{"id":2749,"date":"2016-08-24T18:19:07","date_gmt":"2016-08-24T18:19:07","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/?post_type=chapter&#038;p=2749"},"modified":"2016-08-24T21:36:13","modified_gmt":"2016-08-24T21:36:13","slug":"conversion-of-solubility-to-ksp","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/chapter\/conversion-of-solubility-to-ksp\/","title":{"raw":"Conversion of Solubility to Ksp","rendered":"Conversion of Solubility to Ksp"},"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 molar solubility.<\/li>\r\n \t<li>Perform calculations involving molar solubility and <em>K<sub>sp<\/sub><\/em>.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3><strong>How is baking soda made? <\/strong><\/h3>\r\n<p id=\"x-ck12-OTNkYTQzMDI0YzI1Njk1NWQyZGRlMzQ1NTA3ZWNkZDE.-gsf\">Baking soda (sodium bicarbonate) is prepared by bubbling carbon dioxide gas through a solution of ammonia and sodium chloride. Ammonium carbonate is first formed which then reacts with the NaCl to form sodium bicarbonate and ammonium chloride. The sodium bicarbonate is less soluble than the other materials, so it will precipitate out of solution.<\/p>\r\n<img class=\"aligncenter wp-image-3011\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/515\/2016\/08\/24213526\/24978791011_33068dc09c_z.jpg\" alt=\"a bowl of baking soda\" width=\"500\" height=\"389\" \/>\r\n\r\n<\/div>\r\n<h2>Conversion of Solubility to <em>K<sub>sp<\/sub><\/em><\/h2>\r\n<p id=\"x-ck12-NTc3YjllNjhiNGNhNzViNjY0NTg2ZGMyM2QzZWU0NDY.-cno\">Solubility is normally expressed in g\/L of saturated solution. However, solubility can also be expressed as the moles per liter. Molar solubility is the number of moles of solute in one liter of saturated solution. In other words, the molar solubility of a given compound represents the highest molarity solution that is possible for that compound. The molar mass of a compound is the conversion factor between solubility and molar solubility. Given that the solubility of Zn(OH) <sub> 2 <\/sub> is 4.2\u00a0\u00d7\u00a010 <sup> -4 <\/sup> \u00a0g\/L, the molar solubility can be calculated as shown below:<\/p>\r\n<p id=\"x-ck12-lva\"><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\/19213030\/1b01dd475c8320a4d0cb346fa035a28d.png\" alt=\"frac{4.2 times 10^{-4} cancel{text{g}}}{text{L}} times frac{1 text{mol}}{99.41 cancel{text{g}}}=4.2 times 10^{-6} text{mol\/L} (text{M})\" width=\"373\" height=\"47\" \/><\/p>\r\n<p id=\"x-ck12-OGE2ZTRiYjUwYzI0MjZmNTAxNjIzOWZjNjJhODJjYjg.-vlo\">Solubility data can be used to calculate the\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0for a given compound. The following steps need to be taken.<\/p>\r\n\r\n<ol id=\"x-ck12-NDc5OGQ2MTEwZGJmNGJhNTJkMDIwZjIzMmRmYjRkODM.-w4r\">\r\n \t<li>Convert from solubility to molar solubility.<\/li>\r\n \t<li>Use the dissociation equation to determine the concentration of each of the ions in mol\/L.<\/li>\r\n \t<li>Apply the\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0equation.<\/li>\r\n<\/ol>\r\n<div class=\"textbox shaded\">\r\n<h4>Sample Problem: Calculating\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0from Solubility<\/h4>\r\n<p id=\"x-ck12-MDYyY2UyOWY1NmNjNzMwNWE3MDk1NzIyYjU3OWZlOGQ.-aqd\">The solubility of lead(II) fluoride is found experimentally to be 0.533 g\/L. Calculate the\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0for lead(II) fluoride.