{"id":7631,"date":"2021-06-08T21:57:06","date_gmt":"2021-06-08T21:57:06","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/autoionization-of-water\/"},"modified":"2021-10-06T22:15:01","modified_gmt":"2021-10-06T22:15:01","slug":"autoionization-of-water","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/autoionization-of-water\/","title":{"raw":"Autoionization of Water","rendered":"Autoionization of Water"},"content":{"raw":"[latexpage]\r\n
\r\n

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

\r\n
    \r\n \t
  1. Describe the autoionization of water.<\/li>\r\n \t
  2. Calculate the concentrations of H+<\/sup> and OH\u2212<\/sup> in solutions, knowing the other concentration.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nWe have already seen that H2<\/sub>O can act as an acid or a base:\r\n

    [latex]\\begin{array}{lcll}\r\n\\ce{NH3}+\\ce{H2O}&\\rightarrow&\\ce{NH4^+}+\\ce{OH^-}&\\hspace{5mm}\\ce{H2O}\\text{ acts as an acid} \\\\\r\n\\ce{HCl}+\\ce{H2O}&\\rightarrow&\\ce{H3O^+}+\\ce{Cl^-}&\\hspace{5mm}\\ce{H2O}\\text{ acts as a base}\r\n\\end{array}[\/latex]<\/p>\r\nIt may not surprise you to learn, then, that within any given sample of water, some H2<\/sub>O molecules are acting as acids, and other H2<\/sub>O molecules are acting as bases. The chemical equation is as follows:\r\n

    [latex]\\ce{H2O}+\\ce{H2O}\\rightarrow \\ce{H3O+}+\\ce{OH-}[\/latex]<\/p>\r\nThis occurs only to a very small degree: only about 6 in 108<\/sup> H2<\/sub>O molecules are participating in this process, which is called the autoionization of water<\/strong>. At this level, the concentration of both H+<\/sup>(aq) and OH\u2212<\/sup>(aq) in a sample of pure H2<\/sub>O is about 1.0 \u00d7 10\u22127<\/sup> M. If we use square brackets around a dissolved species to imply the molar concentration of that species, we have:\r\n

    [latex][\\ce{H+}]=[\\ce{OH-}]=1.0\\times 10^{-7}\\text{ M}[\/latex]<\/p>\r\nfor any<\/em> sample of pure water, because H2<\/sub>O can act as both an acid and a base. The product of these two concentrations is 1.0 \u00d7 10\u221214<\/sup>:\r\n

    [latex][\\ce{H+}]\\times [\\ce{OH-}]=(1.0\\times 10^{-7})(1.0\\times 10^{-7})=1.0\\times 10^{-14}[\/latex]<\/p>\r\nIn acids, the concentration of H+<\/sup>(aq) \u2014 written as [H+<\/sup>] \u2014 is greater than 1.0 \u00d7 10\u22127<\/sup> M, while for bases the concentration of OH\u2212<\/sup>(aq) \u2014 [OH\u2212<\/sup>] \u2014 is greater than 1.0 \u00d7 10\u22127<\/sup> M. However, the product<\/em> of the two concentrations \u2014 [H+<\/sup>][OH\u2212<\/sup>] \u2014 is always<\/em> equal to 1.0 \u00d7 10\u221214<\/sup>, no matter whether the aqueous solution is an acid, a base, or neutral:\r\n

    [latex][\\ce{H+}][\\ce{OH-}]=1.0\\times 10^{-14}[\/latex]<\/p>\r\nThis value of the product of concentrations is so important for aqueous solutions that it is called the autoionization constant of water<\/strong>\u00a0and is denoted K<\/em>w<\/sub>:\r\n

    [latex]K_{\\text{w}}=[\\ce{H+}][\\ce{OH-}]=1.0\\times 10^{-14}[\/latex]<\/p>\r\nThis means that if you know [H+<\/sup>] for a solution, you can calculate what [OH\u2212<\/sup>] has to be for the product to equal 1.0 \u00d7 10\u221214<\/sup>, or if you know [OH\u2212<\/sup>], you can calculate [H+<\/sup>]. This also implies that as one concentration goes up, the other must go down to compensate so that their product always equals the value of K<\/em>w<\/sub>.\r\n

    \r\n

    Example 12.9<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nWhat is [OH\u2212<\/sup>] of an aqueous solution if [H+<\/sup>] is 1.0 \u00d7 10\u22124<\/sup> M?\r\n\r\nSolution<\/em>\r\nUsing the expression and known value for K<\/em>w<\/sub>:\r\n

