{"id":7636,"date":"2021-06-08T21:57:07","date_gmt":"2021-06-08T21:57:07","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/the-ph-scale\/"},"modified":"2021-10-06T23:15:23","modified_gmt":"2021-10-06T23:15:23","slug":"the-ph-scale","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/the-ph-scale\/","title":{"raw":"The pH Scale","rendered":"The pH Scale"},"content":{"raw":"[latexpage]\r\n
\r\n

Learning Objectives<\/p>\r\n\r\n<\/header>\r\n

\r\n
    \r\n \t
  1. Define pH<\/em>.<\/li>\r\n \t
  2. Determine the pH of acidic and basic solutions.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nAs we have seen, [H+<\/sup>] and [OH\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.\r\n\r\npH<\/strong>\u00a0is a logarithmic function of [H+<\/sup>]:\r\n

    [latex]\\text{pH}=-\\log[\\ce{H+}][\/latex]<\/p>\r\npH is usually (but not always) between 0 and 14. Knowing the dependence of pH on [H+<\/sup>], we can summarize as follows:\r\n

      \r\n \t
    • If pH < 7, then the solution is acidic.<\/li>\r\n \t
    • If pH = 7, then the solution is neutral.<\/li>\r\n \t
    • If pH > 7, then the solution is basic.<\/li>\r\n<\/ul>\r\n

      This is known as the pH scale<\/strong>. You can use pH to make a quick determination whether a given aqueous solution is acidic, basic, or neutral.<\/p>\r\n\r\n

      \r\n

      Example 12.13<\/p>\r\n\r\n<\/header>\r\n

      \r\n\r\nLabel each solution as acidic, basic, or neutral based only on the stated pH.\r\n
        \r\n \t
      1. milk of magnesia, pH = 10.5<\/li>\r\n \t
      2. pure water, pH = 7<\/li>\r\n \t
      3. wine, pH = 3.0<\/li>\r\n<\/ol>\r\nSolution<\/em>\r\n
          \r\n \t
        1. With a pH greater than 7, milk of magnesia is basic. (Milk of magnesia is largely Mg(OH)2<\/sub>.)<\/li>\r\n \t
        2. Pure water, with a pH of 7, is neutral.<\/li>\r\n \t
        3. With a pH of less than 7, wine is acidic.<\/li>\r\n<\/ol>\r\nTest Yourself<\/em>\r\nIdentify each substance as acidic, basic, or neutral based only on the stated pH.\r\n
            \r\n \t
          1. human blood, pH = 7.4<\/li>\r\n \t
          2. household ammonia, pH = 11.0<\/li>\r\n \t
          3. cherries, pH = 3.6<\/li>\r\n<\/ol>\r\nAnswers<\/em>\r\n
              \r\n \t
            1. basic<\/li>\r\n \t
            2. basic<\/li>\r\n \t
            3. acidic<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nTable 12.3 \"Typical pH Values of Various Substances\" gives the typical pH values of some common substances. Note that several food items are on the list, and most of them are acidic.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
              Table 12.3 Typical pH Values of Various Substances[footnote]Actual values may vary depending on conditions.[\/footnote]<\/caption>\r\n
              Substance<\/th>\r\npH<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
              stomach acid<\/td>\r\n1.7<\/td>\r\n<\/tr>\r\n
              lemon juice<\/td>\r\n2.2<\/td>\r\n<\/tr>\r\n
              vinegar<\/td>\r\n2.9<\/td>\r\n<\/tr>\r\n
              soda<\/td>\r\n3.0<\/td>\r\n<\/tr>\r\n
              wine<\/td>\r\n3.5<\/td>\r\n<\/tr>\r\n
              coffee, black<\/td>\r\n5.0<\/td>\r\n<\/tr>\r\n
              milk<\/td>\r\n6.9<\/td>\r\n<\/tr>\r\n
              pure water<\/td>\r\n7.0<\/td>\r\n<\/tr>\r\n
              blood<\/td>\r\n7.4<\/td>\r\n<\/tr>\r\n
              seawater<\/td>\r\n8.5<\/td>\r\n<\/tr>\r\n
              milk of magnesia<\/td>\r\n10.5<\/td>\r\n<\/tr>\r\n
              ammonia solution<\/td>\r\n12.5<\/td>\r\n<\/tr>\r\n
              1.0 M NaOH<\/td>\r\n14.0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\npH is a logarithmic<\/em> scale. A solution that has a pH of 1.0 has 10 times the [H+<\/sup>] as a solution with a pH of 2.0, which in turn has 10 times the [H+<\/sup>] as a solution with a pH of 3.0 and so forth.\r\n\r\nUsing the definition of pH, it is also possible to calculate [H+<\/sup>] (and [OH\u2212<\/sup>]) from pH and vice versa. The general formula for determining [H+<\/sup>] from pH is as follows:\r\n

