{"id":2826,"date":"2016-08-24T19:55:34","date_gmt":"2016-08-24T19:55:34","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/?post_type=chapter&#038;p=2826"},"modified":"2016-08-24T22:53:43","modified_gmt":"2016-08-24T22:53:43","slug":"calculating-ph-of-salt-solutions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/chapter\/calculating-ph-of-salt-solutions\/","title":{"raw":"Calculating pH of Salt Solutions","rendered":"Calculating pH of Salt Solutions"},"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 to determine pH of salt solutions if\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\/19213301\/23b35b1534be8d9bb64ff42a83c35fd9.png\" alt=\"K_a\" width=\"22\" height=\"15\" \/> or <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\/19212456\/c7189caed27d0f9b02b2f75baf0c8c3f.png\" alt=\"K_b\" width=\"21\" height=\"15\" \/> are known.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Keeping things safe and healthy<\/h3>\r\n<p id=\"x-ck12-ZDBkYWFhYzNkNjcxNjFjN2IxNzE2OTIzYmFlODUxNjc.-ryt\"><span class=\"x-ck12-img-inline\"><img class=\"alignright\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213408\/20140811155917780590.jpeg\" alt=\"A swimming pool is kept around a pH of 7.2 using various acidic and basic salts\" width=\"265\" height=\"199\" \/><\/span><\/p>\r\n<p id=\"x-ck12-MDBkZDg4Yjc3ZmRmNGJlNTlmZWIyNjgxMTc3YmZiMjE.-tr1\">We all enjoy a cool dip in a swimming pool on a hot day, but we may not realize the work needed to keep that water safe and healthy. The ideal pH for a swimming pool is around 7.2. The pH will change as a result of many factors. Adjustment can be accomplished with different chemicals depending on the tested pH. High pH can be lowered with liquid HCl (unsafe material) or sodium bisulfate. The bisulfate anion is a weak acid and can dissociate partially in solution. To increase pH, use sodium carbonate. The carbonate anion forms an equilibrium with protons that results in some formation of carbon dioxide.<\/p>\r\n\r\n<\/div>\r\n<h2>Calculating pH of Salt Solutions<\/h2>\r\n<p id=\"x-ck12-ODI0ZDg0Y2M2M2UyODgxMTEzMmY3M2EwNDdkMWFjMzQ.-9ph\">It is often helpful to be able to predict the effect a salt solution will have on the pH of a certain solution. Knowledge of the relevant acidity or basicity constants allows us to carry out the necessary calculations.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h4>Sample Problem: Salt Hydrolysis<\/h4>\r\n<p id=\"x-ck12-MGVlZDM3ODU0ZTcyYjAyYTExODQwZGE3NmU5MWYyZjQ.-1ag\">If we dissolve NaF in water, we get the following equilibrium:<\/p>\r\n<p id=\"x-ck12-2nu\"><img id=\"x-ck12-MTM3ODMyMDgzMzMwNQ..\" 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\/19213406\/c65fd3e30e6550028e01c8d42203a384.png\" alt=\"text{F}^-(aq)+text{H}_2text{O}(l) rightleftarrows text{HF}(aq)+text{OH}^-(aq)\" width=\"307\" height=\"18\" \/><\/p>\r\n<p id=\"x-ck12-Yjc2OWJlMDRjOTYwMTRlZWI1Y2MwYjdkODZkZWM4YjY.-l3x\">The pH of the resulting solution can be determined if 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\/19212456\/c7189caed27d0f9b02b2f75baf0c8c3f.png\" alt=\"K_b\" width=\"21\" height=\"15\" \/> of the fluoride ion is known.<\/p>\r\n<p id=\"x-ck12-MzU5ZjM4NGRjNjg0N2RlYzY3YzMwNjliMjQyZWZkZmQ.-agc\">20.0 g of sodium fluoride is dissolve in enough water to make 500.0 mL of solution. Calculate the pH of the solution. The\u00a0 <img id=\"x-ck12-MTM3ODMyMDgzMzMwNg..