{"id":609,"date":"2017-12-14T21:42:12","date_gmt":"2017-12-14T21:42:12","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/the-ph-scale\/"},"modified":"2017-12-14T21:42:12","modified_gmt":"2017-12-14T21:42:12","slug":"the-ph-scale","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/chapter\/the-ph-scale\/","title":{"raw":"The pH Scale","rendered":"The pH Scale"},"content":{"raw":"<div class=\"section\" id=\"ball-ch12_s06\" lang=\"en\">\n<div class=\"learning_objectives editable block\" id=\"ball-ch12_s06_n01\">\n<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ol id=\"ball-ch12_s06_l01\"><li>Define <em>pH<\/em>.<\/li>\n\t<li>Determine the pH of acidic and basic solutions.<\/li>\n<\/ol><\/div>\n<\/div>\n<p id=\"ball-ch12_s06_p01\" class=\"para editable block\">As we have seen, [H<sup class=\"superscript\">+<\/sup>] and [OH<sup class=\"superscript\">\u2212<\/sup>] values can be markedly different from one aqueous solution to another. So chemists defined a new scale that succinctly indicates the concentrations of either of these two ions.<\/p>\n<p id=\"ball-ch12_s06_p02\" class=\"para editable block\"><span class=\"margin_term\"><a class=\"glossterm\">pH<\/a><\/span>\u00a0is a logarithmic function of [H<sup class=\"superscript\">+<\/sup>]:<\/p>\n<span class=\"informalequation block\"><span class=\"mathphrase\">pH = \u2212log[H<sup class=\"superscript\">+<\/sup>]<\/span><\/span>\n<p id=\"ball-ch12_s06_p03\" class=\"para editable block\">pH is usually (but not always) between 0 and 14. Knowing the dependence of pH on [H<sup class=\"superscript\">+<\/sup>], we can summarize as follows:<\/p>\n\n<ul id=\"ball-ch12_s06_l02\" class=\"itemizedlist editable block\"><li>If pH &lt; 7, then the solution is acidic.<\/li>\n\t<li>If pH = 7, then the solution is neutral.<\/li>\n\t<li>If pH &gt; 7, then the solution is basic.<\/li>\n<\/ul><p id=\"ball-ch12_s06_p04\" class=\"para editable block\">This is known as the <span class=\"margin_term\"><a class=\"glossterm\">pH scale<\/a><\/span>. You can use pH to make a quick determination whether a given aqueous solution is acidic, basic, or neutral.<\/p>\n\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 12<\/h3>\n<p id=\"ball-ch12_s06_p05\" class=\"para\">Label each solution as acidic, basic, or neutral based only on the stated pH.<\/p>\n\n<ol id=\"ball-ch12_s06_l03\" class=\"orderedlist\"><li>milk of magnesia, pH = 10.5<\/li>\n\t<li>pure water, pH = 7<\/li>\n\t<li>wine, pH = 3.0<\/li>\n<\/ol><p class=\"simpara\">Solution<\/p>\n\n<ol id=\"ball-ch12_s06_l04\" class=\"orderedlist\"><li>With a pH greater than 7, milk of magnesia is basic. (Milk of magnesia is largely Mg(OH)<sub class=\"subscript\">2<\/sub>.)<\/li>\n\t<li>Pure water, with a pH of 7, is neutral.<\/li>\n\t<li>With a pH of less than 7, wine is acidic.<\/li>\n<\/ol><p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch12_s06_p06\" class=\"para\">Identify each substance as acidic, basic, or neutral based only on the stated pH.<\/p>\n\n<ol id=\"ball-ch12_s06_l05\" class=\"orderedlist\"><li>human blood, pH = 7.4<\/li>\n\t<li>household ammonia, pH = 11.0<\/li>\n\t<li>cherries, pH = 3.6<\/li>\n<\/ol><p class=\"simpara\"><em class=\"emphasis\">Answers<\/em><\/p>\n\n<ol id=\"ball-ch12_s06_l06\" class=\"orderedlist\"><li>basic<\/li>\n\t<li>basic<\/li>\n\t<li>acidic<\/li>\n<\/ol><\/div>\n<p id=\"ball-ch12_s06_p07\" class=\"para editable block\"><a class=\"xref\" href=\"#ball-ch12_s06_t01\">Table 12.3 \"Typical pH Values of Various Substances*\"<\/a> gives the typical pH values of some common substances. Note that several food items are on the list, and most of them are acidic.<\/p>\n\n<div class=\"table block\" id=\"ball-ch12_s06_t01\">\n<p class=\"title\"><span class=\"title-prefix\">Table 12.3<\/span> Typical pH Values of Various Substances*<\/p>\n\n<table cellpadding=\"0\" style=\"border-spacing: 0px;\"><thead><tr><th>Substance<\/th>\n<th align=\"right\">pH<\/th>\n<\/tr><\/thead><tbody><tr><td>stomach acid<\/td>\n<td align=\"right\">1.7<\/td>\n<\/tr><tr><td>lemon juice<\/td>\n<td align=\"right\">2.2<\/td>\n<\/tr><tr><td>vinegar<\/td>\n<td align=\"right\">2.9<\/td>\n<\/tr><tr><td>soda<\/td>\n<td align=\"right\">3.0<\/td>\n<\/tr><tr><td>wine<\/td>\n<td align=\"right\">3.5<\/td>\n<\/tr><tr><td>coffee, black<\/td>\n<td align=\"right\">5.0<\/td>\n<\/tr><tr><td>milk<\/td>\n<td align=\"right\">6.9<\/td>\n<\/tr><tr><td>pure water<\/td>\n<td align=\"right\">7.0<\/td>\n<\/tr><tr><td>blood<\/td>\n<td align=\"right\">7.4<\/td>\n<\/tr><tr><td>seawater<\/td>\n<td align=\"right\">8.5<\/td>\n<\/tr><tr><td>milk of magnesia<\/td>\n<td align=\"right\">10.5<\/td>\n<\/tr><tr><td>ammonia solution<\/td>\n<td align=\"right\">12.5<\/td>\n<\/tr><tr><td>1.0 M NaOH<\/td>\n<td align=\"right\">14.0<\/td>\n<\/tr><tr><th colspan=\"2\">*Actual values may vary depending on conditions.