{"id":4902,"date":"2015-08-24T22:57:16","date_gmt":"2015-08-24T22:57:16","guid":{"rendered":"https:\/\/courses.candelalearning.com\/chemistryformajorsxmaster\/?post_type=chapter&#038;p=4902"},"modified":"2016-08-09T18:40:29","modified_gmt":"2016-08-09T18:40:29","slug":"ph-and-poh","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/chapter\/ph-and-poh\/","title":{"raw":"pH and pOH","rendered":"pH and pOH"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>LEARNING OBJECTIVES<\/h3>\r\nBy the end of this module, you will be able to:\r\n<ul>\r\n \t<li>Explain the characterization of aqueous solutions as acidic, basic, or neutral<\/li>\r\n \t<li>Express hydronium and hydroxide ion concentrations on the pH and pOH scales<\/li>\r\n \t<li>Perform calculations relating pH and pOH<\/li>\r\n<\/ul>\r\n<\/div>\r\nAs discussed earlier, hydronium and hydroxide ions are present both in pure water and in all aqueous solutions, and their concentrations are inversely proportional as determined by the ion product of water (<em data-effect=\"italics\">K<\/em><sub>w<\/sub>). The concentrations of these ions in a solution are often critical determinants of the solution\u2019s properties and the chemical behaviors of its other solutes, and specific vocabulary has been developed to describe these concentrations in relative terms. A solution is <strong>neutral<\/strong> if it contains equal concentrations of hydronium and hydroxide ions; <strong>acidic<\/strong> if it contains a greater concentration of hydronium ions than hydroxide ions; and <strong>basic<\/strong> if it contains a lesser concentration of hydronium ions than hydroxide ions.\r\n\r\nA common means of expressing quantities, the values of which may span many orders of magnitude, is to use a logarithmic scale. One such scale that is very popular for chemical concentrations and equilibrium constants is based on the p-function, defined as shown where \u201cX\u201d is the quantity of interest and \u201clog\u201d is the base-10 logarithm:\r\n<p style=\"text-align: center;\">pX = \u2212log X<\/p>\r\nThe pH of a solution is therefore defined as shown here, where [H<sub>3<\/sub>O<sup>+<\/sup>] is the molar concentration of hydronium ion in the solution:\r\n<p style=\"text-align: center;\">pH = \u2212log[H<sub>3<\/sub>O<sup>+<\/sup>]<\/p>\r\nRearranging this equation to isolate the hydronium ion molarity yields the equivalent expression:\r\n<p style=\"text-align: center;\">[H<sub>3<\/sub>O<sup>+<\/sup>] = 10<sup>\u2212pH<\/sup><\/p>\r\nLikewise, the hydroxide ion molarity may be expressed as a p-function, or <b>pOH<\/b>:\r\n<p style=\"text-align: center;\">pOH = \u2212log[OH<sup>\u2212<\/sup>]<\/p>\r\n<p style=\"text-align: center;\">or<\/p>\r\n<p style=\"text-align: center;\">[OH<sup>\u2212<\/sup>] = 10<sup>\u2212pOH<\/sup><\/p>\r\nFinally, the relation between these two ion concentration expressed as p-functions is easily derived from the\u00a0<em data-effect=\"italics\">K<\/em><sub>w<\/sub> expression:\r\n<p style=\"text-align: center;\"><em data-effect=\"italics\">K<\/em><sub>w\u00a0<\/sub>= [H<sub>3<\/sub>O<sup>+<\/sup>][OH<sup>\u2212<\/sup>]\r\n\u2212log<em data-effect=\"italics\">K<\/em><sub>w<\/sub> = \u2212log([H<sub>3<\/sub>O<sup>+<\/sup>][OH<sup>\u2212<\/sup>]) = \u2212log[H<sub>3<\/sub>O<sup>+<\/sup>] + \u2212log[OH<sup>\u2212<\/sup>]\r\np<em data-effect=\"italics\">K<\/em><sub>w\u00a0<\/sub>= pH + pOH<\/p>\r\nAt 25 \u00b0C, the value of\u00a0<em data-effect=\"italics\">K<\/em><sub>w<\/sub>\u00a0is 1.0 \u00d7 10\u221214, and so:\r\n<p style=\"text-align: center;\">14.00 = pH +pOH<\/p>\r\nAs we learned earlier, the hydronium ion molarity in pure water (or any neutral solution) is 1.0 \u00d7 10<sup>\u22127<\/sup> <em data-effect=\"italics\">M<\/em>\u00a0at 25 \u00b0C. The pH and pOH of a neutral solution at this temperature are therefore:\r\n<p style=\"text-align: center;\">pH = \u2212log[H<sub>3<\/sub>O<sup>+<\/sup>] = \u2212log(1.0 \u00d7 10<sup>\u22127<\/sup>) = 7.00<\/p>\r\n<p style=\"text-align: center;\">pOH = \u2212log[OH<sup>\u2212<\/sup>] = \u2212log(1.0 \u00d7 10<sup>\u22127<\/sup>) = 7.00<\/p>\r\nAnd so, <em>at this temperature<\/em>, acidic solutions are those with hydronium ion molarities greater than 1.0 \u00d7 10<sup>\u22127<\/sup> <em data-effect=\"italics\">M<\/em> and hydroxide ion molarities less than 1.0 \u00d7 10<sup>\u22127<\/sup> <em data-effect=\"italics\">M<\/em> (corresponding to pH values less than 7.00 and pOH values greater than 7.00). Basic solutions are those with hydronium ion molarities less than 1.0 \u00d7 10<sup>\u22127<\/sup> <em data-effect=\"italics\">M<\/em> and hydroxide ion molarities greater than 1.0 \u00d7 10<sup>\u22127<\/sup> <em data-effect=\"italics\">M<\/em> (corresponding to pH values greater than 7.00 and pOH values less than 7.00).\r\n\r\nSince the autoionization constant Kw is temperature dependent, these correlations between pH values and the acidic\/neutral\/basic adjectives will be different at temperatures other than 25 \u00b0C. For example, the hydronium molarity of pure water is\u00a080 \u00b0C is 4.9 \u00d7 10<sup>\u22127<\/sup> <em data-effect=\"italics\">M<\/em>, which corresponds to pH and pOH values of:\r\n<p style=\"text-align: center;\">pH = \u2212log[H<sub>3<\/sub>O<sup>+<\/sup>] = \u2212log(4.9 \u00d7 10<sup>\u22127<\/sup>) = 6.31<\/p>\r\n<p style=\"text-align: center;\">pOH = \u2212log[OH<sup>\u2212<\/sup>] = \u2212log(4.9 \u00d7 10<sup>\u22127<\/sup>) = 6.31<\/p>\r\nAt this temperature, then, neutral solutions exhibit pH = pOH = 6.31, acidic solutions exhibit pH less than 6.31 and pOH greater than 6.31, whereas basic solutions exhibit pH greater than 6.31 and pOH less than 6.31. This distinction can be important when studying certain processes that occur at nonstandard temperatures, such as enzyme reactions in warm-blooded organisms. Unless otherwise noted, references to pH values are presumed to be those at standard temperature (25 \u00b0C) (Table 1).\r\n<table>\r\n<thead>\r\n<tr>\r\n<th colspan=\"3\">Table 1.\u00a0Summary of Relations for Acidic, Basic and Neutral Solutions<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Classification<\/th>\r\n<th>\u00a0Relative Ion Concentrations<\/th>\r\n<th>\u00a0pH at 25 \u00b0C<\/th>\r\n<\/tr>\r\n<tr>\r\n<td>acidic<\/td>\r\n<td>\u00a0[H<sub>3<\/sub>O<sup>+<\/sup>] &gt; [OH<sup>\u2212<\/sup>]<\/td>\r\n<td>\u00a0pH &lt; 7<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>neutral<\/td>\r\n<td>\u00a0[H<sub>3<\/sub>O<sup>+<\/sup>] = [OH<sup>\u2212<\/sup>]<\/td>\r\n<td>\u00a0pH = 7<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>basic<\/td>\r\n<td>\u00a0[H<sub>3<\/sub>O<sup>+<\/sup>] &lt; [OH<sup>\u2212<\/sup>]<\/td>\r\n<td>\u00a0pH &gt; 7<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFigure 1 shows the relationships between [H<sub>3<\/sub>O<sup>+<\/sup>], [OH<sup>\u2212<\/sup>], pH, and pOH, and gives values for these properties at standard temperatures for some common substances.\r\n\r\n[caption id=\"attachment_5396\" align=\"aligncenter\" width=\"1024\"]<img class=\"size-large wp-image-5396\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09035250\/CNX_Chem_14_02_phscale-1024x897-1024x897.jpg\" alt=\"A table is provided with 5 columns. The first column is labeled \u201cleft bracket H subscript 3 O superscript plus right bracket (M).