{"id":2750,"date":"2016-08-24T18:21:19","date_gmt":"2016-08-24T18:21:19","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/?post_type=chapter&#038;p=2750"},"modified":"2017-09-06T18:58:54","modified_gmt":"2017-09-06T18:58:54","slug":"conversion-of-ksp-to-solubility","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/chapter\/conversion-of-ksp-to-solubility\/","title":{"raw":"Conversion of Ksp to Solubility","rendered":"Conversion of Ksp to Solubility"},"content":{"raw":"<div class=\"x-ck12-data-objectives\">\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Perform calculations converting\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0to solubility.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3><strong style=\"line-height: 1.5;\">How do you purify water?<\/strong><\/h3>\r\n<p id=\"x-ck12-NTcxMjZiN2FhOGRmMjQyYjdjM2E3YmM5ODM0YzY4NTA.-s7t\">Purification of water for drinking and other uses is a complicated process. Heavy metals need to be removed, a process accomplished by addition of carbonates and sulfates. Lead contamination can present major health problems, especially for younger children. Lead sulfates and carbonates are very insoluble, so will precipitate out of solution very easily.<\/p>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213036\/20140811155758642168.jpeg\" alt=\"Heavy metals can be removed by precipitation with carbonates and sulfates\" width=\"400\" \/>\r\n\r\n<\/div>\r\n<h3>Conversion of\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0to Solubility<\/h3>\r\n<table id=\"x-ck12-OTI5ODAyZDhhMDJlODM3NDEzZjBkMDNlOTgyNDcxNWE.-8hd\" class=\"x-ck12-nofloat\" border=\"1\"><caption>Solubility Product Constants (25\u00b0C)<\/caption>\r\n<tbody>\r\n<tr>\r\n<td><strong> Compound <\/strong><\/td>\r\n<td><em>K<sub>sp<\/sub><\/em><\/td>\r\n<td><strong> Compound <\/strong><\/td>\r\n<td><em>K<sub>sp<\/sub><\/em><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>AgBr<\/td>\r\n<td>5.0\u00a0\u00d7 10 <sup> -13 <\/sup><\/td>\r\n<td>CuS<\/td>\r\n<td>8.0\u00a0\u00d7 10 <sup> -37 <\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>AgCl<\/td>\r\n<td>1.8 \u00d7 10 <sup> -10 <\/sup><\/td>\r\n<td>Fe(OH) <sub> 2 <\/sub><\/td>\r\n<td>7.9 \u00d7 10 <sup> -16 <\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Al(OH) <sub> 3 <\/sub><\/td>\r\n<td>3.0\u00a0\u00d7 10 <sup> -34 <\/sup><\/td>\r\n<td>Mg(OH) <sub> 2 <\/sub><\/td>\r\n<td>7.1 \u00d7 10 <sup> -12 <\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>BaCO <sub> 3 <\/sub><\/td>\r\n<td>5.0\u00a0\u00d7 10 <sup> -9 <\/sup><\/td>\r\n<td>PbCl <sub> 2 <\/sub><\/td>\r\n<td>1.7 \u00d7 10 <sup> -5 <\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>BaSO <sub> 4 <\/sub><\/td>\r\n<td>1.1 \u00d7 10 <sup> -10 <\/sup><\/td>\r\n<td>PbCO <sub> 3 <\/sub><\/td>\r\n<td>7.4 \u00d7 10 <sup> -14 <\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>CaCO <sub> 3 <\/sub><\/td>\r\n<td>4.5 \u00d7 10 <sup> -9 <\/sup><\/td>\r\n<td>PbI <sub> 2 <\/sub><\/td>\r\n<td>7.1 \u00d7 10 <sup> -9 <\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Ca(OH) <sub> 2 <\/sub><\/td>\r\n<td>6.5 \u00d7 10 <sup> -6 <\/sup><\/td>\r\n<td>PbSO <sub> 4 <\/sub><\/td>\r\n<td>6.3 \u00d7 10 <sup> -7 <\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Ca <sub> 3 <\/sub> (PO <sub> 4 <\/sub> ) <sub> 2 <\/sub><\/td>\r\n<td>1.2 \u00d7 10 <sup> -26 <\/sup><\/td>\r\n<td>Zn(OH) <sub> 2 <\/sub><\/td>\r\n<td>3.0\u00a0\u00d7 10 <sup> -16 <\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>CaSO <sub> 4 <\/sub><\/td>\r\n<td>2.4 \u00d7 10 <sup> -5 <\/sup><\/td>\r\n<td>ZnS<\/td>\r\n<td>3.0\u00a0\u00d7 10 <sup> -23 <\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"x-ck12-MGJkN2JhYjU1ZDYyMzgyMjc0MjYyMzFjYzE1Y2U2MTQ.