<\/th>\n<\/tr><\/tbody><\/table><\/div>\n<p id=\"ball-ch12_s06_p08\" class=\"para editable block\">pH is a <em class=\"emphasis\">logarithmic<\/em> scale. A solution that has a pH of 1.0 has 10 times the [H<sup class=\"superscript\">+<\/sup>] as a solution with a pH of 2.0, which in turn has 10 times the [H<sup class=\"superscript\">+<\/sup>] as a solution with a pH of 3.0 and so forth.<\/p>\n<p id=\"ball-ch12_s06_p09\" class=\"para editable block\">Using the definition of pH, it is also possible to calculate [H<sup class=\"superscript\">+<\/sup>] (and [OH<sup class=\"superscript\">\u2212<\/sup>]) from pH and vice versa. The general formula for determining [H<sup class=\"superscript\">+<\/sup>] from pH is as follows:<\/p>\n<span class=\"informalequation block\"><span class=\"mathphrase\">[H<sup class=\"superscript\">+<\/sup>] = 10<sup class=\"superscript\">\u2212pH<\/sup><\/span><\/span>\n<p id=\"ball-ch12_s06_p10\" class=\"para editable block\">You need to determine how to evaluate the above expression on your calculator. Ask your instructor if you have any questions. The other issue that concerns us here is significant figures. Because the number(s) before the decimal point in a logarithm relate to the power on 10, the number of digits <em class=\"emphasis\">after<\/em> the decimal point is what determines the number of significant figures in the final answer:<\/p>\n<p class=\"para editable block\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/09\/X-and-Y.png\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14214211\/X-and-Y-1.png\" alt=\"X and Y\" width=\"600\" height=\"92\" class=\"alignnone size-full wp-image-4755\"\/><\/a><\/p>\n\n<div class=\"informalfigure large block\">\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 13<\/h3>\n<p id=\"ball-ch12_s06_p11\" class=\"para\">What are [H<sup class=\"superscript\">+<\/sup>] and [OH<sup class=\"superscript\">\u2212<\/sup>] for an aqueous solution whose pH is 4.88?<\/p>\n<p class=\"simpara\">Solution<\/p>\n<p id=\"ball-ch12_s06_p12\" class=\"para\">We need to evaluate the expression<\/p>\n<span class=\"informalequation\"><span class=\"mathphrase\">[H<sup class=\"superscript\">+<\/sup>] = 10<sup class=\"superscript\">\u22124.88<\/sup><\/span><\/span>\n<p id=\"ball-ch12_s06_p13\" class=\"para\">Depending on the calculator you use, the method for solving this problem will vary. In some cases, the \u201c\u22124.88\u201d is entered and a \u201c10<sup class=\"superscript\">x<\/sup>\u201d key is pressed; for other calculators, the sequence of keystrokes is reversed. In any case, the correct numerical answer is as follows:<\/p>\n<span class=\"informalequation\"><span class=\"mathphrase\">[H<sup class=\"superscript\">+<\/sup>] = 1.3 \u00d7 10<sup class=\"superscript\">\u22125<\/sup> M<\/span><\/span>\n<p id=\"ball-ch12_s06_p14\" class=\"para\">Because 4.88 has two digits after the decimal point, [H<sup class=\"superscript\">+<\/sup>] is limited to two significant figures. From this, [OH<sup class=\"superscript\">\u2212<\/sup>] can be determined:<\/p>\n<span class=\"informalequation\">[OH<sup>\u2212<\/sup>] = 1\u00d710<sup>\u221214<\/sup> \/\u00a01.3\u00d710<sup>\u22125\u00a0<\/sup>= 7.7\u00d710<sup>\u221210\u2009<\/sup>M<\/span>\n<p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch12_s06_p15\" class=\"para\">What are [H<sup class=\"superscript\">+<\/sup>] and [OH<sup class=\"superscript\">\u2212<\/sup>] for an aqueous solution whose pH is 10.36?<\/p>\n<p class=\"simpara\"><em class=\"emphasis\">Answer<\/em><\/p>\n<p id=\"ball-ch12_s06_p16\" class=\"para\">[H<sup class=\"superscript\">+<\/sup>] = 4.4 \u00d7 10<sup class=\"superscript\">\u221211<\/sup> M; [OH<sup class=\"superscript\">\u2212<\/sup>] = 2.3 \u00d7 10<sup class=\"superscript\">\u22124<\/sup> M<\/p>\n\n<\/div>\n<p id=\"ball-ch12_s06_p17\" class=\"para editable block\">There is an easier way to relate [H<sup class=\"superscript\">+<\/sup>] and [OH<sup class=\"superscript\">\u2212<\/sup>]. We can also define <span class=\"margin_term\"><a class=\"glossterm\">pOH<\/a><\/span>\u00a0similar to pH:<\/p>\n<span class=\"informalequation block\"><span class=\"mathphrase\">pOH = \u2212log[OH<sup class=\"superscript\">\u2212<\/sup>]<\/span><\/span>\n<p id=\"ball-ch12_s06_p18\" class=\"para