{"id":2751,"date":"2016-08-24T18:23:50","date_gmt":"2016-08-24T18:23:50","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/?post_type=chapter&#038;p=2751"},"modified":"2016-08-24T21:38:46","modified_gmt":"2016-08-24T21:38:46","slug":"predicting-precipitates","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/chapter\/predicting-precipitates\/","title":{"raw":"Predicting Precipitates","rendered":"Predicting Precipitates"},"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>Predict solubility of a compound based on the comparison between the ion product and the <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 style=\"line-height: 1.5;\">What do you see?<\/strong><\/h3>\r\n<p id=\"x-ck12-ODQxMmM0YmE3M2M2NWNkMTRjYTAyYTEyOGZiNThiZTg.-bhs\">The invention of the X-ray machine had radically improved medical diagnosis and treatment. For the first time, it was possible to see inside a person\u2019s body to detect broken bones, tumors, obstructions, and other types of problems. Barium sulfate is often used to examine patients with problems of the esophagus, stomach, and intestines. This insoluble compound coats the inside of the tissues and absorbs X-rays, allowing a clear picture of the interior structure of these organs.<\/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\/19213047\/20140811155758736566.jpeg\" alt=\"Barium sulfate is used to obtain clear images of the digestive tract\" width=\"400\" \/>\r\n\r\n<\/div>\r\n<h3>Predicting Precipitates<\/h3>\r\n<p id=\"x-ck12-Yjg2ZTcwMThhN2NlMWE2MjJlODc5N2FjNzQyYWViYWI.-sqi\">Knowledge of\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0values will allow you to be able to predict whether or not a precipitate will form when two solutions are mixed together. For example, suppose that a known solution of barium chloride is mixed with a known solution of sodium sulfate. Barium sulfate is a mostly insoluble compound and so could potentially precipitate from the mixture. However, it is first necessary to calculate the <strong> ion product<\/strong>, [Ba <sup> 2+ <\/sup> ][SO <sub> 4 <\/sub><sup> 2\u2212 <\/sup> ] for the solution. If the value of the ion product is less than the value of the\u00a0<em>K<sub>sp<\/sub><\/em>, then the solution will remain unsaturated. No precipitate will form because the concentrations are not high enough to begin the precipitation process. If the value of the ion product is greater than the value of the\u00a0<em>K<sub>sp<\/sub><\/em>, then a precipitate will form. The formation of the precipitate lowers the concentration of each of the ions until the ion product is exactly equal to the <em>K<sub>sp<\/sub><\/em>, at which point precipitation ceases.<\/p>\r\n\r\n<div id=\"x-ck12-NWY2NWUzZjZhOTRhZmFjYTE0OTVhYjE0MWM0OTZiMjU.-usa\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"500\"]<img id=\"x-ck12-OTgwNDUtMTM2NTQxMDcwMS01OC00MC01LjMuMTQuMg..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213048\/20140811155758842968.png\" alt=\"Barium sulfate is used as a component of white paint and in certain x-ray imaging processes\" width=\"500\" height=\"192\" longdesc=\"Barium%20sulfate%20is%20used%20as%20a%20component%20of%20white%20pigment%20for%20paints%20and%20as%20an%20agent%20in%20certain%20x-ray%20imaging%20processes.\" \/> Figure 1.\u00a0Barium sulfate is used as a component of white pigment for paints and as an agent in certain x-ray imaging processes.[\/caption]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h4>Sample Problem: Predicting Precipitates<\/h4>\r\n<p id=\"x-ck12-OGZhMDE1ZmZjMWJjNDg2Y2ZmNTMwYzAwZWJlMTI1ZmI.-xm9\">Will a precipitate of barium sulfate form when 10.0 mL of 0.0050 M BaCl <sub> 2 <\/sub> is mixed with 20.0 mL of 0.0020 M Na <sub> 2 <\/sub> SO <sub> 4 <\/sub> ?<\/p>\r\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-iy5\"><em> Step 1: List the known quantities and plan the problem <\/em> .