\u0027ve drawn the titration curve."},{"Start":"07:30.740 ","End":"07:33.260","Text":"This is more or less just like a straight line."},{"Start":"07:33.260 ","End":"07:34.565","Text":"What do we see here?"},{"Start":"07:34.565 ","End":"07:38.195","Text":"It rises slowly near the half equivalence point."},{"Start":"07:38.195 ","End":"07:40.970","Text":"Here, we have 1.5 equivalence point, another one,"},{"Start":"07:40.970 ","End":"07:44.690","Text":"and the 3rd one, and more steeply near the equivalence point."},{"Start":"07:44.690 ","End":"07:46.205","Text":"This is equivalence point."},{"Start":"07:46.205 ","End":"07:48.350","Text":"That\u0027s the first, the second,"},{"Start":"07:48.350 ","End":"07:50.300","Text":"you don\u0027t see the 3rd one clearly."},{"Start":"07:50.300 ","End":"07:51.920","Text":"It should be here at 30,"},{"Start":"07:51.920 ","End":"07:54.500","Text":"but you don\u0027t see it clearly. I\u0027ve written that here."},{"Start":"07:54.500 ","End":"07:55.730","Text":"The third equivalence point,"},{"Start":"07:55.730 ","End":"07:58.435","Text":"very basic, doesn\u0027t show up on the graph."},{"Start":"07:58.435 ","End":"08:04.260","Text":"In this video, we completed the discussion of titration of polyprotic acids."}],"ID":29785},{"Watched":false,"Name":"Titration of a Weak Acid with a Strong Base 1","Duration":"6m 45s","ChapterTopicVideoID":28392,"CourseChapterTopicPlaylistID":282562,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.870 ","End":"00:08.265","Text":"The previous video, we discussed the titration the strong acid with a strong base."},{"Start":"00:08.265 ","End":"00:10.560","Text":"In this video and the next,"},{"Start":"00:10.560 ","End":"00:13.455","Text":"we\u0027ll talk about the titration of a weak acid,"},{"Start":"00:13.455 ","End":"00:15.240","Text":"with a strong base."},{"Start":"00:15.240 ","End":"00:20.940","Text":"Let\u0027s talk about the titration of a weak acid with a strong base."},{"Start":"00:20.940 ","End":"00:23.640","Text":"Now the titration of a weak acid with"},{"Start":"00:23.640 ","End":"00:28.575","Text":"a strong base is more complicated than that of a strong acid."},{"Start":"00:28.575 ","End":"00:32.610","Text":"We will consider more points in the titration."},{"Start":"00:32.610 ","End":"00:35.190","Text":"Let\u0027s take an example."},{"Start":"00:35.190 ","End":"00:40.580","Text":"Calculate the pH at each of the following points in the titration of"},{"Start":"00:40.580 ","End":"00:46.775","Text":"50 milliliters of 0.100 molar acetic acid,"},{"Start":"00:46.775 ","End":"00:52.145","Text":"with 0.100 molar sodium hydroxide."},{"Start":"00:52.145 ","End":"00:58.595","Text":"Then plot the pH versus the volume of any NaOH added."},{"Start":"00:58.595 ","End":"01:03.020","Text":"Let\u0027s start with the initial pH of acetic acid."},{"Start":"01:03.020 ","End":"01:09.230","Text":"Then we\u0027ll calculate what happens after addition of 20 milliliters of the base."},{"Start":"01:09.230 ","End":"01:13.780","Text":"Then after addition of 25 milliliters of the base,"},{"Start":"01:13.780 ","End":"01:16.190","Text":"and we\u0027ll call that the half stoichiometric,"},{"Start":"01:16.190 ","End":"01:18.215","Text":"or half equivalence point,"},{"Start":"01:18.215 ","End":"01:26.825","Text":"because we need 50 milliliters to counterbalance the 50 milliliters of the acetic acid."