[{"Name":"Solubility Product","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Solubility Product","Duration":"3m 42s","ChapterTopicVideoID":28484,"CourseChapterTopicPlaylistID":283655,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/28484.jpeg","UploadDate":"2022-02-06T13:26:12.9230000","DurationForVideoObject":"PT3M42S","Description":null,"MetaTitle":"Solubility Product: Video + Workbook | Proprep","MetaDescription":"Solubility and Complex-ion Equilibria - Solubility Product. 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/solubility-and-complex_ion-equilibria/solubility-product/vid31806","VideoComments":[],"Subtitles":[{"Start":"00:00.410 ","End":"00:02.625","Text":"In a previous video,"},{"Start":"00:02.625 ","End":"00:05.745","Text":"we talked about which salts dissolve in water."},{"Start":"00:05.745 ","End":"00:10.875","Text":"In this video, we\u0027ll talk about the solubility product for ionic solids."},{"Start":"00:10.875 ","End":"00:14.265","Text":"We\u0027re going to talk about the solubility product."},{"Start":"00:14.265 ","End":"00:18.030","Text":"The equilibrium constant between an ionic solid"},{"Start":"00:18.030 ","End":"00:21.945","Text":"and it\u0027s dissolved ions is called solubility product."},{"Start":"00:21.945 ","End":"00:27.585","Text":"Sometimes it\u0027s called solubility product constant or solubility constant."},{"Start":"00:27.585 ","End":"00:30.270","Text":"We\u0027re just going to call it the solubility product,"},{"Start":"00:30.270 ","End":"00:32.280","Text":"written as K_sp,"},{"Start":"00:32.280 ","End":"00:34.905","Text":"sp for solubility product."},{"Start":"00:34.905 ","End":"00:40.485","Text":"The solubility product is only used for salts that are sparingly soluble,"},{"Start":"00:40.485 ","End":"00:46.665","Text":"almost insoluble, since ion-ion interactions complicate the interpretation."},{"Start":"00:46.665 ","End":"00:49.445","Text":"We\u0027ll see that even then it\u0027s not terribly accurate."},{"Start":"00:49.445 ","End":"00:51.815","Text":"Even in almost insoluble salts,"},{"Start":"00:51.815 ","End":"00:57.485","Text":"ion clusters occur, making the solubility product only approximate."},{"Start":"00:57.485 ","End":"01:00.465","Text":"Let\u0027s take an example, calcium carbonate."},{"Start":"01:00.465 ","End":"01:04.805","Text":"Here we have it. Calcium carbonate, it\u0027s almost insoluble."},{"Start":"01:04.805 ","End":"01:10.035","Text":"Is an equilibrium with its ions Ca_2 plus and SO_4,"},{"Start":"01:10.035 ","End":"01:14.540","Text":"2 minus, calcium 2 plus and SO_4, 2 minus."},{"Start":"01:14.540 ","End":"01:22.280","Text":"K_sp is written as the activity of the Ca_2 plus times the activity of SO_4,"},{"Start":"01:22.280 ","End":"01:26.930","Text":"2 minus divided by the activity of CaCO_3."},{"Start":"01:26.930 ","End":"01:30.820","Text":"It\u0027s just an equilibrium constant like all the others."},{"Start":"01:30.820 ","End":"01:35.790","Text":"The activity of Ca_2 plus is the concentration of Ca_2 plus"},{"Start":"01:35.790 ","End":"01:40.440","Text":"divided by C_0 times the concentration of SO_4,"},{"Start":"01:40.440 ","End":"01:41.760","Text":"2 minus divided by C_0,"},{"Start":"01:41.760 ","End":"01:47.570","Text":"and the activity of CaSO_4 is just 1 because it\u0027s a solid."},{"Start":"01:47.570 ","End":"01:51.540","Text":"Remember that C_0 is 1 mole per liter,"},{"Start":"01:51.540 ","End":"01:57.830","Text":"so we can write this as a concentration of Ca_2 plus times the concentration of SO_4,"},{"Start":"01:57.830 ","End":"02:02.630","Text":"2 minus and not use any units for the concentrations."},{"Start":"02:02.630 ","End":"02:06.785","Text":"We write them in molarity and then drop the units."},{"Start":"02:06.785 ","End":"02:11.090","Text":"As another equilibrium constants, K_sp is dimensionless."},{"Start":"02:11.090 ","End":"02:15.650","Text":"We use numerical values of the molar concentrations without units."},{"Start":"02:15.650 ","End":"02:22.210","Text":"For calcium sulfate, K_sp is 2.8 times 10 to the power minus 9."},{"Start":"02:22.210 ","End":"02:24.995","Text":"That\u0027s 25 degrees Celsius."},{"Start":"02:24.995 ","End":"02:27.560","Text":"The K_sp is temperature-dependent,"},{"Start":"02:27.560 ","End":"02:30.890","Text":"usually higher temperatures, it\u0027s more soluble."},{"Start":"02:30.890 ","End":"02:34.070","Text":"We have to say at which temperature we\u0027re measuring it."},{"Start":"02:34.070 ","End":"02:37.280","Text":"It\u0027s 2.8 times 10 to the power minus 9."},{"Start":"02:37.280 ","End":"02:41.270","Text":"The solubility products are small since they refer to"},{"Start":"02:41.270 ","End":"02:45.980","Text":"almost insoluble salts that produce low concentrations of ions."},{"Start":"02:45.980 ","End":"02:47.360","Text":"That\u0027s why it\u0027s so small."},{"Start":"02:47.360 ","End":"02:50.675","Text":"We\u0027re going to talk about molar solubility."},{"Start":"02:50.675 ","End":"02:57.035","Text":"The molar solubility is the concentration of a compound in a saturated solution."},{"Start":"02:57.035 ","End":"02:59.270","Text":"It\u0027s that part of the compound,"},{"Start":"02:59.270 ","End":"03:02.570","Text":"that part of the solid that is actually dissolved."},{"Start":"03:02.570 ","End":"03:06.125","Text":"s, which is the molar solubility,"},{"Start":"03:06.125 ","End":"03:08.630","Text":"is in moles per liter."},{"Start":"03:08.630 ","End":"03:12.350","Text":"We use the letter s to indicate molar solubility."},{"Start":"03:12.350 ","End":"03:14.105","Text":"The units are moles per liter,"},{"Start":"03:14.105 ","End":"03:17.360","Text":"and it can be used to calculate the solubility product."},{"Start":"03:17.360 ","End":"03:20.000","Text":"We\u0027ll see in another video how we can go from"},{"Start":"03:20.000 ","End":"03:24.275","Text":"solubility product and molar solubility and vice versa."},{"Start":"03:24.275 ","End":"03:27.830","Text":"If the solubility product is known, for example,"},{"Start":"03:27.830 ","End":"03:30.409","Text":"it can be measured using electrochemical methods,"},{"Start":"03:30.409 ","End":"03:32.329","Text":"we\u0027ll see that in later videos,"},{"Start":"03:32.329 ","End":"03:35.575","Text":"it can be used to calculate the molar solubility."},{"Start":"03:35.575 ","End":"03:41.490","Text":"In this video, we defined the solubility product and molar solubility."}],"ID":31806},{"Watched":false,"Name":"Solubility Product and Molar Solubility","Duration":"4m 22s","ChapterTopicVideoID":28485,"CourseChapterTopicPlaylistID":283655,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.875","Text":"In the previous video,"},{"Start":"00:01.875 ","End":"00:05.820","Text":"we defined the solubility product and molar solubility."},{"Start":"00:05.820 ","End":"00:09.735","Text":"In this video, we\u0027ll show how to calculate one from the other."},{"Start":"00:09.735 ","End":"00:14.940","Text":"How do you determine the solubility product from the molar solubility?"},{"Start":"00:14.940 ","End":"00:16.800","Text":"Let\u0027s take an example."},{"Start":"00:16.800 ","End":"00:24.780","Text":"The molar solubility of calcium sulfate is 0.015 moles per liter at 25 degrees Celsius."},{"Start":"00:24.780 ","End":"00:28.575","Text":"Calculate K_sp for calcium sulfate."