    [latex]K_{\\text{w}}=[\\ce{H+}][\\ce{OH-}]=1.0\\times 10^{-14}=(1.0\\times 10^{-4})[\\ce{OH-}][\/latex]<\/p>\r\nWe solve by dividing both sides of the equation by 1.0 \u00d7 10\u22124<\/sup>:\r\n

    [latex][\\ce{OH-}]=\\dfrac{1.0\\times 10^{-14}}{1.0\\times 10^{-4}}=1.0\\times 10^{-10}\\text{ M}[\/latex]<\/p>\r\nIt is assumed that the concentration unit is molarity, so [OH\u2212<\/sup>] is 1.0 \u00d7 10\u221210<\/sup> M.\r\n\r\nTest Yourself<\/em>\r\nWhat is [H+<\/sup>] of an aqueous solution if [OH\u2212<\/sup>] is 1.0 \u00d7 10\u22129<\/sup> M?\r\n\r\nAnswer<\/em>\r\n1.0 \u00d7 10\u22125<\/sup> M\r\n\r\n<\/div>\r\n<\/div>\r\nWhen you have a solution of a particular acid or base, you need to look at the formula of the acid or base to determine the number of H+<\/sup> or OH\u2212<\/sup> ions in the formula unit because [H+<\/sup>] or [OH\u2212<\/sup>] may not be the same as the concentration of the acid or base itself.\r\n

    \r\n

    Example 12.10<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nWhat is [H+<\/sup>] in a 0.0044 M solution of Ca(OH)2<\/sub>?\r\n\r\nSolution<\/em>\r\nWe begin by determining [OH\u2212<\/sup>].\r\n\r\nThe concentration of the solute is 0.0044 M, but because Ca(OH)2<\/sub> is a strong base, there are two OH\u2212<\/sup> ions in solution for every formula unit dissolved, so the actual [OH\u2212<\/sup>] is two times this, or 2 \u00d7 0.0044 M = 0.0088 M.\r\n\r\nNow we can use the K<\/em>w<\/sub> expression:\r\n

    [latex][\\ce{H+}][\\ce{OH-}]=1.0\\times 10^{-14}=[\\ce{H+}](0.0088\\text{ M})[\/latex]<\/p>\r\nDividing both sides by 0.0088:\r\n

    [latex][\\ce{H+}]=\\dfrac{1.0\\times 10^{-14}}{0.0088}=1.1\\times 10^{-12}\\text{ M}[\/latex]<\/p>\r\n[H+<\/sup>] has decreased significantly in this basic solution.\r\n\r\nTest Yourself<\/em>\r\nWhat is [OH\u2212<\/sup>] in a 0.00032 M solution of H2<\/sub>SO4<\/sub>? (Hint: assume both H+<\/sup> ions ionize.)\r\n\r\nAnswer<\/em>\r\n1.6 \u00d7 10\u221211<\/sup> M\r\n\r\n<\/div>\r\n<\/div>\r\nFor strong acids and bases, [H+<\/sup>] and [OH\u2212<\/sup>] can be determined directly from the concentration of the acid or base itself because these ions are 100% ionized by definition. However, for weak acids and bases, this is not so. The degree, or percentage, of ionization would need to be known before we can determine [H+<\/sup>] and [OH\u2212<\/sup>].\r\n

    \r\n

    Example 12.11<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nA 0.0788 M solution of HC2<\/sub>H3<\/sub>O2<\/sub> is 3.0% ionized into H+<\/sup> ions and C2<\/sub>H3<\/sub>O2<\/sub>\u2212<\/sup> ions. What are [H+<\/sup>] and [OH\u2212<\/sup>] for this solution?\r\n\r\nSolution<\/em>\r\nBecause the acid is only 3.0% ionized, we can determine [H+<\/sup>] from the concentration of the acid. Recall that 3.0% is 0.030 in decimal form:\r\n

    [latex][\\ce{H+}]=0.030\\times 0.0788=0.00236\\text{ M}[\/latex]<\/p>\r\nWith this [H+<\/sup>], then [OH\u2212<\/sup>] can be calculated as follows:\r\n