},{"Start":"01:26.825 ","End":"01:31.670","Text":"Then after addition of 50 milliliters of sodium hydroxide,"},{"Start":"01:31.670 ","End":"01:35.555","Text":"and that\u0027s the stoichiometric or equivalence point."},{"Start":"01:35.555 ","End":"01:41.690","Text":"Then finally, after additional 51 milliliters of sodium hydroxide."},{"Start":"01:41.690 ","End":"01:44.855","Text":"Let\u0027s look at the equation for the titration."},{"Start":"01:44.855 ","End":"01:50.480","Text":"First thing to note is the reaction goes to completion as it involves a strong base."},{"Start":"01:50.480 ","End":"01:52.775","Text":"Whenever there is a strong acid or strong base,"},{"Start":"01:52.775 ","End":"01:55.720","Text":"the reaction goes to completion."},{"Start":"01:55.720 ","End":"01:58.425","Text":"Here\u0027s the full equation."},{"Start":"01:58.425 ","End":"02:06.710","Text":"Acetic acid plus sodium hydroxide to give sodium acetate and water."},{"Start":"02:06.710 ","End":"02:08.810","Text":"That\u0027s the full equation."},{"Start":"02:08.810 ","End":"02:11.390","Text":"Here\u0027s the net equation."},{"Start":"02:11.390 ","End":"02:15.900","Text":"Acetic acid plus hydroxide,"},{"Start":"02:15.900 ","End":"02:19.475","Text":"giving us acetate and water."},{"Start":"02:19.475 ","End":"02:21.395","Text":"That\u0027s the net equation."},{"Start":"02:21.395 ","End":"02:28.685","Text":"We don\u0027t divide the acetic acid into it\u0027s ions because it doesn\u0027t completely dissociate."},{"Start":"02:28.685 ","End":"02:31.910","Text":"Now when we get to the stoichiometric point,"},{"Start":"02:31.910 ","End":"02:36.200","Text":"the millimoles of the acetic acid will be exactly"},{"Start":"02:36.200 ","End":"02:41.470","Text":"equal to the millimoles of the hydroxide, 1:1 ratio."},{"Start":"02:41.470 ","End":"02:46.310","Text":"Let\u0027s first look at the titration points before neutralization."},{"Start":"02:46.310 ","End":"02:49.655","Text":"First, we\u0027ll calculate the initial pH."},{"Start":"02:49.655 ","End":"02:54.920","Text":"Now this is a similar calculation to what we did in a previous video."},{"Start":"02:54.920 ","End":"02:59.390","Text":"K_a is 1.8 times 10^minus 5,"},{"Start":"02:59.390 ","End":"03:04.775","Text":"and that\u0027s = x^2 divide by 0.1."},{"Start":"03:04.775 ","End":"03:09.605","Text":"We\u0027re ignoring the minus x that would be in the denominator."},{"Start":"03:09.605 ","End":"03:15.415","Text":"X is =1.34 times 10^ minus 3,"},{"Start":"03:15.415 ","End":"03:20.350","Text":"and the pH is then =2.87."},{"Start":"03:20.350 ","End":"03:23.195","Text":"We did a similar calculation before."},{"Start":"03:23.195 ","End":"03:27.650","Text":"Before we\u0027ve added any hydroxide, this is the pH."},{"Start":"03:27.650 ","End":"03:31.400","Text":"Now another point before the half stoichiometric,"},{"Start":"03:31.400 ","End":"03:32.720","Text":"or half equivalence point,"},{"Start":"03:32.720 ","End":"03:35.495","Text":"which remember was 25 milliliters."},{"Start":"03:35.495 ","End":"03:44.030","Text":"Now initially the millimoles of the acetic acid is 0.100 m times 50 milliliters."},{"Start":"03:44.030 ","End":"03:45.785","Text":"When we multiply that,"},{"Start":"03:45.785 ","End":"03:48.855","Text":"we get 5.00 Millimoles."},{"Start":"03:48.855 ","End":"03:54.890","Text":"That\u0027s the number of millimoles we have before we add any hydroxide."