-bxd\">The known\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0values from the <strong> Table <\/strong> above can be used to calculate the solubility of a given compound by following the steps listed below.<\/p>\r\n\r\n<ol id=\"x-ck12-NTM3M2I1Y2JmNjQ1YWFhODY5NTZjOTUzYzA4Njk3ZTk.-1yo\">\r\n \t<li>Set up an ICE problem (Initial, Change, Equilibrium) in order to use the\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0value to calculate the concentration of each of the ions.<\/li>\r\n \t<li>The concentration of the ions leads to the molar solubility of the compound.<\/li>\r\n \t<li>Use the molar mass to convert from molar solubility to solubility.<\/li>\r\n<\/ol>\r\n<p id=\"x-ck12-NzZkMjU0YTNlY2YyYWEzZThiMDliMTFmNTUwYmM3OTM.-ymt\">The\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0of calcium carbonate is 4.5\u00a0\u00d7\u00a010 <sup> -9 <\/sup> . We begin by setting up an ICE table showing the dissociation of CaCO <sub> 3 <\/sub> into calcium ions and carbonate ions. The variable\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211145\/1a271d97818de0a9d58667885dc707b3.png\" alt=\"s\" width=\"8\" height=\"8\" \/> will be used to represent the molar solubility of CaCO <sub> 3 <\/sub> . In this case, each formula unit of CaCO <sub> 3 <\/sub> yields one Ca <sup> 2+ <\/sup> ion and one CO <sub> 3 <\/sub><sup> 2\u2212 <\/sup> ion. Therefore, the equilibrium concentrations of each ion are equal to <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211145\/1a271d97818de0a9d58667885dc707b3.png\" alt=\"s\" width=\"8\" height=\"8\" \/> .<\/p>\r\n<p id=\"x-ck12-i6o\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213038\/a2c8dd7fa1970edc56d741206bb79883.png\" alt=\"&amp; text{CaCO}_3(s) quad rightleftarrows quad text{Ca}^{2+}(aq)+ text{CO}_3^{2-}(aq) \\text{Initial }(text{M}) &amp; qquad qquad qquad qquad quad 0.00 qquad quad 0.00 \\text{Change }(text{M}) &amp; qquad qquad qquad qquad quad +s qquad quad +s \\qquad text{Equilibrium }(text{M}) &amp; qquad qquad qquad qquad qquad s qquad qquad s\" width=\"440\" height=\"101\" \/><\/p>\r\n<p id=\"x-ck12-N2FmM2RiNzMzYzk4YzIwODVmOGM3MzI2M2NhYzJlZWU.-beo\">The\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0expression can be written in terms of\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211145\/1a271d97818de0a9d58667885dc707b3.png\" alt=\"s\" width=\"8\" height=\"8\" \/> and then used to solve for <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211145\/1a271d97818de0a9d58667885dc707b3.png\" alt=\"s\" width=\"8\" height=\"8\" \/> .<\/p>\r\n<img id=\"x-ck12-MTM2Nzk1ODQ2ODMwNQ..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213039\/7c7e12bec87657b9ceec66e8d4f3dc2f.png\" alt=\"K_{sp}&amp;=[ text{Ca}^{2+}][ text{CO}_3^{2-}]=(s)(s)=s^2 \\s&amp;=sqrt{K_{sp}}=sqrt{4.5 times 10^{-9}}=6.7 times 10^{-5} text{M}\" width=\"346\" height=\"53\" \/>\r\n<p id=\"x-ck12-ZjQ2NmQzZDY3ODAwYWViYjVkYzc2NDg4M2IwNzUyMzA.-glf\">The concentration of each of the ions at equilibrium is 6.7\u00a0\u00d7\u00a010 <sup> -5 <\/sup> \u00a0M. We can use the molar mass to convert from molar solubility to solubility.<\/p>\r\n<p id=\"x-ck12-eg3\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213040\/10f0727dbce300daa8d80dfbfb7bc102.png\" alt=\"frac{6.7 times 10^{-5} cancel{text{mol}}}{text{L}} times frac{100.09 text{g}}{1 cancel{text{mol}}} = 6.7 times 10^{-3} text{g\/L}\" width=\"347\" height=\"43\" \/><\/p>\r\n<p id=\"x-ck12-ZmM2MTAyMmEwYmU0MjZiODdlODkzZWE4NzdmYzRhY2E.