\u201d Powers of ten are listed in the column beginning at 10 superscript 1, including 10 superscript 0 or 1, 10 superscript negative 1, decreasing by single powers of 10 to 10 superscript negative 15. The second column is labeled \u201cleft bracket O H superscript negative right bracket (M).\u201d Powers of ten are listed in the column beginning at 10 superscript negative 15, increasing by single powers of 10 to including 10 superscript 0 or 1, and 10 superscript 1. The third column is labeled \u201cp H.\u201d Values listed in this column are integers beginning at negative 1, increasing by ones up to 14. The fourth column is labeled \u201cp O H.\u201d Values in this column are integers beginning at 15, decreasing by ones up to negative 1. The fifth column is labeled \u201cSample Solution.\u201d A vertical line at the left of the column has tick marks corresponding to each p H level in the table. Substances are listed next to this line segment with line segments connecting them to the line to show approximate p H and p O H values. 1 M H C l is listed at a p H of 0. Gastric juices are listed at a p H of about 1.5. Lime juice is listed at a p H of about 2, followed by 1 M C H subscript 3 C O subscript 2 H, followed by stomach acid at a p H value of nearly 3. Wine is listed around 3.5. Coffee is listed just past 5. Pure water is listed at a p H of 7. Pure blood is just beyond 7. Milk of Magnesia is listed just past a p H of 10.5. Household ammonia is listed just before a pH of 12. 1 M N a O H is listed at a p H of 0. To the right of this labeled arrow is an arrow that points up and down through the height of the column. A beige strip passes through the table and to this double headed arrow at p H 7. To the left of the double headed arrow in this beige strip is the label \u201cneutral.\u201d A narrow beige strip runs through the arrow. Just above and below this region, the arrow is purple. It gradually turns to a bright red as it extends upward. At the top of the arrow, near the head of the arrow is the label \u201cacidic.\u201d Similarly, the lower region changes color from purple to blue moving to the bottom of the column. The head at this end of the arrow is labeled \u201cbasic.\u201d\" width=\"1024\" height=\"897\" \/> Figure 1. The pH and pOH scales represent concentrations of [H<sub>3<\/sub>O<sup>+<\/sup>] and OH<sup>\u2212<\/sup>, respectively. The pH and pOH values of some common substances at standard temperature (25 \u00b0C) are shown in this chart.[\/caption]\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1<\/h3>\r\n<h4>Calculation of pH from [H<sub>3<\/sub>O<sup>+<\/sup>]<\/h4>\r\nWhat is the pH of stomach acid, a solution of HCl with a hydronium ion concentration of 1.2 \u00d7 10<sup>\u20133\u00a0<\/sup><em>M<\/em>?\r\n<h4>Solution<\/h4>\r\npH = \u2212log[H<sub>3<\/sub>O<sup>+<\/sup>]\r\n\r\n= \u2212log(1.2 \u00d7 10<sup>\u20133<\/sup>)\r\n\r\n= \u2212(\u22122.92) = 2.92\r\n\r\n(The use of logarithms is explained in <a href=\".\/chapter\/essential-mathematics\/\" target=\"_blank\">Essential Mathematics<\/a>. Recall that, as we have done here, when taking the log of a value, keep as many decimal places in the result as there are significant figures in the value.)\r\n<h4>Check Your Learning<\/h4>\r\nWater exposed to air contains carbonic acid, H<sub>2<\/sub>CO<sub>3<\/sub>, due to the reaction between carbon dioxide and water:\r\n\r\nCO<sub>2<\/sub>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>) \u21cc H<sub>2<\/sub>CO<sub>3<\/sub>(<em>aq<\/em>)\r\n\r\nAir-saturated water has a hydronium ion concentration caused by the dissolved CO<sub>2<\/sub> of 2.0 \u00d7 10<sup>\u20136<\/sup> <em>M<\/em>, about 20-times larger than that of pure water. Calculate the pH of the solution at 25 \u00b0C.\r\n\r\n<strong>Answer:\u00a0<\/strong>5.70\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2<\/h3>\r\n<h4>Calculation of Hydronium Ion Concentration from pH<\/h4>\r\nCalculate the hydronium ion concentration of blood, the pH of which is 7.3 (slightly alkaline).\r\n<h4>Solution<\/h4>\r\npH = \u2212log[H<sub>3<\/sub>O<sup>+<\/sup>] = 7.3\r\n\r\nlog[H<sub>3<\/sub>O<sup>+<\/sup>] = \u22127.3\r\n\r\n[H<sub>3<\/sub>O<sup>+<\/sup>] = 10<sup>\u20137.3\u00a0<\/sup>or [H<sub>3<\/sub>O<sup>+<\/sup>] = antilog of \u22127.3\r\n\r\n[H<sub>3<\/sub>O<sup>+<\/sup>] = 5 \u00d7 10<sup>\u20138<\/sup>\u00a0<em>M<\/em>\r\n\r\n(On a calculator take the antilog, or the \u201cinverse\u201d log, of \u22127.3, or calculate 10<sup>\u20137.3<\/sup>.)\r\n<h4>Check Your Learning<\/h4>\r\nCalculate the hydronium ion concentration of a solution with a pH of \u22121.07.\r\n<p style=\"text-align: right;\"><strong>Answer:<\/strong>\u00a012 M<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Environmental Science<\/h3>\r\nNormal rainwater has a pH between 5 and 6 due to the presence of dissolved CO<sub>2<\/sub> which forms carbonic acid:\r\n<p style=\"text-align: center;\">H<sub>2<\/sub>O(<em>l<\/em>)+ CO<sub>2<\/sub>(<em>g<\/em>) \u27f6 H<sub>2<\/sub>CO<sub>3<\/sub>(<em>aq<\/em>)<\/p>\r\n<p style=\"text-align: center;\">H<sub>2<\/sub>CO<sub>3<\/sub>(<em>aq<\/em>) \u21cc H<sup>+<\/sup>(<em>aq<\/em>)+HCO<sub>3<\/sub><sup>\u2212<\/sup>(<em>aq<\/em>)<\/p>\r\nAcid rain is rainwater that has a pH of less than 5, due to a variety of nonmetal oxides, including CO2, SO2, SO3, NO, and NO2 being dissolved in the water and reacting with it to form not only carbonic acid, but sulfuric acid and nitric acid. The formation and subsequent ionization of sulfuric acid are shown here:\r\n<p style=\"text-align: center;\">H<sub>2<\/sub>O(<em>l<\/em>)+SO<sub>3<\/sub>(<em>g<\/em>) \u27f6 H<sub>2<\/sub>SO<sub>4<\/sub>(<em>aq<\/em>)<\/p>\r\n<p style=\"text-align: center;\">H<sub>2<\/sub>SO<sub>4<\/sub>(<em>aq<\/em>) \u27f6 H<sup>+<\/sup>(<em>aq<\/em>) + HSO<sub>2<\/sub><sup>\u2212<\/sup>(<em>aq<\/em>)<\/p>\r\nCarbon dioxide is naturally present in the atmosphere because we and most other organisms produce it as a waste product of metabolism. Carbon dioxide is also formed when fires release carbon stored in vegetation or when we burn wood or fossil fuels. Sulfur trioxide in the atmosphere is naturally produced by volcanic activity, but it also stems from burning fossil fuels, which have traces of sulfur, and from the process of \u201croasting\u201d ores of metal sulfides in metal-refining processes. Oxides of nitrogen are formed in internal combustion engines where the high temperatures make it possible for the nitrogen and oxygen in air to chemically combine.\r\n\r\nAcid rain is a particular problem in industrial areas where the products of combustion and smelting are released into the air without being stripped of sulfur and nitrogen oxides. In North America and Europe until the 1980s, it was responsible for the destruction of forests and freshwater lakes, when the acidity of the rain actually killed trees, damaged soil, and made lakes uninhabitable for all but the most acid-tolerant species. Acid rain also corrodes statuary and building facades that are made of marble and limestone (Figure 2). Regulations limiting the amount of sulfur and nitrogen oxides that can be released into the atmosphere by industry and automobiles have reduced the severity of acid damage to both natural and manmade environments in North America and Europe. It is now a growing problem in industrial areas of China and India.