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212456\/c7189caed27d0f9b02b2f75baf0c8c3f.png\" alt=\"K_b\" width=\"21\" height=\"15\" \/> of the fluoride ion is 1.4 \u00d7 10 <sup> \u221211 <\/sup> .<\/p>\r\n<p id=\"x-ck12-ODFjNTAwNDljM2RhMjljYTgyMTUxYWY2Nzk5ZWFhNGI.-tob\"><em> Step 1: List the known values and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-ilh\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-NmEyMzdmYjA2YTY5MDQ5MDBjZjdkNDQ3NTdjN2YyMzM.-hik\">\r\n \t<li>mass NaF = 20.0 g<\/li>\r\n \t<li>molar mass NaF = 41.99 g\/mol<\/li>\r\n \t<li>volume solution = 0.500 L<\/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\/19212456\/c7189caed27d0f9b02b2f75baf0c8c3f.png\" alt=\"K_b\" width=\"21\" height=\"15\" \/> of F <sup> - <\/sup> = 1.4 \u00d7 10 <sup> \u221211 <\/sup><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-zdg\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-YmEzMGRhNjg2OGQ5ZTRkZTY5ZjkzM2NiYjIyNWE2ODI.-ct9\">\r\n \t<li>pH of solution = ?<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-MzQ4OTUyYjJmNzVlZjAzZTYyYTI2MzNjZGIzZjQwZmQ.-cl5\">The molarity of the F <sup> \u2212 <\/sup> solution can be calculated from the mass, molar mass, and solution volume. Since NaF completely dissociates, the molarity of the NaF is equal to the molarity of the F <sup> \u2212 <\/sup> ion. An ICE <strong> Table <\/strong> (below) can be used to calculate the concentration of OH <sup> \u2212 <\/sup> produced and then the pH of the solution.<\/p>\r\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-5ww\"><em> Step 2: Solve. <\/em><\/p>\r\n<p id=\"x-ck12-tir\"><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\/19213410\/958271f1c98955aa54278fb5548f20ba.png\" alt=\"20.0 cancel{text{g NaF}} times frac{1 cancel{text{mol NaF}}}{41.99 cancel{text{g NaF}}}times frac{1 text{mol F}^-}{1 cancel{text{mol NaF}}}&amp;=0.476 text{mol F}^- \\frac{0.476 text{mol F}^-}{0.5000 text{L}}&amp;=0.953 text{M F}^-\" width=\"444\" height=\"88\" \/><\/p>\r\n<p id=\"x-ck12-NzRhOTBlM2U0ZTVkZTdlNDNmMzIzNmRjNjdmZWY3ZGQ.-hq0\"><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\/19213411\/2f420f2e6d32ddffbafd06ccca8f13e2.png\" alt=\"text{Hydrolysis equation:} qquad text{F}^-(aq)+text{H}_2text{O}(l)rightleftarrows text{HF}(aq)+text{OH}^-(aq)\" width=\"503\" height=\"18\" \/><\/p>\r\n\r\n<table id=\"x-ck12-YTk0NGUzZmRhNjA4MGYzNDcyZTBmOTJmZGJhYTI4M2M.-07i\" class=\"x-ck12-nofloat\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<td><strong> Concentrations <\/strong><\/td>\r\n<td><strong> [F <sup> - <\/sup> ] <\/strong><\/td>\r\n<td><strong> [HF] <\/strong><\/td>\r\n<td><strong> [OH <sup> - <\/sup> ] <\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Initial<\/td>\r\n<td>0.953<\/td>\r\n<td>0<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Change<\/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\/19213315\/128f156bd2177fdb41cca16122d32a69.png\" alt=\"-x\" width=\"23\" 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\/19213315\/ded5a4d8ae585bda875e454ef5fce93d.png\" alt=\"+x\" width=\"24\" 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\/19213315\/ded5a4d8ae585bda875e454ef5fce93d.png\" alt=\"+x\" width=\"24\" height=\"12\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Equilibrium<\/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\/19213413\/1e09476a65467e6df12973e2b4b2a522.png\" alt=\"0.953 - x\" width=\"73\" height=\"13\" \/><\/td>\r\n<td><img id=\"x-ck12-MTM3ODMyMDgzMzMwNw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212123\/0f14991bebd0bc5ca2e82990e029d525.png\" alt=\"x\" width=\"10\" 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\/19212123\/0f14991bebd0bc5ca2e82990e029d525.png\" alt=\"x\" width=\"10\" height=\"8\" \/><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"x-ck12-yv7\"><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\/19213413\/b5d9006fd65c3c7d70daf435f823dc97.png\" alt=\"K_b &amp;=1.4times 10^{-11}=frac{(x)(x)}{0.953-x}=frac{x^2}{0.953-x}approx frac{x^2}{0.953} \\x &amp;=[text{OH}^-]=sqrt{1.4times 10^{-11}(0.953)}=3.65times 10^{-6} text{M} \\text{pOH} &amp;=-log (3.65times 10^{-6})=5.44 \\text{pH} &amp;=14-5.44=8.56\" width=\"428\" height=\"123\" \/><\/p>\r\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-wqr\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-N2NmMTkzYjViMjEwZGI3ZTQwMzYxNzYzMThiYjNjZDk.-v9z\">The solution is slightly basic due to the hydrolysis of the fluoride ion.<\/p>\r\n\r\n<\/div>\r\n<h3>Salts That Form Acidic Solutions<\/h3>\r\n<p id=\"x-ck12-Y2U0NTE3Nzk4NmQ4ZDU4YjQxMjJkMzczMjQ3Njc3OGQ.-h5w\">When the ammonium ion dissolves in water, the following equilibrium exists:<\/p>\r\n<p id=\"x-ck12-h9q\"><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\/19213415\/2ac7e4485859514b1cb6d6d34726e4b5.png\" alt=\"text{NH}_4^+(aq)+text{H}_2text{O}(l)rightleftarrows text{H}_3text{O}^+(aq)+text{NH}_3(aq)\" width=\"339\" height=\"21\" \/><\/p>\r\n<p id=\"x-ck12-MmMwNGM2YWY3YWI4ZjRmMDkwNDZjNWQxNjJmOTIzYzU.-btc\">The production of hydronium ions causes the resulting solution to be acidic. The pH of a solution of ammonium chloride can be found in a very similar way to the sodium fluoride solution in Sample Problem 21.7. However, since the ammonium chloride is acting as an acid, it is necessary to know 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\/19213416\/cbd56ced35c37836d1a05027a2d41284.png\" alt=\"K_a\" width=\"22\" height=\"15\" \/> of NH <sub> 4 <\/sub><sup> + <\/sup> , which is 5.6 \u00d7 10 <sup> \u221210 <\/sup> . We will find the pH of a 2.00 M solution of NH <sub> 4 <\/sub> Cl. Because the NH <sub> 4 <\/sub> Cl completely ionizes, the concentration of the ammonium ion is 2.00 M.<\/p>\r\n<p id=\"x-ck12-l9j\"><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\/19213417\/f98b549229c60c1e4ecaa78908f3d6a8.png\" alt=\"text{NH}_4text{Cl}(s)rightarrow text{NH}_4^+(aq)+text{Cl} ^-(aq)\" width=\"254\" height=\"21\" \/><\/p>\r\n<p id=\"x-ck12-NWRmN2M0NjAzMjljZjg4ZDBiMTVjMThlMDg5MGNkYzg.-xqu\">Again, an ICE <strong> Table <\/strong> (below) is set up in order to solve for the concentration of the hydronium (or H <sup> + <\/sup> ) ion produced.<\/p>\r\n\r\n<table id=\"x-ck12-MTcyNmQxNGE5ZDQwYmE0M2ZlYWFkZGQ1ZDYxNzlkYzU.