{"id":2821,"date":"2016-08-24T19:27:17","date_gmt":"2016-08-24T19:27:17","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/?post_type=chapter&#038;p=2821"},"modified":"2016-08-24T22:39:42","modified_gmt":"2016-08-24T22:39:42","slug":"calculating-ph-of-weak-acid-and-base-solutions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/chapter\/calculating-ph-of-weak-acid-and-base-solutions\/","title":{"raw":"Calculating pH of Weak Acid and Base Solutions","rendered":"Calculating pH of Weak Acid and Base Solutions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Perform calculations to determine the pH of a weak acid or base solution.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Ouch, that hurts!<\/h3>\r\n<p id=\"x-ck12-NzNjM2VlY2Y2NTE3YTg5MDE3NjI4ZDZlN2I2Y2E5MWQ.-srq\"><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\/19213313\/20140811155916233996.jpeg\" alt=\"Sodium bicarbonate can be used to help alleviate a bee sting\" width=\"277\" height=\"208\" \/><\/span><\/p>\r\n<p id=\"x-ck12-NjRlNzY2ZmZlYTlmZGNiYTdjYzE4MDlhNjg0ODQzMDI.-jo3\">Bees are beautiful creatures that help plants flourish. They carry pollen from one plant to another to facilitate plant growth and development. But, they can also be troublesome when they sting you. For people who are allergic to bee venom, this can be a serious, life-threatening problem. For the rest of us, it can be a painful experience.<\/p>\r\nWhen stung by a bee, one first-aid treatment is to apply a paste of baking soda (sodium bicarbonate) to the stung area. This weak base helps with the itching and swelling that accompanies the bee sting.\r\n\r\n<\/div>\r\n<h2>Calculating pH of Weak Acid and Base Solutions<\/h2>\r\n<p id=\"x-ck12-ZTJhOWMzN2I5NmQ2ZWJkNGZiMTg3MjMyY2IyOGFhYWE.-dzu\">The\u00a0<em>K<sub>a<\/sub><\/em>\u00a0and\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\" \/> values have been determined for a great many acids and bases, as shown in Tables 21.5 and 21.6. These can be used to calculate the pH of any solution of a weak acid or base whose ionization constant is known.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3>Sample Problem: Calculating the pH of a Weak Acid<\/h3>\r\n<p id=\"x-ck12-ODdjMjVlMDkyNTBmOGQzOTYyYzNmNzJmOWI4NjgyYzM.-riv\">Calculate the pH of a 2.00 M solution of nitrous acid (HNO <sub> 2 <\/sub> ). The\u00a0<em>K<sub>a<\/sub><\/em>\u00a0for nitrous acid is 4.5\u00a0\u00d7\u00a010<sup>\u22124<\/sup> .<\/p>\r\n<p id=\"x-ck12-OGM1ZTljNzdiMjc4MjQyMjY3OTljNDI4OTk2ZmY0MTE.-4es\"><em> Step 1: List the known values and plan the problem. <\/em><\/p>\r\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-7px\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-YjIzM2EwZjk5NTczNjc3ZDdlZjMxZGY5ZmVkMzUxM2I.-mnw\">\r\n \t<li>initial [HNO <sub> 2 <\/sub> ] = 2.00 M<\/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\/19213315\/fc75365c014920350634a31a8604306d.png\" alt=\"K_a=4.5 times 10^{-4}\" width=\"127\" height=\"18\" \/><\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-f0n\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\r\n\r\n<ul id=\"x-ck12-ZWJkMWRiNDFkOWViNThmZjlhYWM0ZDUzNDhmYjhjMWU.-fzi\">\r\n \t<li>pH = ?<\/li>\r\n<\/ul>\r\n<p id=\"x-ck12-MGJhMWY2N2M2ODZmM2RiYWMxMTYyNjE4OGFlYzdiMTU.