\r\n\r\nFor further information on acid rain, visit this <a href=\"http:\/\/www.epa.gov\/acidrain\/\" target=\"_blank\">US Environmental Protection Agency website<\/a>.\r\n\r\n[caption id=\"attachment_5398\" align=\"aligncenter\" width=\"975\"]<img class=\"size-full wp-image-5398\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09035419\/CNX_Chem_14_02_AcidRain.jpg\" alt=\"Two photos are shown. Photograph a on the left shows the upper portion of trees against a bright blue sky. The tops of several trees at the center of the photograph have bare branches and appear to be dead. Image b shows a statue of a man that appears to from the revolutionary war era in either marble or limestone.\" width=\"975\" height=\"396\" \/> Figure 2. (a) Acid rain makes trees more susceptible to drought and insect infestation, and depletes nutrients in the soil. (b) It also is corrodes statues that are carved from marble or limestone. (credit a: modification of work by Chris M Morris; credit b: modification of work by \u201cEden, Janine and Jim\u201d\/Flickr)[\/caption]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 3<\/h3>\r\n<h4>Calculation of pOH<\/h4>\r\nWhat are the pOH and the pH of a 0.0125<em>-M<\/em> solution of potassium hydroxide, KOH?\r\n<h4>Solution<\/h4>\r\nPotassium hydroxide is a highly soluble ionic compound and completely dissociates when dissolved in dilute solution, yielding\u00a0[OH<sup>\u2212<\/sup>] = 0.0125 <em>M<\/em>:\r\n<p style=\"text-align: center;\">pOH = \u2212log[OH<sup>\u2212<\/sup>] = \u2212log 0.0125<\/p>\r\n<p style=\"text-align: center;\">= \u2212(\u22121.903) = 1.903<\/p>\r\nThe pH can be found from the pOH:\r\n<p style=\"text-align: center;\">pH + pOH = 14.00<\/p>\r\n<p style=\"text-align: center;\">pH = 14.00 \u2212 pOH = 14.00 \u2212 1.903 = 12.10<\/p>\r\n\r\n<h4>Check Your Learning<\/h4>\r\nThe hydronium ion concentration of vinegar is approximately 4 \u00d7 10<sup>\u20133<\/sup> <em>M<\/em>. What are the corresponding values of pOH and pH?\r\n<p style=\"text-align: right;\"><b>Answer:<\/b> pOH = 11.6, pH = 2.4<\/p>\r\n\r\n<\/div>\r\nThe acidity of a solution is typically assessed experimentally by measurement of its pH. The pOH of a solution is not usually measured, as it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter (Figure 3).\r\n\r\n[caption id=\"attachment_5399\" align=\"aligncenter\" width=\"1024\"]<img class=\"size-large wp-image-5399\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09035521\/CNX_Chem_14_02_pHMeter-1024x587-1024x587.jpg\" alt=\"This figure contains two images. The first, image a, is of an analytical digital p H meter on a laboratory counter. The second, image b, is of a portable hand held digital p H meter.\" width=\"1024\" height=\"587\" \/> Figure 3. (a) A research-grade pH meter used in a laboratory can have a resolution of 0.001 pH units, an accuracy of \u00b1 0.002 pH units, and may cost in excess of $1000. (b) A portable pH meter has lower resolution (0.01 pH units), lower accuracy (\u00b1 0.2 pH units), and a far lower price tag. (credit b: modification of work by Jacopo Werther)[\/caption]\r\n\r\nThe pH of a solution may also be visually estimated using colored indicators (Figure 4).\r\n\r\n[caption id=\"attachment_5401\" align=\"aligncenter\" width=\"1024\"]<img class=\"size-large wp-image-5401\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09035614\/CNX_Chem_14_02_indicator-1024x336-1024x336.jpg\" alt=\"This figure contains two images. The first shows a variety of colors of solutions in labeled beakers. A red solution in a beaker is labeled \u201c0.10 M H C l.\u201d An orange solution is labeled \u201c0.10 M C H subscript 3 C O O H.\u201d A yellow-orange solution is labeled \u201c0.1 M N H subscript 4 C l.\u201d A yellow solution is labeled \u201cdeionized water.\u201d A second solution beaker is labeled \u201c0.10 M K C l.\u201d A green solution is labeled \u201c0.10 M aniline.\u201d A blue solution is labeled \u201c0.10 M N H subscript 4 C l (a q).\u201d A final beaker containing a dark blue solution is labeled \u201c0.10 M N a O H.\u201d Image b shows pHydrion paper that is used for measuring pH in the range of p H from 1 to 12. The color scale for identifying p H based on color is shown along with several of the test strips used to evaluate p H.\" width=\"1024\" height=\"336\" \/> Figure 4. (a) A universal indicator assumes a different color in solutions of different pH values. Thus, it can be added to a solution to determine the pH of the solution. The eight vials each contain a universal indicator and 0.1-<em>M<\/em> solutions of progressively weaker acids: HCl (pH = l), CH<sub>3<\/sub>CO<sub>2<\/sub>H (pH = 3), and NH<sub>4<\/sub>Cl (pH = 5), deionized water, a neutral substance (pH = 7); and 0.1-M solutions of the progressively stronger bases: KCl (pH = 7), aniline, C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub> (pH = 9), NH<sub>3<\/sub> (pH = 11), and NaOH (pH = 13). (b) pH paper contains a mixture of indicators that give different colors in solutions of differing pH values. (credit: modification of work by Sahar Atwa)[\/caption]\r\n\r\n<div class=\"bcc-box bcc-success\">\r\n<h2>Key Concepts and Summary<\/h2>\r\nThe concentration of hydronium ion in a solution of an acid in water is greater than 1.0 \u00d7 10<sup>\u22127<\/sup> <em>M<\/em> at 25 \u00b0C. The concentration of hydroxide ion in a solution of a base in water is greater than 1.0 \u00d7 10<sup>\u22127<\/sup> <em>M<\/em> at 25 \u00b0C. The concentration of H<sub>3<\/sub>O<sup>+<\/sup>\u00a0in a solution can be expressed as the pH of the solution; pH = \u2212logH<sub>3<\/sub>O<sup>+<\/sup>. The concentration of OH<sup>\u2212<\/sup> can be expressed as the pOH of the solution: pOH = \u2212log[OH<sup>\u2212<\/sup>]. In pure water, pH = 7.00 and pOH = 7.00\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Key Equations<\/h3>\r\n<ul>\r\n \t<li>pH = \u2212log[H<sub>3<\/sub>O<sup>+<\/sup>]<\/li>\r\n \t<li>pOH = \u2212log[OH<sup>\u2212<\/sup>]<\/li>\r\n \t<li>[H<sub>3<\/sub>O<sup>+<\/sup>] =10<sup>\u2212pH<\/sup><\/li>\r\n \t<li>[OH<sup>\u2212<\/sup>] =10<sup>\u2212pOH<\/sup><\/li>\r\n \t<li>pH + pOH = p<em data-effect=\"italics\">K<\/em><sub>w<\/sub> = 14.00 at 25 \u00b0C<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Chemistry End of Chapter Exercises<\/h3>\r\n<ol>\r\n \t<li>Explain why a sample of pure water at 40 \u00b0C is neutral even though [H<sub>3<\/sub>O<sup>+<\/sup>] = 1.7 \u00d7 10<sup>\u22127<\/sup> <em data-effect=\"italics\">M<\/em>.\u00a0<em data-effect=\"italics\">K<\/em><sub>w<\/sub> is 2.9 \u00d7 10<sup>\u221214<\/sup> at 40 \u00b0C.<\/li>\r\n \t<li>The ionization constant for water (<em data-effect=\"italics\">K<\/em><sub>w<\/sub>) is 2.9 \u00d7 10<sup>\u221214<\/sup> at 40 \u00b0C. Calculate [H<sub>3<\/sub>O<sup>+<\/sup>], [OH<sup>\u2212<\/sup>], pH, and pOH for pure water at 40 \u00b0C.<\/li>\r\n \t<li><span style=\"line-height: 1.5;\">The ionization constant for water (<em data-effect=\"italics\">K<\/em><sub>w<\/sub>) is 9.614 \u00d7 10<sup>\u221214<\/sup> at 60 \u00b0C. Calculate [H<sub>3<\/sub>O<sup>+<\/sup>], [OH<sup>\u2212<\/sup>], pH, and pOH for pure water at 60 \u00b0C.