    [latex][\\ce{OH-}]=\\dfrac{1.0\\times 10^{-14}}{0.00236}=4.2\\times 10^{-12}\\text{ M}[\/latex]<\/p>\r\nThis is about 30 times higher than would be expected for a strong acid of the same concentration.\r\n\r\nTest Yourself<\/em>\r\nA 0.0222 M solution of pyridine (C5<\/sub>H5<\/sub>N) is 0.44% ionized into pyridinium ions (C5<\/sub>H5<\/sub>NH+<\/sup>) and OH\u2212<\/sup> ions. What are [OH\u2212<\/sup>] and [H+<\/sup>] for this solution?\r\n\r\nAnswer<\/em>\r\n[OH\u2212<\/sup>] = 9.77 \u00d7 10\u22125<\/sup> M; [H+<\/sup>] = 1.02 \u00d7 10\u221210<\/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
    • In any aqueous solution, the product of [H+<\/sup>] and [OH\u2212<\/sup>] equals 1.0 \u00d7 10\u221214<\/sup>.<\/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. Does [H+<\/sup>] remain constant in all aqueous solutions? Why or why not?<\/li>\r\n \t
      2. Does [OH\u2212<\/sup>] remain constant in all aqueous solutions? Why or why not?<\/li>\r\n \t
      3. What is the relationship between [H+<\/sup>] and K<\/em>w<\/sub>? Write a mathematical expression that relates them.<\/li>\r\n \t
      4. What is the relationship between [OH\u2212<\/sup>] and K<\/em>w<\/sub>? Write a mathematical expression that relates them.<\/li>\r\n \t
      5. Write the chemical equation for the autoionization of water and label the conjugate acid-base pairs.<\/li>\r\n \t
      6. Write the reverse of the reaction for the autoionization of water. It is still an acid-base reaction? If so, label the acid and base.<\/li>\r\n \t
      7. For a given aqueous solution, if [H+<\/sup>] = 1.0 \u00d7 10\u22123<\/sup> M, what is [OH\u2212<\/sup>]?<\/li>\r\n \t
      8. For a given aqueous solution, if [H+<\/sup>] = 1.0 \u00d7 10\u22129<\/sup> M, what is [OH\u2212<\/sup>]?<\/li>\r\n \t
      9. For a given aqueous solution, if [H+<\/sup>] = 7.92 \u00d7 10\u22125<\/sup> M, what is [OH\u2212<\/sup>]?<\/li>\r\n \t
      10. For a given aqueous solution, if [H+<\/sup>] = 2.07 \u00d7 10\u221211<\/sup> M, what is [H+<\/sup>]?<\/li>\r\n \t
      11. For a given aqueous solution, if [OH\u2212<\/sup>] = 1.0 \u00d7 10\u22125<\/sup> M, what is [H+<\/sup>]?<\/li>\r\n \t
      12. For a given aqueous solution, if [OH\u2212<\/sup>] = 1.0 \u00d7 10\u221212<\/sup> M, what is [H+<\/sup>]?<\/li>\r\n \t
      13. For a given aqueous solution, if [OH\u2212<\/sup>] = 3.77 \u00d7 10\u22124<\/sup> M, what is [H+<\/sup>]?<\/li>\r\n \t
      14. For a given aqueous solution, if [OH\u2212<\/sup>] = 7.11 \u00d7 10\u221210<\/sup> M, what is [H+<\/sup>]?<\/li>\r\n \t
      15. What are [H+<\/sup>] and [OH\u2212<\/sup>] in a 0.344 M solution of HNO3<\/sub>?<\/li>\r\n \t
      16. What are [H+<\/sup>] and [OH\u2212<\/sup>] in a 2.86 M solution of HBr?<\/li>\r\n \t
      17. What are [H+<\/sup>] and [OH\u2212<\/sup>] in a 0.00338 M solution of KOH?<\/li>\r\n \t
      18. What are [H+<\/sup>] and [OH\u2212<\/sup>] in a 6.02 \u00d7 10\u22124<\/sup> M solution of Ca(OH)2<\/sub>?<\/li>\r\n \t
      19. If HNO2<\/sub> is dissociated only to an extent of 0.445%, what are [H+<\/sup>] and [OH\u2212<\/sup>] in a 0.307 M solution of HNO2<\/sub>?<\/li>\r\n \t
      20. If (C2<\/sub>H5<\/sub>)2<\/sub>NH is dissociated only to an extent of 0.077%, what are [H+<\/sup>] and [OH\u2212<\/sup>] in a 0.0955 M solution of (C2<\/sub>H5<\/sub>)2<\/sub>NH?<\/li>\r\n<\/ol>\r\n