},{"Start":"00:28.575 ","End":"00:32.115","Text":"We always start by writing the chemical equation."},{"Start":"00:32.115 ","End":"00:35.625","Text":"That\u0027s calcium sulfate solid,"},{"Start":"00:35.625 ","End":"00:42.480","Text":"there\u0027s an equilibrium with Ca^2+ and sulfate ions."},{"Start":"00:42.480 ","End":"00:46.615","Text":"They are the ones dissolved in the water."},{"Start":"00:46.615 ","End":"00:49.860","Text":"Now we\u0027ll write the expression for the K_sp."},{"Start":"00:49.860 ","End":"00:54.060","Text":"K_sp is the concentration of Ca^2+,"},{"Start":"00:54.060 ","End":"00:57.855","Text":"times the concentration of SO_4^2-."},{"Start":"00:57.855 ","End":"01:03.575","Text":"Now, we see here the molar solubility of calcium sulfate is 0.015."},{"Start":"01:03.575 ","End":"01:09.080","Text":"That\u0027s the amount of the calcium sulfate that\u0027s dissolved in the water,"},{"Start":"01:09.080 ","End":"01:11.800","Text":"and it\u0027s 1 mole to 1 mole to 1 mole."},{"Start":"01:11.800 ","End":"01:17.100","Text":"That means the concentration of Ca^2+ is 0.015,"},{"Start":"01:17.100 ","End":"01:19.980","Text":"and so is the concentration of the sulfates."},{"Start":"01:19.980 ","End":"01:26.440","Text":"We have 0.015 squared and that\u0027s 2.3 times 10 to the power minus 4."},{"Start":"01:26.440 ","End":"01:31.730","Text":"Now when we measure it accurately by electrochemistry,"},{"Start":"01:31.730 ","End":"01:36.530","Text":"if K_sp is 9.1 times 10 to the power minus 6,"},{"Start":"01:36.530 ","End":"01:42.600","Text":"you see it\u0027s much smaller than 2.3 times 10 to the power minus 4."},{"Start":"01:42.600 ","End":"01:47.555","Text":"The difference is due to the use of concentrations instead of activities."},{"Start":"01:47.555 ","End":"01:51.215","Text":"We should really calculate everything using activities,"},{"Start":"01:51.215 ","End":"01:54.603","Text":"but for simplicity, we\u0027re just using concentrations."},{"Start":"01:54.603 ","End":"01:56.900","Text":"That\u0027s the reason for the difference."},{"Start":"01:56.900 ","End":"02:00.950","Text":"Now, supposing we know the solubility product,"},{"Start":"02:00.950 ","End":"02:04.220","Text":"how can we calculate the molar solubility?"},{"Start":"02:04.220 ","End":"02:05.915","Text":"Here\u0027s an example."},{"Start":"02:05.915 ","End":"02:09.695","Text":"Calculate the molar solubility of magnesium phosphate"},{"Start":"02:09.695 ","End":"02:13.609","Text":"at 25 degrees Celsius from its solubility constant,"},{"Start":"02:13.609 ","End":"02:20.735","Text":"K_sp equal to 1 times 10 to the power minus 25, something extremely small."},{"Start":"02:20.735 ","End":"02:26.830","Text":"We start by writing the chemical equation, Mg_3 (PO4)_2."},{"Start":"02:26.830 ","End":"02:29.830","Text":"That\u0027s our solid magnesium phosphate,"},{"Start":"02:29.830 ","End":"02:34.655","Text":"and here are the ions that dissolve in water."},{"Start":"02:34.655 ","End":"02:36.905","Text":"3 moles of Mg^2+,"},{"Start":"02:36.905 ","End":"02:41.940","Text":"plus 2 of PO_4^3-."},{"Start":"02:41.940 ","End":"02:48.680","Text":"If S is the number of moles of magnesium phosphate dissolved in 1 liter of solution,"},{"Start":"02:48.680 ","End":"02:54.395","Text":"then the concentration of Mg^2+ will be 3 times that."},{"Start":"02:54.395 ","End":"02:56.455","Text":"For every 1 mole of this,"},{"Start":"02:56.455 ","End":"02:58.545","Text":"we get 3 of that,"},{"Start":"02:58.545 ","End":"03:08.205","Text":"so this concentration of Mg^2+ is 3s and the concentration of PO_4^3- is 2s,"},{"Start":"03:08.205 ","End":"03:11.100","Text":"because here we have 2 moles."},{"Start":"03:11.100 ","End":"03:13.530","Text":"Now we can write the K_sp,"},{"Start":"03:13.530 ","End":"03:19.500","Text":"so concentration of Mg^2+ cubed because of this 3,"},{"Start":"03:19.500 ","End":"03:25.730","Text":"times the concentration of the phosphate squared because of this 2."},{"Start":"03:25.730 ","End":"03:32.015","Text":"Then we can write for the concentration of Mg^2+ 3s,"},{"Start":"03:32.015 ","End":"03:34.285","Text":"so that\u0027s 3s^3,"},{"Start":"03:34.285 ","End":"03:38.115","Text":"and for the concentration PO_4^3- 2s,"},{"Start":"03:38.115 ","End":"03:40.830","Text":"so we have 2s^2."},{"Start":"03:40.830 ","End":"03:43.560","Text":"Now if we multiply 3 to the power of 3,"},{"Start":"03:43.560 ","End":"03:46.950","Text":"which is 27, times 2^2,"},{"Start":"03:46.950 ","End":"03:48.270","Text":"which is 4,"},{"Start":"03:48.270 ","End":"03:52.815","Text":"we get a 108, and s to the power 5,"},{"Start":"03:52.815 ","End":"03:56.880","Text":"and that\u0027s equal to 1 times 10 to the power of minus 25."},{"Start":"03:56.880 ","End":"04:00.300","Text":"We can work out the s to the power of 5 is equal to"},{"Start":"04:00.300 ","End":"04:04.320","Text":"9.26 times 10 to the power of minus 28,"},{"Start":"04:04.320 ","End":"04:12.165","Text":"and s is equal to 3.92 times 10 to the power minus 6 molar."},{"Start":"04:12.165 ","End":"04:17.720","Text":"In this video, we calculated the solubility product from the molar solubility,"},{"Start":"04:17.720 ","End":"04:22.110","Text":"and the molar solubility from the solubility product."}],"ID":31807},{"Watched":false,"Name":"Exercise 1","Duration":"4m 14s","ChapterTopicVideoID":30088,"CourseChapterTopicPlaylistID":283655,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31808},{"Watched":false,"Name":"Exercise 2","Duration":"2m 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10","Duration":"2m 40s","ChapterTopicVideoID":30097,"CourseChapterTopicPlaylistID":283655,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":31817},{"Watched":false,"Name":"Predicting precipitation","Duration":"4m 42s","ChapterTopicVideoID":28482,"CourseChapterTopicPlaylistID":283655,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.830","Text":"In previous videos, we talked about the solubility product."},{"Start":"00:04.830 ","End":"00:07.890","Text":"In this video, we\u0027ll learn how to predict where"},{"Start":"00:07.890 ","End":"00:12.150","Text":"the precipitation will occur in an ionic solution."},{"Start":"00:12.150 ","End":"00:17.010","Text":"We\u0027re going to talk about the solubility product and the ion product."},{"Start":"00:17.010 ","End":"00:19.965","Text":"Solubility product, K_sp,"},{"Start":"00:19.965 ","End":"00:23.520","Text":"is the equilibrium constant between an ionic"},{"Start":"00:23.520 ","End":"00:28.050","Text":"solid and it\u0027s dissolved ions in a saturated solution."},{"Start":"00:28.050 ","End":"00:31.859","Text":"The 2 important points are that it\u0027s an equilibrium"},{"Start":"00:31.859 ","End":"00:37.205","Text":"constant and that it relates to a saturated solution."},{"Start":"00:37.205 ","End":"00:40.910","Text":"Now, the ion product Q_sp is"},{"Start":"00:40.910 ","End":"00:46.840","Text":"the reaction quotient between an ionic solid and it\u0027s dissolved ions."},{"Start":"00:46.840 ","End":"00:49.705","Text":"It\u0027s similar to the K_sp,"},{"Start":"00:49.705 ","End":"00:54.590","Text":"but it can be calculated for any concentration of the ions."},{"Start":"00:54.590 ","End":"00:58.030","Text":"Now, if Q_sp is equal to K_sp,"},{"Start":"00:58.030 ","End":"01:02.945","Text":"the solution is saturated and no precipitation can occur."