-lqj\">So the maximum amount of calcium carbonate that is capable of dissolving in 1 liter of water at 25\u00b0C is 6.7\u00a0\u00d7\u00a010 <sup> -3 <\/sup> \u00a0grams. Note that in the case above, the 1:1 ratio of the ions upon dissociation led to the\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0being equal to <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213041\/a13a0f5d5e1213e2adc9f2c9f7caea61.png\" alt=\"s^2\" width=\"14\" height=\"15\" \/> . This is referred to as a formula of the type <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213042\/81c6b8fb0f89e65473814916c64fb053.png\" alt=\"AB\" width=\"27\" height=\"12\" \/> , where\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/d5ebd507fac84fdb3364e6593d198a76.png\" alt=\"A\" width=\"13\" height=\"12\" \/> is the cation and\u00a0 <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/00025e2ee78e7b712fbb42f74f2d6cb7.png\" alt=\"B\" width=\"14\" height=\"12\" \/> is the anion. Now let\u2019s consider a formula of the type <img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213042\/8735d5f5e52109132e5d695fbea001a6.png\" alt=\"AB_2\" width=\"33\" height=\"15\" \/> , such as Fe(OH) <sub> 2 <\/sub> . In this case the setup of the ICE table would look like the following:<\/p>\r\n<img id=\"x-ck12-MTM2Nzk1ODQ2ODMwNg..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213042\/6c4a62889a1377d890fef7ce071965ff.png\" alt=\"&amp; text{Fe(OH)}_2(s) quad rightleftarrows quad text{Fe}^{2+}(aq)+2 text{OH}^-(aq) \\text{Initial }(text{M}) &amp; qquad qquad qquad qquad qquad 0.00 qquad quad 0.00 \\text{Change }(text{M}) &amp; qquad qquad qquad qquad qquad +s qquad quad +2s \\text{Equilibrium }(text{M}) &amp; qquad qquad qquad qquad qquad quad s qquad qquad 2s\" width=\"450\" height=\"101\" \/>\r\n<p id=\"x-ck12-YTA2YTBhZDIyNWEyOTVlMDQ1ZjU3OWUzMDU4MWQ5N2E.-uob\">When the\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0expression is written in terms of <img id=\"x-ck12-MTM2Nzk1ODQ2ODMwNw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211145\/1a271d97818de0a9d58667885dc707b3.png\" alt=\"s\" width=\"8\" height=\"8\" \/> , we get the following result for the molar solubility.<\/p>\r\n<p id=\"x-ck12-vf5\"><img class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213044\/6e7eb2e6e313f45004ef0e05e340cf01.png\" alt=\"K_{sp}&amp;=[ text{Fe}^{2+}][ text{OH}^-]^2=(s)(2s)^2=4s^3 \\s&amp;=sqrt [3]{frac{K_{sp}}{4}}=sqrt [3]{frac{7.9 times 10^{-16}}{4}}=5.8 times 10^{-6} text{M}\" width=\"364\" height=\"75\" \/><\/p>\r\n<p id=\"x-ck12-MmY4ZTk1ZDc1ZjdkY2ZkZmE1YzQ2MDgxMGY2MTBiZTk.-tgl\">The <strong> Table <\/strong> below shows the relationship between\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0and molar solubility based on the formula.<\/p>\r\n\r\n<table id=\"x-ck12-NWIxNDYxYjdjMzdhNDZiZmY5NDAzN2RmNDI0MjI5NmY.-ygi\" class=\"x-ck12-nofloat\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<td><strong> Compound Type <\/strong><\/td>\r\n<td><strong> Example <\/strong><\/td>\r\n<td><strong><em>K<sub>sp<\/sub><\/em> Expression <\/strong><\/td>\r\n<td><strong> Cation <\/strong><\/td>\r\n<td><strong> Anion <\/strong><\/td>\r\n<td><strong><em>K<sub>sp<\/sub><\/em> in Terms of <em>s<\/em><\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>AB<\/td>\r\n<td>CuS<\/td>\r\n<td>[Cu <sup> 2+ <\/sup> ][S <sup> 2\u2212 <\/sup> ]<\/td>\r\n<td><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211145\/1a271d97818de0a9d58667885dc707b3.png\" alt=\"s\" width=\"8\" height=\"8\" \/><\/td>\r\n<td><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211145\/1a271d97818de0a9d58667885dc707b3.png\" alt=\"s\" width=\"8\" height=\"8\" \/><\/td>\r\n<td><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213041\/a13a0f5d5e1213e2adc9f2c9f7caea61.png\" alt=\"s^2\" width=\"14\" height=\"15\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>AB <sub> 2 <\/sub> or A <sub> 2 <\/sub> B<\/td>\r\n<td>Ag <sub> 2 <\/sub> CrO <sub> 4 <\/sub><\/td>\r\n<td>[Ag <sup> + <\/sup> ] <sup> 2 <\/sup> [CrO <sub> 4 <\/sub><sup> 2\u2212 <\/sup> ]<\/td>\r\n<td><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213045\/db8f7016281c16f48682bf562d92c014.png\" alt=\"2s\" width=\"17\" height=\"12\" \/><\/td>\r\n<td><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211145\/1a271d97818de0a9d58667885dc707b3.png\" alt=\"s\" width=\"8\" height=\"8\" \/><\/td>\r\n<td><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213046\/65f08f7c5ac9e2182f02f48bcc54bded.png\" alt=\"4s^3\" width=\"23\" height=\"16\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>AB <sub> 3 <\/sub> or A <sub> 3 <\/sub> B<\/td>\r\n<td>Al(OH) <sub> 3 <\/sub><\/td>\r\n<td>[Al <sup> 3+ <\/sup> ][OH <sup> \u2212 <\/sup> ] <sup> 3 <\/sup><\/td>\r\n<td><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211145\/1a271d97818de0a9d58667885dc707b3.png\" alt=\"s\" width=\"8\" height=\"8\" \/><\/td>\r\n<td><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213046\/f5e43a5cd04d2ab074dd4284c0e038d8.png\" alt=\"3s\" width=\"17\" height=\"12\" \/><\/td>\r\n<td><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213046\/d98d174c59bfee1af9d0100e5dd6616a.png\" alt=\"27s^4\" width=\"33\" height=\"15\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>A <sub> 2 <\/sub> B <sub> 3 <\/sub> or A <sub> 3 <\/sub> B <sub> 2 <\/sub><\/td>\r\n<td>Ba <sub> 3 <\/sub> (PO <sub> 4 <\/sub> ) <sub> 2 <\/sub><\/td>\r\n<td>[Ba <sup> 2+ <\/sup> ] <sup> 3 <\/sup> [PO <sub> 4 <\/sub><sup> 3\u2212 <\/sup> ] <sup> 2 <\/sup><\/td>\r\n<td><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213046\/f5e43a5cd04d2ab074dd4284c0e038d8.png\" alt=\"3s\" width=\"17\" height=\"12\" \/><\/td>\r\n<td><img id=\"x-ck12-MTM2Nzk1ODQ2ODMwOA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213045\/db8f7016281c16f48682bf562d92c014.png\" alt=\"2s\" width=\"17\" height=\"12\" \/><\/td>\r\n<td><img class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213046\/b227085fddc8ef1036ebd5256e042b49.png\" alt=\"108s^5\" width=\"40\" height=\"16\" \/><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"x-ck12-MzFkMzBlMWM3NTBjYjU5Mzg1YzgwMGVjMTgwMTQxMTM.-tle\">The\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0expressions in terms of\u00a0 <em>s<\/em>\u00a0can be used to solve problems in which the\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0is used to calculate the molar solubility as in the examples above. Molar solubility can then be converted to solubility.<\/p>\r\n\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Summary<\/h3>\r\n<ul id=\"x-ck12-Mjg4ZWViOGEzMzdhOTNkOTg5ZjliOGJhYjRlODkxMzU.-dhy\">\r\n \t<li>The process of determining solubilities using\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0values is described.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Review<\/h3>\r\n<ol id=\"x-ck12-ZGI3NTAyMjkyMjE3NDkwMjA0ZDdlNjcwZTYzOTc2ZWQ.-pdo\">\r\n \t<li>What information is needed to carry out these calculations?<\/li>\r\n \t<li>What allows the calculation of molar solubility?<\/li>\r\n \t<li>How is solubility determined?<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"836080\"]Show References[\/reveal-answer]\r\n[hidden-answer a=\"836080\"]\r\n<h2>References<\/h2>\r\n<ol>\r\n \t<li>Courtesy of Ken Hackman, US Air Force.<a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Luke_AFB_waste_water_treatment_plant_1982.JPEG\">http:\/\/commons.wikimedia.org\/wiki\/File:Luke_AFB_waste_water_treatment_plant_1982.JPEG <\/a>.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]","rendered":"<div class=\"x-ck12-data-objectives\">\n<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Perform calculations converting\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0to solubility.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3><strong style=\"line-height: 1.5;\">How do you purify water?