        Answers<\/h1>\r\n
          \r\n \t
        1. [H+<\/sup>] varies with the amount of acid or base in a solution.<\/li>\r\n<\/ol>\r\n
            \r\n \t
          1. [H+<\/sup>] = K<\/em>w<\/sub>[OH\u2212<\/sup>]<\/li>\r\n<\/ol>\r\n
              \r\n \t
            1. H2<\/sub>O +\u00a0H2<\/sub>O \u2192\u00a0H3<\/sub>O+<\/sup> +\u00a0OH\u2212<\/sup>; H2<\/sub>O\/H3<\/sub>O+<\/sup> and H2<\/sub>O\/OH\u2212<\/sup><\/li>\r\n<\/ol>\r\n
                \r\n \t
              1. 1.0 \u00d7 10\u221211<\/sup> M<\/li>\r\n<\/ol>\r\n
                  \r\n \t
                1. 1.26 \u00d7 10\u221210<\/sup> M<\/li>\r\n<\/ol>\r\n
                    \r\n \t
                  1. 1.0 \u00d7 10\u22129<\/sup> M<\/li>\r\n<\/ol>\r\n
                      \r\n \t
                    1. 2.65 \u00d7 10\u221211<\/sup> M<\/li>\r\n<\/ol>\r\n
                        \r\n \t
                      1. [H+<\/sup>] = 0.344 M; [OH\u2212<\/sup>] = 2.91 \u00d7 10\u221214<\/sup> M<\/li>\r\n<\/ol>\r\n
                          \r\n \t
                        1. [OH\u2212<\/sup>] = 0.00338 M; [H+<\/sup>] = 2.96 \u00d7 10\u221212<\/sup> M<\/li>\r\n<\/ol>\r\n
                            \r\n \t
                          1. [H+<\/sup>] = 0.00137 M; [OH\u2212<\/sup>] = 7.32 \u00d7 10\u221212<\/sup> M<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>","rendered":"
                            \n
                            \n

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

                            \n
                              \n
                            1. Describe the autoionization of water.<\/li>\n
                            2. Calculate the concentrations of H+<\/sup> and OH\u2212<\/sup> in solutions, knowing the other concentration.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n

                              We have already seen that H2<\/sub>O can act as an acid or a base:<\/p>\n

                              \"\begin{array}{lcll}<\/p>\n

                              It may not surprise you to learn, then, that within any given sample of water, some H2<\/sub>O molecules are acting as acids, and other H2<\/sub>O molecules are acting as bases. The chemical equation is as follows:<\/p>\n

                              \"\ce{H2O}+\ce{H2O}\rightarrow \ce{H3O+}+\ce{OH-}\"<\/p>\n

                              This occurs only to a very small degree: only about 6 in 108<\/sup> H2<\/sub>O molecules are participating in this process, which is called the autoionization of water<\/strong>. At this level, the concentration of both H+<\/sup>(aq) and OH\u2212<\/sup>(aq) in a sample of pure H2<\/sub>O is about 1.0 \u00d7 10\u22127<\/sup> M. If we use square brackets around a dissolved species to imply the molar concentration of that species, we have:<\/p>\n

                              \"[\ce{H+}]=[\ce{OH-}]=1.0\times 10^{-7}\text{ M}\"<\/p>\n

                              for any<\/em> sample of pure water, because H2<\/sub>O can act as both an acid and a base. The product of these two concentrations is 1.0 \u00d7 10\u221214<\/sup>:<\/p>\n

                              \"[\ce{H+}]\times [\ce{OH-}]=(1.0\times 10^{-7})(1.0\times 10^{-7})=1.0\times 10^{-14}\"<\/p>\n

                              In acids, the concentration of H+<\/sup>(aq) \u2014 written as [H+<\/sup>] \u2014 is greater than 1.0 \u00d7 10\u22127<\/sup> M, while for bases the concentration of OH\u2212<\/sup>(aq) \u2014 [OH\u2212<\/sup>] \u2014 is greater than 1.0 \u00d7 10\u22127<\/sup> M. However, the product<\/em> of the two concentrations \u2014 [H+<\/sup>][OH\u2212<\/sup>] \u2014 is always<\/em> equal to 1.0 \u00d7 10\u221214<\/sup>, no matter whether the aqueous solution is an acid, a base, or neutral:<\/p>\n