<\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-zug\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-NzExMTllNTllMDIyMzRiMmQ0MDQ3MjNmYWM1ODlkYzY.-go8\">\r\n \t<li>concentration of BaCl <sub> 2 <\/sub> = 0.0050 M<\/li>\r\n \t<li>volume of BaCl <sub> 2 <\/sub> = 10.0 mL<\/li>\r\n \t<li>concentration of Na <sub> 2 <\/sub> SO <sub> 4 <\/sub> = 0.0020 M<\/li>\r\n \t<li>volume of Na <sub> 2 <\/sub> SO <sub> 4 <\/sub> = 20.0 mL<\/li>\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\/19213027\/152ea892d77eb7f4d21f1d84974b9c17.png\" alt=\"K_{sp}\" width=\"28\" height=\"18\" \/> of BaSO <sub> 4 <\/sub> = 1.1\u00a0\u00d7\u00a010 <sup> -10 <\/sup><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-en4\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-N2FhNDMyNDBkMTNiN2ZlY2ZjZGZlY2YxMmMzYTcxNTM.-w23\">\r\n \t<li>ion product [Ba <sup> 2+ <\/sup> ][SO <sub> 4 <\/sub><sup> 2- <\/sup> ]<\/li>\r\n \t<li>if a precipitate forms<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-MmQxMjE5NzIxN2M5ZDkyNDczMzFjYzc5YThjYzUxZmI.-cua\">The concentration and volume of each solution that is mixed together must be used to calculate the [Ba <sup> 2+ <\/sup> ] and the [SO <sub> 4 <\/sub><sup> 2\u2212 <\/sup> ]. Each individual solution is diluted when they are mixed together. The ion product is calculated and compared to the\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0to determine if a precipitate forms.<\/p>\r\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-tq3\"><em> Step 2: Solve <\/em> .<\/p>\r\n<p id=\"x-ck12-NzFmNmY2MWZiN2NlYWM5ZTczMjZhMmYzNGI2N2YzNmI.-otj\">The moles of each ion from the original solutions are calculated by multiplying the molarity by the volume in liters.<\/p>\r\n<p id=\"x-ck12-d7b\"><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\/19213050\/7574b796d00a90bde69ee33191ec010d.png\" alt=\"text{mol Ba}^{2+}&amp;=0.0050 text{M} times 0.010 text{L}=5.0 times 10^{-5} text{mol Ba}^{2+} \\text{mol SO}_4^{2-}&amp;=0.0020 text{M} times 0.020 text{L}=4.0 times 10^{-5} text{mol SO}_4^{2-}\" width=\"436\" height=\"50\" \/><\/p>\r\n<p id=\"x-ck12-MTIxODljYjJkOTkzOTUxMmRiYzIzOTM1Y2Q4MDI5NjE.-0v8\">The concentration of each ion after dilution is then calculated by dividing the moles by the final solution volume of 0.030 L.<\/p>\r\n<p id=\"x-ck12-bd7\"><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\/19213051\/c663a205fc771869f29398109eced4d1.png\" alt=\"[text{Ba}^{2+}] &amp;=frac{5.0 times 10^{-5} text{mol}}{0.030 text{L}}=1.7 times 10^{-3} text{M} \\left [text{SO}_4^{2-}right ] &amp;=frac{4.0 times 10^{-5} text{mol}}{0.030 text{L}}=1.3 times 10^{-3} text{M}\" width=\"323\" height=\"86\" \/><\/p>\r\n<p id=\"x-ck12-ODZlODU0MzQ4NTk4MzdmZWQ4NzY2NTBlZGE0MWI3YTQ.-hqv\">Now the ion product is calculated.<\/p>\r\n<p id=\"x-ck12-hkf\"><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\/19213052\/f0cb54b372c1159476c03f332cc3926b.png\" alt=\"[text{Ba}^{2+}][text{SO}_4^{2-}]=(1.7 times 10^{-3})(1.3 times 10^{-3})=2.2 times 10^{-6}\" width=\"413\" height=\"22\" \/><\/p>\r\n<p id=\"x-ck12-YTNjZDdkOGYzYzg2Y2U2MzAxYTFlYmMyNTA1Mzg3NWQ.-zt6\">Since the ion product is greater than the\u00a0<em>K<sub>sp<\/sub><\/em>, a precipitate of barium sulfate will form.<\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-eqk\"><em> Step 3: Think about your result <\/em> .<\/p>\r\n<p id=\"x-ck12-ZjM5MTllNzU1MTVjZmIxODBmNDBlMGYwNTBmMjIzMDQ.-cv4\">Two significant figures are appropriate for the calculated value of the ion product.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Summary<\/h3>\r\n<ul id=\"x-ck12-ZmIwZWFlM2ZkZmJhZTY3MDA1ZDkyYmM2MjA4ZmVhYjM.-rwo\">\r\n \t<li>Calculations are shown which allow the prediction of precipitate formation based on\u00a0<em>K<sub>sp<\/sub><\/em>.<\/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.-bv1\">Work the problems at the following link:\u00a0<a style=\"line-height: 1.5;\" href=\"http:\/\/misterguch.brinkster.net\/jan2002.pdf\">http:\/\/misterguch.brinkster.net\/jan2002.pdf<\/a><\/p>\r\n<p id=\"x-ck12-ZDFiYzcyZjUzNjY0Nzg4ZTI2MDFjYmFkMjE0ZDBjMzY.