},{"Start":"01:02.945 ","End":"01:06.635","Text":"If Q_sp is smaller than K_sp,"},{"Start":"01:06.635 ","End":"01:09.065","Text":"the solution is unsaturated."},{"Start":"01:09.065 ","End":"01:12.230","Text":"Again, no precipitation can occur."},{"Start":"01:12.230 ","End":"01:16.200","Text":"However, if Q_sp is greater than K_sp,"},{"Start":"01:16.200 ","End":"01:19.635","Text":"the solution is said to be supersaturated."},{"Start":"01:19.635 ","End":"01:22.365","Text":"The salt will precipitate,"},{"Start":"01:22.365 ","End":"01:28.415","Text":"Q_sp will decrease until equilibrium is achieved."},{"Start":"01:28.415 ","End":"01:32.765","Text":"Q_sp will decrease until it\u0027s equal to K_sp."},{"Start":"01:32.765 ","End":"01:35.000","Text":"Let\u0027s consider an example."},{"Start":"01:35.000 ","End":"01:41.780","Text":"If 250 milliliters of 0.20 molar lead nitrate is mixed with"},{"Start":"01:41.780 ","End":"01:49.415","Text":"350 milliliters of 0.20 molar KI, will precipitation occur?"},{"Start":"01:49.415 ","End":"01:51.050","Text":"Here\u0027s the reaction."},{"Start":"01:51.050 ","End":"01:59.735","Text":"Lead nitrate plus 2 moles of KI react to give us lead iodide,"},{"Start":"01:59.735 ","End":"02:01.385","Text":"that\u0027s a solid,"},{"Start":"02:01.385 ","End":"02:07.030","Text":"almost insoluble, plus 2 moles of KNO_3."},{"Start":"02:07.030 ","End":"02:13.645","Text":"The K_sp of lead iodide is 7.1 times 10 to the power minus 9."},{"Start":"02:13.645 ","End":"02:17.630","Text":"Let\u0027s see how many millimoles of lead nitrate we have."},{"Start":"02:17.630 ","End":"02:23.420","Text":"The 250 milliliters and the concentration is 0.20 molar."},{"Start":"02:23.420 ","End":"02:24.830","Text":"When we multiply these 2,"},{"Start":"02:24.830 ","End":"02:27.140","Text":"we get 50 millimoles."},{"Start":"02:27.140 ","End":"02:30.555","Text":"Now the concentration of lead 2 plus,"},{"Start":"02:30.555 ","End":"02:31.910","Text":"Pb^ 2 plus,"},{"Start":"02:31.910 ","End":"02:35.134","Text":"is the same as the concentration of lead nitrate."},{"Start":"02:35.134 ","End":"02:39.190","Text":"We have 50 millimoles divided by the volume."},{"Start":"02:39.190 ","End":"02:45.080","Text":"The volume is 600 milliliters because we have 250 milliliters of"},{"Start":"02:45.080 ","End":"02:51.610","Text":"lead nitrate and 350 milliliters of KI, potassium iodide."},{"Start":"02:51.610 ","End":"02:57.210","Text":"In total, we have 600 milliliters, 250 plus 350."},{"Start":"02:57.210 ","End":"03:00.120","Text":"When we divide 50 by 600,"},{"Start":"03:00.120 ","End":"03:04.760","Text":"we get 0.083 molar."},{"Start":"03:04.760 ","End":"03:11.390","Text":"That\u0027s the concentration of the Pb_^2 plus 0.083 molar."},{"Start":"03:11.390 ","End":"03:19.495","Text":"Now the millimoles of the KI is given by 350 milliliters times 0.20 molar,"},{"Start":"03:19.495 ","End":"03:21.950","Text":"and that 70 millimoles."},{"Start":"03:21.950 ","End":"03:26.754","Text":"Now the concentration of the iodide is the same as the concentration of KI,"},{"Start":"03:26.754 ","End":"03:31.745","Text":"and that\u0027s 70 millimoles divided by 600 milliliters,"},{"Start":"03:31.745 ","End":"03:35.360","Text":"and that\u0027s 0.117 molar."},{"Start":"03:35.360 ","End":"03:38.105","Text":"We can calculate the Q_sp."},{"Start":"03:38.105 ","End":"03:41.930","Text":"It\u0027s equal to the concentration of Pb^ 2 plus,"},{"Start":"03:41.930 ","End":"03:45.440","Text":"times the concentration of I minus squared,"},{"Start":"03:45.440 ","End":"03:48.395","Text":"squared because we have this two."},{"Start":"03:48.395 ","End":"03:55.080","Text":"That\u0027s equal to 0.083 from here,"},{"Start":"03:55.080 ","End":"03:59.220","Text":"times 0.117 from here."},{"Start":"03:59.220 ","End":"04:01.005","Text":"That has to be squared."},{"Start":"04:01.005 ","End":"04:02.670","Text":"We multiply these 2,"},{"Start":"04:02.670 ","End":"04:07.005","Text":"we get 1.14 times 10^minus 3."},{"Start":"04:07.005 ","End":"04:16.770","Text":"Now we can see that Q_sp 1.14 times 10^minus 3 is much greater than K_sp."},{"Start":"04:16.770 ","End":"04:22.000","Text":"K_sp is 7.1 times 10^minus 9."},{"Start":"04:22.000 ","End":"04:25.475","Text":"That means that precipitation will occur."},{"Start":"04:25.475 ","End":"04:31.010","Text":"The ion product will decrease until it\u0027s equal to the K_sp,"},{"Start":"04:31.010 ","End":"04:34.445","Text":"until it\u0027s equal to the solubility product."},{"Start":"04:34.445 ","End":"04:37.250","Text":"In this video, we learned how to predict where"},{"Start":"04:37.250 ","End":"04:42.180","Text":"the precipitation will occur in an ionic solution."}],"ID":31818},{"Watched":false,"Name":"Common-ion Effect in Solubility Equilibria","Duration":"5m 8s","ChapterTopicVideoID":28483,"CourseChapterTopicPlaylistID":283655,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.860","Text":"In the previous videos,"},{"Start":"00:01.860 ","End":"00:04.289","Text":"we learned about the solubility equilibria."},{"Start":"00:04.289 ","End":"00:05.790","Text":"In this video,"},{"Start":"00:05.790 ","End":"00:09.735","Text":"we\u0027ll talk about the common ion effect in these equilibria."},{"Start":"00:09.735 ","End":"00:12.840","Text":"We\u0027re going to talk about the common ion effect,"},{"Start":"00:12.840 ","End":"00:18.840","Text":"and this is similar to the effect we observed in acids."},{"Start":"00:18.840 ","End":"00:22.530","Text":"Remember we had common ion effect between strong and weak acids and"},{"Start":"00:22.530 ","End":"00:26.430","Text":"also between weak acids and its conjugate base,"},{"Start":"00:26.430 ","End":"00:29.235","Text":"and that was the basis of buffers."},{"Start":"00:29.235 ","End":"00:30.750","Text":"Let\u0027s take an example,"},{"Start":"00:30.750 ","End":"00:34.815","Text":"PbI_2 lead iodide, and potassium iodide."},{"Start":"00:34.815 ","End":"00:40.490","Text":"Lead iodide is insoluble and potassium iodide is soluble."},{"Start":"00:40.490 ","End":"00:45.200","Text":"First, you prepare a saturated solution of PbI_2."},{"Start":"00:45.200 ","End":"00:52.085","Text":"The little that dissolves gives us Pb^2 plus, plus 2 I^minus."},{"Start":"00:52.085 ","End":"00:54.260","Text":"Then we\u0027re dissolving this,"},{"Start":"00:54.260 ","End":"01:03.170","Text":"the soluble salt KI and we get KI completely dissociate into K^plus and I^minus."},{"Start":"01:03.170 ","End":"01:09.330","Text":"It completely ionizes, whereas the PbI_2 almost doesn\u0027t ionize."},{"Start":"01:09.330 ","End":"01:13.160","Text":"Clearly, the common ion is the iodide anion."},{"Start":"01:13.160 ","End":"01:15.845","Text":"Now according to Le Chatelier\u0027s principle,"},{"Start":"01:15.845 ","End":"01:22.780","Text":"the common ion I^minus pushes the equilibrium to the left to form more PbI_2,"},{"Start":"01:22.780 ","End":"01:27.065","Text":"and the PBI_2 then precipitates out of the solution."},{"Start":"01:27.065 ","End":"01:31.180","Text":"Here we have the main source of I^minus,"},{"Start":"01:31.180 ","End":"01:34.130","Text":"and we\u0027re considering this equilibrium."},{"Start":"01:34.130 ","End":"01:39.365","Text":"The extra I^minus pushes the equilibrium to the left."},{"Start":"01:39.365 ","End":"01:45.570","Text":"This results in more precipitation of PbI_2,"},{"Start":"01:45.570 ","End":"01:48.905","Text":"more of the PbI_2 comes out of the solution."