                              \"[\ce{H+}][\ce{OH-}]=1.0\times 10^{-14}\"<\/p>\n

                              This value of the product of concentrations is so important for aqueous solutions that it is called the autoionization constant of water<\/strong>\u00a0and is denoted K<\/em>w<\/sub>:<\/p>\n

                              \"K_{\text{w}}=[\ce{H+}][\ce{OH-}]=1.0\times 10^{-14}\"<\/p>\n

                              This means that if you know [H+<\/sup>] for a solution, you can calculate what [OH\u2212<\/sup>] has to be for the product to equal 1.0 \u00d7 10\u221214<\/sup>, or if you know [OH\u2212<\/sup>], you can calculate [H+<\/sup>]. This also implies that as one concentration goes up, the other must go down to compensate so that their product always equals the value of K<\/em>w<\/sub>.<\/p>\n

                              \n
                              \n

                              Example 12.9<\/p>\n<\/header>\n

                              \n

                              What is [OH\u2212<\/sup>] of an aqueous solution if [H+<\/sup>] is 1.0 \u00d7 10\u22124<\/sup> M?<\/p>\n

                              Solution<\/em>
                              \nUsing the expression and known value for K<\/em>w<\/sub>:<\/p>\n

                              \"K_{\text{w}}=[\ce{H+}][\ce{OH-}]=1.0\times 10^{-14}=(1.0\times 10^{-4})[\ce{OH-}]\"<\/p>\n

                              We solve by dividing both sides of the equation by 1.0 \u00d7 10\u22124<\/sup>:<\/p>\n

                              \"[\ce{OH-}]=\dfrac{1.0\times 10^{-14}}{1.0\times 10^{-4}}=1.0\times 10^{-10}\text{ M}\"<\/p>\n

                              It is assumed that the concentration unit is molarity, so [OH\u2212<\/sup>] is 1.0 \u00d7 10\u221210<\/sup> M.<\/p>\n

                              Test Yourself<\/em>
                              \nWhat is [H+<\/sup>] of an aqueous solution if [OH\u2212<\/sup>] is 1.0 \u00d7 10\u22129<\/sup> M?<\/p>\n

                              Answer<\/em>
                              \n1.0 \u00d7 10\u22125<\/sup> M<\/p>\n<\/div>\n<\/div>\n

                              When you have a solution of a particular acid or base, you need to look at the formula of the acid or base to determine the number of H+<\/sup> or OH\u2212<\/sup> ions in the formula unit because [H+<\/sup>] or [OH\u2212<\/sup>] may not be the same as the concentration of the acid or base itself.<\/p>\n

                              \n
                              \n

                              Example 12.10<\/p>\n<\/header>\n

                              \n

                              What is [H+<\/sup>] in a 0.0044 M solution of Ca(OH)2<\/sub>?<\/p>\n

                              Solution<\/em>
                              \nWe begin by determining [OH\u2212<\/sup>].<\/p>\n

                              The concentration of the solute is 0.0044 M, but because Ca(OH)2<\/sub> is a strong base, there are two OH\u2212<\/sup> ions in solution for every formula unit dissolved, so the actual [OH\u2212<\/sup>] is two times this, or 2 \u00d7 0.0044 M = 0.0088 M.<\/p>\n

                              Now we can use the K<\/em>w<\/sub> expression:<\/p>\n

                              \"[\ce{H+}][\ce{OH-}]=1.0\times 10^{-14}=[\ce{H+}](0.0088\text{ M})\"<\/p>\n

                              Dividing both sides by 0.0088:<\/p>\n

                              \"[\ce{H+}]=\dfrac{1.0\times 10^{-14}}{0.0088}=1.1\times 10^{-12}\text{ M}\"<\/p>\n

                              [H+<\/sup>] has decreased significantly in this basic solution.<\/p>\n

                              Test Yourself<\/em>
                              \nWhat is [OH\u2212<\/sup>] in a 0.00032 M solution of H2<\/sub>SO4<\/sub>? (Hint: assume both H+<\/sup> ions ionize.)<\/p>\n

                              Answer<\/em>
                              \n1.6 \u00d7 10\u221211<\/sup> M<\/p>\n<\/div>\n<\/div>\n

                              For strong acids and bases, [H+<\/sup>] and [OH\u2212<\/sup>] can be determined directly from the concentration of the acid or base itself because these ions are 100% ionized by definition. However, for weak acids and bases, this is not so. The degree, or percentage, of ionization would need to be known before we can determine [H+<\/sup>] and [OH\u2212<\/sup>].<\/p>\n