},{"Start":"03:54.890 ","End":"03:59.390","Text":"And at this point it\u0027s 20 milliliters of hydroxide that we\u0027re adding,"},{"Start":"03:59.390 ","End":"04:07.340","Text":"so the number of millimoles of hydroxide is 0.1M multiplied by 20 milliliters,"},{"Start":"04:07.340 ","End":"04:10.085","Text":"and that\u0027s 2.00 Millimoles."},{"Start":"04:10.085 ","End":"04:15.088","Text":"We have 55 millimoles of acetic acid,"},{"Start":"04:15.088 ","End":"04:18.440","Text":"and 2 millimoles of hydroxide."},{"Start":"04:18.440 ","End":"04:24.500","Text":"When these 2 react we\u0027ll only need 2 millimoles of the acetic acid,"},{"Start":"04:24.500 ","End":"04:27.664","Text":"and we\u0027ll be left with 3 millimoles of acetic acid."},{"Start":"04:27.664 ","End":"04:33.230","Text":"Once the 2 millimoles of hydroxide is reacted,"},{"Start":"04:33.230 ","End":"04:37.295","Text":"so we have 2 millimoles of acetate,"},{"Start":"04:37.295 ","End":"04:39.980","Text":"and 3 millimoles of acetic acid."},{"Start":"04:39.980 ","End":"04:44.540","Text":"Now as both the acid and it\u0027s conjugate base are present at the same time,"},{"Start":"04:44.540 ","End":"04:46.810","Text":"the solution is like a buffer solution."},{"Start":"04:46.810 ","End":"04:49.534","Text":"We can use the Henderson-Hasselbach equation."},{"Start":"04:49.534 ","End":"04:57.635","Text":"We have pH=pK_a plus the log of the ratio of the acetate and acetic acid."},{"Start":"04:57.635 ","End":"05:02.270","Text":"That\u0027s 4.74 plus the log of 2 divided by 3."},{"Start":"05:02.270 ","End":"05:04.100","Text":"We can just use the millimoles."},{"Start":"05:04.100 ","End":"05:06.050","Text":"We don\u0027t need to worry about the volume"},{"Start":"05:06.050 ","End":"05:09.694","Text":"because it\u0027s the same in the numerator and denominator."},{"Start":"05:09.694 ","End":"05:14.190","Text":"That gives us 4.74 minus 0.18=4.56."},{"Start":"05:14.770 ","End":"05:22.850","Text":"That means that when we\u0027ve added 20 milliliters of any OH,"},{"Start":"05:22.850 ","End":"05:25.310","Text":"the pH is 4.56."},{"Start":"05:25.310 ","End":"05:27.361","Text":"Let\u0027s look at the half stoichiometric,"},{"Start":"05:27.361 ","End":"05:29.135","Text":"or half equivalence point."},{"Start":"05:29.135 ","End":"05:35.615","Text":"Again, initially the amount of millimoles of acetic acid is 5 millimoles."},{"Start":"05:35.615 ","End":"05:42.290","Text":"Now we\u0027re adding 25 milliliters of hydroxide and that\u0027s 2.50 millimoles."},{"Start":"05:42.290 ","End":"05:44.360","Text":"Once these 2 have reacted,"},{"Start":"05:44.360 ","End":"05:48.625","Text":"only 2.50 millimoles of"},{"Start":"05:48.625 ","End":"05:54.035","Text":"the acetic acid will be able to react with 2.50 millimoles of the hydroxide,"},{"Start":"05:54.035 ","End":"05:59.735","Text":"so we\u0027re left with 2.50 millimoles of the acetic acid,"},{"Start":"05:59.735 ","End":"06:05.190","Text":"and we\u0027ll produce 2.50 millimoles of acetate."},{"Start":"06:05.190 ","End":"06:10.040","Text":"Now we have equal amount of acetic acid and acetate."},{"Start":"06:10.040 ","End":"06:13.505","Text":"Again, we could use the Henderson-Hasselbach equation,"},{"Start":"06:13.505 ","End":"06:17.530","Text":"and this time the 2 concentrations are the same."