<\/span><\/li>\r\n \t<li>Calculate the pH and the pOH of each of the following solutions at 25 \u00b0C for which the substances ionize completely:\r\n<ol>\r\n \t<li><span style=\"line-height: 1.5;\">0.200 <em>M<\/em> HCl<\/span><\/li>\r\n \t<li><span style=\"line-height: 1.5;\">0.0143 <em>M<\/em> NaOH<\/span><\/li>\r\n \t<li><span style=\"line-height: 1.5;\">3.0 <em>M<\/em> HNO<sub>3<\/sub><\/span><\/li>\r\n \t<li><span style=\"line-height: 1.5;\">0.0031 <em>M\u00a0<\/em>Ca(OH)<sub>2<\/sub><\/span><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li><span style=\"line-height: 1.5;\">Calculate the pH and the pOH of each of the following solutions at 25 \u00b0C for which the substances ionize completely:<\/span>\r\n<ol>\r\n \t<li><span style=\"line-height: 1.5;\">0.000259\u00a0<em data-effect=\"italics\">M<\/em> HClO<sub>4<\/sub><\/span><\/li>\r\n \t<li><span style=\"line-height: 1.5;\">0.21 <em>M<\/em> NaOH<\/span><\/li>\r\n \t<li><span style=\"line-height: 1.5;\">0.000071 <em>M<\/em>\u00a0Ba(OH)<sub>2<\/sub><\/span><\/li>\r\n \t<li><span style=\"line-height: 1.5;\">2.5 <em>M<\/em> KOH<\/span><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>What are the pH and pOH of a solution of 2.0 <em>M<\/em> HCl, which ionizes completely?<\/li>\r\n \t<li><span style=\"line-height: 1.5;\">What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52?<\/span><\/li>\r\n \t<li>Calculate the hydrogen ion concentration and the hydroxide ion concentration in wine from its pH. See Figure 1 for useful information.<\/li>\r\n \t<li><span style=\"line-height: 1.5;\">Calculate the hydronium ion concentration and the hydroxide ion concentration in lime juice from its pH. See Figure 1 for useful information.<\/span><\/li>\r\n \t<li>The hydronium ion concentration in a sample of rainwater is found to be 1.7 \u00d7 10<sup>\u22126<\/sup> <em data-effect=\"italics\">M<\/em> at 25 \u00b0C. What is the concentration of hydroxide ions in the rainwater?<\/li>\r\n \t<li><span style=\"line-height: 1.5;\">The hydroxide ion concentration in household ammonia is 3.2 \u00d7 10<sup>\u22123<\/sup> <em data-effect=\"italics\">M<\/em> at 25 \u00b0C. What is the concentration of hydronium ions in the solution?<\/span><\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h4>Selected Answers<\/h4>\r\n1.\u00a0In a neutral solution [H<sub>3<\/sub>O<sup>+<\/sup>] = [OH<sup>\u2212<\/sup>]. At 40 \u00b0C,\r\n\r\n[H<sub>3<\/sub>O<sup>+<\/sup>] = [OH<sup>\u2212<\/sup>] = (2.910<sup>\u221214<\/sup>)<sup>1\/2<\/sup> = 1.7 <span id=\"MathJax-Element-1285-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-34312\" class=\"math\"><span id=\"MathJax-Span-34313\" class=\"mrow\"><span id=\"MathJax-Span-34314\" class=\"semantics\"><span id=\"MathJax-Span-34315\" class=\"mrow\"><span id=\"MathJax-Span-34316\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span> 10<sup>\u22127<\/sup>.\r\n\r\n3. \u00a0<em data-effect=\"italics\">x<\/em> = 3.101 <span id=\"MathJax-Element-1288-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-34327\" class=\"math\"><span id=\"MathJax-Span-34328\" class=\"mrow\"><span id=\"MathJax-Span-34329\" class=\"semantics\"><span id=\"MathJax-Span-34330\" class=\"mrow\"><span id=\"MathJax-Span-34331\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span> 10<sup>\u22127<\/sup> <em data-effect=\"italics\">M<\/em> = [H<sub>3<\/sub>O<sup>+<\/sup>] = [OH<sup>\u2212<\/sup>]\r\n\r\npH = \u2212log3.101 <span id=\"MathJax-Element-1289-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-34332\" class=\"math\"><span id=\"MathJax-Span-34333\" class=\"mrow\"><span id=\"MathJax-Span-34334\" class=\"semantics\"><span id=\"MathJax-Span-34335\" class=\"mrow\"><span id=\"MathJax-Span-34336\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span> 10<sup>\u22127<\/sup> = \u2212(\u22126.5085) = 6.5085\r\n\r\npOH = pH = 6.5085\r\n\r\n5.\u00a0(a) pH = 3.587; pOH = 10.413;\r\n\r\n(b) pH = 0.68; pOH = 13.32;\r\n\r\n(c) pOH = 3.85; pH = 10.15;\r\n\r\n(d) pH = \u22120.40; pOH = 14.4\r\n\r\n7.\u00a0[H<sub>3<\/sub>O<sup>+<\/sup>] = 3.0 <span id=\"MathJax-Element-1290-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-34337\" class=\"math\"><span id=\"MathJax-Span-34338\" class=\"mrow\"><span id=\"MathJax-Span-34339\" class=\"semantics\"><span id=\"MathJax-Span-34340\" class=\"mrow\"><span id=\"MathJax-Span-34341\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span> 10<sup>\u22127<\/sup> <em data-effect=\"italics\">M<\/em>; [OH<sup>\u2212<\/sup>] = 3.3 <span id=\"MathJax-Element-1291-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-34342\" class=\"math\"><span id=\"MathJax-Span-34343\" class=\"mrow\"><span id=\"MathJax-Span-34344\" class=\"semantics\"><span id=\"MathJax-Span-34345\" class=\"mrow\"><span id=\"MathJax-Span-34346\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span> 10<sup>\u22128<\/sup> <em data-effect=\"italics\">M<\/em>\r\n\r\n9.\u00a0[H<sub>3<\/sub>O<sup>+<\/sup>] = 1 <span id=\"MathJax-Element-1292-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-34347\" class=\"math\"><span id=\"MathJax-Span-34348\" class=\"mrow\"><span id=\"MathJax-Span-34349\" class=\"semantics\"><span id=\"MathJax-Span-34350\" class=\"mrow\"><span id=\"MathJax-Span-34351\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span> 10<sup>\u22122<\/sup> <em data-effect=\"italics\">M<\/em>; [OH<sup>\u2212<\/sup>] = 1 <span id=\"MathJax-Element-1293-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-34352\" class=\"math\"><span id=\"MathJax-Span-34353\" class=\"mrow\"><span id=\"MathJax-Span-34354\" class=\"semantics\"><span id=\"MathJax-Span-34355\" class=\"mrow\"><span id=\"MathJax-Span-34356\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span> 10<sup>\u221212<\/sup> <em data-effect=\"italics\">M<\/em>\r\n\r\n11.\u00a0[OH<sup>\u2212<\/sup>] = 3.1 <span id=\"MathJax-Element-1296-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-34367\" class=\"math\"><span id=\"MathJax-Span-34368\" class=\"mrow\"><span id=\"MathJax-Span-34369\" class=\"semantics\"><span id=\"MathJax-Span-34370\" class=\"mrow\"><span id=\"MathJax-Span-34371\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span> 10<sup>\u221212<\/sup> <em data-effect=\"italics\">M<\/em>\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Glossary<\/h3>\r\n<strong>acidic<\/strong>\r\ndescribes a solution in which [H<sub>3<\/sub>O<sup>+<\/sup>] &gt; [OH<sup>\u2212<\/sup>]\r\n\r\n<strong>basic<\/strong>\r\ndescribes a solution in which [H<sub>3<\/sub>O<sup>+<\/sup>] &lt; [OH<sup>\u2212<\/sup>]\r\n\r\n<strong>neutral<\/strong>\r\ndescribes a solution in which [H<sub>3<\/sub>O<sup>+<\/sup>] = [OH<sup>\u2212<\/sup>]\r\n\r\n<strong>pH<\/strong>\r\nlogarithmic measure of the concentration of hydronium ions in a solution\r\n\r\n<strong>pOH<\/strong>\r\nlogarithmic measure of the concentration of hydroxide ions in a solution\r\n\r\n<\/div>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>LEARNING OBJECTIVES<\/h3>\n<p>By the end of this module, you will be able to:<\/p>\n<ul>\n<li>Explain the characterization of aqueous solutions as acidic, basic, or neutral<\/li>\n<li>Express hydronium and hydroxide ion concentrations on the pH and pOH scales<\/li>\n<li>Perform calculations relating pH and pOH<\/li>\n<\/ul>\n<\/div>\n<p>As discussed earlier, hydronium and hydroxide ions are present both in pure water and in all aqueous solutions, and their concentrations are inversely proportional as determined by the ion product of water (<em data-effect=\"italics\">K<\/em><sub>w<\/sub>). The concentrations of these ions in a solution are often critical determinants of the solution\u2019s properties and the chemical behaviors of its other solutes, and specific vocabulary has been developed to describe these concentrations in relative terms. A solution is <strong>neutral<\/strong> if it contains equal concentrations of hydronium and hydroxide ions; <strong>acidic<\/strong> if it contains a greater concentration of hydronium ions than hydroxide ions; and <strong>basic<\/strong> if it contains a lesser concentration of hydronium ions than hydroxide ions.