<\/strong><\/h3>\n<p id=\"x-ck12-NTcxMjZiN2FhOGRmMjQyYjdjM2E3YmM5ODM0YzY4NTA.-s7t\">Purification of water for drinking and other uses is a complicated process. Heavy metals need to be removed, a process accomplished by addition of carbonates and sulfates. Lead contamination can present major health problems, especially for younger children. Lead sulfates and carbonates are very insoluble, so will precipitate out of solution very easily.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213036\/20140811155758642168.jpeg\" alt=\"Heavy metals can be removed by precipitation with carbonates and sulfates\" width=\"400\" \/><\/p>\n<\/div>\n<h3>Conversion of\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0to Solubility<\/h3>\n<table id=\"x-ck12-OTI5ODAyZDhhMDJlODM3NDEzZjBkMDNlOTgyNDcxNWE.-8hd\" class=\"x-ck12-nofloat\">\n<caption>Solubility Product Constants (25\u00b0C)<\/caption>\n<tbody>\n<tr>\n<td><strong> Compound <\/strong><\/td>\n<td><em>K<sub>sp<\/sub><\/em><\/td>\n<td><strong> Compound <\/strong><\/td>\n<td><em>K<sub>sp<\/sub><\/em><\/td>\n<\/tr>\n<tr>\n<td>AgBr<\/td>\n<td>5.0\u00a0\u00d7 10 <sup> -13 <\/sup><\/td>\n<td>CuS<\/td>\n<td>8.0\u00a0\u00d7 10 <sup> -37 <\/sup><\/td>\n<\/tr>\n<tr>\n<td>AgCl<\/td>\n<td>1.8 \u00d7 10 <sup> -10 <\/sup><\/td>\n<td>Fe(OH) <sub> 2 <\/sub><\/td>\n<td>7.9 \u00d7 10 <sup> -16 <\/sup><\/td>\n<\/tr>\n<tr>\n<td>Al(OH) <sub> 3 <\/sub><\/td>\n<td>3.0\u00a0\u00d7 10 <sup> -34 <\/sup><\/td>\n<td>Mg(OH) <sub> 2 <\/sub><\/td>\n<td>7.1 \u00d7 10 <sup> -12 <\/sup><\/td>\n<\/tr>\n<tr>\n<td>BaCO <sub> 3 <\/sub><\/td>\n<td>5.0\u00a0\u00d7 10 <sup> -9 <\/sup><\/td>\n<td>PbCl <sub> 2 <\/sub><\/td>\n<td>1.7 \u00d7 10 <sup> -5 <\/sup><\/td>\n<\/tr>\n<tr>\n<td>BaSO <sub> 4 <\/sub><\/td>\n<td>1.1 \u00d7 10 <sup> -10 <\/sup><\/td>\n<td>PbCO <sub> 3 <\/sub><\/td>\n<td>7.4 \u00d7 10 <sup> -14 <\/sup><\/td>\n<\/tr>\n<tr>\n<td>CaCO <sub> 3 <\/sub><\/td>\n<td>4.5 \u00d7 10 <sup> -9 <\/sup><\/td>\n<td>PbI <sub> 2 <\/sub><\/td>\n<td>7.1 \u00d7 10 <sup> -9 <\/sup><\/td>\n<\/tr>\n<tr>\n<td>Ca(OH) <sub> 2 <\/sub><\/td>\n<td>6.5 \u00d7 10 <sup> -6 <\/sup><\/td>\n<td>PbSO <sub> 4 <\/sub><\/td>\n<td>6.3 \u00d7 10 <sup> -7 <\/sup><\/td>\n<\/tr>\n<tr>\n<td>Ca <sub> 3 <\/sub> (PO <sub> 4 <\/sub> ) <sub> 2 <\/sub><\/td>\n<td>1.2 \u00d7 10 <sup> -26 <\/sup><\/td>\n<td>Zn(OH) <sub> 2 <\/sub><\/td>\n<td>3.0\u00a0\u00d7 10 <sup> -16 <\/sup><\/td>\n<\/tr>\n<tr>\n<td>CaSO <sub> 4 <\/sub><\/td>\n<td>2.4 \u00d7 10 <sup> -5 <\/sup><\/td>\n<td>ZnS<\/td>\n<td>3.0\u00a0\u00d7 10 <sup> -23 <\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"x-ck12-MGJkN2JhYjU1ZDYyMzgyMjc0MjYyMzFjYzE1Y2U2MTQ.-bxd\">The known\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0values from the <strong> Table <\/strong> above can be used to calculate the solubility of a given compound by following the steps listed below.<\/p>\n<ol id=\"x-ck12-NTM3M2I1Y2JmNjQ1YWFhODY5NTZjOTUzYzA4Njk3ZTk.-1yo\">\n<li>Set up an ICE problem (Initial, Change, Equilibrium) in order to use the\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0value to calculate the concentration of each of the ions.<\/li>\n<li>The concentration of the ions leads to the molar solubility of the compound.<\/li>\n<li>Use the molar mass to convert from molar solubility to solubility.<\/li>\n<\/ol>\n<p id=\"x-ck12-NzZkMjU0YTNlY2YyYWEzZThiMDliMTFmNTUwYmM3OTM.