<\/p>\r\n<p id=\"x-ck12-OGU2YWEyYzY0NzAwMDUxZjI4NjFjY2E4MjYyNmNhN2I.-rly\"><em> Step 1: List the known quantities and plan the problem <\/em> .<\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-5xl\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-MzYyMmQ2YzI5OGVlM2Y3MzJhOTEyNmFhZmUyYWMyNWY.-4em\">\r\n \t<li>solubility of PbF <sub> 2 <\/sub> = 0.533 g\/L<\/li>\r\n \t<li>molar mass = 245.20 g\/mol<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-rs7\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-YjIwNDBkMzMxYTcyMTg4NTNlMTZhMjlkMmE0ZTJjMmU.-d6k\">\r\n \t<li><em>K<sub>sp<\/sub><\/em> of PbF<sub>2 <\/sub> = ?<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ZDg2MTRiZWUxMDQ1MDZkMDUwYTZmZmVmNGY2ZmQ1MDA.-xfa\">The dissociation equation for PbF <sub> 2 <\/sub> and the corresponding\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0expression<\/p>\r\n<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\/19213031\/1682b1bd53c37e4b8b14c60db113f4eb.png\" alt=\"text{PbF}_2(s) rightleftarrows text{Pb}^{2+}(aq)+2text{F}^-(aq) &amp;&amp; K_{sp}=[text{Pb}^{2+}][text{F}^-]^2\" width=\"457\" height=\"23\" \/>\r\n<p id=\"x-ck12-YjM2YTBmNTIyNTBmZTRjOGExMzFlYjAxZjk0ODZkZmU.-tsc\">The steps above will be followed to calculate the\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0for PbF <sub> 2 <\/sub> .<\/p>\r\n<p id=\"x-ck12-ZjZhN2M2ZWIzYjU5NWY1NDY2Yzg4ZDMxMDBjN2VkODc.-b0p\"><em> Step 2: Solve <\/em> .<\/p>\r\n<p id=\"x-ck12-OTg4ZjJkMGQwNWY4YjEyMTA4MGYzZmUxNDk4MmE4MmY.-kp5\"><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\/19213033\/82258380623f5cf138f36ae543c17365.png\" alt=\"text{molar solubility} qquad frac{0.533 cancel{text{g}}}{text{L}} times frac{1 text{mol}}{245.20 cancel{text{g}}}=2.17 times 10^{-3} text{M}\" width=\"444\" height=\"44\" \/><\/p>\r\n<p id=\"x-ck12-NjhlZWUwN2YyMmUzNGI4NmQ5MWZiNzAwMWVhN2IyY2I.-irq\">The dissociation equation shows that for every mole of PbF <sub> 2 <\/sub> that dissociates, 1 mol of Pb <sup> 2+ <\/sup> and 2 mol of F <sup> \u2212 <\/sup> are produced. Therefore, at equilibrium the concentrations of the ions are:<\/p>\r\n<p id=\"x-ck12-awi\"><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\/19213034\/fe369d746d277575d4f29770a7f6ec6b.png\" alt=\"[text{Pb}^{2+}]=2.17 times 10^{-3} text{M} quad text{and} quad [text{F}^-]=2 times 2.17 times 10^{-3}=4.35 times 10^{-3} text{M}\" width=\"560\" height=\"22\" \/><\/p>\r\n<p id=\"x-ck12-NmY5NDliNjFiYmFkODhhYTViNjkyOGEwNmVkYTE1NzY.-jv4\">Substitute into the expression and solve for 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\" \/> .<\/p>\r\n<p id=\"x-ck12-u9p\"><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\/19213035\/a4dc347960b0f0f0a841f78a12e9e08a.png\" alt=\"K_{sp}=(2.17 times 10^{-3})(4.35 times 10^{-3})^2=4.11 times 10^{-8}\" width=\"379\" height=\"23\" \/><\/p>\r\n<p id=\"x-ck12-ZGJmYWE4NWM0YmIyZGJiOTdhZTZmY2NmYmJmYjM1MjI.-8ua\"><em> Step 3: Think about your result <\/em> .<\/p>\r\n<p id=\"x-ck12-NmQ4NjRmYTFkY2VlYjA1OTQwMzE0NzUxN2E4MmRkZDU.-hyd\">The solubility product constant is significantly less than 1 for a nearly insoluble compound such as PbF <sub> 2 <\/sub> .<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Summary<\/h3>\r\n<ul id=\"x-ck12-NDg0M2Y2MmZhM2I3MDdmNDI1ZmVkYjBhOWZiMjAwNmQ.