<\/p>\n<p>A common means of expressing quantities, the values of which may span many orders of magnitude, is to use a logarithmic scale. One such scale that is very popular for chemical concentrations and equilibrium constants is based on the p-function, defined as shown where \u201cX\u201d is the quantity of interest and \u201clog\u201d is the base-10 logarithm:<\/p>\n<p style=\"text-align: center;\">pX = \u2212log X<\/p>\n<p>The pH of a solution is therefore defined as shown here, where [H<sub>3<\/sub>O<sup>+<\/sup>] is the molar concentration of hydronium ion in the solution:<\/p>\n<p style=\"text-align: center;\">pH = \u2212log[H<sub>3<\/sub>O<sup>+<\/sup>]<\/p>\n<p>Rearranging this equation to isolate the hydronium ion molarity yields the equivalent expression:<\/p>\n<p style=\"text-align: center;\">[H<sub>3<\/sub>O<sup>+<\/sup>] = 10<sup>\u2212pH<\/sup><\/p>\n<p>Likewise, the hydroxide ion molarity may be expressed as a p-function, or <b>pOH<\/b>:<\/p>\n<p style=\"text-align: center;\">pOH = \u2212log[OH<sup>\u2212<\/sup>]<\/p>\n<p style=\"text-align: center;\">or<\/p>\n<p style=\"text-align: center;\">[OH<sup>\u2212<\/sup>] = 10<sup>\u2212pOH<\/sup><\/p>\n<p>Finally, the relation between these two ion concentration expressed as p-functions is easily derived from the\u00a0<em data-effect=\"italics\">K<\/em><sub>w<\/sub> expression:<\/p>\n<p style=\"text-align: center;\"><em data-effect=\"italics\">K<\/em><sub>w\u00a0<\/sub>= [H<sub>3<\/sub>O<sup>+<\/sup>][OH<sup>\u2212<\/sup>]<br \/>\n\u2212log<em data-effect=\"italics\">K<\/em><sub>w<\/sub> = \u2212log([H<sub>3<\/sub>O<sup>+<\/sup>][OH<sup>\u2212<\/sup>]) = \u2212log[H<sub>3<\/sub>O<sup>+<\/sup>] + \u2212log[OH<sup>\u2212<\/sup>]<br \/>\np<em data-effect=\"italics\">K<\/em><sub>w\u00a0<\/sub>= pH + pOH<\/p>\n<p>At 25 \u00b0C, the value of\u00a0<em data-effect=\"italics\">K<\/em><sub>w<\/sub>\u00a0is 1.0 \u00d7 10\u221214, and so:<\/p>\n<p style=\"text-align: center;\">14.00 = pH +pOH<\/p>\n<p>As we learned earlier, the hydronium ion molarity in pure water (or any neutral solution) is 1.0 \u00d7 10<sup>\u22127<\/sup> <em data-effect=\"italics\">M<\/em>\u00a0at 25 \u00b0C. The pH and pOH of a neutral solution at this temperature are therefore:<\/p>\n<p style=\"text-align: center;\">pH = \u2212log[H<sub>3<\/sub>O<sup>+<\/sup>] = \u2212log(1.0 \u00d7 10<sup>\u22127<\/sup>) = 7.00<\/p>\n<p style=\"text-align: center;\">pOH = \u2212log[OH<sup>\u2212<\/sup>] = \u2212log(1.0 \u00d7 10<sup>\u22127<\/sup>) = 7.00<\/p>\n<p>And so, <em>at this temperature<\/em>, acidic solutions are those with hydronium ion molarities greater than 1.0 \u00d7 10<sup>\u22127<\/sup> <em data-effect=\"italics\">M<\/em> and hydroxide ion molarities less than 1.0 \u00d7 10<sup>\u22127<\/sup> <em data-effect=\"italics\">M<\/em> (corresponding to pH values less than 7.00 and pOH values greater than 7.00). Basic solutions are those with hydronium ion molarities less than 1.0 \u00d7 10<sup>\u22127<\/sup> <em data-effect=\"italics\">M<\/em> and hydroxide ion molarities greater than 1.0 \u00d7 10<sup>\u22127<\/sup> <em data-effect=\"italics\">M<\/em> (corresponding to pH values greater than 7.00 and pOH values less than 7.00).<\/p>\n<p>Since the autoionization constant Kw is temperature dependent, these correlations between pH values and the acidic\/neutral\/basic adjectives will be different at temperatures other than 25 \u00b0C. For example, the hydronium molarity of pure water is\u00a080 \u00b0C is 4.9 \u00d7 10<sup>\u22127<\/sup> <em data-effect=\"italics\">M<\/em>, which corresponds to pH and pOH values of:<\/p>\n<p style=\"text-align: center;\">pH = \u2212log[H<sub>3<\/sub>O<sup>+<\/sup>] = \u2212log(4.9 \u00d7 10<sup>\u22127<\/sup>) = 6.31<\/p>\n<p style=\"text-align: center;\">pOH = \u2212log[OH<sup>\u2212<\/sup>] = \u2212log(4.9 \u00d7 10<sup>\u22127<\/sup>) = 6.31<\/p>\n<p>At this temperature, then, neutral solutions exhibit pH = pOH = 6.31, acidic solutions exhibit pH less than 6.31 and pOH greater than 6.31, whereas basic solutions exhibit pH greater than 6.31 and pOH less than 6.31. This distinction can be important when studying certain processes that occur at nonstandard temperatures, such as enzyme reactions in warm-blooded organisms. Unless otherwise noted, references to pH values are presumed to be those at standard temperature (25 \u00b0C) (Table 1).<\/p>\n<table>\n<thead>\n<tr>\n<th colspan=\"3\">Table 1.\u00a0Summary of Relations for Acidic, Basic and Neutral Solutions<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Classification<\/th>\n<th>\u00a0Relative Ion Concentrations<\/th>\n<th>\u00a0pH at 25 \u00b0C<\/th>\n<\/tr>\n<tr>\n<td>acidic<\/td>\n<td>\u00a0[H<sub>3<\/sub>O<sup>+<\/sup>] &gt; [OH<sup>\u2212<\/sup>]<\/td>\n<td>\u00a0pH &lt; 7<\/td>\n<\/tr>\n<tr>\n<td>neutral<\/td>\n<td>\u00a0[H<sub>3<\/sub>O<sup>+<\/sup>] = [OH<sup>\u2212<\/sup>]<\/td>\n<td>\u00a0pH = 7<\/td>\n<\/tr>\n<tr>\n<td>basic<\/td>\n<td>\u00a0[H<sub>3<\/sub>O<sup>+<\/sup>] &lt; [OH<sup>\u2212<\/sup>]<\/td>\n<td>\u00a0pH &gt; 7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Figure 1 shows the relationships between [H<sub>3<\/sub>O<sup>+<\/sup>], [OH<sup>\u2212<\/sup>], pH, and pOH, and gives values for these properties at standard temperatures for some common substances.<\/p>\n<div id=\"attachment_5396\" style=\"width: 1034px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5396\" class=\"size-large wp-image-5396\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09035250\/CNX_Chem_14_02_phscale-1024x897-1024x897.jpg\" alt=\"A table is provided with 5 columns. The first column is labeled \u201cleft bracket H subscript 3 O superscript plus right bracket (M).\u201d Powers of ten are listed in the column beginning at 10 superscript 1, including 10 superscript 0 or 1, 10 superscript negative 1, decreasing by single powers of 10 to 10 superscript negative 15. The second column is labeled \u201cleft bracket O H superscript negative right bracket (M).\u201d Powers of ten are listed in the column beginning at 10 superscript negative 15, increasing by single powers of 10 to including 10 superscript 0 or 1, and 10 superscript 1. The third column is labeled \u201cp H.\u201d Values listed in this column are integers beginning at negative 1, increasing by ones up to 14. The fourth column is labeled \u201cp O H.\u201d Values in this column are integers beginning at 15, decreasing by ones up to negative 1. The fifth column is labeled \u201cSample Solution.\u201d A vertical line at the left of the column has tick marks corresponding to each p H level in the table. Substances are listed next to this line segment with line segments connecting them to the line to show approximate p H and p O H values. 