-ymt\">The\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0of calcium carbonate is 4.5\u00a0\u00d7\u00a010 <sup> -9 <\/sup> . We begin by setting up an ICE table showing the dissociation of CaCO <sub> 3 <\/sub> into calcium ions and carbonate ions. The variable\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211145\/1a271d97818de0a9d58667885dc707b3.png\" alt=\"s\" width=\"8\" height=\"8\" \/> will be used to represent the molar solubility of CaCO <sub> 3 <\/sub> . In this case, each formula unit of CaCO <sub> 3 <\/sub> yields one Ca <sup> 2+ <\/sup> ion and one CO <sub> 3 <\/sub><sup> 2\u2212 <\/sup> ion. Therefore, the equilibrium concentrations of each ion are equal to <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211145\/1a271d97818de0a9d58667885dc707b3.png\" alt=\"s\" width=\"8\" height=\"8\" \/> .<\/p>\n<p id=\"x-ck12-i6o\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213038\/a2c8dd7fa1970edc56d741206bb79883.png\" alt=\"&amp; text{CaCO}_3(s) quad rightleftarrows quad text{Ca}^{2+}(aq)+ text{CO}_3^{2-}(aq) \\text{Initial }(text{M}) &amp; qquad qquad qquad qquad quad 0.00 qquad quad 0.00 \\text{Change }(text{M}) &amp; qquad qquad qquad qquad quad +s qquad quad +s \\qquad text{Equilibrium }(text{M}) &amp; qquad qquad qquad qquad qquad s qquad qquad s\" width=\"440\" height=\"101\" \/><\/p>\n<p id=\"x-ck12-N2FmM2RiNzMzYzk4YzIwODVmOGM3MzI2M2NhYzJlZWU.-beo\">The\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0expression can be written in terms of\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211145\/1a271d97818de0a9d58667885dc707b3.png\" alt=\"s\" width=\"8\" height=\"8\" \/> and then used to solve for <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211145\/1a271d97818de0a9d58667885dc707b3.png\" alt=\"s\" width=\"8\" height=\"8\" \/> .<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2Nzk1ODQ2ODMwNQ..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213039\/7c7e12bec87657b9ceec66e8d4f3dc2f.png\" alt=\"K_{sp}&amp;=[ text{Ca}^{2+}][ text{CO}_3^{2-}]=(s)(s)=s^2 \\s&amp;=sqrt{K_{sp}}=sqrt{4.5 times 10^{-9}}=6.7 times 10^{-5} text{M}\" width=\"346\" height=\"53\" \/><\/p>\n<p id=\"x-ck12-ZjQ2NmQzZDY3ODAwYWViYjVkYzc2NDg4M2IwNzUyMzA.-glf\">The concentration of each of the ions at equilibrium is 6.7\u00a0\u00d7\u00a010 <sup> -5 <\/sup> \u00a0M. We can use the molar mass to convert from molar solubility to solubility.<\/p>\n<p id=\"x-ck12-eg3\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213040\/10f0727dbce300daa8d80dfbfb7bc102.png\" alt=\"frac{6.7 times 10^{-5} cancel{text{mol}}}{text{L}} times frac{100.09 text{g}}{1 cancel{text{mol}}} = 6.7 times 10^{-3} text{g\/L}\" width=\"347\" height=\"43\" \/><\/p>\n<p id=\"x-ck12-ZmM2MTAyMmEwYmU0MjZiODdlODkzZWE4NzdmYzRhY2E.-lqj\">So the maximum amount of calcium carbonate that is capable of dissolving in 1 liter of water at 25\u00b0C is 6.7\u00a0\u00d7\u00a010 <sup> -3 <\/sup> \u00a0grams. Note that in the case above, the 1:1 ratio of the ions upon dissociation led to the\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0being equal to <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213041\/a13a0f5d5e1213e2adc9f2c9f7caea61.png\" alt=\"s^2\" width=\"14\" height=\"15\" \/> . This is referred to as a formula of the type <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213042\/81c6b8fb0f89e65473814916c64fb053.png\" alt=\"AB\" width=\"27\" height=\"12\" \/> , where\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/d5ebd507fac84fdb3364e6593d198a76.png\" alt=\"A\" width=\"13\" height=\"12\" \/> is the cation and\u00a0 <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211811\/00025e2ee78e7b712fbb42f74f2d6cb7.png\" alt=\"B\" width=\"14\" height=\"12\" \/> is the anion. Now let\u2019s consider a formula of the type <img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213042\/8735d5f5e52109132e5d695fbea001a6.png\" alt=\"AB_2\" width=\"33\" height=\"15\" \/> , such as Fe(OH) <sub> 2 <\/sub> . In this case the setup of the ICE table would look like the following:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2Nzk1ODQ2ODMwNg..