},{"Start":"06:17.530 ","End":"06:21.290","Text":"We have plus log of 2.5 divide by 2.5,"},{"Start":"06:21.290 ","End":"06:22.670","Text":"which just log of 1,"},{"Start":"06:22.670 ","End":"06:25.385","Text":"and the log of 1, as you know, is 0."},{"Start":"06:25.385 ","End":"06:30.110","Text":"The pH is exactly 4.74 is very important."},{"Start":"06:30.110 ","End":"06:32.360","Text":"The pH is equal to the pK_A,"},{"Start":"06:32.360 ","End":"06:34.040","Text":"the half equivalence point,"},{"Start":"06:34.040 ","End":"06:37.400","Text":"the pH is equal to the pK_a."},{"Start":"06:37.400 ","End":"06:43.190","Text":"In this video, we began to talk about the titration for weak acid with a strong base,"},{"Start":"06:43.190 ","End":"06:46.140","Text":"and we\u0027ll continue in the next video."}],"ID":29786},{"Watched":false,"Name":"Titration of a Weak Acid with a Strong Base 2","Duration":"5m 59s","ChapterTopicVideoID":28393,"CourseChapterTopicPlaylistID":282562,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.850","Text":"In the previous video we began to study the titration of a weak acid with a strong base."},{"Start":"00:05.850 ","End":"00:09.750","Text":"In this video we continue to study this titration."},{"Start":"00:09.750 ","End":"00:11.250","Text":"Just to remind you,"},{"Start":"00:11.250 ","End":"00:12.750","Text":"here\u0027s the net equation for"},{"Start":"00:12.750 ","End":"00:15.836","Text":"neutralization of a weak acid, acetic acid plus hydroxide to give us acetate and water."},{"Start":"00:15.836 ","End":"00:21.840","Text":"That\u0027s the net equation."},{"Start":"00:21.840 ","End":"00:27.375","Text":"We have a 1 to 1 ratio of the hydroxide and acetic acid."},{"Start":"00:27.375 ","End":"00:34.740","Text":"When these react, for every mole of acetic acid that reacts we get 1 mole of acetate."},{"Start":"00:34.740 ","End":"00:37.415","Text":"Now, at the stoichiometric point,"},{"Start":"00:37.415 ","End":"00:42.880","Text":"the millimoles of acetic acid are equal to the millimoles of hydroxide."},{"Start":"00:42.880 ","End":"00:45.750","Text":"They\u0027re exactly 1 to 1."},{"Start":"00:45.750 ","End":"00:49.365","Text":"That\u0027s called the stoichiometric point."},{"Start":"00:49.365 ","End":"00:53.240","Text":"Let\u0027s look at the stoichiometric or equivalence point."},{"Start":"00:53.240 ","End":"00:57.980","Text":"Now, again, initially we have 5 millimoles of acetic acid"},{"Start":"00:57.980 ","End":"01:03.170","Text":"and now we have 5 millimoles also of hydroxide,"},{"Start":"01:03.170 ","End":"01:05.560","Text":"so there is a 1 to 1 ratio."},{"Start":"01:05.560 ","End":"01:10.835","Text":"That means that all the acetic acid will react this time."},{"Start":"01:10.835 ","End":"01:13.820","Text":"After the reaction, remember we had"},{"Start":"01:13.820 ","End":"01:18.170","Text":"the same number of millimoles of acetic acid with hydroxide."},{"Start":"01:18.170 ","End":"01:21.140","Text":"At the end there will be none of them left,"},{"Start":"01:21.140 ","End":"01:27.755","Text":"there will be 0 and the millimoles of acetate will be 5 millimoles."},{"Start":"01:27.755 ","End":"01:31.925","Text":"All the acetic acid has turned into acetate."},{"Start":"01:31.925 ","End":"01:35.360","Text":"No acetic acid left, just acetate."},{"Start":"01:35.360 ","End":"01:39.295","Text":"Now, what\u0027s the concentration of this acetate?"