-aps\">A useful table for constants can be found here:\u00a0 <a href=\"http:\/\/users.stlcc.edu\/gkrishnan\/ksptable.html\"> http:\/\/users.stlcc.edu\/gkrishnan\/ksptable.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-ZDk1NzBmOTQ0NjNiMTNmZmQ5NDgwNGY3MzliNzZjZDM.-bjy\">\r\n \t<li>What would be the equation for the ion product of BaCl <sub> 2 <\/sub> ?<\/li>\r\n \t<li>What happens if the ion product is less 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\" \/> ?<\/li>\r\n \t<li>Why did we not need to calculate an ion product for NaCl?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<h3 class=\"x-ck12-data-problem-set\">Glossary<\/h3>\r\n<div class=\"x-ck12-data-vocabulary\">\r\n<ul id=\"x-ck12-NTgxMmFjOWFjZDZiNjc5M2E2MjM2MTdkNTEwNmJmM2I.-kza\">\r\n \t<li><strong> ion product: <\/strong> The product of the concentrations of ions in a solution.<\/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 Journalist Seaman Apprentice Mike Leporati, US Navy.<a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:US_Navy_060505-N-2832L-010_An_X-Ray_machine_located_aboard_Military_Sealift_Command_%28MSC%29_hospital_ship_USNS_Mercy_%28T-AH_19%29.jpg\">http:\/\/commons.wikimedia.org\/wiki\/File:US_Navy_060505-N-2832L-010_An_X-Ray_machine_located_aboard_Military_Sealift_Command_%28MSC%29_hospital_ship_USNS_Mercy_%28T-AH_19%29.jpg <\/a>.<\/li>\r\n \t<li>Ond\u0159ej Mangl. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:S%C3%ADran_barnat%C3%BD.PNG\">http:\/\/commons.wikimedia.org\/wiki\/File:S%C3%ADran_barnat%C3%BD.PNG <\/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>Predict solubility of a compound based on the comparison between the ion product and the <em>K<sub>sp<\/sub><\/em>.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3><strong style=\"line-height: 1.5;\">What do you see?<\/strong><\/h3>\n<p id=\"x-ck12-ODQxMmM0YmE3M2M2NWNkMTRjYTAyYTEyOGZiNThiZTg.-bhs\">The invention of the X-ray machine had radically improved medical diagnosis and treatment. For the first time, it was possible to see inside a person\u2019s body to detect broken bones, tumors, obstructions, and other types of problems. Barium sulfate is often used to examine patients with problems of the esophagus, stomach, and intestines. This insoluble compound coats the inside of the tissues and absorbs X-rays, allowing a clear picture of the interior structure of these organs.<\/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\/19213047\/20140811155758736566.jpeg\" alt=\"Barium sulfate is used to obtain clear images of the digestive tract\" width=\"400\" \/><\/p>\n<\/div>\n<h3>Predicting Precipitates<\/h3>\n<p id=\"x-ck12-Yjg2ZTcwMThhN2NlMWE2MjJlODc5N2FjNzQyYWViYWI.-sqi\">Knowledge of\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0values will allow you to be able to predict whether or not a precipitate will form when two solutions are mixed together. For example, suppose that a known solution of barium chloride is mixed with a known solution of sodium sulfate. Barium sulfate is a mostly insoluble compound and so could potentially precipitate from the mixture. However, it is first necessary to calculate the <strong> ion product<\/strong>, [Ba <sup> 2+ <\/sup> ][SO <sub> 4 <\/sub><sup> 2\u2212 <\/sup> ] for the solution. If the value of the ion product is less than the value of the\u00a0<em>K<sub>sp<\/sub><\/em>, then the solution will remain unsaturated. No precipitate will form because the concentrations are not high enough to begin the precipitation process. If the value of the ion product is greater than the value of the\u00a0<em>K<sub>sp<\/sub><\/em>, then a precipitate will form. The formation of the precipitate lowers the concentration of each of the ions until the ion product is exactly equal to the <em>K<sub>sp<\/sub><\/em>, at which point precipitation ceases.