{"id":2752,"date":"2016-08-24T18:24:59","date_gmt":"2016-08-24T18:24:59","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/?post_type=chapter&#038;p=2752"},"modified":"2016-08-24T21:42:29","modified_gmt":"2016-08-24T21:42:29","slug":"common-ion-effect","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/chapter\/common-ion-effect\/","title":{"raw":"Common Ion Effect","rendered":"Common Ion Effect"},"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>Define common ion.<\/li>\r\n \t<li>Define common ion effect.<\/li>\r\n \t<li>Perform calculations involving the common ion effect.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3><strong>Quite a charge <\/strong><\/h3>\r\n<p id=\"x-ck12-NjYxY2E4ZmRjZWVhZDdjYThlMTA2YjgxMmZlNGQ1Mzk.-gvb\">Lithium carbonate is an essential component of lithium batteries, which tend to be longer-lasting than regular alkaline batteries. The material is obtained from lithium ores by adding CO<sub>2 <\/sub> under high pressure to form the more soluble LiHCO<sub>3 <\/sub> . The mixture is then depressurized to remove the carbon dioxide and the lithium carbonate precipitates out of solution.<\/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\/19213053\/20140811155759002589.jpeg\" alt=\"Lithium batteries are enhanced by the common ion effect\" width=\"300\" \/>\r\n\r\n<\/div>\r\n<h2>Common Ion Effect<\/h2>\r\n<p id=\"x-ck12-MGNkM2YzZDQyNTYwNWE4ZGI5MmQ3YWE2N2JjYzNlNGQ.-dyj\">In a saturated solution of calcium sulfate, an equilibrium exists between the solid calcium sulfate and its ions in solution.<\/p>\r\n<p id=\"x-ck12-tax\"><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\/19213055\/1ecc1236819f0684cc61c626b716d9c0.png\" alt=\"text{CaSO}_4(s) rightleftarrows text{Ca}^{2+}(aq)+text{SO}_4^{2-}(aq) qquad K_{sp}=2.4 times 10^{-5}\" width=\"437\" height=\"23\" \/><\/p>\r\n<p id=\"x-ck12-MTgwZThiMTM5ZmUzYTc4MjcwOWJjMzY3Mzk0YTY5Mzk.-vto\">Suppose that some calcium nitrate were added to this saturated solution. Immediately, the concentration of the calcium ion in the solution would increase. As a result, the ion product of the [Ca <sup> 2+ <\/sup> ] times the [SO <sub> 4 <\/sub><sup> 2\u2212 <\/sup> ] would increase and now be greater than the <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\/19213027\/152ea892d77eb7f4d21f1d84974b9c17.png\" alt=\"K_{sp}\" width=\"28\" height=\"18\" \/> . According to LeCh\u00e2telier\u2019s principle, the equilibrium above would shift to the left in order to relieve the stress of the added calcium ion. Additional calcium sulfate would precipitate out of the solution until the ion product once again becomes equal to the <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\/19213027\/152ea892d77eb7f4d21f1d84974b9c17.png\" alt=\"K_{sp}\" width=\"28\" height=\"18\" \/> . Note that in the new equilibrium the concentrations of the calcium ion and the sulfate ion would no longer be equal to each other. The calcium ion concentration would be larger than the sulfate ion concentration.<\/p>\r\n<p id=\"x-ck12-MmVjZTdhZTEwMTY0NjMyMGZlMzM0NTUxYjI5ZmNlZDY.-oqj\">This situation describes the common ion effect. A <strong> common ion <\/strong> is an ion that is in common to both salts in a solution. In the above example, the common ion is Ca <sup> 2+ <\/sup> . The <strong> common ion effect <\/strong> is a decrease in the solubility of an ionic compound as a result of the addition of a common ion. Adding calcium ion to the saturated solution of calcium sulfate causes additional CaSO <sub> 4 <\/sub> to precipitate from the solution, lowering its solubility. The addition of a solution containing sulfate ion, such as potassium sulfate, would result in the same common ion effect.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3>Sample Problem: The Common Ion Effect<\/h3>\r\n<p id=\"x-ck12-YzZlNTIxODZkZTI2ZmUzN2I0ZWY5NTVkZmQzNmE4ZmU.