},{"Start":"01:39.295 ","End":"01:41.520","Text":"The number of millimoles is 5,"},{"Start":"01:41.520 ","End":"01:45.170","Text":"so it\u0027s 5 millimoles and the volume is"},{"Start":"01:45.170 ","End":"01:49.655","Text":"100 milliliters because we had 50 milliliters of acid,"},{"Start":"01:49.655 ","End":"01:51.470","Text":"50 milliliters of base."},{"Start":"01:51.470 ","End":"01:54.785","Text":"The total volume is 100 milliliters."},{"Start":"01:54.785 ","End":"01:59.005","Text":"We divide this we get 0.05 M,"},{"Start":"01:59.005 ","End":"02:05.225","Text":"0.05 molar, so that\u0027s the concentration of the acetate."},{"Start":"02:05.225 ","End":"02:08.705","Text":"Now, the acetate reacts with water,"},{"Start":"02:08.705 ","End":"02:16.785","Text":"it becomes hydrolyzed and reacts to give us acetic acid and hydroxide."},{"Start":"02:16.785 ","End":"02:19.250","Text":"This is an equilibrium reaction."},{"Start":"02:19.250 ","End":"02:23.490","Text":"Now, k_b for the acetate is k_w divide by k_a, that\u0027s 1.0 times to 10"},{"Start":"02:34.520 ","End":"02:35.520","Text":"to the power of"},{"Start":"02:35.520 ","End":"02:36.062","Text":"minus 14 divide by"},{"Start":"02:36.062 ","End":"02:36.063","Text":"1.8 times 10 to the power of minus 5 and that\u0027s 5.6 times 10 to the power of minus 10."},{"Start":"02:36.063 ","End":"02:39.380","Text":"It\u0027s a very weak base."},{"Start":"02:39.380 ","End":"02:46.610","Text":"Now, we found the pH of the acetate in a similar calculation in a previous video."},{"Start":"02:46.610 ","End":"02:52.310","Text":"So Kb is equal to 5.6 times 10^-10 and that\u0027s x^2 divide by 0.05."},{"Start":"02:52.310 ","End":"02:59.055","Text":"0.05 is the concentration of the acetate."},{"Start":"02:59.055 ","End":"03:01.455","Text":"We\u0027re ignoring here -x."},{"Start":"03:01.455 ","End":"03:06.265","Text":"That means that x is equal to 5.3 times 10^-6."},{"Start":"03:06.265 ","End":"03:07.865","Text":"Now, in this case,"},{"Start":"03:07.865 ","End":"03:11.350","Text":"the x is the concentration of the OH minus."},{"Start":"03:11.350 ","End":"03:14.340","Text":"This is the concentration of OH minus."},{"Start":"03:14.340 ","End":"03:20.670","Text":"From that we can calculate that the pOH is 5.28 and the pH which"},{"Start":"03:20.670 ","End":"03:27.660","Text":"is 14 minus 5.28 is 8.72."},{"Start":"03:27.660 ","End":"03:32.030","Text":"The stoichiometric point occurs at a basic pH,"},{"Start":"03:32.030 ","End":"03:41.315","Text":"8.72 and this you would expect because we have a weak acid versus a strong base."},{"Start":"03:41.315 ","End":"03:46.040","Text":"Now, when we get beyond the stoichiometric point, then again,"},{"Start":"03:46.040 ","End":"03:54.650","Text":"initially we had 5 millimoles of acetic acid and now we have 51 milliliters of hydroxide,"},{"Start":"03:54.650 ","End":"03:57.685","Text":"so that\u0027s 5.10 millimoles."},{"Start":"03:57.685 ","End":"04:04.115","Text":"After these 2 react we\u0027re left with 0.10 millimoles of hydroxide."},{"Start":"04:04.115 ","End":"04:06.635","Text":"All the acetic acid has reacted."},{"Start":"04:06.635 ","End":"04:12.185","Text":"The millimoles of hydroxide is 0.10 millimoles,"},{"Start":"04:12.185 ","End":"04:13.700","Text":"so the concentration is"},{"Start":"04:13.700 ","End":"04:16.230","Text":"0.10 millimoles divided by the volume which is now 101 milliliters."