1 M H C l is listed at a p H of 0. Gastric juices are listed at a p H of about 1.5. Lime juice is listed at a p H of about 2, followed by 1 M C H subscript 3 C O subscript 2 H, followed by stomach acid at a p H value of nearly 3. Wine is listed around 3.5. Coffee is listed just past 5. Pure water is listed at a p H of 7. Pure blood is just beyond 7. Milk of Magnesia is listed just past a p H of 10.5. Household ammonia is listed just before a pH of 12. 1 M N a O H is listed at a p H of 0. To the right of this labeled arrow is an arrow that points up and down through the height of the column. A beige strip passes through the table and to this double headed arrow at p H 7. To the left of the double headed arrow in this beige strip is the label \u201cneutral.\u201d A narrow beige strip runs through the arrow. Just above and below this region, the arrow is purple. It gradually turns to a bright red as it extends upward. At the top of the arrow, near the head of the arrow is the label \u201cacidic.\u201d Similarly, the lower region changes color from purple to blue moving to the bottom of the column. The head at this end of the arrow is labeled \u201cbasic.\u201d\" width=\"1024\" height=\"897\" \/><\/p>\n<p id=\"caption-attachment-5396\" class=\"wp-caption-text\">Figure 1. The pH and pOH scales represent concentrations of [H<sub>3<\/sub>O<sup>+<\/sup>] and OH<sup>\u2212<\/sup>, respectively. The pH and pOH values of some common substances at standard temperature (25 \u00b0C) are shown in this chart.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1<\/h3>\n<h4>Calculation of pH from [H<sub>3<\/sub>O<sup>+<\/sup>]<\/h4>\n<p>What is the pH of stomach acid, a solution of HCl with a hydronium ion concentration of 1.2 \u00d7 10<sup>\u20133\u00a0<\/sup><em>M<\/em>?<\/p>\n<h4>Solution<\/h4>\n<p>pH = \u2212log[H<sub>3<\/sub>O<sup>+<\/sup>]<\/p>\n<p>= \u2212log(1.2 \u00d7 10<sup>\u20133<\/sup>)<\/p>\n<p>= \u2212(\u22122.92) = 2.92<\/p>\n<p>(The use of logarithms is explained in <a href=\".\/chapter\/essential-mathematics\/\" target=\"_blank\">Essential Mathematics<\/a>. Recall that, as we have done here, when taking the log of a value, keep as many decimal places in the result as there are significant figures in the value.)<\/p>\n<h4>Check Your Learning<\/h4>\n<p>Water exposed to air contains carbonic acid, H<sub>2<\/sub>CO<sub>3<\/sub>, due to the reaction between carbon dioxide and water:<\/p>\n<p>CO<sub>2<\/sub>(<em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>) \u21cc H<sub>2<\/sub>CO<sub>3<\/sub>(<em>aq<\/em>)<\/p>\n<p>Air-saturated water has a hydronium ion concentration caused by the dissolved CO<sub>2<\/sub> of 2.0 \u00d7 10<sup>\u20136<\/sup> <em>M<\/em>, about 20-times larger than that of pure water. Calculate the pH of the solution at 25 \u00b0C.<\/p>\n<p><strong>Answer:\u00a0<\/strong>5.70<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2<\/h3>\n<h4>Calculation of Hydronium Ion Concentration from pH<\/h4>\n<p>Calculate the hydronium ion concentration of blood, the pH of which is 7.3 (slightly alkaline).<\/p>\n<h4>Solution<\/h4>\n<p>pH = \u2212log[H<sub>3<\/sub>O<sup>+<\/sup>] = 7.3<\/p>\n<p>log[H<sub>3<\/sub>O<sup>+<\/sup>] = \u22127.3<\/p>\n<p>[H<sub>3<\/sub>O<sup>+<\/sup>] = 10<sup>\u20137.3\u00a0<\/sup>or [H<sub>3<\/sub>O<sup>+<\/sup>] = antilog of \u22127.3<\/p>\n<p>[H<sub>3<\/sub>O<sup>+<\/sup>] = 5 \u00d7 10<sup>\u20138<\/sup>\u00a0<em>M<\/em><\/p>\n<p>(On a calculator take the antilog, or the \u201cinverse\u201d log, of \u22127.3, or calculate 10<sup>\u20137.3<\/sup>.)<\/p>\n<h4>Check Your Learning<\/h4>\n<p>Calculate the hydronium ion concentration of a solution with a pH of \u22121.07.<\/p>\n<p style=\"text-align: right;\"><strong>Answer:<\/strong>\u00a012 M<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Environmental Science<\/h3>\n<p>Normal rainwater has a pH between 5 and 6 due to the presence of dissolved CO<sub>2<\/sub> which forms carbonic acid:<\/p>\n<p style=\"text-align: center;\">H<sub>2<\/sub>O(<em>l<\/em>)+ CO<sub>2<\/sub>(<em>g<\/em>) \u27f6 H<sub>2<\/sub>CO<sub>3<\/sub>(<em>aq<\/em>)<\/p>\n<p style=\"text-align: center;\">H<sub>2<\/sub>CO<sub>3<\/sub>(<em>aq<\/em>) \u21cc H<sup>+<\/sup>(<em>aq<\/em>)+HCO<sub>3<\/sub><sup>\u2212<\/sup>(<em>aq<\/em>)<\/p>\n<p>Acid rain is rainwater that has a pH of less than 5, due to a variety of nonmetal oxides, including CO2, SO2, SO3, NO, and NO2 being dissolved in the water and reacting with it to form not only carbonic acid, but sulfuric acid and nitric acid. The formation and subsequent ionization of sulfuric acid are shown here:<\/p>\n<p style=\"text-align: center;\">H<sub>2<\/sub>O(<em>l<\/em>)+SO<sub>3<\/sub>(<em>g<\/em>) \u27f6 H<sub>2<\/sub>SO<sub>4<\/sub>(<em>aq<\/em>)<\/p>\n<p style=\"text-align: center;\">H<sub>2<\/sub>SO<sub>4<\/sub>(<em>aq<\/em>) \u27f6 H<sup>+<\/sup>(<em>aq<\/em>) + HSO<sub>2<\/sub><sup>\u2212<\/sup>(<em>aq<\/em>)<\/p>\n<p>Carbon dioxide is naturally present in the atmosphere because we and most other organisms produce it as a waste product of metabolism. Carbon dioxide is also formed when fires release carbon stored in vegetation or when we burn wood or fossil fuels. Sulfur trioxide in the atmosphere is naturally produced by volcanic activity, but it also stems from burning fossil fuels, which have traces of sulfur, and from the process of \u201croasting\u201d ores of metal sulfides in metal-refining processes. Oxides of nitrogen are formed in internal combustion engines where the high temperatures make it possible for the nitrogen and oxygen in air to chemically combine.<\/p>\n<p>Acid rain is a particular problem in industrial areas where the products of combustion and smelting are released into the air without being stripped of sulfur and nitrogen oxides. In North America and Europe until the 1980s, it was responsible for the destruction of forests and freshwater lakes, when the acidity of the rain actually killed trees, damaged soil, and made lakes uninhabitable for all but the most acid-tolerant species. Acid rain also corrodes statuary and building facades that are made of marble and limestone (Figure 2). Regulations limiting the amount of sulfur and nitrogen oxides that can be released into the atmosphere by industry and automobiles have reduced the severity of acid damage to both natural and manmade environments in North America and Europe. It is now a growing problem in industrial areas of China and India.<\/p>\n<p>For further information on acid rain, visit this <a href=\"http:\/\/www.epa.gov\/acidrain\/\" target=\"_blank\">US Environmental Protection Agency website<\/a>.<\/p>\n<div id=\"attachment_5398\" style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5398\" class=\"size-full wp-image-5398\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09035419\/CNX_Chem_14_02_AcidRain.jpg\" alt=\"Two photos are shown. Photograph a on the left shows the upper portion of trees against a bright blue sky. The tops of several trees at the center of the photograph have bare branches and appear to be dead. Image b shows a statue of a man that appears to from the revolutionary war era in either marble or limestone.\" width=\"975\" height=\"396\" \/><\/p>\n<p id=\"caption-attachment-5398\" class=\"wp-caption-text\">Figure 2. (a) Acid rain makes trees more susceptible to drought and insect infestation, and depletes nutrients in the soil. (b) It also is corrodes statues that are carved from marble or limestone. (credit a: modification of work by Chris M Morris; credit b: modification of work by \u201cEden, Janine and Jim\u201d\/Flickr)<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3<\/h3>\n<h4>Calculation of pOH<\/h4>\n<p>What are the pOH and the pH of a 0.0125<em>-M<\/em> solution of potassium hydroxide, KOH?