-bxx\">What is the concentration of zinc ion in 1.00 L of a saturated solution of zinc hydroxide to which 0.040 mol of NaOH has been added?<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-8q2\"><em> Step 1: List the known quantities and plan the problem <\/em> .<\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-082\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-NDllMGUwMDcyMzhhOGMzNmIxNWY4MGZlNjM2MGQyYjU.-f1z\">\r\n \t<li><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\/19213056\/dfb675c540cd60ebfcb8315bbb6d33a0.png\" alt=\"K_{sp}= 3.0 times 10^{-16}\" width=\"138\" height=\"21\" \/> (from table in\u00a0 \"Conversion of Ksp to Solubility\" )<\/li>\r\n \t<li>moles of added NaOH = 0.040 mol<\/li>\r\n \t<li>volumes of solution = 1.00 L<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-2sb\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-YTVhYTgwMDJkYWY4MTgyZWE4MjE1MzE2YzI0MGUzNmI.-d0z\">\r\n \t<li>[Zn <sup> 2+ <\/sup> ] = ? M<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-MjlmYTIzMTVkMjc2ZTE2MzU5ZjBiMWVjMWVhZDc1YzA.-brd\">Express the concentrations of the two ions relative to the variable <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\" \/> . The concentration of the zinc ion will be 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\" \/> , while the concentration of the hydroxide ion will be 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\/19213057\/07f64c525e5f1fd895fee0510c230e91.png\" alt=\"0.040 + 2s\" width=\"80\" height=\"13\" \/> .<\/p>\r\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-iut\"><em> Step 2: Solve <\/em> .<\/p>\r\n<p id=\"x-ck12-YTMyNzNhZTJmMzQ1ZjNkZTYwYzI4YTg0NjhkMjA4Nzk.-9k0\">The\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\/19213027\/152ea892d77eb7f4d21f1d84974b9c17.png\" alt=\"K_{sp}\" width=\"28\" height=\"18\" \/> expression can be written in terms of the variable <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-gfb\"><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\/19213058\/9766fca082ac4820fd944b3a2631ae75.png\" alt=\"K_{sp}=[text{Zn}^{2+}][text{OH}^-]^2=(s)(0.040+2s)^2\" width=\"302\" height=\"23\" \/><\/p>\r\n<p id=\"x-ck12-MzBiMzY1YWZmOTJmMGQzMzg5ZDgwYzVlZGYzNTU0N2Q.-hbn\">Because the value of the\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\/19213027\/152ea892d77eb7f4d21f1d84974b9c17.png\" alt=\"K_{sp}\" width=\"28\" height=\"18\" \/> is so small, we can make the assumption that the value 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\" \/> will be very small compared to 0.040. This simplifies the mathematics involved in solving 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<p id=\"x-ck12-pih\"><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\/19213059\/551ab63daa1e4d0b586ed9d699808598.png\" alt=\"K_{sp}&amp;=(s)(0.040)^2=0.0016 s=3.0 times 10^{-16} \\s&amp;=frac{K_{sp}}{left [text{OH}^- right ]^2}=frac{3.0 times 10^{-16}}{0.0016}=1.9 times 10^{-13} text{M}\" width=\"364\" height=\"79\" \/><\/p>\r\n<p id=\"x-ck12-YzhkYTc3M2YxMDc1YWYxZmViYTQ0M2EzZjIxY2E1ODY.-mqo\">The concentration of the zinc ion is equal to\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 so [Zn<sup>2+<\/sup>]\u00a0=\u00a01.9\u00a0\u00d7\u00a010<sup>\u221213 <\/sup> \u00a0M.<\/p>\r\n<p id=\"x-ck12-ZGJmYWE4NWM0YmIyZGJiOTdhZTZmY2NmYmJmYjM1MjI.-sfw\"><em> Step 3: Think about your result <\/em> .<\/p>\r\n<p id=\"x-ck12-ZjNkMjVjYThjMzYwZmFiODYyMmZmYjg0NDQ2ZmNiYTQ.-fef\">The relatively high concentration of the common ion, OH<sup>\u2212 <\/sup> , results in a very low concentration of zinc ion. The molar solubility of the zinc hydroxide is less in the presence of the common ion than it would be in water.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Summary<\/h3>\r\n<ul id=\"x-ck12-YTkxY2I3OWVlYmNjYmE5NTE0NTE1NWZlOWU4ZTVhYjA.