},{"Start":"01:48.905 ","End":"01:53.330","Text":"It\u0027s often indicated as an arrow going downwards."},{"Start":"01:53.330 ","End":"01:58.590","Text":"That\u0027s precipitation, which is often abbreviated as ppt."},{"Start":"01:58.590 ","End":"02:04.750","Text":"That means that the solubility of PbI_2 is reduced by the addition of the common ion."},{"Start":"02:04.750 ","End":"02:06.985","Text":"Estimate the effect numerically."},{"Start":"02:06.985 ","End":"02:14.505","Text":"What\u0027s the molar solubility of PbI_2 in 0.20 molar NaI?"},{"Start":"02:14.505 ","End":"02:21.070","Text":"We\u0027re given the K_sp of PbI_2 is 7.1 times 10 to the power minus 9."},{"Start":"02:21.070 ","End":"02:25.580","Text":"Now we\u0027re going to assume that all the iodide comes from the NaI."},{"Start":"02:25.580 ","End":"02:30.905","Text":"The concentration of I^minus will just be 0.20 molar,"},{"Start":"02:30.905 ","End":"02:34.445","Text":"just like the concentration of the NaI."},{"Start":"02:34.445 ","End":"02:37.170","Text":"Here\u0027s the initial concentration of the NaI,"},{"Start":"02:37.170 ","End":"02:40.080","Text":"it\u0027s all going to dissociate and it will give"},{"Start":"02:40.080 ","End":"02:46.080","Text":"us concentration of I^minus equal to 0.20 molar."},{"Start":"02:46.080 ","End":"02:50.210","Text":"The K_sp is 7.1 times 10 to power minus 9."},{"Start":"02:50.210 ","End":"02:53.780","Text":"We can write that as the concentration of Pb^2 plus times"},{"Start":"02:53.780 ","End":"02:58.885","Text":"the concentration of I^minus squared because of the 2 here."},{"Start":"02:58.885 ","End":"03:03.600","Text":"The concentration of Pb^2 plus is going to be s,"},{"Start":"03:03.600 ","End":"03:06.150","Text":"that\u0027s the molar solubility."},{"Start":"03:06.150 ","End":"03:09.270","Text":"The concentration of I^minus is 0.20,"},{"Start":"03:09.270 ","End":"03:14.050","Text":"so we square that, we get 0.04s."},{"Start":"03:14.050 ","End":"03:18.140","Text":"Now we can calculate s will be equal to K_sp is"},{"Start":"03:18.140 ","End":"03:22.880","Text":"7.1 times 10 to power minus 9 divided by 0.04."},{"Start":"03:22.880 ","End":"03:27.910","Text":"That gives us 1.8 times 10 to the power minus 7."},{"Start":"03:27.910 ","End":"03:36.405","Text":"That\u0027s our value of s when the PbI_2 is in a solution of NaI."},{"Start":"03:36.405 ","End":"03:43.925","Text":"Now the molar solubility in water is 1.2 times 10 to the power minus 3."},{"Start":"03:43.925 ","End":"03:48.830","Text":"We can see a considerable reduction in solubility,"},{"Start":"03:48.830 ","End":"03:51.500","Text":"some 5 orders of magnitude."},{"Start":"03:51.500 ","End":"03:54.920","Text":"Again, we should point out that the numerical value of s is"},{"Start":"03:54.920 ","End":"04:01.115","Text":"only an estimate since we haven\u0027t taken into account the interaction between the ions."},{"Start":"04:01.115 ","End":"04:05.285","Text":"We should really use activities and not concentrations."},{"Start":"04:05.285 ","End":"04:11.840","Text":"Now there\u0027s another effect related to salts and that\u0027s when we have a non-common ion."},{"Start":"04:11.840 ","End":"04:18.575","Text":"Now non-common ions have the opposite effect on the solubility equilibria to common ions."},{"Start":"04:18.575 ","End":"04:21.890","Text":"That means that non-common ions increase the solubility,"},{"Start":"04:21.890 ","End":"04:25.505","Text":"but the fact is smaller than the common ion effect."},{"Start":"04:25.505 ","End":"04:28.370","Text":"Here\u0027s a graph illustrating the effect."},{"Start":"04:28.370 ","End":"04:30.530","Text":"It\u0027s just qualitative."},{"Start":"04:30.530 ","End":"04:37.040","Text":"The solubility is plotted against the concentration of the added salt."},{"Start":"04:37.040 ","End":"04:40.180","Text":"Here\u0027s a common ion effect,"},{"Start":"04:40.180 ","End":"04:44.720","Text":"the solubility decreases as you add more salt."},{"Start":"04:44.720 ","End":"04:50.585","Text":"Here\u0027s the salt effect on non-common ion."},{"Start":"04:50.585 ","End":"04:54.710","Text":"We see that the solubility increases,"},{"Start":"04:54.710 ","End":"05:00.075","Text":"but it\u0027s not as steep a curve as the one for the common ion."},{"Start":"05:00.075 ","End":"05:07.120","Text":"In this video, we talked about the common ion effect and also about the salt effect."}],"ID":31819}],"Thumbnail":null,"ID":283655},{"Name":"Precipitation","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Fractional or Selective Precipitation","Duration":"6m 10s","ChapterTopicVideoID":28480,"CourseChapterTopicPlaylistID":283656,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.040","Text":"In the previous video,"},{"Start":"00:02.040 ","End":"00:05.370","Text":"we learned how to predict where the precipitation will occur,"},{"Start":"00:05.370 ","End":"00:10.275","Text":"and in this video we\u0027ll discuss fractional or selective precipitation."},{"Start":"00:10.275 ","End":"00:14.520","Text":"We\u0027re going to talk about fractional or selective precipitation,"},{"Start":"00:14.520 ","End":"00:16.845","Text":"2 names for the same thing."},{"Start":"00:16.845 ","End":"00:19.830","Text":"Fractional or selective precipitation is"},{"Start":"00:19.830 ","End":"00:23.820","Text":"a method of separating 2 or more ions in the solution."},{"Start":"00:23.820 ","End":"00:25.190","Text":"How do we do this?"},{"Start":"00:25.190 ","End":"00:30.485","Text":"We add a reagent that can potentially cause precipitation of each of the ions."},{"Start":"00:30.485 ","End":"00:32.690","Text":"Now as the reagent is dripped in,"},{"Start":"00:32.690 ","End":"00:36.466","Text":"it will cause precipitation of each of the ions in turn,"},{"Start":"00:36.466 ","End":"00:41.255","Text":"and we choose the conditions so the first ion completely precipitates,"},{"Start":"00:41.255 ","End":"00:44.435","Text":"that means up to 99.9 percent,"},{"Start":"00:44.435 ","End":"00:47.750","Text":"before the second begins to precipitate, and so on."},{"Start":"00:47.750 ","End":"00:51.420","Text":"We can never get complete precipitation,"},{"Start":"00:51.420 ","End":"00:53.330","Text":"there\u0027s always a little left."},{"Start":"00:53.330 ","End":"00:56.330","Text":"Here\u0027s an example, AgNO_3,"},{"Start":"00:56.330 ","End":"00:57.920","Text":"that\u0027s our reagent,"},{"Start":"00:57.920 ","End":"01:03.455","Text":"is gradually added to bromide and chromate ions in the same solution and"},{"Start":"01:03.455 ","End":"01:09.920","Text":"both have concentrations 0.02 M. Here\u0027s the question,"},{"Start":"01:09.920 ","End":"01:11.870","Text":"which will precipitate first?"},{"Start":"01:11.870 ","End":"01:15.130","Text":"Silver bromide or silver chromate,"},{"Start":"01:15.130 ","End":"01:18.330","Text":"then how much of the first precipitate will remain in"},{"Start":"01:18.330 ","End":"01:21.593","Text":"the solution when the second one begins to precipitate,"},{"Start":"01:21.593 ","End":"01:24.950","Text":"and finally is complete separation possible?"},{"Start":"01:24.950 ","End":"01:31.775","Text":"First, we write the equations for silver bromide and silver chromate."},{"Start":"01:31.775 ","End":"01:34.640","Text":"Here\u0027s silver chromate."