editable block\">(In fact, p\u201canything\u201d is defined as the negative logarithm of that anything.) This also implies that<\/p>\n<span class=\"informalequation block\"><span class=\"mathphrase\">[OH<sup class=\"superscript\">\u2212<\/sup>] = 10<sup class=\"superscript\">\u2212pOH<\/sup><\/span><\/span>\n<p id=\"ball-ch12_s06_p19\" class=\"para editable block\">A simple and useful relationship is that for any aqueous solution,<\/p>\n<span class=\"informalequation block\"><span class=\"mathphrase\">pH +\u00a0pOH = 14<\/span><\/span>\n<p id=\"ball-ch12_s06_p20\" class=\"para editable block\">This relationship makes it simple to determine pH from pOH or pOH from pH and then calculate the resulting ion concentration.<\/p>\n\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 14<\/h3>\n<p id=\"ball-ch12_s06_p21\" class=\"para\">The pH of a solution is 8.22. What are pOH, [H<sup class=\"superscript\">+<\/sup>], and [OH<sup class=\"superscript\">\u2212<\/sup>]?<\/p>\n<p class=\"simpara\">Solution<\/p>\n<p id=\"ball-ch12_s06_p22\" class=\"para\">Because the sum of pH and pOH equals 14, we have<\/p>\n<span class=\"informalequation\"><span class=\"mathphrase\">8.22 +\u00a0pOH = 14<\/span><\/span>\n<p id=\"ball-ch12_s06_p23\" class=\"para\">Subtracting 8.22 from 14, we get<\/p>\n<span class=\"informalequation\"><span class=\"mathphrase\">pOH = 5.78<\/span><\/span>\n<p id=\"ball-ch12_s06_p24\" class=\"para\">Now we evaluate the following two expressions:<\/p>\n<span class=\"informalequation\"><span class=\"mathphrase\">[H<sup class=\"superscript\">+<\/sup>] = 10<sup class=\"superscript\">\u22128.22<\/sup><\/span><\/span>\n<span class=\"informalequation\"><span class=\"mathphrase\">[OH<sup class=\"superscript\">\u2212<\/sup>] = 10<sup class=\"superscript\">\u22125.78<\/sup><\/span><\/span>\n<p id=\"ball-ch12_s06_p25\" class=\"para\">So<\/p>\n<span class=\"informalequation\"><span class=\"mathphrase\">[H<sup class=\"superscript\">+<\/sup>] = 6.0 \u00d7 10<sup class=\"superscript\">\u22129<\/sup> M<\/span><\/span>\n<span class=\"informalequation\"><span class=\"mathphrase\">[OH<sup class=\"superscript\">\u2212<\/sup>] = 1.7 \u00d7 10<sup class=\"superscript\">\u22126<\/sup> M<\/span><\/span>\n<p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch12_s06_p26\" class=\"para\">The pOH of a solution is 12.04. What are pH, [H<sup class=\"superscript\">+<\/sup>], and [OH<sup class=\"superscript\">\u2212<\/sup>]?<\/p>\n<p class=\"simpara\"><em class=\"emphasis\">Answer<\/em><\/p>\n<p id=\"ball-ch12_s06_p27\" class=\"para\">pH = 1.96; [H<sup class=\"superscript\">+<\/sup>] = 1.1 \u00d7 10<sup class=\"superscript\">\u22122<\/sup> M; [OH<sup class=\"superscript\">\u2212<\/sup>] = 9.1 \u00d7 10<sup class=\"superscript\">\u221213<\/sup> M<\/p>\n\n<\/div>\n<div class=\"key_takeaways editable block\" id=\"ball-ch12_s06_n05\">\n<div class=\"bcc-box bcc-success\">\n<h3>Key Takeaways<\/h3>\n<ul id=\"ball-ch12_s06_l07\" class=\"itemizedlist\"><li>pH is a logarithmic function of [H<sup class=\"superscript\">+<\/sup>].<\/li>\n\t<li>[H<sup class=\"superscript\">+<\/sup>] can be calculated directly from pH.<\/li>\n\t<li>pOH is related to pH and can be easily calculated from pH.<\/li>\n<\/ul><\/div>\n\u00a0\n<div class=\"bcc-box bcc-info\">\n<h3>Exercises<\/h3>\n<div class=\"qandaset block\" id=\"ball-ch12_s06_qs01\">\n<ol id=\"ball-ch12_s06_qs01_qd01\" class=\"qandadiv\"><li id=\"ball-ch12_s06_qs01_qd01_qa01\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p01\" class=\"para\">Define <em class=\"emphasis\">pH<\/em>. How is it related to pOH?<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch12_s06_qs01_qd01_qa02\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p03\" class=\"para\">Define <em class=\"emphasis\">pOH<\/em>. How is it related to pH?<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch12_s06_qs01_qd01_qa03\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p05\" class=\"para\">What is the pH range for an acidic solution?<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch12_s06_qs01_qd01_qa04\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p07\" class=\"para\">What is the pH range for a basic solution?<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch12_s06_qs01_qd01_qa05\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p09\" class=\"para\">What is [H<sup class=\"superscript\">+<\/sup>] for a neutral solution?