-y3d\">First, an ICE table is set up with the variable\u00a0<em>x<\/em>\u00a0used to signify the change in concentration of the substance due to ionization of the acid. Then the\u00a0<em>K<sub>a<\/sub><\/em>\u00a0expression is used to solve for\u00a0<em>x<\/em>\u00a0and calculate the pH.<\/p>\r\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-dbm\"><em> Step 2: Solve. <\/em><\/p>\r\n\r\n<table id=\"x-ck12-ODYxNzY0ZDc1OWQ4MWViNDgwYmQ5MjkxN2Q5MjZkMzM.-qhp\" class=\"x-ck12-nofloat\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<td><strong> Concentrations <\/strong><\/td>\r\n<td><strong> [HNO <sub> 2 <\/sub> ] <\/strong><\/td>\r\n<td><strong> [H <sup> + <\/sup> ] <\/strong><\/td>\r\n<td><strong> [NO <sub> 2 <\/sub><sup> \u2212 <\/sup> ] <\/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\/19213316\/3e351a4be6827d9c67edb5fba73f1f64.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 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-YjgyZjFjOTFlYTc0YmI2MzQ2NTZjNzFhMGYzNjJhNmQ.-zue\">The\u00a0<em>K<sub>a<\/sub><\/em>\u00a0expression and value is used to set up an equation to solve for<em>x<\/em>\u00a0.<\/p>\r\n<p id=\"x-ck12-ZTZiOTJiYThkYTgxNTA0MjRiYTY2MGFkNmJjZTkxMDM.-y3x\"><img id=\"x-ck12-MTM2ODAxMzk2NTI3Nw..\" 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\/19213316\/a6cafb0d1e8ed0adf3312ccd0f9bbe76.png\" alt=\"K_a=4.5 times 10^{-4}=frac{(x)(x)}{2.00 - x}=frac{x^2}{2.00 - x}\" width=\"310\" height=\"41\" \/><\/p>\r\n<p id=\"x-ck12-ZTZiOTJiYThkYTgxNTA0MjRiYTY2MGFkNmJjZTkxMDM.-fho\">The quadratic equation is required to solve this equation for <em>x<\/em>. However, a simplification can be made because of the fact that the extent of ionization of weak acids is small. The value of\u00a0<em>x<\/em>\u00a0will be significantly less than 2.00, so the\u00a0\u2212<em>x<\/em>\u00a0in the denominator can be dropped.<\/p>\r\n<p id=\"x-ck12-NzZmZjEyOTczMTljYTkyMzZkY2I4MzA4MTIyY2MxNGQ.-0mk\"><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\/19213317\/979ad15c13f7eaa966de95b86cc83170.png\" alt=\"4.5 times 10^{-4}&amp;=frac{x^2}{2.00 - x} approx frac{x^2}{2.00} \\x&amp;=sqrt{4.5 times 10^{-4}(2.00)}=2.9 times 10^{-2} text{M}= left [ H^+ right ]\" width=\"438\" height=\"72\" \/><\/p>\r\n<p id=\"x-ck12-NzZmZjEyOTczMTljYTkyMzZkY2I4MzA4MTIyY2MxNGQ.-s1q\">Since the variable\u00a0<em>x<\/em>\u00a0represents the hydrogen-ion concentration, the pH of the solution can now be calculated.<\/p>\r\n<p id=\"x-ck12-oi9\"><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\/19213318\/e621b682ea3a0310abb4872faab35c32.png\" alt=\"pH=- log[text{H}^+]=- log[2.9 times 10^{-2}]=1.54\" width=\"336\" height=\"22\" \/><\/p>\r\n<p id=\"x-ck12-ZGJmYWE4NWM0YmIyZGJiOTdhZTZmY2NmYmJmYjM1MjI.-2e3\"><em> Step 3: Think about your result. <\/em><\/p>\r\n<p id=\"x-ck12-NGRlMjg3MDA0MTMxZDY3OTViYWJkOTAwYmYxMjQ3NjY.-rpu\">The pH of a 2.00 M solution of a strong acid would be equal to \u2212log (2.00) = \u22120.30\u00a0. The higher pH of the 2.00 M nitrous acid is consistent with it being a weak acid and therefore not as acidic as a strong acid would be.<\/p>\r\n<p id=\"x-ck12-MTQ5ZjhhZTJkNTc1ZjRkMTk2NzQ5YzE5OGIzYjZlNWI.-alg\">The procedure for calculating the pH of a solution of a weak base is similar to that of the weak acid in the sample problem. However, the variable\u00a0<em>x<\/em>\u00a0will represent the concentration of the hydroxide ion. The pH is found by taking the negative logarithm to get the pOH, followed by subtracting from 14 to get the pH.