},{"Start":"01:34.640 ","End":"01:42.050","Text":"Now the solid is in equilibrium with 2 Ag plus, plus CrO_4^2 minus."},{"Start":"01:42.050 ","End":"01:47.000","Text":"The K_sp is 1.1 times 10^minus 12."},{"Start":"01:47.000 ","End":"01:49.675","Text":"Now let\u0027s turn to AgBr."},{"Start":"01:49.675 ","End":"01:53.752","Text":"It\u0027s an equilibrium with Ag plus and Br minus and"},{"Start":"01:53.752 ","End":"01:59.315","Text":"the K_sp for this is 5 times 10^minus 13."},{"Start":"01:59.315 ","End":"02:02.405","Text":"They\u0027re both fairly insoluble."},{"Start":"02:02.405 ","End":"02:06.140","Text":"Now let\u0027s calculate the concentration of silver ion;"},{"Start":"02:06.140 ","End":"02:07.580","Text":"that\u0027s what we\u0027re dripping in,"},{"Start":"02:07.580 ","End":"02:10.400","Text":"required to precipitate each of the anion."},{"Start":"02:10.400 ","End":"02:13.540","Text":"We start with silver chromate."},{"Start":"02:13.540 ","End":"02:17.030","Text":"The K_sp of silver chromate is 1.1 times"},{"Start":"02:17.030 ","End":"02:22.085","Text":"10^minus 12 and that\u0027s the concentration of Ag plus squared."},{"Start":"02:22.085 ","End":"02:24.860","Text":"The 2 is because of this 2 here,"},{"Start":"02:24.860 ","End":"02:28.690","Text":"times the concentration of Cr_4^2 minus."},{"Start":"02:28.690 ","End":"02:31.380","Text":"That\u0027s Ag plus squared,"},{"Start":"02:31.380 ","End":"02:35.490","Text":"and we know the concentration of CrO_4^2 minus,"},{"Start":"02:35.490 ","End":"02:39.950","Text":"we know it\u0027s 0.02 because that was given in the question."},{"Start":"02:39.950 ","End":"02:46.775","Text":"We can see that Ag plus squared is equal to 5.5 times 10^minus 11."},{"Start":"02:46.775 ","End":"02:52.600","Text":"That\u0027s 1.1 times 10^minus 12 divided by 0.02."},{"Start":"02:52.600 ","End":"02:57.656","Text":"Ag plus squared is 5.5 times 10^minus 11,"},{"Start":"02:57.656 ","End":"03:01.580","Text":"and that means the concentration of Ag plus is a square root"},{"Start":"03:01.580 ","End":"03:06.160","Text":"at 7.4 times 10^minus 6 molar."},{"Start":"03:06.160 ","End":"03:10.370","Text":"That\u0027s the amount, that\u0027s the concentration of Ag plus that needs to be"},{"Start":"03:10.370 ","End":"03:15.770","Text":"dripped in to get precipitation of silver chromate."},{"Start":"03:15.770 ","End":"03:17.120","Text":"We need this amount."},{"Start":"03:17.120 ","End":"03:19.895","Text":"Let\u0027s turn to the silver bromide."},{"Start":"03:19.895 ","End":"03:24.710","Text":"K_sp for silver bromide is 5 times 10^minus 13,"},{"Start":"03:24.710 ","End":"03:30.160","Text":"and that\u0027s equal to concentration of Ag plus times the concentration of Br minus."},{"Start":"03:30.160 ","End":"03:34.370","Text":"We want to find the concentration of Ag plus, we\u0027ll write that here,"},{"Start":"03:34.370 ","End":"03:37.100","Text":"Ag plus times,"},{"Start":"03:37.100 ","End":"03:42.415","Text":"and we know the concentration Br minus because it was given in the problem; 0.02."},{"Start":"03:42.415 ","End":"03:47.780","Text":"We can work out that the concentration of Ag plus required to give"},{"Start":"03:47.780 ","End":"03:55.460","Text":"precipitation is 5.0 times 10^minus 13 divided by 0.02,"},{"Start":"03:55.460 ","End":"04:00.600","Text":"and that\u0027s 2.5 times 10^minus 11."},{"Start":"04:00.600 ","End":"04:03.590","Text":"In order to precipitate the chromate,"},{"Start":"04:03.590 ","End":"04:09.320","Text":"we require that the concentration of Ag plus is 7.4 times 10^minus 6,"},{"Start":"04:09.320 ","End":"04:15.890","Text":"and to precipitate the bromide it needs to be 2.5 times 10^minus 11."},{"Start":"04:15.890 ","End":"04:17.420","Text":"As we\u0027re dripping it in gradually,"},{"Start":"04:17.420 ","End":"04:21.320","Text":"we\u0027ll get to this number long before we get to the other."},{"Start":"04:21.320 ","End":"04:25.885","Text":"That means that the silver bromide will precipitate first."},{"Start":"04:25.885 ","End":"04:30.659","Text":"We need less silver plus to precipitate AgBr,"},{"Start":"04:30.659 ","End":"04:32.995","Text":"so it will precipitate first."},{"Start":"04:32.995 ","End":"04:39.080","Text":"As we drip in the silver plus AgBr precipitate"},{"Start":"04:39.080 ","End":"04:44.930","Text":"and then we get to the concentration required to precipitate the silver chromate,"},{"Start":"04:44.930 ","End":"04:48.080","Text":"that\u0027s 7.4 times 10^minus 6."},{"Start":"04:48.080 ","End":"04:49.865","Text":"We ask the question,"},{"Start":"04:49.865 ","End":"04:56.660","Text":"how much of this bromide will be left when the chromate starts to precipitate?"},{"Start":"04:56.660 ","End":"05:01.340","Text":"Ag_2CrO_4 starts to precipitate"},{"Start":"05:01.340 ","End":"05:06.080","Text":"when the concentration of Ag plus is 7.4 times 10^minus 6."},{"Start":"05:06.080 ","End":"05:07.670","Text":"We work that out before."},{"Start":"05:07.670 ","End":"05:11.675","Text":"That means at that point the concentration of bromide will be"},{"Start":"05:11.675 ","End":"05:18.270","Text":"the K_sp of bromide that\u0027s divided by 7.4 times 10^minus 6,"},{"Start":"05:18.270 ","End":"05:23.420","Text":"and that gives us 6.8 times 10^minus 8."},{"Start":"05:23.420 ","End":"05:27.020","Text":"That\u0027s the concentration of bromide left in"},{"Start":"05:27.020 ","End":"05:30.595","Text":"the solution when the chromate starts to precipitate."},{"Start":"05:30.595 ","End":"05:35.720","Text":"The percentage of the bromide remaining in the solution is this concentration,"},{"Start":"05:35.720 ","End":"05:41.696","Text":"6.8 times 10^minus 8 divided by the initial concentration,"},{"Start":"05:41.696 ","End":"05:46.205","Text":"2 times 10^minus 2 times 100 percent."},{"Start":"05:46.205 ","End":"05:51.545","Text":"If you work that out, it comes to 3.4 times 10^minus 8 percent."},{"Start":"05:51.545 ","End":"05:57.598","Text":"There\u0027s virtually no bromide left when the chromate precipitates,"},{"Start":"05:57.598 ","End":"06:00.290","Text":"so we can say that complete separation is"},{"Start":"06:00.290 ","End":"06:04.610","Text":"possible by fractional precipitation in this case."},{"Start":"06:04.610 ","End":"06:09.210","Text":"In this video, we talked about fractional precipitation."}],"ID":29946},{"Watched":false,"Name":"Dissolving Almost Insoluble Salts","Duration":"4m 5s","ChapterTopicVideoID":28481,"CourseChapterTopicPlaylistID":283656,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.350","Text":"In the previous video,"},{"Start":"00:01.350 ","End":"00:04.470","Text":"we talked about fractional or selective precipitation."},{"Start":"00:04.470 ","End":"00:06.840","Text":"In this video we\u0027ll talk about methods of"},{"Start":"00:06.840 ","End":"00:10.860","Text":"dissolving salts that are almost insoluble in water."},{"Start":"00:10.860 ","End":"00:16.140","Text":"The first method we\u0027re going to talk about is dissolving salts by adding an acid."},{"Start":"00:16.140 ","End":"00:19.560","Text":"We can increase the solubility of a hydroxide,"},{"Start":"00:19.560 ","End":"00:21.699","Text":"for example, by adding an acid."},{"Start":"00:21.699 ","End":"00:23.135","Text":"2 examples."