<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch12_s06_qs01_qd01_qa06\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p11\" class=\"para\">What is [OH<sup class=\"superscript\">\u2212<\/sup>] for a neutral solution? Compare your answer to Exercise 6. Does this make sense?<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch12_s06_qs01_qd01_qa07\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p13\" class=\"para\">Which substances in <a class=\"xref\" href=\"#ball-ch12_s06_t01\">Table 12.3 \"Typical pH Values of Various Substances*\"<\/a> are acidic?<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch12_s06_qs01_qd01_qa08\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p15\" class=\"para\">Which substances in <a class=\"xref\" href=\"#ball-ch12_s06_t01\">Table 12.3 \"Typical pH Values of Various Substances*\"<\/a> are basic?<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch12_s06_qs01_qd01_qa09\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p17\" class=\"para\">What is the pH of a solution when [H<sup class=\"superscript\">+<\/sup>] is 3.44 \u00d7 10<sup class=\"superscript\">\u22124<\/sup> M?<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch12_s06_qs01_qd01_qa10\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p19\" class=\"para\">What is the pH of a solution when [H<sup class=\"superscript\">+<\/sup>] is 9.04 \u00d7 10<sup class=\"superscript\">\u221213<\/sup> M?<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch12_s06_qs01_qd01_qa11\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p21\" class=\"para\">What is the pH of a solution when [OH<sup class=\"superscript\">\u2212<\/sup>] is 6.22 \u00d7 10<sup class=\"superscript\">\u22127<\/sup> M?<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch12_s06_qs01_qd01_qa12\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p23\" class=\"para\">What is the pH of a solution when [OH<sup class=\"superscript\">\u2212<\/sup>] is 0.0222 M?<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch12_s06_qs01_qd01_qa13\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p25\" class=\"para\">What is the pOH of a solution when [H<sup class=\"superscript\">+<\/sup>] is 3.44 \u00d7 10<sup class=\"superscript\">\u22124<\/sup> M?<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch12_s06_qs01_qd01_qa14\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p27\" class=\"para\">What is the pOH of a solution when [H<sup class=\"superscript\">+<\/sup>] is 9.04 \u00d7 10<sup class=\"superscript\">\u221213<\/sup> M?<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch12_s06_qs01_qd01_qa15\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p29\" class=\"para\">What is the pOH of a solution when [OH<sup class=\"superscript\">\u2212<\/sup>] is 6.22 \u00d7 10<sup class=\"superscript\">\u22127<\/sup> M?<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch12_s06_qs01_qd01_qa16\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p31\" class=\"para\">What is the pOH of a solution when [OH<sup class=\"superscript\">\u2212<\/sup>] is 0.0222 M?<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch12_s06_qs01_qd01_qa17\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p33\" class=\"para\">If a solution has a pH of 0.77, what is its pOH, [H<sup class=\"superscript\">+<\/sup>], and [OH<sup class=\"superscript\">\u2212<\/sup>]?<\/p>\n\n<\/div><\/li>\n\t<li id=\"ball-ch12_s06_qs01_qd01_qa18\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p35\" class=\"para\">If a solution has a pOH of 13.09, what is its pH, [H<sup class=\"superscript\">+<\/sup>], and [OH<sup class=\"superscript\">\u2212<\/sup>]?<\/p>\n\n<\/div><\/li>\n<\/ol><\/div>\n<b>Answers<\/b>\n\n<strong>1.<\/strong>\n\npH is the negative logarithm of [H<sup class=\"superscript\">+<\/sup>] and is equal to 14 \u2212 pOH.\n\n<strong>3.<\/strong>\n\npH &lt; 7\n\n<strong>5.<\/strong>\n\n1.0 \u00d7 10<sup class=\"superscript\">\u22127<\/sup> M\n\n<strong>7.<\/strong>\n\nEvery entry above pure water is acidic.\n\n<strong>9.<\/strong>\n\n3.46\n\n<strong>11.<\/strong>\n\n7.79\n\n<strong>13.<\/strong>\n\n10.54\n\n<strong>15.<\/strong>\n\n6.21\n\n<strong>17.<\/strong>\n\npOH = 13.23; [H<sup class=\"superscript\">+<\/sup>] = 1.70 \u00d7 10<sup class=\"superscript\">\u22121<\/sup> M; [OH<sup class=\"superscript\">\u2212<\/sup>] = 5.89 \u00d7 10<sup class=\"superscript\">\u221214<\/sup> M\n\n<\/div>\n<\/div>\n\u00a0\n\n<\/div><\/div>","rendered":"<div class=\"section\" id=\"ball-ch12_s06\" lang=\"en\">\n<div class=\"learning_objectives editable block\" id=\"ball-ch12_s06_n01\">\n<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ol id=\"ball-ch12_s06_l01\">\n<li>Define <em>pH<\/em>.