-ioq\" class=\"x-ck12-nofloat\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<td><strong> Concentrations <\/strong><\/td>\r\n<td><strong> [NH <sub> 4 <\/sub><sup> + <\/sup> ] <\/strong><\/td>\r\n<td><strong> [H <sup> + <\/sup> ] <\/strong><\/td>\r\n<td><strong> [NH <sub> 3 <\/sub> ] <\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Initial<\/td>\r\n<td>2.00<\/td>\r\n<td>0<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Change<\/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\/19213315\/128f156bd2177fdb41cca16122d32a69.png\" alt=\"-x\" width=\"23\" 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\/19213315\/ded5a4d8ae585bda875e454ef5fce93d.png\" alt=\"+x\" width=\"24\" 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\/19213315\/ded5a4d8ae585bda875e454ef5fce93d.png\" alt=\"+x\" width=\"24\" height=\"12\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Equilibrium<\/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\/19213418\/cc5e2c32d5fee93312254fc0c1cf64c6.png\" alt=\"2.00 - x\" width=\"64\" 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\/19212123\/0f14991bebd0bc5ca2e82990e029d525.png\" alt=\"x\" width=\"10\" height=\"8\" \/><\/td>\r\n<td><img id=\"x-ck12-MTM3ODMyMDgzMzMwOA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212123\/0f14991bebd0bc5ca2e82990e029d525.png\" alt=\"x\" width=\"10\" height=\"8\" \/><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"x-ck12-OTc4MTEzOTlkZDFlYzk5ODkyNjFhOTU0NWI0ZWUxYjU.-slo\">Now substituting into 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\/19213301\/23b35b1534be8d9bb64ff42a83c35fd9.png\" alt=\"K_a\" width=\"22\" height=\"15\" \/> expression gives:<\/p>\r\n<p id=\"x-ck12-dya\"><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\/19213418\/0b26f10589ea4b56af8dce6e78f6ae45.png\" alt=\"K_a &amp;=5.6 times 10^{-10}=frac{x^2}{2.00-x} approx frac{x^2}{2.00} \\x &amp;=[text{H}^+]=sqrt{5.6 times 10^{-10}(2.00)}=3.3 times 10^{-5} text{M} \\text{pH} &amp;=- log (3.3 times 10^{-5})=4.48\" width=\"382\" height=\"97\" \/><\/p>\r\n<p id=\"x-ck12-ZDg4NmViOTBiZWVjODA2YTIzMzA5M2ZjNzc2ZTc3MmE.-42x\">A salt produced from a strong acid and a weak base yields a solution that is acidic.<\/p>\r\n\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Summary<\/h3>\r\n<ul id=\"x-ck12-N2QzM2RkMzU3MzFhY2U1MWM1MzM5YjEyZTU1ZjI1NmQ.-rip\">\r\n \t<li>Calculations to determine pH of salt solutions 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.-bcy\">Work the problems at the link below:<\/p>\r\n<p id=\"x-ck12-Y2VhY2RlNjExZjM0ZGU1YmQ3YWNjYmI2MzJhNmNkMmE.-qke\"><a href=\"http:\/\/www.sparknotes.com\/chemistry\/acidsbases\/phcalc\/problems.html\"> http:\/\/www.sparknotes.com\/chemistry\/acidsbases\/phcalc\/problems.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-Y2MzZGRhNjAyMmFiZGRlZDgzNGRjZDNlM2I4YzgwMjc.-iot\">\r\n \t<li>In the first example, how do we know that we can ignore\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\/19212123\/0f14991bebd0bc5ca2e82990e029d525.png\" alt=\"x\" width=\"10\" height=\"8\" \/> when determining [F <sup> - <\/sup> ]?<\/li>\r\n \t<li>In example two, how do we know the ammonium ion concentration?<\/li>\r\n \t<li>Could we write the equilibrium in example two as <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\/19213419\/7d3ca23d4749ad242fe6fa6397f2e824.png\" alt=\"text{NH}_4^+ rightleftarrows text{H}^+ + text{NH}_3\" width=\"145\" height=\"21\" \/> ?