-bh5\">\r\n \t<li>The common ion and common ion effect are described.<\/li>\r\n \t<li>Calculations involving the common ion effect are described.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Practice<\/h3>\r\n<p id=\"x-ck12-ZGIyN2QxMzI1ODI1MWRlNWVlMjE5YTlhNmM5NDJlN2M.-rks\">Work the problems at the link below:<\/p>\r\n<p id=\"x-ck12-ZjEzMjJmYTVmMzg3MzA5MjJmNDZhNWU1YWY0ZmRjZGM.-cai\"><a href=\"http:\/\/science.widener.edu\/svb\/tutorial\/saltcomioncsn7.html\"> http:\/\/science.widener.edu\/svb\/tutorial\/saltcomioncsn7.html<\/a><\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Review<\/h3>\r\n<ol id=\"x-ck12-ZTNhZTljYjNkOTBkODU5ZjI3Y2IxMWQ4NjJhZmMxNDg.-rsq\">\r\n \t<li>How is Le Ch\u00e2telier\u2019s principle involved in the common-ion effect?<\/li>\r\n \t<li>Could the common ion effect ever increase the solubility of a compound?<\/li>\r\n \t<li>In the sample problem, what would the effect be of adding Zn(NO<sub>3<\/sub>)<sub>2<\/sub>?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<h2 class=\"x-ck12-data-problem-set\">Glossary<\/h2>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-NmM5YjkxZmFiM2ZlMTk0YWEyNDZlOWIwYzIxMTcxNDg.-zfj\">\r\n \t<li><strong> common ion: <\/strong> An ion that is in common to both salts in a solution.<\/li>\r\n \t<li><strong> common ion effect: <\/strong> A decrease in the solubility of an ionic compound as a result of the addition of a common ion.<\/li>\r\n<\/ul>\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 NASA. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:NASA_Lithium_Ion_Polymer_Battery.jpg\">http:\/\/commons.wikimedia.org\/wiki\/File:NASA_Lithium_Ion_Polymer_Battery.jpg <\/a>.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"x-ck12-data-objectives\">\n<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Define common ion.<\/li>\n<li>Define common ion effect.<\/li>\n<li>Perform calculations involving the common ion effect.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3><strong>Quite a charge <\/strong><\/h3>\n<p id=\"x-ck12-NjYxY2E4ZmRjZWVhZDdjYThlMTA2YjgxMmZlNGQ1Mzk.-gvb\">Lithium carbonate is an essential component of lithium batteries, which tend to be longer-lasting than regular alkaline batteries. The material is obtained from lithium ores by adding CO<sub>2 <\/sub> under high pressure to form the more soluble LiHCO<sub>3 <\/sub> . The mixture is then depressurized to remove the carbon dioxide and the lithium carbonate precipitates out of solution.<\/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\/19213053\/20140811155759002589.jpeg\" alt=\"Lithium batteries are enhanced by the common ion effect\" width=\"300\" \/><\/p>\n<\/div>\n<h2>Common Ion Effect<\/h2>\n<p id=\"x-ck12-MGNkM2YzZDQyNTYwNWE4ZGI5MmQ3YWE2N2JjYzNlNGQ.-dyj\">In a saturated solution of calcium sulfate, an equilibrium exists between the solid calcium sulfate and its ions in solution.<\/p>\n<p id=\"x-ck12-tax\"><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\/19213055\/1ecc1236819f0684cc61c626b716d9c0.png\" alt=\"text{CaSO}_4(s) rightleftarrows text{Ca}^{2+}(aq)+text{SO}_4^{2-}(aq) qquad K_{sp}=2.4 times 10^{-5}\" width=\"437\" height=\"23\" \/><\/p>\n<p id=\"x-ck12-MTgwZThiMTM5ZmUzYTc4MjcwOWJjMzY3Mzk0YTY5Mzk.-vto\">Suppose that some calcium nitrate were added to this saturated solution. Immediately, the concentration of the calcium ion in the solution would increase. As a result, the ion product of the [Ca <sup> 2+ <\/sup> ] times the [SO <sub> 4 <\/sub><sup> 2\u2212 <\/sup> ] would increase and now be greater than the <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\/19213027\/152ea892d77eb7f4d21f1d84974b9c17.png\" alt=\"K_{sp}\" width=\"28\" height=\"18\" \/> . According to LeCh\u00e2telier\u2019s principle, the equilibrium above would shift to the left in order to relieve the stress of the added calcium ion. Additional calcium sulfate would precipitate out of the solution until the ion product once again becomes equal to the <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\/19213027\/152ea892d77eb7f4d21f1d84974b9c17.png\" alt=\"K_{sp}\" width=\"28\" height=\"18\" \/> . Note that in the new equilibrium the concentrations of the calcium ion and the sulfate ion would no longer be equal to each other. The calcium ion concentration would be larger than the sulfate ion concentration.