<\/li>\n<li>Determine the pH of acidic and basic solutions.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p id=\"ball-ch12_s06_p01\" class=\"para editable block\">As we have seen, [H<sup class=\"superscript\">+<\/sup>] and [OH<sup class=\"superscript\">\u2212<\/sup>] values can be markedly different from one aqueous solution to another. So chemists defined a new scale that succinctly indicates the concentrations of either of these two ions.<\/p>\n<p id=\"ball-ch12_s06_p02\" class=\"para editable block\"><span class=\"margin_term\"><a class=\"glossterm\">pH<\/a><\/span>\u00a0is a logarithmic function of [H<sup class=\"superscript\">+<\/sup>]:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">pH = \u2212log[H<sup class=\"superscript\">+<\/sup>]<\/span><\/span><\/p>\n<p id=\"ball-ch12_s06_p03\" class=\"para editable block\">pH is usually (but not always) between 0 and 14. Knowing the dependence of pH on [H<sup class=\"superscript\">+<\/sup>], we can summarize as follows:<\/p>\n<ul id=\"ball-ch12_s06_l02\" class=\"itemizedlist editable block\">\n<li>If pH &lt; 7, then the solution is acidic.<\/li>\n<li>If pH = 7, then the solution is neutral.<\/li>\n<li>If pH &gt; 7, then the solution is basic.<\/li>\n<\/ul>\n<p id=\"ball-ch12_s06_p04\" class=\"para editable block\">This is known as the <span class=\"margin_term\"><a class=\"glossterm\">pH scale<\/a><\/span>. You can use pH to make a quick determination whether a given aqueous solution is acidic, basic, or neutral.<\/p>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 12<\/h3>\n<p id=\"ball-ch12_s06_p05\" class=\"para\">Label each solution as acidic, basic, or neutral based only on the stated pH.<\/p>\n<ol id=\"ball-ch12_s06_l03\" class=\"orderedlist\">\n<li>milk of magnesia, pH = 10.5<\/li>\n<li>pure water, pH = 7<\/li>\n<li>wine, pH = 3.0<\/li>\n<\/ol>\n<p class=\"simpara\">Solution<\/p>\n<ol id=\"ball-ch12_s06_l04\" class=\"orderedlist\">\n<li>With a pH greater than 7, milk of magnesia is basic. (Milk of magnesia is largely Mg(OH)<sub class=\"subscript\">2<\/sub>.)<\/li>\n<li>Pure water, with a pH of 7, is neutral.<\/li>\n<li>With a pH of less than 7, wine is acidic.<\/li>\n<\/ol>\n<p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch12_s06_p06\" class=\"para\">Identify each substance as acidic, basic, or neutral based only on the stated pH.<\/p>\n<ol id=\"ball-ch12_s06_l05\" class=\"orderedlist\">\n<li>human blood, pH = 7.4<\/li>\n<li>household ammonia, pH = 11.0<\/li>\n<li>cherries, pH = 3.6<\/li>\n<\/ol>\n<p class=\"simpara\"><em class=\"emphasis\">Answers<\/em><\/p>\n<ol id=\"ball-ch12_s06_l06\" class=\"orderedlist\">\n<li>basic<\/li>\n<li>basic<\/li>\n<li>acidic<\/li>\n<\/ol>\n<\/div>\n<p id=\"ball-ch12_s06_p07\" class=\"para editable block\"><a class=\"xref\" href=\"#ball-ch12_s06_t01\">Table 12.3 &#8220;Typical pH Values of Various Substances*&#8221;<\/a> gives the typical pH values of some common substances. Note that several food items are on the list, and most of them are acidic.<\/p>\n<div class=\"table block\" id=\"ball-ch12_s06_t01\">\n<p class=\"title\"><span class=\"title-prefix\">Table 12.3<\/span> Typical pH Values of Various Substances*<\/p>\n<table cellpadding=\"0\" style=\"border-spacing: 0px;\">\n<thead>\n<tr>\n<th>Substance<\/th>\n<th align=\"right\">pH<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>stomach acid<\/td>\n<td align=\"right\">1.7<\/td>\n<\/tr>\n<tr>\n<td>lemon juice<\/td>\n<td align=\"right\">2.2<\/td>\n<\/tr>\n<tr>\n<td>vinegar<\/td>\n<td align=\"right\">2.9<\/td>\n<\/tr>\n<tr>\n<td>soda<\/td>\n<td align=\"right\">3.0<\/td>\n<\/tr>\n<tr>\n<td>wine<\/td>\n<td align=\"right\">3.5<\/td>\n<\/tr>\n<tr>\n<td>coffee, black<\/td>\n<td align=\"right\">5.0<\/td>\n<\/tr>\n<tr>\n<td>milk<\/td>\n<td align=\"right\">6.9<\/td>\n<\/tr>\n<tr>\n<td>pure water<\/td>\n<td align=\"right\">7.0<\/td>\n<\/tr>\n<tr>\n<td>blood<\/td>\n<td align=\"right\">7.4<\/td>\n<\/tr>\n<tr>\n<td>seawater<\/td>\n<td align=\"right\">8.5<\/td>\n<\/tr>\n<tr>\n<td>milk of magnesia<\/td>\n<td align=\"right\">10.5<\/td>\n<\/tr>\n<tr>\n<td>ammonia solution<\/td>\n<td align=\"right\">12.5<\/td>\n<\/tr>\n<tr>\n<td>1.0 M NaOH<\/td>\n<td align=\"right\">14.0<\/td>\n<\/tr>\n<tr>\n<th colspan=\"2\">*Actual values may vary depending on conditions.<\/th>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p id=\"ball-ch12_s06_p08\" class=\"para editable block\">pH is a <em class=\"emphasis\">logarithmic<\/em> scale. A solution that has a pH of 1.0 has 10 times the [H<sup class=\"superscript\">+<\/sup>] as a solution with a pH of 2.0, which in turn has 10 times the [H<sup class=\"superscript\">+<\/sup>] as a solution with a pH of 3.0 and so forth.