},{"Start":"00:23.135 ","End":"00:30.645","Text":"Magnesium hydroxide is an equilibrium with Mg_2 plus and 2 OH minus,"},{"Start":"00:30.645 ","End":"00:35.325","Text":"that\u0027s milk of magnesia which can be used to settle your stomach."},{"Start":"00:35.325 ","End":"00:43.360","Text":"Or iron hydroxide, which is in equilibrium with Fe^3 plus and 3 OH minus."},{"Start":"00:43.360 ","End":"00:47.665","Text":"That\u0027s Fe^3 hydroxide."},{"Start":"00:47.665 ","End":"00:50.000","Text":"Now, if we add acid,"},{"Start":"00:50.000 ","End":"00:52.310","Text":"it will react with the hydroxide."},{"Start":"00:52.310 ","End":"00:55.790","Text":"Could be adding acid or the acid in your stomach,"},{"Start":"00:55.790 ","End":"00:58.070","Text":"that\u0027s how the milk of magnesia works."},{"Start":"00:58.070 ","End":"01:00.425","Text":"It reacts with the acid in the stomach."},{"Start":"01:00.425 ","End":"01:08.165","Text":"We get the acid HCl plus a hydroxide OH minus giving us 2 molecules of water."},{"Start":"01:08.165 ","End":"01:14.630","Text":"Now that pushes the equilibrium to the right so that more hydroxide dissolves."},{"Start":"01:14.630 ","End":"01:21.290","Text":"We\u0027re removing the hydroxide by reacting it with the acid and then more will dissolve."},{"Start":"01:21.290 ","End":"01:25.625","Text":"The equilibrium is pushed to the right."},{"Start":"01:25.625 ","End":"01:29.150","Text":"Now we can also improve the solubility of"},{"Start":"01:29.150 ","End":"01:32.990","Text":"conjugate bases of weak acids by adding an acid."},{"Start":"01:32.990 ","End":"01:35.435","Text":"For example, if we have zinc carbonate,"},{"Start":"01:35.435 ","End":"01:40.535","Text":"that\u0027s an equilibrium with Zn_2 plus CO_3 2 minus."},{"Start":"01:40.535 ","End":"01:42.035","Text":"If we add an acid,"},{"Start":"01:42.035 ","End":"01:45.935","Text":"it will react with the carbonate to produce CO_2 gas."},{"Start":"01:45.935 ","End":"01:54.485","Text":"Here\u0027s the reaction, carbonate plus 2 molecules of the acid to give us CO_2 and 3H2O."},{"Start":"01:54.485 ","End":"02:01.300","Text":"Again, we\u0027re removing the CO_3 2 minus and pushing the equilibrium to the right."},{"Start":"02:01.300 ","End":"02:04.265","Text":"This is the basis of acid rain."},{"Start":"02:04.265 ","End":"02:11.690","Text":"Acid rain comes down and reacts with a carbonate in statues and buildings and so"},{"Start":"02:11.690 ","End":"02:19.565","Text":"on and then we get damage of famous statues and buildings and so on."},{"Start":"02:19.565 ","End":"02:23.975","Text":"Now we can also dissolve ions by changing the oxidation number."},{"Start":"02:23.975 ","End":"02:30.295","Text":"For example, an insoluble sulfide can be dissolved by converting it to elemental sulfur."},{"Start":"02:30.295 ","End":"02:32.955","Text":"Here\u0027s copper sulfide,"},{"Start":"02:32.955 ","End":"02:37.910","Text":"and in equilibrium with CO_2 plus and S_2 minus."},{"Start":"02:37.910 ","End":"02:42.065","Text":"If we add nitric acid to oxidize the sulfide,"},{"Start":"02:42.065 ","End":"02:45.630","Text":"this is a redox reaction."},{"Start":"02:45.630 ","End":"02:50.550","Text":"The 3 S_2 minus plus 8 HNO_3 in"},{"Start":"02:50.550 ","End":"02:57.570","Text":"equilibrium with 3 S sulfur and 2 NO,"},{"Start":"02:57.570 ","End":"03:02.655","Text":"4 H2O, and 6 NO_3 minus, that\u0027s nitrate."},{"Start":"03:02.655 ","End":"03:07.459","Text":"We are taking the sulfide and converting it to sulfur,"},{"Start":"03:07.459 ","End":"03:12.080","Text":"so the sulfide is going out of the reaction,"},{"Start":"03:12.080 ","End":"03:16.525","Text":"it\u0027s being removed, pushing the equilibrium to the right."},{"Start":"03:16.525 ","End":"03:22.235","Text":"The copper ions are dissolved as copper nitrate because we\u0027ve produced here nitrate."},{"Start":"03:22.235 ","End":"03:28.820","Text":"Another method is changing the temperature and that can be used to purify precipitates."},{"Start":"03:28.820 ","End":"03:30.695","Text":"Sometimes when we heat,"},{"Start":"03:30.695 ","End":"03:32.855","Text":"we can dissolve the precipitate."},{"Start":"03:32.855 ","End":"03:37.320","Text":"Preferably it dissolves 1 as it precipitates, but not always."},{"Start":"03:37.320 ","End":"03:38.810","Text":"Then we filter to remove"},{"Start":"03:38.810 ","End":"03:42.965","Text":"the insoluble impurities after we have dissolved the precipitate,"},{"Start":"03:42.965 ","End":"03:47.095","Text":"and then we can cool it in the fridge often overnight."},{"Start":"03:47.095 ","End":"03:50.840","Text":"Often the desired compound will recrystallize leaving"},{"Start":"03:50.840 ","End":"03:54.815","Text":"impurities in the solution and then we can filter again."},{"Start":"03:54.815 ","End":"03:58.340","Text":"You must be aware of this method from your labs."},{"Start":"03:58.340 ","End":"04:01.400","Text":"In this video, we talked about methods of dissolving"},{"Start":"04:01.400 ","End":"04:05.970","Text":"salts that are almost insoluble in water"}],"ID":29947},{"Watched":false,"Name":"Exercise 1","Duration":"6m 25s","ChapterTopicVideoID":31687,"CourseChapterTopicPlaylistID":283656,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33925},{"Watched":false,"Name":"Exercise 2","Duration":"5m 21s","ChapterTopicVideoID":31686,"CourseChapterTopicPlaylistID":283656,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33926}],"Thumbnail":null,"ID":283656},{"Name":"Equilibria Involving Complex Ions","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Complex-ion Formation","Duration":"5m 13s","ChapterTopicVideoID":28479,"CourseChapterTopicPlaylistID":283657,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.175","Text":"In the previous video,"},{"Start":"00:02.175 ","End":"00:05.865","Text":"we showed how to increase solubility and in this video,"},{"Start":"00:05.865 ","End":"00:12.285","Text":"we\u0027ll show how forming complex ions of metal cations can increase solubility."},{"Start":"00:12.285 ","End":"00:19.020","Text":"We\u0027re going to talk about the effect of complex ion formation on almost insoluble salts."},{"Start":"00:19.020 ","End":"00:22.290","Text":"We\u0027ll show that forming a complex ion can lead to"},{"Start":"00:22.290 ","End":"00:26.670","Text":"increased solubility of almost insoluble compounds."},{"Start":"00:26.670 ","End":"00:28.155","Text":"Here\u0027s an example."},{"Start":"00:28.155 ","End":"00:31.410","Text":"Silver chloride is almost insoluble in water."},{"Start":"00:31.410 ","End":"00:37.770","Text":"Here\u0027s the equilibrium, AgCl(s) in equilibrium with Ag plus and Cl minus,"},{"Start":"00:37.770 ","End":"00:42.280","Text":"and the K_sp is 1.8 times 10^ minus 10,"},{"Start":"00:42.280 ","End":"00:44.250","Text":"which is very small."},{"Start":"00:44.250 ","End":"00:54.515","Text":"Now, Ag plus can react with an ammonia solution to form the complex ion, Ag(NH_3)_2 plus."},{"Start":"00:54.515 ","End":"00:57.410","Text":"We talked about this in a previous video."},{"Start":"00:57.410 ","End":"01:02.570","Text":"Now we\u0027re going to sometimes write it like this with the brackets and sometimes"},{"Start":"01:02.570 ","End":"01:08.080","Text":"without the brackets to make it convenient to talk about concentrations."},{"Start":"01:08.080 ","End":"01:15.680","Text":"Ag plus reacts with 2 moles of ammonia to give us Ag(NH_3)2 plus,"},{"Start":"01:15.680 ","End":"01:17.825","Text":"that\u0027s our complex ion."