<\/p>\n<p id=\"x-ck12-MmVjZTdhZTEwMTY0NjMyMGZlMzM0NTUxYjI5ZmNlZDY.-oqj\">This situation describes the common ion effect. A <strong> common ion <\/strong> is an ion that is in common to both salts in a solution. In the above example, the common ion is Ca <sup> 2+ <\/sup> . The <strong> common ion effect <\/strong> is a decrease in the solubility of an ionic compound as a result of the addition of a common ion. Adding calcium ion to the saturated solution of calcium sulfate causes additional CaSO <sub> 4 <\/sub> to precipitate from the solution, lowering its solubility. The addition of a solution containing sulfate ion, such as potassium sulfate, would result in the same common ion effect.<\/p>\n<div class=\"textbox shaded\">\n<h3>Sample Problem: The Common Ion Effect<\/h3>\n<p id=\"x-ck12-YzZlNTIxODZkZTI2ZmUzN2I0ZWY5NTVkZmQzNmE4ZmU.-bxx\">What is the concentration of zinc ion in 1.00 L of a saturated solution of zinc hydroxide to which 0.040 mol of NaOH has been added?<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-8q2\"><em> Step 1: List the known quantities and plan the problem <\/em> .<\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-082\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-NDllMGUwMDcyMzhhOGMzNmIxNWY4MGZlNjM2MGQyYjU.-f1z\">\n<li><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\/19213056\/dfb675c540cd60ebfcb8315bbb6d33a0.png\" alt=\"K_{sp}= 3.0 times 10^{-16}\" width=\"138\" height=\"21\" \/> (from table in\u00a0 &#8220;Conversion of Ksp to Solubility&#8221; )<\/li>\n<li>moles of added NaOH = 0.040 mol<\/li>\n<li>volumes of solution = 1.00 L<\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-2sb\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-YTVhYTgwMDJkYWY4MTgyZWE4MjE1MzE2YzI0MGUzNmI.-d0z\">\n<li>[Zn <sup> 2+ <\/sup> ] = ? M<\/li>\n<\/ul>\n<p id=\"x-ck12-MjlmYTIzMTVkMjc2ZTE2MzU5ZjBiMWVjMWVhZDc1YzA.-brd\">Express the concentrations of the two ions relative to the variable <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\" \/> . The concentration of the zinc ion will be 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\" \/> , while the concentration of the hydroxide ion will be 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\/19213057\/07f64c525e5f1fd895fee0510c230e91.png\" alt=\"0.040 + 2s\" width=\"80\" height=\"13\" \/> .<\/p>\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-iut\"><em> Step 2: Solve <\/em> .<\/p>\n<p id=\"x-ck12-YTMyNzNhZTJmMzQ1ZjNkZTYwYzI4YTg0NjhkMjA4Nzk.-9k0\">The\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\/19213027\/152ea892d77eb7f4d21f1d84974b9c17.png\" alt=\"K_{sp}\" width=\"28\" height=\"18\" \/> expression can be written in terms of the variable <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-gfb\"><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\/19213058\/9766fca082ac4820fd944b3a2631ae75.png\" alt=\"K_{sp}=[text{Zn}^{2+}][text{OH}^-]^2=(s)(0.040+2s)^2\" width=\"302\" height=\"23\" \/><\/p>\n<p id=\"x-ck12-MzBiMzY1YWZmOTJmMGQzMzg5ZDgwYzVlZGYzNTU0N2Q.