<\/p>\n<p id=\"ball-ch12_s06_p09\" class=\"para editable block\">Using the definition of pH, it is also possible to calculate [H<sup class=\"superscript\">+<\/sup>] (and [OH<sup class=\"superscript\">\u2212<\/sup>]) from pH and vice versa. The general formula for determining [H<sup class=\"superscript\">+<\/sup>] from pH is as follows:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">[H<sup class=\"superscript\">+<\/sup>] = 10<sup class=\"superscript\">\u2212pH<\/sup><\/span><\/span><\/p>\n<p id=\"ball-ch12_s06_p10\" class=\"para editable block\">You need to determine how to evaluate the above expression on your calculator. Ask your instructor if you have any questions. The other issue that concerns us here is significant figures. Because the number(s) before the decimal point in a logarithm relate to the power on 10, the number of digits <em class=\"emphasis\">after<\/em> the decimal point is what determines the number of significant figures in the final answer:<\/p>\n<p class=\"para editable block\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/09\/X-and-Y.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2017\/12\/14214211\/X-and-Y-1.png\" alt=\"X and Y\" width=\"600\" height=\"92\" class=\"alignnone size-full wp-image-4755\" \/><\/a><\/p>\n<div class=\"informalfigure large block\">\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 13<\/h3>\n<p id=\"ball-ch12_s06_p11\" class=\"para\">What are [H<sup class=\"superscript\">+<\/sup>] and [OH<sup class=\"superscript\">\u2212<\/sup>] for an aqueous solution whose pH is 4.88?<\/p>\n<p class=\"simpara\">Solution<\/p>\n<p id=\"ball-ch12_s06_p12\" class=\"para\">We need to evaluate the expression<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">[H<sup class=\"superscript\">+<\/sup>] = 10<sup class=\"superscript\">\u22124.88<\/sup><\/span><\/span><\/p>\n<p id=\"ball-ch12_s06_p13\" class=\"para\">Depending on the calculator you use, the method for solving this problem will vary. In some cases, the \u201c\u22124.88\u201d is entered and a \u201c10<sup class=\"superscript\">x<\/sup>\u201d key is pressed; for other calculators, the sequence of keystrokes is reversed. In any case, the correct numerical answer is as follows:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">[H<sup class=\"superscript\">+<\/sup>] = 1.3 \u00d7 10<sup class=\"superscript\">\u22125<\/sup> M<\/span><\/span><\/p>\n<p id=\"ball-ch12_s06_p14\" class=\"para\">Because 4.88 has two digits after the decimal point, [H<sup class=\"superscript\">+<\/sup>] is limited to two significant figures. From this, [OH<sup class=\"superscript\">\u2212<\/sup>] can be determined:<\/p>\n<p><span class=\"informalequation\">[OH<sup>\u2212<\/sup>] = 1\u00d710<sup>\u221214<\/sup> \/\u00a01.3\u00d710<sup>\u22125\u00a0<\/sup>= 7.7\u00d710<sup>\u221210\u2009<\/sup>M<\/span><\/p>\n<p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch12_s06_p15\" class=\"para\">What are [H<sup class=\"superscript\">+<\/sup>] and [OH<sup class=\"superscript\">\u2212<\/sup>] for an aqueous solution whose pH is 10.36?<\/p>\n<p class=\"simpara\"><em class=\"emphasis\">Answer<\/em><\/p>\n<p id=\"ball-ch12_s06_p16\" class=\"para\">[H<sup class=\"superscript\">+<\/sup>] = 4.4 \u00d7 10<sup class=\"superscript\">\u221211<\/sup> M; [OH<sup class=\"superscript\">\u2212<\/sup>] = 2.3 \u00d7 10<sup class=\"superscript\">\u22124<\/sup> M<\/p>\n<\/div>\n<p id=\"ball-ch12_s06_p17\" class=\"para editable block\">There is an easier way to relate [H<sup class=\"superscript\">+<\/sup>] and [OH<sup class=\"superscript\">\u2212<\/sup>]. We can also define <span class=\"margin_term\"><a class=\"glossterm\">pOH<\/a><\/span>\u00a0similar to pH:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">pOH = \u2212log[OH<sup class=\"superscript\">\u2212<\/sup>]<\/span><\/span><\/p>\n<p id=\"ball-ch12_s06_p18\" class=\"para editable block\">(In fact, p\u201canything\u201d is defined as the negative logarithm of that anything.) This also implies that<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">[OH<sup class=\"superscript\">\u2212<\/sup>] = 10<sup class=\"superscript\">\u2212pOH<\/sup><\/span><\/span><\/p>\n<p id=\"ball-ch12_s06_p19\" class=\"para editable block\">A simple and useful relationship is that for any aqueous solution,<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">pH +\u00a0pOH = 14<\/span><\/span><\/p>\n<p id=\"ball-ch12_s06_p20\" class=\"para editable block\">This relationship makes it simple to determine pH from pOH or pOH from pH and then calculate the resulting ion concentration.