},{"Start":"01:17.825 ","End":"01:22.665","Text":"The K of formation is 1.6 times 10^7,"},{"Start":"01:22.665 ","End":"01:24.600","Text":"so it\u0027s very large."},{"Start":"01:24.600 ","End":"01:30.275","Text":"Recall, the Ag plus is a Lewis acid and NH_3 is a Lewis base."},{"Start":"01:30.275 ","End":"01:33.005","Text":"Here\u0027s our Cl minus ion."},{"Start":"01:33.005 ","End":"01:36.425","Text":"If we remove Ag plus from equilibrium,"},{"Start":"01:36.425 ","End":"01:38.660","Text":"the equilibrium will shift to"},{"Start":"01:38.660 ","End":"01:39.890","Text":"the right so that"},{"Start":"01:39.890 ","End":"01:41.600","Text":"more AgCl dissolves. By removing the Ag plus,"},{"Start":"01:41.600 ","End":"01:41.601","Text":"the equilibrium will shift to the right, and more and more AgCl will dissolve."},{"Start":"01:41.601 ","End":"01:42.930","Text":"Let\u0027s"},{"Start":"01:51.340 ","End":"01:53.870","Text":"calculate the equilibrium"},{"Start":"01:53.870 ","End":"01:56.170","Text":"constant for the overall reaction."},{"Start":"01:56.170 ","End":"02:05.950","Text":"AgCl reacting with 2 moles of ammonia to give us the complex Ag(NH_3)2 plus and Cl minus."},{"Start":"02:05.950 ","End":"02:10.865","Text":"Now, this reaction is the sum of the reactions we\u0027ve talked about before,"},{"Start":"02:10.865 ","End":"02:13.940","Text":"this one and this one."},{"Start":"02:13.940 ","End":"02:16.085","Text":"If we add these two,"},{"Start":"02:16.085 ","End":"02:19.610","Text":"we see the Ag plus cancels."},{"Start":"02:19.610 ","End":"02:26.880","Text":"We\u0027re left with AgCl plus 2NH_3 in the left-hand side,"},{"Start":"02:26.880 ","End":"02:28.580","Text":"and in the right-hand side,"},{"Start":"02:28.580 ","End":"02:32.359","Text":"the complex plus Cl minus."},{"Start":"02:32.359 ","End":"02:34.425","Text":"If we add 2 reactions,"},{"Start":"02:34.425 ","End":"02:44.450","Text":"then we multiply the Ks to get the overall K. K for this reaction is K_sp times K_f."},{"Start":"02:44.450 ","End":"02:50.830","Text":"That\u0027s 1.8 times 10^minus 10 times 1.6 times 10^7."},{"Start":"02:50.830 ","End":"02:55.890","Text":"That\u0027s equal to 2.9 times 10^minus 3."},{"Start":"02:55.890 ","End":"02:59.785","Text":"That\u0027s much larger than the K_sp."},{"Start":"02:59.785 ","End":"03:01.985","Text":"Here\u0027s a numerical example."},{"Start":"03:01.985 ","End":"03:07.940","Text":"Calculate the molar solubility of AgCl in 0.20 molar ammonia."},{"Start":"03:07.940 ","End":"03:11.195","Text":"Now, for every mole of AgCl that dissolves,"},{"Start":"03:11.195 ","End":"03:13.520","Text":"1 mole of Cl minus is produced,"},{"Start":"03:13.520 ","End":"03:19.625","Text":"so the molar solubility S is equal to the concentration of Cl minus."},{"Start":"03:19.625 ","End":"03:22.070","Text":"Here\u0027s our ICE table."},{"Start":"03:22.070 ","End":"03:25.415","Text":"We start with 0.20 molar ammonia,"},{"Start":"03:25.415 ","End":"03:29.620","Text":"0 of the complex and 0 Cl minus."},{"Start":"03:29.620 ","End":"03:32.030","Text":"On the way to equilibrium,"},{"Start":"03:32.030 ","End":"03:37.070","Text":"the amount of NH_3 decreases by minus 2S."},{"Start":"03:37.070 ","End":"03:40.430","Text":"Here we have 2 in the equation."},{"Start":"03:40.430 ","End":"03:44.155","Text":"The amount of the complex increases by plus s,"},{"Start":"03:44.155 ","End":"03:46.290","Text":"just 1 mole here,"},{"Start":"03:46.290 ","End":"03:50.220","Text":"and Cl minus also plus s. At equilibrium,"},{"Start":"03:50.220 ","End":"03:53.475","Text":"we have 0.20 minus 2s for the ammonia,"},{"Start":"03:53.475 ","End":"03:55.995","Text":"plus s for the complex,"},{"Start":"03:55.995 ","End":"03:58.305","Text":"and plus s for Cl minus."},{"Start":"03:58.305 ","End":"04:03.920","Text":"K, which is the concentration of the complex,"},{"Start":"04:03.920 ","End":"04:05.975","Text":"times the concentration of Cl minus,"},{"Start":"04:05.975 ","End":"04:13.155","Text":"divided by the concentration of ammonia^2=S^2 because s times s,"},{"Start":"04:13.155 ","End":"04:17.850","Text":"divided by 0.20 minus 2s and that\u0027s squared."},{"Start":"04:17.850 ","End":"04:20.010","Text":"That\u0027s equal to K,"},{"Start":"04:20.010 ","End":"04:22.840","Text":"which is 2.9 time 10^minus 3."},{"Start":"04:22.840 ","End":"04:25.460","Text":"Now, we take the square root of each side,"},{"Start":"04:25.460 ","End":"04:29.630","Text":"we get s divided by 0.20 minus 2s equal to"},{"Start":"04:29.630 ","End":"04:35.995","Text":"the square root of 2.9 times 10^minus 3, which is 0.053."},{"Start":"04:35.995 ","End":"04:38.730","Text":"From that, we can work out that"},{"Start":"04:38.730 ","End":"04:47.265","Text":"1.106s=0.0106, so that s=0.91."},{"Start":"04:47.265 ","End":"04:49.655","Text":"We can add the M to that."},{"Start":"04:49.655 ","End":"04:54.380","Text":"The molar solubility of AgCl in ammonia is 0.91"},{"Start":"04:54.380 ","End":"05:00.435","Text":"molar compared to 1.3 times 10^minus 5 in water."},{"Start":"05:00.435 ","End":"05:02.390","Text":"By forming the complex,"},{"Start":"05:02.390 ","End":"05:05.080","Text":"we\u0027ve increased the solubility."},{"Start":"05:05.080 ","End":"05:08.220","Text":"In this video, we showed that forming a complex ion of"},{"Start":"05:08.220 ","End":"05:13.170","Text":"a metal ion can increase solubility considerably."}],"ID":29944},{"Watched":false,"Name":"Qualitative Analysis","Duration":"5m 28s","ChapterTopicVideoID":28478,"CourseChapterTopicPlaylistID":283657,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.200 ","End":"00:04.425","Text":"In this video, we will talk about qualitative analysis,"},{"Start":"00:04.425 ","End":"00:08.639","Text":"which involves many of the topics discussed in recent videos."},{"Start":"00:08.639 ","End":"00:12.135","Text":"We\u0027re talking about qualitative analysis."},{"Start":"00:12.135 ","End":"00:15.090","Text":"This is a method of identifying the presence or"},{"Start":"00:15.090 ","End":"00:18.435","Text":"absence of metal cations in aqueous solutions."},{"Start":"00:18.435 ","End":"00:20.850","Text":"It\u0027s often taught in first year lab courses,"},{"Start":"00:20.850 ","End":"00:24.465","Text":"but not widely used in real analytical laboratories."},{"Start":"00:24.465 ","End":"00:27.390","Text":"It\u0027s taught because it involves complex formation,"},{"Start":"00:27.390 ","End":"00:30.405","Text":"selective precipitation, and control of pH,"},{"Start":"00:30.405 ","End":"00:32.565","Text":"all of which are important topics."},{"Start":"00:32.565 ","End":"00:35.790","Text":"Now we\u0027re going to discuss a scheme which is often used."},{"Start":"00:35.790 ","End":"00:39.130","Text":"It\u0027s not the only scheme, there are many variations."},{"Start":"00:39.130 ","End":"00:42.800","Text":"Basically the ions are divided into 5 groups and"},{"Start":"00:42.800 ","End":"00:46.895","Text":"the order in which we perform the various actions is important."},{"Start":"00:46.895 ","End":"00:49.020","Text":"Let\u0027s start with Group 1."