<\/li>\r\n<\/ol>\r\n<\/div>\r\n&nbsp;","rendered":"<div class=\"x-ck12-data-objectives\">\n<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Perform calculations to determine pH of salt solutions if\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\/19213301\/23b35b1534be8d9bb64ff42a83c35fd9.png\" alt=\"K_a\" width=\"22\" height=\"15\" \/> or <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\/19212456\/c7189caed27d0f9b02b2f75baf0c8c3f.png\" alt=\"K_b\" width=\"21\" height=\"15\" \/> are known.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Keeping things safe and healthy<\/h3>\n<p id=\"x-ck12-ZDBkYWFhYzNkNjcxNjFjN2IxNzE2OTIzYmFlODUxNjc.-ryt\"><span class=\"x-ck12-img-inline\"><img loading=\"lazy\" decoding=\"async\" class=\"alignright\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19213408\/20140811155917780590.jpeg\" alt=\"A swimming pool is kept around a pH of 7.2 using various acidic and basic salts\" width=\"265\" height=\"199\" \/><\/span><\/p>\n<p id=\"x-ck12-MDBkZDg4Yjc3ZmRmNGJlNTlmZWIyNjgxMTc3YmZiMjE.-tr1\">We all enjoy a cool dip in a swimming pool on a hot day, but we may not realize the work needed to keep that water safe and healthy. The ideal pH for a swimming pool is around 7.2. The pH will change as a result of many factors. Adjustment can be accomplished with different chemicals depending on the tested pH. High pH can be lowered with liquid HCl (unsafe material) or sodium bisulfate. The bisulfate anion is a weak acid and can dissociate partially in solution. To increase pH, use sodium carbonate. The carbonate anion forms an equilibrium with protons that results in some formation of carbon dioxide.<\/p>\n<\/div>\n<h2>Calculating pH of Salt Solutions<\/h2>\n<p id=\"x-ck12-ODI0ZDg0Y2M2M2UyODgxMTEzMmY3M2EwNDdkMWFjMzQ.-9ph\">It is often helpful to be able to predict the effect a salt solution will have on the pH of a certain solution. Knowledge of the relevant acidity or basicity constants allows us to carry out the necessary calculations.<\/p>\n<div class=\"textbox shaded\">\n<h4>Sample Problem: Salt Hydrolysis<\/h4>\n<p id=\"x-ck12-MGVlZDM3ODU0ZTcyYjAyYTExODQwZGE3NmU5MWYyZjQ.-1ag\">If we dissolve NaF in water, we get the following equilibrium:<\/p>\n<p id=\"x-ck12-2nu\"><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM3ODMyMDgzMzMwNQ..\" 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\/19213406\/c65fd3e30e6550028e01c8d42203a384.png\" alt=\"text{F}^-(aq)+text{H}_2text{O}(l) rightleftarrows text{HF}(aq)+text{OH}^-(aq)\" width=\"307\" height=\"18\" \/><\/p>\n<p id=\"x-ck12-Yjc2OWJlMDRjOTYwMTRlZWI1Y2MwYjdkODZkZWM4YjY.-l3x\">The pH of the resulting solution can be determined if 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\/19212456\/c7189caed27d0f9b02b2f75baf0c8c3f.png\" alt=\"K_b\" width=\"21\" height=\"15\" \/> of the fluoride ion is known.<\/p>\n<p id=\"x-ck12-MzU5ZjM4NGRjNjg0N2RlYzY3YzMwNjliMjQyZWZkZmQ.-agc\">20.0 g of sodium fluoride is dissolve in enough water to make 500.0 mL of solution. Calculate the pH of the solution. The\u00a0 <img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM3ODMyMDgzMzMwNg..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212456\/c7189caed27d0f9b02b2f75baf0c8c3f.png\" alt=\"K_b\" width=\"21\" height=\"15\" \/> of the fluoride ion is 1.4 \u00d7 10 <sup> \u221211 <\/sup> .<\/p>\n<p id=\"x-ck12-ODFjNTAwNDljM2RhMjljYTgyMTUxYWY2Nzk5ZWFhNGI.-tob\"><em> Step 1: List the known values and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-ilh\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-NmEyMzdmYjA2YTY5MDQ5MDBjZjdkNDQ3NTdjN2YyMzM.