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Summary<\/h3>\r\n<ul id=\"x-ck12-ZDljNTlhYWYxZGMxZjJiMmEyMmE4ZjZhNWFmODM1ZjA.-mgh\">\r\n \t<li>The procedure for calculating the pH of a weak acid or base is illustrated.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Practice<\/h3>\r\n<p id=\"x-ck12-ZWRkOGNjYmQ4ZTQ1YmVkY2JkNmEyNDlkNDcwM2FjNjg.-19s\">Perform the calculations at the site below:<\/p>\r\n<p id=\"x-ck12-YmIyYjJlZTcxOTE2ZjNjNTAyYjJhOTM4NWI1NjlmZmI.-kto\"><a href=\"http:\/\/www.sciencegeek.net\/APchemistry\/APtaters\/pHcalculations.htm\"> http:\/\/www.sciencegeek.net\/APchemistry\/APtaters\/pHcalculations.htm<\/a><\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Review<\/h3>\r\n<ol id=\"x-ck12-YTY5NTFhMDUzZWJlYTlmMDA5MDY2NGJhZjM5MDZlMGE.-m5x\">\r\n \t<li>What does\u00a0<em>x<\/em>\u00a0stand for in the equation?<\/li>\r\n \t<li>What simplifying assumption is made?<\/li>\r\n \t<li>What would\u00a0<em>x<\/em>\u00a0stand for if we were calculating pOH?<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"836080\"]Show References[\/reveal-answer]\r\n[hidden-answer a=\"836080\"]\r\n<h2>References<\/h2>\r\n<ol>\r\n \t<li>Jon Sullivan. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Bees_pollenating_basil.jpg\">http:\/\/commons.wikimedia.org\/wiki\/File:Bees_pollenating_basil.jpg <\/a>.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Perform calculations to determine the pH of a weak acid or base solution.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Ouch, that hurts!<\/h3>\n<p id=\"x-ck12-NzNjM2VlY2Y2NTE3YTg5MDE3NjI4ZDZlN2I2Y2E5MWQ.-srq\"><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\/19213313\/20140811155916233996.jpeg\" alt=\"Sodium bicarbonate can be used to help alleviate a bee sting\" width=\"277\" height=\"208\" \/><\/span><\/p>\n<p id=\"x-ck12-NjRlNzY2ZmZlYTlmZGNiYTdjYzE4MDlhNjg0ODQzMDI.-jo3\">Bees are beautiful creatures that help plants flourish. They carry pollen from one plant to another to facilitate plant growth and development. But, they can also be troublesome when they sting you. For people who are allergic to bee venom, this can be a serious, life-threatening problem. For the rest of us, it can be a painful experience.<\/p>\n<p>When stung by a bee, one first-aid treatment is to apply a paste of baking soda (sodium bicarbonate) to the stung area. This weak base helps with the itching and swelling that accompanies the bee sting.<\/p>\n<\/div>\n<h2>Calculating pH of Weak Acid and Base Solutions<\/h2>\n<p id=\"x-ck12-ZTJhOWMzN2I5NmQ2ZWJkNGZiMTg3MjMyY2IyOGFhYWE.-dzu\">The\u00a0<em>K<sub>a<\/sub><\/em>\u00a0and\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\" \/> values have been determined for a great many acids and bases, as shown in Tables 21.5 and 21.6. These can be used to calculate the pH of any solution of a weak acid or base whose ionization constant is known.<\/p>\n<div class=\"textbox shaded\">\n<h3>Sample Problem: Calculating the pH of a Weak Acid<\/h3>\n<p id=\"x-ck12-ODdjMjVlMDkyNTBmOGQzOTYyYzNmNzJmOWI4NjgyYzM.-riv\">Calculate the pH of a 2.00 M solution of nitrous acid (HNO <sub> 2 <\/sub> ). The\u00a0<em>K<sub>a<\/sub><\/em>\u00a0for nitrous acid is 4.5\u00a0\u00d7\u00a010<sup>\u22124<\/sup> .