              [latex][\\ce{H+}]=10^{-\\text{pH}}[\/latex]<\/p>\r\nYou need to determine how to evaluate the above expression on your calculator. Ask your instructor if you have any questions.\r\n\r\nThe 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 after<\/em> the decimal point is what determines the number of significant figures in the final answer:\r\n\r\n\"X\r\n

              \r\n

              Example 12.14<\/p>\r\n\r\n<\/header>\r\n

              \r\n\r\nWhat are [H+<\/sup>] and [OH\u2212<\/sup>] for an aqueous solution whose pH is 4.88?\r\n\r\nSolution<\/em>\r\nWe need to evaluate the following expression:\r\n

              [latex][\\ce{H+}]=10^{-4.88}[\/latex]<\/p>\r\nDepending 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 \u201c10x<\/sup>\u201d key is pressed; for other calculators, the sequence of keystrokes is reversed. In any case, the correct numerical answer is as follows:\r\n

              [latex][\\ce{H+}]=1.3\\times 10^{-5}\\text{ M}[\/latex]<\/p>\r\nBecause 4.88 has two digits after the decimal point, [H+<\/sup>] is limited to two significant figures. From this, [OH\u2212<\/sup>] can be determined:\r\n

              [latex][\\ce{OH-}]=\\dfrac{1\\times 10^{-14}}{1.3\\times 10^{-5}}=7.7\\times 10^{-10}\\text{ M}[\/latex]<\/p>\r\nTest Yourself<\/em>\r\nWhat are [H+<\/sup>] and [OH\u2212<\/sup>] for an aqueous solution whose pH is 10.36?\r\n\r\nAnswer<\/em>\r\n[H+<\/sup>] = 4.4 \u00d7 10\u221211<\/sup> M; [OH\u2212<\/sup>] = 2.3 \u00d7 10\u22124<\/sup> M\r\n\r\n<\/div>\r\n<\/div>\r\nThere is an easier way to relate [H+<\/sup>] and [OH\u2212<\/sup>]. We can also define pOH<\/strong>\u00a0similar to pH:\r\n

              [latex]\\text{pOH}=-\\log[\\ce{OH-}][\/latex]<\/p>\r\n(In fact, p\u201canything\u201d is defined as the negative logarithm of that anything.) This also implies that:\r\n

              [latex][\\ce{OH-}]=10^{-\\text{pOH}}[\/latex]<\/p>\r\nA simple and useful relationship is that, for any aqueous solution:\r\n

              [latex]\\text{pH}+\\text{pOH}=14[\/latex]<\/p>\r\nThis relationship makes it simple to determine pH from pOH or pOH from pH and then calculate the resulting ion concentration.\r\n

              \r\n

              Example 12.15<\/p>\r\n\r\n<\/header>\r\n

              \r\n\r\nThe pH of a solution is 8.22. What are pOH, [H+<\/sup>], and [OH\u2212<\/sup>]?\r\n\r\nSolution<\/em>\r\nBecause the sum of pH and pOH equals 14, we have:\r\n

              [latex]8.22+\\text{pOH}=14[\/latex]<\/p>\r\nSubtracting 8.22 from 14, we get:\r\n

              [latex]\\text{pOH}=5.78[\/latex]<\/p>\r\nNow we evaluate the following two expressions:\r\n

              [latex]\\begin{array}{rcl}\r\n\\left[\\ce{H+}\\right]&=&10^{-8.22} \\\\ \\\\\r\n\\left[\\ce{OH-}\\right]&=&10^{-5.78}\r\n\\end{array}[\/latex]<\/p>\r\nSo:\r\n

              [latex]\\begin{array}{rcl}\r\n\\left[\\ce{H+}\\right]&=&6.0\\times 10^{-9}\\text{ M} \\\\ \\\\\r\n\\left[\\ce{OH-}\\right]&=&1.7\\times 10^{-6}\\text{ M}\r\n\\end{array}[\/latex]<\/p>\r\nTest Yourself<\/em>\r\nThe pOH of a solution is 12.04. What are pH, [H+<\/sup>], and [OH\u2212<\/sup>]?\r\n\r\nAnswer<\/em>\r\npH = 1.96; [H+<\/sup>] = 1.1 \u00d7 10\u22122<\/sup> M; [OH\u2212<\/sup>] = 9.1 \u00d7 10\u221213<\/sup> M\r\n\r\n<\/div>\r\n<\/div>\r\n