-hik\">\n<li>mass NaF = 20.0 g<\/li>\n<li>molar mass NaF = 41.99 g\/mol<\/li>\n<li>volume solution = 0.500 L<\/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\/19212456\/c7189caed27d0f9b02b2f75baf0c8c3f.png\" alt=\"K_b\" width=\"21\" height=\"15\" \/> of F <sup> &#8211; <\/sup> = 1.4 \u00d7 10 <sup> \u221211 <\/sup><\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-zdg\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-YmEzMGRhNjg2OGQ5ZTRkZTY5ZjkzM2NiYjIyNWE2ODI.-ct9\">\n<li>pH of solution = ?<\/li>\n<\/ul>\n<p id=\"x-ck12-MzQ4OTUyYjJmNzVlZjAzZTYyYTI2MzNjZGIzZjQwZmQ.-cl5\">The molarity of the F <sup> \u2212 <\/sup> solution can be calculated from the mass, molar mass, and solution volume. Since NaF completely dissociates, the molarity of the NaF is equal to the molarity of the F <sup> \u2212 <\/sup> ion. An ICE <strong> Table <\/strong> (below) can be used to calculate the concentration of OH <sup> \u2212 <\/sup> produced and then the pH of the solution.<\/p>\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-5ww\"><em> Step 2: Solve. <\/em><\/p>\n<p id=\"x-ck12-tir\"><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\/19213410\/958271f1c98955aa54278fb5548f20ba.png\" alt=\"20.0 cancel{text{g NaF}} times frac{1 cancel{text{mol NaF}}}{41.99 cancel{text{g NaF}}}times frac{1 text{mol F}^-}{1 cancel{text{mol NaF}}}&amp;=0.476 text{mol F}^- \\frac{0.476 text{mol F}^-}{0.5000 text{L}}&amp;=0.953 text{M F}^-\" width=\"444\" height=\"88\" \/><\/p>\n<p id=\"x-ck12-NzRhOTBlM2U0ZTVkZTdlNDNmMzIzNmRjNjdmZWY3ZGQ.-hq0\"><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\/19213411\/2f420f2e6d32ddffbafd06ccca8f13e2.png\" alt=\"text{Hydrolysis equation:} qquad text{F}^-(aq)+text{H}_2text{O}(l)rightleftarrows text{HF}(aq)+text{OH}^-(aq)\" width=\"503\" height=\"18\" \/><\/p>\n<table id=\"x-ck12-YTk0NGUzZmRhNjA4MGYzNDcyZTBmOTJmZGJhYTI4M2M.-07i\" class=\"x-ck12-nofloat\">\n<tbody>\n<tr>\n<td><strong> Concentrations <\/strong><\/td>\n<td><strong> [F <sup> &#8211; <\/sup> ] <\/strong><\/td>\n<td><strong> [HF] <\/strong><\/td>\n<td><strong> [OH <sup> &#8211; <\/sup> ] <\/strong><\/td>\n<\/tr>\n<tr>\n<td>Initial<\/td>\n<td>0.953<\/td>\n<td>0<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<td>Change<\/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\/19213315\/128f156bd2177fdb41cca16122d32a69.png\" alt=\"-x\" width=\"23\" 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\/19213315\/ded5a4d8ae585bda875e454ef5fce93d.png\" alt=\"+x\" width=\"24\" 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\/19213315\/ded5a4d8ae585bda875e454ef5fce93d.png\" alt=\"+x\" width=\"24\" height=\"12\" \/><\/td>\n<\/tr>\n<tr>\n<td>Equilibrium<\/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\/19213413\/1e09476a65467e6df12973e2b4b2a522.png\" alt=\"0.953 - x\" width=\"73\" height=\"13\" \/><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM3ODMyMDgzMzMwNw..