<\/p>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 14<\/h3>\n<p id=\"ball-ch12_s06_p21\" class=\"para\">The pH of a solution is 8.22. What are pOH, [H<sup class=\"superscript\">+<\/sup>], and [OH<sup class=\"superscript\">\u2212<\/sup>]?<\/p>\n<p class=\"simpara\">Solution<\/p>\n<p id=\"ball-ch12_s06_p22\" class=\"para\">Because the sum of pH and pOH equals 14, we have<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">8.22 +\u00a0pOH = 14<\/span><\/span><\/p>\n<p id=\"ball-ch12_s06_p23\" class=\"para\">Subtracting 8.22 from 14, we get<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">pOH = 5.78<\/span><\/span><\/p>\n<p id=\"ball-ch12_s06_p24\" class=\"para\">Now we evaluate the following two expressions:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">[H<sup class=\"superscript\">+<\/sup>] = 10<sup class=\"superscript\">\u22128.22<\/sup><\/span><\/span><br \/>\n<span class=\"informalequation\"><span class=\"mathphrase\">[OH<sup class=\"superscript\">\u2212<\/sup>] = 10<sup class=\"superscript\">\u22125.78<\/sup><\/span><\/span><\/p>\n<p id=\"ball-ch12_s06_p25\" class=\"para\">So<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">[H<sup class=\"superscript\">+<\/sup>] = 6.0 \u00d7 10<sup class=\"superscript\">\u22129<\/sup> M<\/span><\/span><br \/>\n<span class=\"informalequation\"><span class=\"mathphrase\">[OH<sup class=\"superscript\">\u2212<\/sup>] = 1.7 \u00d7 10<sup class=\"superscript\">\u22126<\/sup> M<\/span><\/span><\/p>\n<p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch12_s06_p26\" class=\"para\">The pOH of a solution is 12.04. What are pH, [H<sup class=\"superscript\">+<\/sup>], and [OH<sup class=\"superscript\">\u2212<\/sup>]?<\/p>\n<p class=\"simpara\"><em class=\"emphasis\">Answer<\/em><\/p>\n<p id=\"ball-ch12_s06_p27\" class=\"para\">pH = 1.96; [H<sup class=\"superscript\">+<\/sup>] = 1.1 \u00d7 10<sup class=\"superscript\">\u22122<\/sup> M; [OH<sup class=\"superscript\">\u2212<\/sup>] = 9.1 \u00d7 10<sup class=\"superscript\">\u221213<\/sup> M<\/p>\n<\/div>\n<div class=\"key_takeaways editable block\" id=\"ball-ch12_s06_n05\">\n<div class=\"bcc-box bcc-success\">\n<h3>Key Takeaways<\/h3>\n<ul id=\"ball-ch12_s06_l07\" class=\"itemizedlist\">\n<li>pH is a logarithmic function of [H<sup class=\"superscript\">+<\/sup>].<\/li>\n<li>[H<sup class=\"superscript\">+<\/sup>] can be calculated directly from pH.<\/li>\n<li>pOH is related to pH and can be easily calculated from pH.<\/li>\n<\/ul>\n<\/div>\n<p>\u00a0<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Exercises<\/h3>\n<div class=\"qandaset block\" id=\"ball-ch12_s06_qs01\">\n<ol id=\"ball-ch12_s06_qs01_qd01\" class=\"qandadiv\">\n<li id=\"ball-ch12_s06_qs01_qd01_qa01\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p01\" class=\"para\">Define <em class=\"emphasis\">pH<\/em>. How is it related to pOH?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s06_qs01_qd01_qa02\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p03\" class=\"para\">Define <em class=\"emphasis\">pOH<\/em>. How is it related to pH?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s06_qs01_qd01_qa03\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p05\" class=\"para\">What is the pH range for an acidic solution?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s06_qs01_qd01_qa04\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p07\" class=\"para\">What is the pH range for a basic solution?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s06_qs01_qd01_qa05\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p09\" class=\"para\">What is [H<sup class=\"superscript\">+<\/sup>] for a neutral solution?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s06_qs01_qd01_qa06\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p11\" class=\"para\">What is [OH<sup class=\"superscript\">\u2212<\/sup>] for a neutral solution? Compare your answer to Exercise 6. Does this make sense?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s06_qs01_qd01_qa07\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p13\" class=\"para\">Which substances in <a class=\"xref\" href=\"#ball-ch12_s06_t01\">Table 12.3 &#8220;Typical pH Values of Various Substances*&#8221;<\/a> are acidic?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s06_qs01_qd01_qa08\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p15\" class=\"para\">Which substances in <a class=\"xref\" href=\"#ball-ch12_s06_t01\">Table 12.3 &#8220;Typical pH Values of Various Substances*&#8221;<\/a> are basic?