-jyu\">\r\n \t<li>Molar solubility calculations are described.<\/li>\r\n \t<li>Calculations of\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0using molar solubility 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-ZDY4YjkyZjQ1NjFmOWJlZDVkZTk4YWM0ZGI4YTJjMmI.-smc\">Read the material at <a href=\"http:\/\/www.chemteam.info\/Equilibrium\/Calc-Ksp-FromMolSolub.html\" target=\"_blank\">ChemTeam.info<\/a>\u00a0and do the problems at the end.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Review<\/h3>\r\n<ol id=\"x-ck12-ZWU5OTI0Zjk0ODQ0MGYxZGM4N2ExMzEzNDQ5NDNlMmM.-qlc\">\r\n \t<li>What are the solution requirements for determining molar solubility?<\/li>\r\n \t<li>Why do we need to convert mass to molarity to determine\u00a0<em>K<sub>sp<\/sub><\/em>?<\/li>\r\n \t<li>What\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0values would you expect for very insoluble compounds?<\/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-YjIxMzc3MzYwNDVkYWFiMTAzNWU3NDk4ZmQ1Yjg5ZGU.-ukb\">\r\n \t<li><strong> molar solubility: <\/strong> The number of moles of solute in one liter of saturated solution.<\/li>\r\n<\/ul>\r\n&nbsp;\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 molar solubility.<\/li>\n<li>Perform calculations involving molar solubility and <em>K<sub>sp<\/sub><\/em>.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3><strong>How is baking soda made? <\/strong><\/h3>\n<p id=\"x-ck12-OTNkYTQzMDI0YzI1Njk1NWQyZGRlMzQ1NTA3ZWNkZDE.-gsf\">Baking soda (sodium bicarbonate) is prepared by bubbling carbon dioxide gas through a solution of ammonia and sodium chloride. Ammonium carbonate is first formed which then reacts with the NaCl to form sodium bicarbonate and ammonium chloride. The sodium bicarbonate is less soluble than the other materials, so it will precipitate out of solution.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3011\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/515\/2016\/08\/24213526\/24978791011_33068dc09c_z.jpg\" alt=\"a bowl of baking soda\" width=\"500\" height=\"389\" \/><\/p>\n<\/div>\n<h2>Conversion of Solubility to <em>K<sub>sp<\/sub><\/em><\/h2>\n<p id=\"x-ck12-NTc3YjllNjhiNGNhNzViNjY0NTg2ZGMyM2QzZWU0NDY.-cno\">Solubility is normally expressed in g\/L of saturated solution. However, solubility can also be expressed as the moles per liter. Molar solubility is the number of moles of solute in one liter of saturated solution. In other words, the molar solubility of a given compound represents the highest molarity solution that is possible for that compound. The molar mass of a compound is the conversion factor between solubility and molar solubility. Given that the solubility of Zn(OH) <sub> 2 <\/sub> is 4.2\u00a0\u00d7\u00a010 <sup> -4 <\/sup> \u00a0g\/L, the molar solubility can be calculated as shown below:<\/p>\n<p id=\"x-ck12-lva\"><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\/19213030\/1b01dd475c8320a4d0cb346fa035a28d.png\" alt=\"frac{4.2 times 10^{-4} cancel{text{g}}}{text{L}} times frac{1 text{mol}}{99.41 cancel{text{g}}}=4.2 times 10^{-6} text{mol\/L} (text{M})\" width=\"373\" height=\"47\" \/><\/p>\n<p id=\"x-ck12-OGE2ZTRiYjUwYzI0MjZmNTAxNjIzOWZjNjJhODJjYjg.-vlo\">Solubility data can be used to calculate the\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0for a given compound. The following steps need to be taken.