\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213042\/6c4a62889a1377d890fef7ce071965ff.png\" alt=\"&amp; text{Fe(OH)}_2(s) quad rightleftarrows quad text{Fe}^{2+}(aq)+2 text{OH}^-(aq) \\text{Initial }(text{M}) &amp; qquad qquad qquad qquad qquad 0.00 qquad quad 0.00 \\text{Change }(text{M}) &amp; qquad qquad qquad qquad qquad +s qquad quad +2s \\text{Equilibrium }(text{M}) &amp; qquad qquad qquad qquad qquad quad s qquad qquad 2s\" width=\"450\" height=\"101\" \/><\/p>\n<p id=\"x-ck12-YTA2YTBhZDIyNWEyOTVlMDQ1ZjU3OWUzMDU4MWQ5N2E.-uob\">When the\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0expression is written in terms of <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2Nzk1ODQ2ODMwNw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211145\/1a271d97818de0a9d58667885dc707b3.png\" alt=\"s\" width=\"8\" height=\"8\" \/> , we get the following result for the molar solubility.<\/p>\n<p id=\"x-ck12-vf5\"><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-block-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213044\/6e7eb2e6e313f45004ef0e05e340cf01.png\" alt=\"K_{sp}&amp;=[ text{Fe}^{2+}][ text{OH}^-]^2=(s)(2s)^2=4s^3 \\s&amp;=sqrt [3]{frac{K_{sp}}{4}}=sqrt [3]{frac{7.9 times 10^{-16}}{4}}=5.8 times 10^{-6} text{M}\" width=\"364\" height=\"75\" \/><\/p>\n<p id=\"x-ck12-MmY4ZTk1ZDc1ZjdkY2ZkZmE1YzQ2MDgxMGY2MTBiZTk.-tgl\">The <strong> Table <\/strong> below shows the relationship between\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0and molar solubility based on the formula.<\/p>\n<table id=\"x-ck12-NWIxNDYxYjdjMzdhNDZiZmY5NDAzN2RmNDI0MjI5NmY.-ygi\" class=\"x-ck12-nofloat\">\n<tbody>\n<tr>\n<td><strong> Compound Type <\/strong><\/td>\n<td><strong> Example <\/strong><\/td>\n<td><strong><em>K<sub>sp<\/sub><\/em> Expression <\/strong><\/td>\n<td><strong> Cation <\/strong><\/td>\n<td><strong> Anion <\/strong><\/td>\n<td><strong><em>K<sub>sp<\/sub><\/em> in Terms of <em>s<\/em><\/strong><\/td>\n<\/tr>\n<tr>\n<td>AB<\/td>\n<td>CuS<\/td>\n<td>[Cu <sup> 2+ <\/sup> ][S <sup> 2\u2212 <\/sup> ]<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211145\/1a271d97818de0a9d58667885dc707b3.png\" alt=\"s\" width=\"8\" height=\"8\" \/><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211145\/1a271d97818de0a9d58667885dc707b3.png\" alt=\"s\" width=\"8\" height=\"8\" \/><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213041\/a13a0f5d5e1213e2adc9f2c9f7caea61.png\" alt=\"s^2\" width=\"14\" height=\"15\" \/><\/td>\n<\/tr>\n<tr>\n<td>AB <sub> 2 <\/sub> or A <sub> 2 <\/sub> B<\/td>\n<td>Ag <sub> 2 <\/sub> CrO <sub> 4 <\/sub><\/td>\n<td>[Ag <sup> + <\/sup> ] <sup> 2 <\/sup> [CrO <sub> 4 <\/sub><sup> 2\u2212 <\/sup> ]<\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213045\/db8f7016281c16f48682bf562d92c014.png\" alt=\"2s\" width=\"17\" height=\"12\" \/><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211145\/1a271d97818de0a9d58667885dc707b3.png\" alt=\"s\" width=\"8\" height=\"8\" \/><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213046\/65f08f7c5ac9e2182f02f48bcc54bded.png\" alt=\"4s^3\" width=\"23\" height=\"16\" \/><\/td>\n<\/tr>\n<tr>\n<td>AB <sub> 3 <\/sub> or A <sub> 3 <\/sub> B<\/td>\n<td>Al(OH) <sub> 3 <\/sub><\/td>\n<td>[Al <sup> 3+ <\/sup> ][OH <sup> \u2212 <\/sup> ] <sup> 3 <\/sup><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19211145\/1a271d97818de0a9d58667885dc707b3.png\" alt=\"s\" width=\"8\" height=\"8\" \/><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213046\/f5e43a5cd04d2ab074dd4284c0e038d8.png\" alt=\"3s\" width=\"17\" height=\"12\" \/><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213046\/d98d174c59bfee1af9d0100e5dd6616a.png\" alt=\"27s^4\" width=\"33\" height=\"15\" \/><\/td>\n<\/tr>\n<tr>\n<td>A <sub> 2 <\/sub> B <sub> 3 <\/sub> or A <sub> 3 <\/sub> B <sub> 2 <\/sub><\/td>\n<td>Ba <sub> 3 <\/sub> (PO <sub> 4 <\/sub> ) <sub> 2 <\/sub><\/td>\n<td>[Ba <sup> 2+ <\/sup> ] <sup> 3 <\/sup> [PO <sub> 4 <\/sub><sup> 3\u2212 <\/sup> ] <sup> 2 <\/sup><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213046\/f5e43a5cd04d2ab074dd4284c0e038d8.png\" alt=\"3s\" width=\"17\" height=\"12\" \/><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2Nzk1ODQ2ODMwOA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213045\/db8f7016281c16f48682bf562d92c014.png\" alt=\"2s\" width=\"17\" height=\"12\" \/><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213046\/b227085fddc8ef1036ebd5256e042b49.png\" alt=\"108s^5\" width=\"40\" height=\"16\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"x-ck12-MzFkMzBlMWM3NTBjYjU5Mzg1YzgwMGVjMTgwMTQxMTM.-tle\">The\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0expressions in terms of\u00a0 <em>s<\/em>\u00a0can be used to solve problems in which the\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0is used to calculate the molar solubility as in the examples above. Molar solubility can then be converted to solubility.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Summary<\/h3>\n<ul id=\"x-ck12-Mjg4ZWViOGEzMzdhOTNkOTg5ZjliOGJhYjRlODkxMzU.-dhy\">\n<li>The process of determining solubilities using\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0values is described.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Review<\/h3>\n<ol id=\"x-ck12-ZGI3NTAyMjkyMjE3NDkwMjA0ZDdlNjcwZTYzOTc2ZWQ.-pdo\">\n<li>What information is needed to carry out these calculations?<\/li>\n<li>What allows the calculation of molar solubility?<\/li>\n<li>How is solubility determined?<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q836080\">Show References<\/span><\/p>\n<div id=\"q836080\" class=\"hidden-answer\" style=\"display: none\">\n<h2>References<\/h2>\n<ol>\n<li>Courtesy of Ken Hackman, US Air Force.<a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Luke_AFB_waste_water_treatment_plant_1982.JPEG\">http:\/\/commons.wikimedia.org\/wiki\/File:Luke_AFB_waste_water_treatment_plant_1982.JPEG <\/a>.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2750\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Chemistry Concepts Intermediate. <strong>Authored by<\/strong>: Calbreath, Baxter, et al.. <strong>Provided by<\/strong>: CK12.org. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/\">http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc\/4.0\/\">CC BY-NC: Attribution-NonCommercial<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17,"menu_order":13,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry Concepts Intermediate\",\"author\":\"Calbreath, Baxter, et al.\",\"organization\":\"CK12.org\",\"url\":\"http:\/\/www.ck12.org\/book\/CK-12-Chemistry-Concepts-Intermediate\/\",\"project\":\"\",\"license\":\"cc-by-nc\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2750","chapter","type-chapter","status-publish","hentry"],"part":2340,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2750","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/users\/17"}],"version-history":[{"count":4,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2750\/revisions"}],"predecessor-version":[{"id":3695,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2750\/revisions\/3695"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/parts\/2340"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2750\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/media?parent=2750"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapter-type?post=2750"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/contributor?post=2750"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/license?post=2750"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}