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s06_qs01_qd01_qa09\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p17\" class=\"para\">What is the pH of a solution when [H<sup class=\"superscript\">+<\/sup>] is 3.44 \u00d7 10<sup class=\"superscript\">\u22124<\/sup> M?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s06_qs01_qd01_qa10\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p19\" class=\"para\">What is the pH of a solution when [H<sup class=\"superscript\">+<\/sup>] is 9.04 \u00d7 10<sup class=\"superscript\">\u221213<\/sup> M?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s06_qs01_qd01_qa11\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p21\" class=\"para\">What is the pH of a solution when [OH<sup class=\"superscript\">\u2212<\/sup>] is 6.22 \u00d7 10<sup class=\"superscript\">\u22127<\/sup> M?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s06_qs01_qd01_qa12\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p23\" class=\"para\">What is the pH of a solution when [OH<sup class=\"superscript\">\u2212<\/sup>] is 0.0222 M?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s06_qs01_qd01_qa13\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p25\" class=\"para\">What is the pOH of a solution when [H<sup class=\"superscript\">+<\/sup>] is 3.44 \u00d7 10<sup class=\"superscript\">\u22124<\/sup> M?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s06_qs01_qd01_qa14\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p27\" class=\"para\">What is the pOH of a solution when [H<sup class=\"superscript\">+<\/sup>] is 9.04 \u00d7 10<sup class=\"superscript\">\u221213<\/sup> M?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s06_qs01_qd01_qa15\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p29\" class=\"para\">What is the pOH of a solution when [OH<sup class=\"superscript\">\u2212<\/sup>] is 6.22 \u00d7 10<sup class=\"superscript\">\u22127<\/sup> M?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s06_qs01_qd01_qa16\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p31\" class=\"para\">What is the pOH of a solution when [OH<sup class=\"superscript\">\u2212<\/sup>] is 0.0222 M?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s06_qs01_qd01_qa17\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p33\" class=\"para\">If a solution has a pH of 0.77, what is its pOH, [H<sup class=\"superscript\">+<\/sup>], and [OH<sup class=\"superscript\">\u2212<\/sup>]?<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s06_qs01_qd01_qa18\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s06_qs01_p35\" class=\"para\">If a solution has a pOH of 13.09, what is its pH, [H<sup class=\"superscript\">+<\/sup>], and [OH<sup class=\"superscript\">\u2212<\/sup>]?<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<p><b>Answers<\/b><\/p>\n<p><strong>1.<\/strong><\/p>\n<p>pH is the negative logarithm of [H<sup class=\"superscript\">+<\/sup>] and is equal to 14 \u2212 pOH.<\/p>\n<p><strong>3.<\/strong><\/p>\n<p>pH &lt; 7<\/p>\n<p><strong>5.<\/strong><\/p>\n<p>1.0 \u00d7 10<sup class=\"superscript\">\u22127<\/sup> M<\/p>\n<p><strong>7.<\/strong><\/p>\n<p>Every entry above pure water is acidic.<\/p>\n<p><strong>9.<\/strong><\/p>\n<p>3.46<\/p>\n<p><strong>11.<\/strong><\/p>\n<p>7.79<\/p>\n<p><strong>13.<\/strong><\/p>\n<p>10.54<\/p>\n<p><strong>15.<\/strong><\/p>\n<p>6.21<\/p>\n<p><strong>17.<\/strong><\/p>\n<p>pOH = 13.23; [H<sup class=\"superscript\">+<\/sup>] = 1.70 \u00d7 10<sup class=\"superscript\">\u22121<\/sup> M; [OH<sup class=\"superscript\">\u2212<\/sup>] = 5.89 \u00d7 10<sup class=\"superscript\">\u221214<\/sup> M<\/p>\n<\/div>\n<\/div>\n<p>\u00a0<\/p>\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-609\">\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>Introductory Chemistry- 1st Canadian Edition . <strong>Authored by<\/strong>: Jessie A. Key and David W. Ball. <strong>Provided by<\/strong>: BCCampus. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/opentextbc.ca\/introductorychemistry\/\">https:\/\/opentextbc.ca\/introductorychemistry\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Download this book for free at http:\/\/open.bccampus.ca<\/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":23485,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Introductory Chemistry- 1st Canadian Edition \",\"author\":\"Jessie A. Key and David W. 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