<\/p>\n<ol id=\"x-ck12-NDc5OGQ2MTEwZGJmNGJhNTJkMDIwZjIzMmRmYjRkODM.-w4r\">\n<li>Convert from solubility to molar solubility.<\/li>\n<li>Use the dissociation equation to determine the concentration of each of the ions in mol\/L.<\/li>\n<li>Apply the\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0equation.<\/li>\n<\/ol>\n<div class=\"textbox shaded\">\n<h4>Sample Problem: Calculating\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0from Solubility<\/h4>\n<p id=\"x-ck12-MDYyY2UyOWY1NmNjNzMwNWE3MDk1NzIyYjU3OWZlOGQ.-aqd\">The solubility of lead(II) fluoride is found experimentally to be 0.533 g\/L. Calculate the\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0for lead(II) fluoride.<\/p>\n<p id=\"x-ck12-OGU2YWEyYzY0NzAwMDUxZjI4NjFjY2E4MjYyNmNhN2I.-rly\"><em> Step 1: List the known quantities and plan the problem <\/em> .<\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-5xl\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-MzYyMmQ2YzI5OGVlM2Y3MzJhOTEyNmFhZmUyYWMyNWY.-4em\">\n<li>solubility of PbF <sub> 2 <\/sub> = 0.533 g\/L<\/li>\n<li>molar mass = 245.20 g\/mol<\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-rs7\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-YjIwNDBkMzMxYTcyMTg4NTNlMTZhMjlkMmE0ZTJjMmU.-d6k\">\n<li><em>K<sub>sp<\/sub><\/em> of PbF<sub>2 <\/sub> = ?<\/li>\n<\/ul>\n<p id=\"x-ck12-ZDg2MTRiZWUxMDQ1MDZkMDUwYTZmZmVmNGY2ZmQ1MDA.-xfa\">The dissociation equation for PbF <sub> 2 <\/sub> and the corresponding\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0expression<\/p>\n<p><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\/19213031\/1682b1bd53c37e4b8b14c60db113f4eb.png\" alt=\"text{PbF}_2(s) rightleftarrows text{Pb}^{2+}(aq)+2text{F}^-(aq) &amp;&amp; K_{sp}=[text{Pb}^{2+}][text{F}^-]^2\" width=\"457\" height=\"23\" \/><\/p>\n<p id=\"x-ck12-YjM2YTBmNTIyNTBmZTRjOGExMzFlYjAxZjk0ODZkZmU.-tsc\">The steps above will be followed to calculate the\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0for PbF <sub> 2 <\/sub> .<\/p>\n<p id=\"x-ck12-ZjZhN2M2ZWIzYjU5NWY1NDY2Yzg4ZDMxMDBjN2VkODc.-b0p\"><em> Step 2: Solve <\/em> .<\/p>\n<p id=\"x-ck12-OTg4ZjJkMGQwNWY4YjEyMTA4MGYzZmUxNDk4MmE4MmY.-kp5\"><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\/19213033\/82258380623f5cf138f36ae543c17365.png\" alt=\"text{molar solubility} qquad frac{0.533 cancel{text{g}}}{text{L}} times frac{1 text{mol}}{245.20 cancel{text{g}}}=2.17 times 10^{-3} text{M}\" width=\"444\" height=\"44\" \/><\/p>\n<p id=\"x-ck12-NjhlZWUwN2YyMmUzNGI4NmQ5MWZiNzAwMWVhN2IyY2I.-irq\">The dissociation equation shows that for every mole of PbF <sub> 2 <\/sub> that dissociates, 1 mol of Pb <sup> 2+ <\/sup> and 2 mol of F <sup> \u2212 <\/sup> are produced. Therefore, at equilibrium the concentrations of the ions are:<\/p>\n<p id=\"x-ck12-awi\"><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\/19213034\/fe369d746d277575d4f29770a7f6ec6b.png\" alt=\"[text{Pb}^{2+}]=2.17 times 10^{-3} text{M} quad text{and} quad [text{F}^-]=2 times 2.17 times 10^{-3}=4.35 times 10^{-3} text{M}\" width=\"560\" height=\"22\" \/><\/p>\n<p id=\"x-ck12-NmY5NDliNjFiYmFkODhhYTViNjkyOGEwNmVkYTE1NzY.-jv4\">Substitute into the expression and solve for 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\" \/> .