},{"Start":"00:49.020 ","End":"00:55.280","Text":"To aqueous solution that contains the unknowns that we wish to identify,"},{"Start":"00:55.280 ","End":"00:57.440","Text":"we add 6 molar HCl,"},{"Start":"00:57.440 ","End":"01:00.620","Text":"then we get insoluble chlorides precipitate,"},{"Start":"01:00.620 ","End":"01:03.260","Text":"and these are the chlorides of silver,"},{"Start":"01:03.260 ","End":"01:06.080","Text":"lead, and mercury (I)."},{"Start":"01:06.080 ","End":"01:08.750","Text":"Now Group 2. To the supernatant,"},{"Start":"01:08.750 ","End":"01:14.225","Text":"that\u0027s a liquid remaining after we filter off the insoluble chlorides,"},{"Start":"01:14.225 ","End":"01:18.790","Text":"we add H_2S gas and 0.2 molar HCl."},{"Start":"01:18.790 ","End":"01:22.920","Text":"Then we get the acid insoluble sulfides precipitate,"},{"Start":"01:22.920 ","End":"01:25.640","Text":"and these are the sulfides of copper,"},{"Start":"01:25.640 ","End":"01:28.849","Text":"cadmium, bismuth, lead, mercury,"},{"Start":"01:28.849 ","End":"01:33.266","Text":"now mercury(II), arsenic, antimony and tin."},{"Start":"01:33.266 ","End":"01:34.450","Text":"Now Group 3."},{"Start":"01:34.450 ","End":"01:38.000","Text":"To the supernatant that remains after Group 2,"},{"Start":"01:38.000 ","End":"01:41.495","Text":"we add ammonium sulfide at pH 4,"},{"Start":"01:41.495 ","End":"01:46.505","Text":"and then we get base-soluble sulfides and hydroxides precipitate."},{"Start":"01:46.505 ","End":"01:50.740","Text":"We get aluminum hydroxide, iron hydroxide,"},{"Start":"01:50.740 ","End":"01:53.570","Text":"chromium hydroxide, zinc sulfide,"},{"Start":"01:53.570 ","End":"01:57.980","Text":"nickel sulfide, manganese sulfide, and cobalt sulfide."},{"Start":"01:57.980 ","End":"02:03.012","Text":"All these ions are contained in Group 3."},{"Start":"02:03.012 ","End":"02:04.635","Text":"Now Group 4."},{"Start":"02:04.635 ","End":"02:07.600","Text":"To the supernatant after Group 3,"},{"Start":"02:07.600 ","End":"02:12.410","Text":"we add ammonium hydrogen phosphate and ammonia."},{"Start":"02:12.410 ","End":"02:16.230","Text":"Then we get insoluble phosphates precipitate."},{"Start":"02:16.230 ","End":"02:20.810","Text":"These are the phosphates of calcium, Ca_3(PO_4)_2,"},{"Start":"02:20.810 ","End":"02:23.315","Text":"strontium, barium,"},{"Start":"02:23.315 ","End":"02:26.755","Text":"and magnesium ammonium phosphate."},{"Start":"02:26.755 ","End":"02:29.160","Text":"We can identify calcium,"},{"Start":"02:29.160 ","End":"02:30.675","Text":"strontium, barium,"},{"Start":"02:30.675 ","End":"02:33.768","Text":"or magnesium in Group 4."},{"Start":"02:33.768 ","End":"02:35.940","Text":"Now Group 5."},{"Start":"02:35.940 ","End":"02:42.675","Text":"This is the final supernatant and it contains alkali metal ions and NH_4+."},{"Start":"02:42.675 ","End":"02:45.415","Text":"These are all soluble in water."},{"Start":"02:45.415 ","End":"02:50.900","Text":"Now how can we identify each of the insoluble chlorides of Group 1?"},{"Start":"02:50.900 ","End":"02:52.970","Text":"Let\u0027s start with lead chloride."},{"Start":"02:52.970 ","End":"02:54.740","Text":"Lead chloride dissolves in"},{"Start":"02:54.740 ","End":"03:00.140","Text":"hot water because it has a higher K_sp than the other 2 chlorides."},{"Start":"03:00.140 ","End":"03:05.555","Text":"So we can identify it by adding potassium chromate to the hot water."},{"Start":"03:05.555 ","End":"03:09.305","Text":"Let\u0027s dissolve the lead chloride and we get lead 2+"},{"Start":"03:09.305 ","End":"03:15.275","Text":"reacting with chromate to give us lead chromate, which is yellow."},{"Start":"03:15.275 ","End":"03:17.690","Text":"If we get a yellow precipitate,"},{"Start":"03:17.690 ","End":"03:19.730","Text":"we know we have lead."},{"Start":"03:19.730 ","End":"03:22.070","Text":"Now, what about the other 2?"},{"Start":"03:22.070 ","End":"03:25.615","Text":"To the remaining precipitate, we add ammonia."},{"Start":"03:25.615 ","End":"03:28.220","Text":"Now, if there is silver present,"},{"Start":"03:28.220 ","End":"03:32.630","Text":"the silver chloride will react with 2 moles of ammonia,"},{"Start":"03:32.630 ","End":"03:37.790","Text":"giving us this complex we\u0027ve met several times,"},{"Start":"03:37.790 ","End":"03:39.980","Text":"silver ammonia plus, plus Cl minus."},{"Start":"03:39.980 ","End":"03:42.200","Text":"Now, it is soluble in water."},{"Start":"03:42.200 ","End":"03:44.330","Text":"When add nitric acid to it,"},{"Start":"03:44.330 ","End":"03:47.630","Text":"we get AgCl and that\u0027s white."},{"Start":"03:47.630 ","End":"03:49.160","Text":"If we get a white precipitate,"},{"Start":"03:49.160 ","End":"03:55.715","Text":"we know we have Ag+ present in our original solution."},{"Start":"03:55.715 ","End":"03:59.165","Text":"Now, what about mercury (I) chloride?"},{"Start":"03:59.165 ","End":"04:05.720","Text":"We can perform a redox reaction by adding ammonia to Hg_2Cl_2,"},{"Start":"04:05.720 ","End":"04:09.005","Text":"that gives us mercury,"},{"Start":"04:09.005 ","End":"04:15.815","Text":"Hg, in the liquid form and that\u0027s black, and HgNH_2Cl,"},{"Start":"04:15.815 ","End":"04:17.480","Text":"which is white,"},{"Start":"04:17.480 ","End":"04:20.150","Text":"and it\u0027s a solid,"},{"Start":"04:20.150 ","End":"04:21.920","Text":"that\u0027s a liquid and a solid,"},{"Start":"04:21.920 ","End":"04:25.505","Text":"plus ammonium and chloride."},{"Start":"04:25.505 ","End":"04:27.350","Text":"Now, because we have black and white,"},{"Start":"04:27.350 ","End":"04:28.985","Text":"we get dark gray."},{"Start":"04:28.985 ","End":"04:31.220","Text":"If we get a dark gray precipitate,"},{"Start":"04:31.220 ","End":"04:33.995","Text":"we know we have Hg present,"},{"Start":"04:33.995 ","End":"04:35.780","Text":"and if we get a white precipitate,"},{"Start":"04:35.780 ","End":"04:37.820","Text":"we know we have silver present."},{"Start":"04:37.820 ","End":"04:41.795","Text":"Now how can we identify the soluble salts of Group 5?"},{"Start":"04:41.795 ","End":"04:43.895","Text":"If we use a flame test,"},{"Start":"04:43.895 ","End":"04:49.058","Text":"then Na+ gives a yellow flame and K+ gives a violet flame."},{"Start":"04:49.058 ","End":"04:57.245","Text":"We can add a hydroxide that will react with NH_4+ to give us NH_3, ammonia and water."},{"Start":"04:57.245 ","End":"05:02.615","Text":"Then we identify the ammonia by its significant pungent smell."},{"Start":"05:02.615 ","End":"05:08.255","Text":"By adding hydroxide, we can identify that we had ammonium present."},{"Start":"05:08.255 ","End":"05:09.785","Text":"Using the flame test,"},{"Start":"05:09.785 ","End":"05:13.730","Text":"we can see whether we have sodium plus or potassium plus."},{"Start":"05:13.730 ","End":"05:17.690","Text":"Now the ions in other groups are identified using similar principles;"},{"Start":"05:17.690 ","End":"05:21.875","Text":"complex formation, selective precipitation, and control of pH."},{"Start":"05:21.875 ","End":"05:28.590","Text":"In this video, we talked a little about qualitative analysis of metal cations."}],"ID":29945},{"Watched":false,"Name":"Exercise 1","Duration":"3m 30s","ChapterTopicVideoID":31689,"CourseChapterTopicPlaylistID":283657,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33927},{"Watched":false,"Name":"Exercise 2","Duration":"6m 1s","ChapterTopicVideoID":31688,"CourseChapterTopicPlaylistID":283657,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33928}],"Thumbnail":null,"ID":283657}]