<\/p>\n<h4>Solution<\/h4>\n<p>Potassium hydroxide is a highly soluble ionic compound and completely dissociates when dissolved in dilute solution, yielding\u00a0[OH<sup>\u2212<\/sup>] = 0.0125 <em>M<\/em>:<\/p>\n<p style=\"text-align: center;\">pOH = \u2212log[OH<sup>\u2212<\/sup>] = \u2212log 0.0125<\/p>\n<p style=\"text-align: center;\">= \u2212(\u22121.903) = 1.903<\/p>\n<p>The pH can be found from the pOH:<\/p>\n<p style=\"text-align: center;\">pH + pOH = 14.00<\/p>\n<p style=\"text-align: center;\">pH = 14.00 \u2212 pOH = 14.00 \u2212 1.903 = 12.10<\/p>\n<h4>Check Your Learning<\/h4>\n<p>The hydronium ion concentration of vinegar is approximately 4 \u00d7 10<sup>\u20133<\/sup> <em>M<\/em>. What are the corresponding values of pOH and pH?<\/p>\n<p style=\"text-align: right;\"><b>Answer:<\/b> pOH = 11.6, pH = 2.4<\/p>\n<\/div>\n<p>The acidity of a solution is typically assessed experimentally by measurement of its pH. The pOH of a solution is not usually measured, as it is easily calculated from an experimentally determined pH value. The pH of a solution can be directly measured using a pH meter (Figure 3).<\/p>\n<div id=\"attachment_5399\" style=\"width: 1034px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5399\" class=\"size-large wp-image-5399\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09035521\/CNX_Chem_14_02_pHMeter-1024x587-1024x587.jpg\" alt=\"This figure contains two images. The first, image a, is of an analytical digital p H meter on a laboratory counter. The second, image b, is of a portable hand held digital p H meter.\" width=\"1024\" height=\"587\" \/><\/p>\n<p id=\"caption-attachment-5399\" class=\"wp-caption-text\">Figure 3. (a) A research-grade pH meter used in a laboratory can have a resolution of 0.001 pH units, an accuracy of \u00b1 0.002 pH units, and may cost in excess of $1000. (b) A portable pH meter has lower resolution (0.01 pH units), lower accuracy (\u00b1 0.2 pH units), and a far lower price tag. (credit b: modification of work by Jacopo Werther)<\/p>\n<\/div>\n<p>The pH of a solution may also be visually estimated using colored indicators (Figure 4).<\/p>\n<div id=\"attachment_5401\" style=\"width: 1034px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5401\" class=\"size-large wp-image-5401\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09035614\/CNX_Chem_14_02_indicator-1024x336-1024x336.jpg\" alt=\"This figure contains two images. The first shows a variety of colors of solutions in labeled beakers. A red solution in a beaker is labeled \u201c0.10 M H C l.\u201d An orange solution is labeled \u201c0.10 M C H subscript 3 C O O H.\u201d A yellow-orange solution is labeled \u201c0.1 M N H subscript 4 C l.\u201d A yellow solution is labeled \u201cdeionized water.\u201d A second solution beaker is labeled \u201c0.10 M K C l.\u201d A green solution is labeled \u201c0.10 M aniline.\u201d A blue solution is labeled \u201c0.10 M N H subscript 4 C l (a q).\u201d A final beaker containing a dark blue solution is labeled \u201c0.10 M N a O H.\u201d Image b shows pHydrion paper that is used for measuring pH in the range of p H from 1 to 12. The color scale for identifying p H based on color is shown along with several of the test strips used to evaluate p H.\" width=\"1024\" height=\"336\" \/><\/p>\n<p id=\"caption-attachment-5401\" class=\"wp-caption-text\">Figure 4. (a) A universal indicator assumes a different color in solutions of different pH values. Thus, it can be added to a solution to determine the pH of the solution. The eight vials each contain a universal indicator and 0.1-<em>M<\/em> solutions of progressively weaker acids: HCl (pH = l), CH<sub>3<\/sub>CO<sub>2<\/sub>H (pH = 3), and NH<sub>4<\/sub>Cl (pH = 5), deionized water, a neutral substance (pH = 7); and 0.1-M solutions of the progressively stronger bases: KCl (pH = 7), aniline, C<sub>6<\/sub>H<sub>5<\/sub>NH<sub>2<\/sub> (pH = 9), NH<sub>3<\/sub> (pH = 11), and NaOH (pH = 13). (b) pH paper contains a mixture of indicators that give different colors in solutions of differing pH values. (credit: modification of work by Sahar Atwa)<\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h2>Key Concepts and Summary<\/h2>\n<p>The concentration of hydronium ion in a solution of an acid in water is greater than 1.0 \u00d7 10<sup>\u22127<\/sup> <em>M<\/em> at 25 \u00b0C. The concentration of hydroxide ion in a solution of a base in water is greater than 1.0 \u00d7 10<sup>\u22127<\/sup> <em>M<\/em> at 25 \u00b0C. The concentration of H<sub>3<\/sub>O<sup>+<\/sup>\u00a0in a solution can be expressed as the pH of the solution; pH = \u2212logH<sub>3<\/sub>O<sup>+<\/sup>. The concentration of OH<sup>\u2212<\/sup> can be expressed as the pOH of the solution: pOH = \u2212log[OH<sup>\u2212<\/sup>]. In pure water, pH = 7.00 and pOH = 7.00<\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Key Equations<\/h3>\n<ul>\n<li>pH = \u2212log[H<sub>3<\/sub>O<sup>+<\/sup>]<\/li>\n<li>pOH = \u2212log[OH<sup>\u2212<\/sup>]<\/li>\n<li>[H<sub>3<\/sub>O<sup>+<\/sup>] =10<sup>\u2212pH<\/sup><\/li>\n<li>[OH<sup>\u2212<\/sup>] =10<sup>\u2212pOH<\/sup><\/li>\n<li>pH + pOH = p<em data-effect=\"italics\">K<\/em><sub>w<\/sub> = 14.00 at 25 \u00b0C<\/li>\n<\/ul>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Chemistry End of Chapter Exercises<\/h3>\n<ol>\n<li>Explain why a sample of pure water at 40 \u00b0C is neutral even though [H<sub>3<\/sub>O<sup>+<\/sup>] = 1.7 \u00d7 10<sup>\u22127<\/sup> <em data-effect=\"italics\">M<\/em>.\u00a0<em data-effect=\"italics\">K<\/em><sub>w<\/sub> is 2.9 \u00d7 10<sup>\u221214<\/sup> at 40 \u00b0C.<\/li>\n<li>The ionization constant for water (<em data-effect=\"italics\">K<\/em><sub>w<\/sub>) is 2.9 \u00d7 10<sup>\u221214<\/sup> at 40 \u00b0C. Calculate [H<sub>3<\/sub>O<sup>+<\/sup>], [OH<sup>\u2212<\/sup>], pH, and pOH for pure water at 40 \u00b0C.<\/li>\n<li><span style=\"line-height: 1.5;\">The ionization constant for water (<em data-effect=\"italics\">K<\/em><sub>w<\/sub>) is 9.614 \u00d7 10<sup>\u221214<\/sup> at 60 \u00b0C. Calculate [H<sub>3<\/sub>O<sup>+<\/sup>], [OH<sup>\u2212<\/sup>], pH, and pOH for pure water at 60 \u00b0C.<\/span><\/li>\n<li>Calculate the pH and the pOH of each of the following solutions at 25 \u00b0C for which the substances ionize completely:\n<ol>\n<li><span style=\"line-height: 1.5;\">0.200 <em>M<\/em> HCl<\/span><\/li>\n<li><span style=\"line-height: 1.5;\">0.0143 <em>M<\/em> NaOH<\/span><\/li>\n<li><span style=\"line-height: 1.5;\">3.0 <em>M<\/em> HNO<sub>3<\/sub><\/span><\/li>\n<li><span style=\"line-height: 1.5;\">0.0031 <em>M\u00a0<\/em>Ca(OH)<sub>2<\/sub><\/span><\/li>\n<\/ol>\n<\/li>\n<li><span style=\"line-height: 1.5;\">Calculate the pH and the pOH of each of the following solutions at 25 \u00b0C for which the substances ionize completely:<\/span>\n<ol>\n<li><span style=\"line-height: 1.5;\">0.000259\u00a0<em data-effect=\"italics\">M<\/em> HClO<sub>4<\/sub><\/span><\/li>\n<li><span style=\"line-height: 1.5;\">0.21 <em>M<\/em> NaOH<\/span><\/li>\n<li><span style=\"line-height: 1.5;\">0.000071 <em>M<\/em>\u00a0Ba(OH)<sub>2<\/sub><\/span><\/li>\n<li><span style=\"line-height: 1.5;\">2.5 <em>M<\/em> KOH<\/span><\/li>\n<\/ol>\n<\/li>\n<li>What are the pH and pOH of a solution of 2.0 <em>M<\/em> HCl, which ionizes completely?