-hbn\">Because the value of the\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\/19213027\/152ea892d77eb7f4d21f1d84974b9c17.png\" alt=\"K_{sp}\" width=\"28\" height=\"18\" \/> is so small, we can make the assumption that the value 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\" \/> will be very small compared to 0.040. This simplifies the mathematics involved in solving 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 id=\"x-ck12-pih\"><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\/19213059\/551ab63daa1e4d0b586ed9d699808598.png\" alt=\"K_{sp}&amp;=(s)(0.040)^2=0.0016 s=3.0 times 10^{-16} \\s&amp;=frac{K_{sp}}{left [text{OH}^- right ]^2}=frac{3.0 times 10^{-16}}{0.0016}=1.9 times 10^{-13} text{M}\" width=\"364\" height=\"79\" \/><\/p>\n<p id=\"x-ck12-YzhkYTc3M2YxMDc1YWYxZmViYTQ0M2EzZjIxY2E1ODY.-mqo\">The concentration of the zinc ion is equal to\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 so [Zn<sup>2+<\/sup>]\u00a0=\u00a01.9\u00a0\u00d7\u00a010<sup>\u221213 <\/sup> \u00a0M.<\/p>\n<p id=\"x-ck12-ZGJmYWE4NWM0YmIyZGJiOTdhZTZmY2NmYmJmYjM1MjI.-sfw\"><em> Step 3: Think about your result <\/em> .<\/p>\n<p id=\"x-ck12-ZjNkMjVjYThjMzYwZmFiODYyMmZmYjg0NDQ2ZmNiYTQ.-fef\">The relatively high concentration of the common ion, OH<sup>\u2212 <\/sup> , results in a very low concentration of zinc ion. The molar solubility of the zinc hydroxide is less in the presence of the common ion than it would be in water.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Summary<\/h3>\n<ul id=\"x-ck12-YTkxY2I3OWVlYmNjYmE5NTE0NTE1NWZlOWU4ZTVhYjA.-bh5\">\n<li>The common ion and common ion effect are described.<\/li>\n<li>Calculations involving the common ion effect are described.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Practice<\/h3>\n<p id=\"x-ck12-ZGIyN2QxMzI1ODI1MWRlNWVlMjE5YTlhNmM5NDJlN2M.-rks\">Work the problems at the link below:<\/p>\n<p id=\"x-ck12-ZjEzMjJmYTVmMzg3MzA5MjJmNDZhNWU1YWY0ZmRjZGM.-cai\"><a href=\"http:\/\/science.widener.edu\/svb\/tutorial\/saltcomioncsn7.html\"> http:\/\/science.widener.edu\/svb\/tutorial\/saltcomioncsn7.html<\/a><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Review<\/h3>\n<ol id=\"x-ck12-ZTNhZTljYjNkOTBkODU5ZjI3Y2IxMWQ4NjJhZmMxNDg.-rsq\">\n<li>How is Le Ch\u00e2telier\u2019s principle involved in the common-ion effect?<\/li>\n<li>Could the common ion effect ever increase the solubility of a compound?<\/li>\n<li>In the sample problem, what would the effect be of adding Zn(NO<sub>3<\/sub>)<sub>2<\/sub>?<\/li>\n<\/ol>\n<\/div>\n<h2 class=\"x-ck12-data-problem-set\">Glossary<\/h2>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-NmM5YjkxZmFiM2ZlMTk0YWEyNDZlOWIwYzIxMTcxNDg.-zfj\">\n<li><strong> common ion: <\/strong> An ion that is in common to both salts in a solution.<\/li>\n<li><strong> common ion effect: <\/strong> A decrease in the solubility of an ionic compound as a result of the addition of a common ion.<\/li>\n<\/ul>\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 NASA. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:NASA_Lithium_Ion_Polymer_Battery.jpg\">http:\/\/commons.wikimedia.org\/wiki\/File:NASA_Lithium_Ion_Polymer_Battery.jpg <\/a>.<\/li>\n<\/ol>\n<\/div>\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-2752\">\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":15,"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-2752","chapter","type-chapter","status-publish","hentry"],"part":2340,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2752","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":3,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2752\/revisions"}],"predecessor-version":[{"id":3017,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2752\/revisions\/3017"}],"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\/2752\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/media?parent=2752"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapter-type?post=2752"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/contributor?post=2752"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/license?post=2752"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}