<\/p>\n<p id=\"x-ck12-OGM1ZTljNzdiMjc4MjQyMjY3OTljNDI4OTk2ZmY0MTE.-4es\"><em> Step 1: List the known values and plan the problem. <\/em><\/p>\n<p id=\"x-ck12-MmY1NjUwZTQ4NGZhMTk1OTQ5YWM2YzhkMTFkY2E0ZmQ.-7px\"><span class=\"x-ck12-underline\"> Known <\/span><\/p>\n<ul id=\"x-ck12-YjIzM2EwZjk5NTczNjc3ZDdlZjMxZGY5ZmVkMzUxM2I.-mnw\">\n<li>initial [HNO <sub> 2 <\/sub> ] = 2.00 M<\/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\/19213315\/fc75365c014920350634a31a8604306d.png\" alt=\"K_a=4.5 times 10^{-4}\" width=\"127\" height=\"18\" \/><\/li>\n<\/ul>\n<p id=\"x-ck12-ODgxODNiOTQ2Y2M1ZjBlOGM5NmIyZTY2ZTFjNzRhN2U.-f0n\"><span class=\"x-ck12-underline\"> Unknown <\/span><\/p>\n<ul id=\"x-ck12-ZWJkMWRiNDFkOWViNThmZjlhYWM0ZDUzNDhmYjhjMWU.-fzi\">\n<li>pH = ?<\/li>\n<\/ul>\n<p id=\"x-ck12-MGJhMWY2N2M2ODZmM2RiYWMxMTYyNjE4OGFlYzdiMTU.-y3d\">First, an ICE table is set up with the variable\u00a0<em>x<\/em>\u00a0used to signify the change in concentration of the substance due to ionization of the acid. Then the\u00a0<em>K<sub>a<\/sub><\/em>\u00a0expression is used to solve for\u00a0<em>x<\/em>\u00a0and calculate the pH.<\/p>\n<p id=\"x-ck12-ZmYxNjkxNzA3ODcxNDhkYjhmZGUyOTFlNDk1NmEzZjE.-dbm\"><em> Step 2: Solve. <\/em><\/p>\n<table id=\"x-ck12-ODYxNzY0ZDc1OWQ4MWViNDgwYmQ5MjkxN2Q5MjZkMzM.-qhp\" class=\"x-ck12-nofloat\">\n<tbody>\n<tr>\n<td><strong> Concentrations <\/strong><\/td>\n<td><strong> [HNO <sub> 2 <\/sub> ] <\/strong><\/td>\n<td><strong> [H <sup> + <\/sup> ] <\/strong><\/td>\n<td><strong> [NO <sub> 2 <\/sub><sup> \u2212 <\/sup> ] <\/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\/19213316\/3e351a4be6827d9c67edb5fba73f1f64.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\" 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-YjgyZjFjOTFlYTc0YmI2MzQ2NTZjNzFhMGYzNjJhNmQ.-zue\">The\u00a0<em>K<sub>a<\/sub><\/em>\u00a0expression and value is used to set up an equation to solve for<em>x<\/em>\u00a0.<\/p>\n<p id=\"x-ck12-ZTZiOTJiYThkYTgxNTA0MjRiYTY2MGFkNmJjZTkxMDM.-y3x\"><img loading=\"lazy\" decoding=\"async\" id=\"x-ck12-MTM2ODAxMzk2NTI3Nw..\" 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\/19213316\/a6cafb0d1e8ed0adf3312ccd0f9bbe76.png\" alt=\"K_a=4.5 times 10^{-4}=frac{(x)(x)}{2.00 - x}=frac{x^2}{2.00 - x}\" width=\"310\" height=\"41\" \/><\/p>\n<p id=\"x-ck12-ZTZiOTJiYThkYTgxNTA0MjRiYTY2MGFkNmJjZTkxMDM.-fho\">The quadratic equation is required to solve this equation for <em>x<\/em>. However, a simplification can be made because of the fact that the extent of ionization of weak acids is small. The value of\u00a0<em>x<\/em>\u00a0will be significantly less than 2.00, so the\u00a0\u2212<em>x<\/em>\u00a0in the denominator can be dropped.<\/p>\n<p id=\"x-ck12-NzZmZjEyOTczMTljYTkyMzZkY2I4MzA4MTIyY2MxNGQ.-0mk\"><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\/19213317\/979ad15c13f7eaa966de95b86cc83170.png\" alt=\"4.5 times 10^{-4}&amp;=frac{x^2}{2.00 - x} approx frac{x^2}{2.00} \\x&amp;=sqrt{4.5 times 10^{-4}(2.00)}=2.9 times 10^{-2} text{M}= left [ H^+ right ]\" width=\"438\" height=\"72\" \/><\/p>\n<p id=\"x-ck12-NzZmZjEyOTczMTljYTkyMzZkY2I4MzA4MTIyY2MxNGQ.-s1q\">Since the variable\u00a0<em>x<\/em>\u00a0represents the hydrogen-ion concentration, the pH of the solution can now be calculated.