<\/li>\n<li><span style=\"line-height: 1.5;\">What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52?<\/span><\/li>\n<li>Calculate the hydrogen ion concentration and the hydroxide ion concentration in wine from its pH. See Figure 1 for useful information.<\/li>\n<li><span style=\"line-height: 1.5;\">Calculate the hydronium ion concentration and the hydroxide ion concentration in lime juice from its pH. See Figure 1 for useful information.<\/span><\/li>\n<li>The hydronium ion concentration in a sample of rainwater is found to be 1.7 \u00d7 10<sup>\u22126<\/sup> <em data-effect=\"italics\">M<\/em> at 25 \u00b0C. What is the concentration of hydroxide ions in the rainwater?<\/li>\n<li><span style=\"line-height: 1.5;\">The hydroxide ion concentration in household ammonia is 3.2 \u00d7 10<sup>\u22123<\/sup> <em data-effect=\"italics\">M<\/em> at 25 \u00b0C. What is the concentration of hydronium ions in the solution?<\/span><\/li>\n<\/ol>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h4>Selected Answers<\/h4>\n<p>1.\u00a0In a neutral solution [H<sub>3<\/sub>O<sup>+<\/sup>] = [OH<sup>\u2212<\/sup>]. At 40 \u00b0C,<\/p>\n<p>[H<sub>3<\/sub>O<sup>+<\/sup>] = [OH<sup>\u2212<\/sup>] = (2.910<sup>\u221214<\/sup>)<sup>1\/2<\/sup> = 1.7 <span id=\"MathJax-Element-1285-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-34312\" class=\"math\"><span id=\"MathJax-Span-34313\" class=\"mrow\"><span id=\"MathJax-Span-34314\" class=\"semantics\"><span id=\"MathJax-Span-34315\" class=\"mrow\"><span id=\"MathJax-Span-34316\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span> 10<sup>\u22127<\/sup>.<\/p>\n<p>3. \u00a0<em data-effect=\"italics\">x<\/em> = 3.101 <span id=\"MathJax-Element-1288-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-34327\" class=\"math\"><span id=\"MathJax-Span-34328\" class=\"mrow\"><span id=\"MathJax-Span-34329\" class=\"semantics\"><span id=\"MathJax-Span-34330\" class=\"mrow\"><span id=\"MathJax-Span-34331\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span> 10<sup>\u22127<\/sup> <em data-effect=\"italics\">M<\/em> = [H<sub>3<\/sub>O<sup>+<\/sup>] = [OH<sup>\u2212<\/sup>]<\/p>\n<p>pH = \u2212log3.101 <span id=\"MathJax-Element-1289-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-34332\" class=\"math\"><span id=\"MathJax-Span-34333\" class=\"mrow\"><span id=\"MathJax-Span-34334\" class=\"semantics\"><span id=\"MathJax-Span-34335\" class=\"mrow\"><span id=\"MathJax-Span-34336\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span> 10<sup>\u22127<\/sup> = \u2212(\u22126.5085) = 6.5085<\/p>\n<p>pOH = pH = 6.5085<\/p>\n<p>5.\u00a0(a) pH = 3.587; pOH = 10.413;<\/p>\n<p>(b) pH = 0.68; pOH = 13.32;<\/p>\n<p>(c) pOH = 3.85; pH = 10.15;<\/p>\n<p>(d) pH = \u22120.40; pOH = 14.4<\/p>\n<p>7.\u00a0[H<sub>3<\/sub>O<sup>+<\/sup>] = 3.0 <span id=\"MathJax-Element-1290-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-34337\" class=\"math\"><span id=\"MathJax-Span-34338\" class=\"mrow\"><span id=\"MathJax-Span-34339\" class=\"semantics\"><span id=\"MathJax-Span-34340\" class=\"mrow\"><span id=\"MathJax-Span-34341\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span> 10<sup>\u22127<\/sup> <em data-effect=\"italics\">M<\/em>; [OH<sup>\u2212<\/sup>] = 3.3 <span id=\"MathJax-Element-1291-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-34342\" class=\"math\"><span id=\"MathJax-Span-34343\" class=\"mrow\"><span id=\"MathJax-Span-34344\" class=\"semantics\"><span id=\"MathJax-Span-34345\" class=\"mrow\"><span id=\"MathJax-Span-34346\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span> 10<sup>\u22128<\/sup> <em data-effect=\"italics\">M<\/em><\/p>\n<p>9.\u00a0[H<sub>3<\/sub>O<sup>+<\/sup>] = 1 <span id=\"MathJax-Element-1292-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-34347\" class=\"math\"><span id=\"MathJax-Span-34348\" class=\"mrow\"><span id=\"MathJax-Span-34349\" class=\"semantics\"><span id=\"MathJax-Span-34350\" class=\"mrow\"><span id=\"MathJax-Span-34351\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span> 10<sup>\u22122<\/sup> <em data-effect=\"italics\">M<\/em>; [OH<sup>\u2212<\/sup>] = 1 <span id=\"MathJax-Element-1293-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-34352\" class=\"math\"><span id=\"MathJax-Span-34353\" class=\"mrow\"><span id=\"MathJax-Span-34354\" class=\"semantics\"><span id=\"MathJax-Span-34355\" class=\"mrow\"><span id=\"MathJax-Span-34356\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span> 10<sup>\u221212<\/sup> <em data-effect=\"italics\">M<\/em><\/p>\n<p>11.\u00a0[OH<sup>\u2212<\/sup>] = 3.1 <span id=\"MathJax-Element-1296-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-34367\" class=\"math\"><span id=\"MathJax-Span-34368\" class=\"mrow\"><span id=\"MathJax-Span-34369\" class=\"semantics\"><span id=\"MathJax-Span-34370\" class=\"mrow\"><span id=\"MathJax-Span-34371\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span> 10<sup>\u221212<\/sup> <em data-effect=\"italics\">M<\/em><\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Glossary<\/h3>\n<p><strong>acidic<\/strong><br \/>\ndescribes a solution in which [H<sub>3<\/sub>O<sup>+<\/sup>] &gt; [OH<sup>\u2212<\/sup>]<\/p>\n<p><strong>basic<\/strong><br \/>\ndescribes a solution in which [H<sub>3<\/sub>O<sup>+<\/sup>] &lt; [OH<sup>\u2212<\/sup>]<\/p>\n<p><strong>neutral<\/strong><br \/>\ndescribes a solution in which [H<sub>3<\/sub>O<sup>+<\/sup>] = [OH<sup>\u2212<\/sup>]<\/p>\n<p><strong>pH<\/strong><br \/>\nlogarithmic measure of the concentration of hydronium ions in a solution<\/p>\n<p><strong>pOH<\/strong><br \/>\nlogarithmic measure of the concentration of hydroxide ions in a solution<\/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-4902\">\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. <strong>Provided by<\/strong>: OpenStax College. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/openstaxcollege.org\">http:\/\/openstaxcollege.org<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get<\/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":78,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry\",\"author\":\"\",\"organization\":\"OpenStax College\",\"url\":\"http:\/\/openstaxcollege.org\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-4902","chapter","type-chapter","status-publish","hentry"],"part":2988,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/4902","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/users\/78"}],"version-history":[{"count":5,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/4902\/revisions"}],"predecessor-version":[{"id":5542,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/4902\/revisions\/5542"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/parts\/2988"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapters\/4902\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/media?parent=4902"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/pressbooks\/v2\/chapter-type?post=4902"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/contributor?post=4902"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/wp-json\/wp\/v2\/license?post=4902"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}