<\/p>\n<p id=\"x-ck12-u9p\"><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\/19213035\/a4dc347960b0f0f0a841f78a12e9e08a.png\" alt=\"K_{sp}=(2.17 times 10^{-3})(4.35 times 10^{-3})^2=4.11 times 10^{-8}\" width=\"379\" height=\"23\" \/><\/p>\n<p id=\"x-ck12-ZGJmYWE4NWM0YmIyZGJiOTdhZTZmY2NmYmJmYjM1MjI.-8ua\"><em> Step 3: Think about your result <\/em> .<\/p>\n<p id=\"x-ck12-NmQ4NjRmYTFkY2VlYjA1OTQwMzE0NzUxN2E4MmRkZDU.-hyd\">The solubility product constant is significantly less than 1 for a nearly insoluble compound such as PbF <sub> 2 <\/sub> .<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Summary<\/h3>\n<ul id=\"x-ck12-NDg0M2Y2MmZhM2I3MDdmNDI1ZmVkYjBhOWZiMjAwNmQ.-jyu\">\n<li>Molar solubility calculations are described.<\/li>\n<li>Calculations of\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0using molar solubility are described.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Practice<\/h3>\n<p id=\"x-ck12-ZDY4YjkyZjQ1NjFmOWJlZDVkZTk4YWM0ZGI4YTJjMmI.-smc\">Read the material at <a href=\"http:\/\/www.chemteam.info\/Equilibrium\/Calc-Ksp-FromMolSolub.html\" target=\"_blank\">ChemTeam.info<\/a>\u00a0and do the problems at the end.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Review<\/h3>\n<ol id=\"x-ck12-ZWU5OTI0Zjk0ODQ0MGYxZGM4N2ExMzEzNDQ5NDNlMmM.-qlc\">\n<li>What are the solution requirements for determining molar solubility?<\/li>\n<li>Why do we need to convert mass to molarity to determine\u00a0<em>K<sub>sp<\/sub><\/em>?<\/li>\n<li>What\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0values would you expect for very insoluble compounds?<\/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-YjIxMzc3MzYwNDVkYWFiMTAzNWU3NDk4ZmQ1Yjg5ZGU.-ukb\">\n<li><strong> molar solubility: <\/strong> The number of moles of solute in one liter of saturated solution.<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\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-2749\">\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><li>Close-up Of Baking Soda On Spoon.. <strong>Authored by<\/strong>: Aqua Mechanical. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/flic.kr\/p\/E4hQjx\">https:\/\/flic.kr\/p\/E4hQjx<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/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":12,"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\":\"\"},{\"type\":\"cc\",\"description\":\"Close-up Of Baking Soda On Spoon.\",\"author\":\"Aqua Mechanical\",\"organization\":\"\",\"url\":\"https:\/\/flic.kr\/p\/E4hQjx\",\"project\":\"\",\"license\":\"cc-by\",\"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-2749","chapter","type-chapter","status-publish","hentry"],"part":2340,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2749","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\/2749\/revisions"}],"predecessor-version":[{"id":3012,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2749\/revisions\/3012"}],"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\/2749\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/media?parent=2749"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapter-type?post=2749"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/contributor?post=2749"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/license?post=2749"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}