<\/p>\n<p id=\"x-ck12-oi9\"><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\/19213318\/e621b682ea3a0310abb4872faab35c32.png\" alt=\"pH=- log[text{H}^+]=- log[2.9 times 10^{-2}]=1.54\" width=\"336\" height=\"22\" \/><\/p>\n<p id=\"x-ck12-ZGJmYWE4NWM0YmIyZGJiOTdhZTZmY2NmYmJmYjM1MjI.-2e3\"><em> Step 3: Think about your result. <\/em><\/p>\n<p id=\"x-ck12-NGRlMjg3MDA0MTMxZDY3OTViYWJkOTAwYmYxMjQ3NjY.-rpu\">The pH of a 2.00 M solution of a strong acid would be equal to \u2212log (2.00) = \u22120.30\u00a0. The higher pH of the 2.00 M nitrous acid is consistent with it being a weak acid and therefore not as acidic as a strong acid would be.<\/p>\n<p id=\"x-ck12-MTQ5ZjhhZTJkNTc1ZjRkMTk2NzQ5YzE5OGIzYjZlNWI.-alg\">The procedure for calculating the pH of a solution of a weak base is similar to that of the weak acid in the sample problem. However, the variable\u00a0<em>x<\/em>\u00a0will represent the concentration of the hydroxide ion. The pH is found by taking the negative logarithm to get the pOH, followed by subtracting from 14 to get the pH.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Summary<\/h3>\n<ul id=\"x-ck12-ZDljNTlhYWYxZGMxZjJiMmEyMmE4ZjZhNWFmODM1ZjA.-mgh\">\n<li>The procedure for calculating the pH of a weak acid or base is illustrated.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Practice<\/h3>\n<p id=\"x-ck12-ZWRkOGNjYmQ4ZTQ1YmVkY2JkNmEyNDlkNDcwM2FjNjg.-19s\">Perform the calculations at the site below:<\/p>\n<p id=\"x-ck12-YmIyYjJlZTcxOTE2ZjNjNTAyYjJhOTM4NWI1NjlmZmI.-kto\"><a href=\"http:\/\/www.sciencegeek.net\/APchemistry\/APtaters\/pHcalculations.htm\"> http:\/\/www.sciencegeek.net\/APchemistry\/APtaters\/pHcalculations.htm<\/a><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Review<\/h3>\n<ol id=\"x-ck12-YTY5NTFhMDUzZWJlYTlmMDA5MDY2NGJhZjM5MDZlMGE.-m5x\">\n<li>What does\u00a0<em>x<\/em>\u00a0stand for in the equation?<\/li>\n<li>What simplifying assumption is made?<\/li>\n<li>What would\u00a0<em>x<\/em>\u00a0stand for if we were calculating pOH?<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q836080\">Show References<\/span><\/p>\n<div id=\"q836080\" class=\"hidden-answer\" style=\"display: none\">\n<h2>References<\/h2>\n<ol>\n<li>Jon Sullivan. <a href=\"http:\/\/commons.wikimedia.org\/wiki\/File:Bees_pollenating_basil.jpg\">http:\/\/commons.wikimedia.org\/wiki\/File:Bees_pollenating_basil.jpg <\/a>.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2821\">\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":11,"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-2821","chapter","type-chapter","status-publish","hentry"],"part":2342,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2821","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\/2821\/revisions"}],"predecessor-version":[{"id":3042,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2821\/revisions\/3042"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/parts\/2342"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapters\/2821\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/media?parent=2821"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/pressbooks\/v2\/chapter-type?post=2821"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/contributor?post=2821"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/umes-cheminter\/wp-json\/wp\/v2\/license?post=2821"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}