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212123\/0f14991bebd0bc5ca2e82990e029d525.png\" alt=\"x\" width=\"10\" 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\/19212123\/0f14991bebd0bc5ca2e82990e029d525.png\" alt=\"x\" width=\"10\" height=\"8\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"x-ck12-yv7\"><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\/19213413\/b5d9006fd65c3c7d70daf435f823dc97.png\" alt=\"K_b &amp;=1.4times 10^{-11}=frac{(x)(x)}{0.953-x}=frac{x^2}{0.953-x}approx frac{x^2}{0.953} \\x &amp;=[text{OH}^-]=sqrt{1.4times 10^{-11}(0.953)}=3.65times 10^{-6} text{M} \\text{pOH} &amp;=-log (3.65times 10^{-6})=5.44 \\text{pH} &amp;=14-5.44=8.56\" width=\"428\" height=\"123\" \/><\/p>\n<p id=\"x-ck12-MGZhMzc5YjY2OWU0YzY0ZWM4YjYzZWNhYzc4OTQ3ZDA.-wqr\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-N2NmMTkzYjViMjEwZGI3ZTQwMzYxNzYzMThiYjNjZDk.-v9z\">The solution is slightly basic due to the hydrolysis of the fluoride ion.<\/p>\n<\/div>\n<h3>Salts That Form Acidic Solutions<\/h3>\n<p id=\"x-ck12-Y2U0NTE3Nzk4NmQ4ZDU4YjQxMjJkMzczMjQ3Njc3OGQ.-h5w\">When the ammonium ion dissolves in water, the following equilibrium exists:<\/p>\n<p id=\"x-ck12-h9q\"><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\/19213415\/2ac7e4485859514b1cb6d6d34726e4b5.png\" alt=\"text{NH}_4^+(aq)+text{H}_2text{O}(l)rightleftarrows text{H}_3text{O}^+(aq)+text{NH}_3(aq)\" width=\"339\" height=\"21\" \/><\/p>\n<p id=\"x-ck12-MmMwNGM2YWY3YWI4ZjRmMDkwNDZjNWQxNjJmOTIzYzU.-btc\">The production of hydronium ions causes the resulting solution to be acidic. The pH of a solution of ammonium chloride can be found in a very similar way to the sodium fluoride solution in Sample Problem 21.7. However, since the ammonium chloride is acting as an acid, it is necessary to know 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\/19213416\/cbd56ced35c37836d1a05027a2d41284.png\" alt=\"K_a\" width=\"22\" height=\"15\" \/> of NH <sub> 4 <\/sub><sup> + <\/sup> , which is 5.6 \u00d7 10 <sup> \u221210 <\/sup> . We will find the pH of a 2.00 M solution of NH <sub> 4 <\/sub> Cl. Because the NH <sub> 4 <\/sub> Cl completely ionizes, the concentration of the ammonium ion is 2.00 M.<\/p>\n<p id=\"x-ck12-l9j\"><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\/19213417\/f98b549229c60c1e4ecaa78908f3d6a8.png\" alt=\"text{NH}_4text{Cl}(s)rightarrow text{NH}_4^+(aq)+text{Cl} ^-(aq)\" width=\"254\" height=\"21\" \/><\/p>\n<p id=\"x-ck12-NWRmN2M0NjAzMjljZjg4ZDBiMTVjMThlMDg5MGNkYzg.-xqu\">Again, an ICE <strong> Table <\/strong> (below) is set up in order to solve for the concentration of the hydronium (or H <sup> + <\/sup> ) ion produced.<\/p>\n<table id=\"x-ck12-MTcyNmQxNGE5ZDQwYmE0M2ZlYWFkZGQ1ZDYxNzlkYzU.-ioq\" class=\"x-ck12-nofloat\">\n<tbody>\n<tr>\n<td><strong> Concentrations <\/strong><\/td>\n<td><strong> [NH <sub> 4 <\/sub><sup> + <\/sup> ] <\/strong><\/td>\n<td><strong> [H <sup> + <\/sup> ] <\/strong><\/td>\n<td><strong> [NH <sub> 3 <\/sub> ] <\/strong><\/td>\n<\/tr>\n<tr>\n<td>Initial<\/td>\n<td>2.00<\/td>\n<td>0<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<td>Change<\/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\/19213315\/128f156bd2177fdb41cca16122d32a69.png\" alt=\"-x\" width=\"23\" 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\/19213315\/ded5a4d8ae585bda875e454ef5fce93d.png\" alt=\"+x\" width=\"24\" 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\/19213315\/ded5a4d8ae585bda875e454ef5fce93d.png\" alt=\"+x\" width=\"24\" height=\"12\" \/><\/td>\n<\/tr>\n<tr>\n<td>Equilibrium<\/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\/19213418\/cc5e2c32d5fee93312254fc0c1cf64c6.png\" alt=\"2.00 - x\" width=\"64\" 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\/19212123\/0f14991bebd0bc5ca2e82990e029d525.png\" alt=\"x\" width=\"10\" height=\"8\" \/><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM3ODMyMDgzMzMwOA..