<\/p>\n<div id=\"x-ck12-NWY2NWUzZjZhOTRhZmFjYTE0OTVhYjE0MWM0OTZiMjU.-usa\" class=\"x-ck12-img-postcard x-ck12-nofloat\">\n<div style=\"width: 510px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-OTgwNDUtMTM2NTQxMDcwMS01OC00MC01LjMuMTQuMg..\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213048\/20140811155758842968.png\" alt=\"Barium sulfate is used as a component of white paint and in certain x-ray imaging processes\" width=\"500\" height=\"192\" longdesc=\"Barium%20sulfate%20is%20used%20as%20a%20component%20of%20white%20pigment%20for%20paints%20and%20as%20an%20agent%20in%20certain%20x-ray%20imaging%20processes.\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1.\u00a0Barium sulfate is used as a component of white pigment for paints and as an agent in certain x-ray imaging processes.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h4>Sample Problem: Predicting Precipitates<\/h4>\n<p id=\"x-ck12-OGZhMDE1ZmZjMWJjNDg2Y2ZmNTMwYzAwZWJlMTI1ZmI.-xm9\">Will a precipitate of barium sulfate form when 10.0 mL of 0.0050 M BaCl <sub> 2 <\/sub> is mixed with 20.0 mL of 0.0020 M Na <sub> 2 <\/sub> SO <sub> 4 <\/sub> ?<\/p>\n<p id=\"x-ck12-Y2VlNGQ1OTZkZWQ3ZDY1MzMyNmI1ODhhNWUzY2Q3NmU.-iy5\"><em> Step 1: List the known quantities and plan the problem <\/em> .<\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-zug\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-NzExMTllNTllMDIyMzRiMmQ0MDQ3MjNmYWM1ODlkYzY.-go8\">\n<li>concentration of BaCl <sub> 2 <\/sub> = 0.0050 M<\/li>\n<li>volume of BaCl <sub> 2 <\/sub> = 10.0 mL<\/li>\n<li>concentration of Na <sub> 2 <\/sub> SO <sub> 4 <\/sub> = 0.0020 M<\/li>\n<li>volume of Na <sub> 2 <\/sub> SO <sub> 4 <\/sub> = 20.0 mL<\/li>\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\/19213027\/152ea892d77eb7f4d21f1d84974b9c17.png\" alt=\"K_{sp}\" width=\"28\" height=\"18\" \/> of BaSO <sub> 4 <\/sub> = 1.1\u00a0\u00d7\u00a010 <sup> -10 <\/sup><\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-en4\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-N2FhNDMyNDBkMTNiN2ZlY2ZjZGZlY2YxMmMzYTcxNTM.-w23\">\n<li>ion product [Ba <sup> 2+ <\/sup> ][SO <sub> 4 <\/sub><sup> 2- <\/sup> ]<\/li>\n<li>if a precipitate forms<\/li>\n<\/ul>\n<p id=\"x-ck12-MmQxMjE5NzIxN2M5ZDkyNDczMzFjYzc5YThjYzUxZmI.-cua\">The concentration and volume of each solution that is mixed together must be used to calculate the [Ba <sup> 2+ <\/sup> ] and the [SO <sub> 4 <\/sub><sup> 2\u2212 <\/sup> ]. Each individual solution is diluted when they are mixed together. The ion product is calculated and compared to the\u00a0<em>K<sub>sp<\/sub><\/em>\u00a0to determine if a precipitate forms.<\/p>\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-tq3\"><em> Step 2: Solve <\/em> .<\/p>\n<p id=\"x-ck12-NzFmNmY2MWZiN2NlYWM5ZTczMjZhMmYzNGI2N2YzNmI.-otj\">The moles of each ion from the original solutions are calculated by multiplying the molarity by the volume in liters.<\/p>\n<p id=\"x-ck12-d7b\"><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\/19213050\/7574b796d00a90bde69ee33191ec010d.png\" alt=\"text{mol Ba}^{2+}&amp;=0.0050 text{M} times 0.010 text{L}=5.0 times 10^{-5} text{mol Ba}^{2+} \\text{mol SO}_4^{2-}&amp;=0.0020 text{M} times 0.020 text{L}=4.0 times 10^{-5} text{mol SO}_4^{2-}\" width=\"436\" height=\"50\" \/><\/p>\n<p id=\"x-ck12-MTIxODljYjJkOTkzOTUxMmRiYzIzOTM1Y2Q4MDI5NjE.-0v8\">The concentration of each ion after dilution is then calculated by dividing the moles by the final solution volume of 0.030 L.<\/p>\n<p id=\"x-ck12-bd7\"><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\/19213051\/c663a205fc771869f29398109eced4d1.png\" alt=\"[text{Ba}^{2+}] &amp;=frac{5.0 times 10^{-5} text{mol}}{0.030 text{L}}=1.7 times 10^{-3} text{M} \\left [text{SO}_4^{2-}right ] &amp;=frac{4.0 times 10^{-5} text{mol}}{0.030 text{L}}=1.3 times 10^{-3} text{M}\" width=\"323\" height=\"86\" \/><\/p>\n<p id=\"x-ck12-ODZlODU0MzQ4NTk4MzdmZWQ4NzY2NTBlZGE0MWI3YTQ.