},{"Start":"04:19.390 ","End":"04:26.640","Text":"That gives us 1 times 10^-3 for the concentration of the hydroxide."},{"Start":"04:26.640 ","End":"04:31.815","Text":"The pOH is 3 and the pH is 11."},{"Start":"04:31.815 ","End":"04:35.240","Text":"This is after the equivalence point."},{"Start":"04:35.240 ","End":"04:38.480","Text":"Now we\u0027re going to draw the titration curve and I\u0027ve"},{"Start":"04:38.480 ","End":"04:42.020","Text":"added a few extra points to make the curve smoother."},{"Start":"04:42.020 ","End":"04:50.465","Text":"We see it rises slowly and then more steeply and then slowly again."},{"Start":"04:50.465 ","End":"04:56.875","Text":"Here at 25, this is the half equivalence point and here\u0027s the equivalence point."},{"Start":"04:56.875 ","End":"05:01.960","Text":"Now, let\u0027s compare this with the titration curve for the strong acid."},{"Start":"05:01.960 ","End":"05:04.735","Text":"First of all, the pH starts at a higher volume,"},{"Start":"05:04.735 ","End":"05:09.040","Text":"starts at 2 point something instead of 1 in the previous when we had"},{"Start":"05:09.040 ","End":"05:15.730","Text":"the strong acid and it rises more slowly until near the stoichiometric point."},{"Start":"05:15.730 ","End":"05:21.175","Text":"The steep slope near the stoichiometric point is less steep than for strong acids."},{"Start":"05:21.175 ","End":"05:25.824","Text":"It\u0027s less steep and it covers a smaller range of pH."},{"Start":"05:25.824 ","End":"05:29.200","Text":"After the stoichiometric point both titration curves"},{"Start":"05:29.200 ","End":"05:32.990","Text":"are the same because they\u0027re just related to the hydroxide."},{"Start":"05:32.990 ","End":"05:36.660","Text":"Now, what indicator could we use for this?"},{"Start":"05:36.660 ","End":"05:41.070","Text":"Now, phenolphthalein has a range of pH equal to"},{"Start":"05:41.070 ","End":"05:48.850","Text":"8.2-10 and it covers the stoichiometric point which is pH equal to 8.72."},{"Start":"05:48.850 ","End":"05:52.070","Text":"This is a possible indicator."},{"Start":"05:52.070 ","End":"05:54.230","Text":"In this video and the previous one,"},{"Start":"05:54.230 ","End":"05:59.399","Text":"we talked about the titration of a weak acid with a strong base"}],"ID":29787},{"Watched":false,"Name":"Exercise 1","Duration":"3m 56s","ChapterTopicVideoID":31911,"CourseChapterTopicPlaylistID":282562,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34181},{"Watched":false,"Name":"Exercise 2 part a","Duration":"9m 2s","ChapterTopicVideoID":31912,"CourseChapterTopicPlaylistID":282562,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34182},{"Watched":false,"Name":"Exercise 2 part b","Duration":"7m 30s","ChapterTopicVideoID":31907,"CourseChapterTopicPlaylistID":282562,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34183},{"Watched":false,"Name":"Exercise 3","Duration":"6m 52s","ChapterTopicVideoID":31915,"CourseChapterTopicPlaylistID":282562,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34184},{"Watched":false,"Name":"Exercise 4 part a","Duration":"7m 9s","ChapterTopicVideoID":31913,"CourseChapterTopicPlaylistID":282562,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34185},{"Watched":false,"Name":"Exercise 4 part b","Duration":"5m 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