              \r\n

              Key Takeaways<\/p>\r\n\r\n<\/header>\r\n

              \r\n
                \r\n \t
              • pH is a logarithmic function of [H+<\/sup>].<\/li>\r\n \t
              • [H+<\/sup>] can be calculated directly from pH.<\/li>\r\n \t
              • pOH is related to pH and can be easily calculated from pH.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n
                \r\n

                Exercises<\/p>\r\n\r\n<\/header>\r\n

                \r\n

                Questions<\/h1>\r\n
                  \r\n \t
                1. Define pH<\/em>. How is it related to pOH?<\/li>\r\n \t
                2. Define pOH<\/em>. How is it related to pH?<\/li>\r\n \t
                3. What is the pH range for an acidic solution?<\/li>\r\n \t
                4. What is the pH range for a basic solution?<\/li>\r\n \t
                5. What is [H+<\/sup>] for a neutral solution?<\/li>\r\n \t
                6. What is [OH\u2212<\/sup>] for a neutral solution? Compare your answer to Exercise 5. Does this make sense?<\/li>\r\n \t
                7. Which substances in Table 12.3 \u201cTypical pH Values of Various Substances\u201d are acidic?<\/li>\r\n \t
                8. Which substances in Table 12.3 are basic?<\/li>\r\n \t
                9. What is the pH of a solution when [H+<\/sup>] is 3.44 \u00d7 10\u22124<\/sup> M?<\/li>\r\n \t
                10. What is the pH of a solution when [H+<\/sup>] is 9.04 \u00d7 10\u221213<\/sup> M?<\/li>\r\n \t
                11. What is the pH of a solution when [OH\u2212<\/sup>] is 6.22 \u00d7 10\u22127<\/sup> M?<\/li>\r\n \t
                12. What is the pH of a solution when [OH\u2212<\/sup>] is 0.0222 M?<\/li>\r\n \t
                13. What is the pOH of a solution when [H+<\/sup>] is 3.44 \u00d7 10\u22124<\/sup> M?<\/li>\r\n \t
                14. What is the pOH of a solution when [H+<\/sup>] is 9.04 \u00d7 10\u221213<\/sup> M?<\/li>\r\n \t
                15. What is the pOH of a solution when [OH\u2212<\/sup>] is 6.22 \u00d7 10\u22127<\/sup> M?<\/li>\r\n \t
                16. What is the pOH of a solution when [OH\u2212<\/sup>] is 0.0222 M?<\/li>\r\n \t
                17. If a solution has a pH of 0.77, what is its pOH, [H+<\/sup>], and [OH\u2212<\/sup>]?<\/li>\r\n \t
                18. If a solution has a pOH of 13.09, what is its pH, [H+<\/sup>], and [OH\u2212<\/sup>]?<\/li>\r\n<\/ol>\r\n

                  Answers<\/h1>\r\n
                    \r\n \t
                  1. pH is the negative logarithm of [H+<\/sup>] and is equal to 14 \u2212 pOH.<\/li>\r\n<\/ol>\r\n
                      \r\n \t
                    1. pH < 7<\/li>\r\n<\/ol>\r\n
                        \r\n \t
                      1. 1.0 \u00d7 10\u22127<\/sup> M<\/li>\r\n<\/ol>\r\n
                          \r\n \t
                        1. Every entry above pure water is acidic.<\/li>\r\n<\/ol>\r\n
                            \r\n \t
                          1. pH of 3.46<\/li>\r\n<\/ol>\r\n
                              \r\n \t
                            1. pH of 7.79<\/li>\r\n<\/ol>\r\n
                                \r\n \t
                              1. pOH of 10.54<\/li>\r\n<\/ol>\r\n
                                  \r\n \t
                                1. pOH of 6.21<\/li>\r\n<\/ol>\r\n
                                    \r\n \t
                                  1. pOH = 13.23; [H+<\/sup>] = 1.70 \u00d7 10\u22121<\/sup> M; [OH\u2212<\/sup>] = 5.89 \u00d7 10\u221214<\/sup> M<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>","rendered":"
                                    \n
                                    \n

                                    Learning Objectives<\/p>\n<\/header>\n

                                    \n
                                      \n
                                    1. Define pH<\/em>.<\/li>\n
                                    2. Determine the pH of acidic and basic solutions.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n

                                      As we have seen, [H+<\/sup>] and [OH\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

                                      pH<\/strong>\u00a0is a logarithmic function of [H+<\/sup>]:<\/p>\n

                                      \"\text{pH}=-\log[\ce{H+}]\"<\/p>\n

                                      pH is usually (but not always) between 0 and 14. Knowing the dependence of pH on [H+<\/sup>], we can summarize as follows:<\/p>\n