-hqv\">Now the ion product is calculated.<\/p>\n<p id=\"x-ck12-hkf\"><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\/19213052\/f0cb54b372c1159476c03f332cc3926b.png\" alt=\"[text{Ba}^{2+}][text{SO}_4^{2-}]=(1.7 times 10^{-3})(1.3 times 10^{-3})=2.2 times 10^{-6}\" width=\"413\" height=\"22\" \/><\/p>\n<p id=\"x-ck12-YTNjZDdkOGYzYzg2Y2U2MzAxYTFlYmMyNTA1Mzg3NWQ.-zt6\">Since the ion product is greater than the\u00a0<em>K<sub>sp<\/sub><\/em>, a precipitate of barium sulfate will form.<\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-eqk\"><em> Step 3: Think about your result <\/em> .<\/p>\n<p id=\"x-ck12-ZjM5MTllNzU1MTVjZmIxODBmNDBlMGYwNTBmMjIzMDQ.-cv4\">Two significant figures are appropriate for the calculated value of the ion product.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Summary<\/h3>\n<ul id=\"x-ck12-ZmIwZWFlM2ZkZmJhZTY3MDA1ZDkyYmM2MjA4ZmVhYjM.-rwo\">\n<li>Calculations are shown which allow the prediction of precipitate formation based on\u00a0<em>K<sub>sp<\/sub><\/em>.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Practice<\/h3>\n<p id=\"x-ck12-ZGIyN2QxMzI1ODI1MWRlNWVlMjE5YTlhNmM5NDJlN2M.-bv1\">Work the problems at the following link:\u00a0<a style=\"line-height: 1.5;\" href=\"http:\/\/misterguch.brinkster.net\/jan2002.pdf\">http:\/\/misterguch.brinkster.net\/jan2002.pdf<\/a><\/p>\n<p id=\"x-ck12-ZDFiYzcyZjUzNjY0Nzg4ZTI2MDFjYmFkMjE0ZDBjMzY.-aps\">A useful table for constants can be found here:\u00a0 <a href=\"http:\/\/users.stlcc.edu\/gkrishnan\/ksptable.html\"> http:\/\/users.stlcc.edu\/gkrishnan\/ksptable.html<\/a><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Review<\/h3>\n<ol id=\"x-ck12-ZDk1NzBmOTQ0NjNiMTNmZmQ5NDgwNGY3MzliNzZjZDM.-bjy\">\n<li>What would be the equation for the ion product of BaCl <sub> 2 <\/sub> ?<\/li>\n<li>What happens if the ion product is less 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\" \/> ?<\/li>\n<li>Why did we not need to calculate an ion product for NaCl?<\/li>\n<\/ol>\n<\/div>\n<h3 class=\"x-ck12-data-problem-set\">Glossary<\/h3>\n<div class=\"x-ck12-data-vocabulary\">\n<ul id=\"x-ck12-NTgxMmFjOWFjZDZiNjc5M2E2MjM2MTdkNTEwNmJmM2I.-kza\">\n<li><strong> ion product: <\/strong> The product of the concentrations of ions in a solution.<\/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 Journalist Seaman Apprentice Mike Leporati, US Navy.<a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:US_Navy_060505-N-2832L-010_An_X-Ray_machine_located_aboard_Military_Sealift_Command_%28MSC%29_hospital_ship_USNS_Mercy_%28T-AH_19%29.jpg\">http:\/\/commons.wikimedia.org\/wiki\/File:US_Navy_060505-N-2832L-010_An_X-Ray_machine_located_aboard_Military_Sealift_Command_%28MSC%29_hospital_ship_USNS_Mercy_%28T-AH_19%29.jpg <\/a>.<\/li>\n<li>Ond\u0159ej Mangl. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:S%C3%ADran_barnat%C3%BD.PNG\">http:\/\/commons.wikimedia.org\/wiki\/File:S%C3%ADran_barnat%C3%BD.PNG <\/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-2751\">\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":14,"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-2751","chapter","type-chapter","status-publish","hentry"],"part":2340,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2751","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":2,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2751\/revisions"}],"predecessor-version":[{"id":3015,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2751\/revisions\/3015"}],"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\/2751\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/media?parent=2751"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapter-type?post=2751"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/contributor?post=2751"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/license?post=2751"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}