\" class=\"x-ck12-math\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/53\/2014\/08\/19212123\/0f14991bebd0bc5ca2e82990e029d525.png\" alt=\"x\" width=\"10\" height=\"8\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"x-ck12-OTc4MTEzOTlkZDFlYzk5ODkyNjFhOTU0NWI0ZWUxYjU.-slo\">Now substituting into 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\/19213301\/23b35b1534be8d9bb64ff42a83c35fd9.png\" alt=\"K_a\" width=\"22\" height=\"15\" \/> expression gives:<\/p>\n<p id=\"x-ck12-dya\"><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\/19213418\/0b26f10589ea4b56af8dce6e78f6ae45.png\" alt=\"K_a &amp;=5.6 times 10^{-10}=frac{x^2}{2.00-x} approx frac{x^2}{2.00} \\x &amp;=[text{H}^+]=sqrt{5.6 times 10^{-10}(2.00)}=3.3 times 10^{-5} text{M} \\text{pH} &amp;=- log (3.3 times 10^{-5})=4.48\" width=\"382\" height=\"97\" \/><\/p>\n<p id=\"x-ck12-ZDg4NmViOTBiZWVjODA2YTIzMzA5M2ZjNzc2ZTc3MmE.-42x\">A salt produced from a strong acid and a weak base yields a solution that is acidic.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Summary<\/h3>\n<ul id=\"x-ck12-N2QzM2RkMzU3MzFhY2U1MWM1MzM5YjEyZTU1ZjI1NmQ.-rip\">\n<li>Calculations to determine pH of salt solutions are described.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Practice<\/h3>\n<p id=\"x-ck12-ZGIyN2QxMzI1ODI1MWRlNWVlMjE5YTlhNmM5NDJlN2M.-bcy\">Work the problems at the link below:<\/p>\n<p id=\"x-ck12-Y2VhY2RlNjExZjM0ZGU1YmQ3YWNjYmI2MzJhNmNkMmE.-qke\"><a href=\"http:\/\/www.sparknotes.com\/chemistry\/acidsbases\/phcalc\/problems.html\"> http:\/\/www.sparknotes.com\/chemistry\/acidsbases\/phcalc\/problems.html<\/a><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Review<\/h3>\n<ol id=\"x-ck12-Y2MzZGRhNjAyMmFiZGRlZDgzNGRjZDNlM2I4YzgwMjc.-iot\">\n<li>In the first example, how do we know that we can ignore\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\/19212123\/0f14991bebd0bc5ca2e82990e029d525.png\" alt=\"x\" width=\"10\" height=\"8\" \/> when determining [F <sup> &#8211; <\/sup> ]?<\/li>\n<li>In example two, how do we know the ammonium ion concentration?<\/li>\n<li>Could we write the equilibrium in example two as <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\/19213419\/7d3ca23d4749ad242fe6fa6397f2e824.png\" alt=\"text{NH}_4^+ rightleftarrows text{H}^+ + text{NH}_3\" width=\"145\" height=\"21\" \/> ?<\/li>\n<\/ol>\n<\/div>\n<p>&nbsp;<\/p>\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-2826\">\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 class=\"license-attribution-dropdown-subheading\">Public domain content<\/div><ul class=\"citation-list\"><li>Fessenden School. <strong>Authored by<\/strong>: Daderot. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Pool_2,_Fessenden_School_-_IMG_0262.JPG\">https:\/\/commons.wikimedia.org\/wiki\/File:Pool_2,_Fessenden_School_-_IMG_0262.JPG<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/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":16,"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\":\"pd\",\"description\":\"Fessenden 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