                              \n
                              \n

                              Example 12.11<\/p>\n<\/header>\n

                              \n

                              A 0.0788 M solution of HC2<\/sub>H3<\/sub>O2<\/sub> is 3.0% ionized into H+<\/sup> ions and C2<\/sub>H3<\/sub>O2<\/sub>\u2212<\/sup> ions. What are [H+<\/sup>] and [OH\u2212<\/sup>] for this solution?<\/p>\n

                              Solution<\/em>
                              \nBecause the acid is only 3.0% ionized, we can determine [H+<\/sup>] from the concentration of the acid. Recall that 3.0% is 0.030 in decimal form:<\/p>\n

                              \"[\ce{H+}]=0.030\times 0.0788=0.00236\text{ M}\"<\/p>\n

                              With this [H+<\/sup>], then [OH\u2212<\/sup>] can be calculated as follows:<\/p>\n

                              \"[\ce{OH-}]=\dfrac{1.0\times 10^{-14}}{0.00236}=4.2\times 10^{-12}\text{ M}\"<\/p>\n

                              This is about 30 times higher than would be expected for a strong acid of the same concentration.<\/p>\n

                              Test Yourself<\/em>
                              \nA 0.0222 M solution of pyridine (C5<\/sub>H5<\/sub>N) is 0.44% ionized into pyridinium ions (C5<\/sub>H5<\/sub>NH+<\/sup>) and OH\u2212<\/sup> ions. What are [OH\u2212<\/sup>] and [H+<\/sup>] for this solution?<\/p>\n

                              Answer<\/em>
                              \n[OH\u2212<\/sup>] = 9.77 \u00d7 10\u22125<\/sup> M; [H+<\/sup>] = 1.02 \u00d7 10\u221210<\/sup> M<\/p>\n<\/div>\n<\/div>\n

                              \n
                              \n

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

                              \n
                                \n
                              • In any aqueous solution, the product of [H+<\/sup>] and [OH\u2212<\/sup>] equals 1.0 \u00d7 10\u221214<\/sup>.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n
                                \n
                                \n

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

                                \n

                                Questions<\/h1>\n
                                  \n
                                1. Does [H+<\/sup>] remain constant in all aqueous solutions? Why or why not?<\/li>\n
                                2. Does [OH\u2212<\/sup>] remain constant in all aqueous solutions? Why or why not?<\/li>\n
                                3. What is the relationship between [H+<\/sup>] and K<\/em>w<\/sub>? Write a mathematical expression that relates them.<\/li>\n
                                4. What is the relationship between [OH\u2212<\/sup>] and K<\/em>w<\/sub>? Write a mathematical expression that relates them.<\/li>\n
                                5. Write the chemical equation for the autoionization of water and label the conjugate acid-base pairs.<\/li>\n
                                6. Write the reverse of the reaction for the autoionization of water. It is still an acid-base reaction? If so, label the acid and base.<\/li>\n
                                7. For a given aqueous solution, if [H+<\/sup>] = 1.0 \u00d7 10\u22123<\/sup> M, what is [OH\u2212<\/sup>]?<\/li>\n
                                8. For a given aqueous solution, if [H+<\/sup>] = 1.0 \u00d7 10\u22129<\/sup> M, what is [OH\u2212<\/sup>]?<\/li>\n
                                9. For a given aqueous solution, if [H+<\/sup>] = 7.92 \u00d7 10\u22125<\/sup> M, what is [OH\u2212<\/sup>]?<\/li>\n
                                10. For a given aqueous solution, if [H+<\/sup>] = 2.07 \u00d7 10\u221211<\/sup> M, what is [H+<\/sup>]?<\/li>\n
                                11. For a given aqueous solution, if [OH\u2212<\/sup>] = 1.0 \u00d7 10\u22125<\/sup> M, what is [H+<\/sup>]?<\/li>\n
                                12. For a given aqueous solution, if [OH\u2212<\/sup>] = 1.0 \u00d7 10\u221212<\/sup> M, what is [H+<\/sup>]?<\/li>\n
                                13. For a given aqueous solution, if [OH\u2212<\/sup>] = 3.77 \u00d7 10\u22124<\/sup> M, what is [H+<\/sup>]?<\/li>\n
                                14. For a given aqueous solution, if [OH\u2212<\/sup>] = 7.11 \u00d7 10\u221210<\/sup> M, what is [H+<\/sup>]?<\/li>\n
                                15. What are [H+<\/sup>] and [OH\u2212<\/sup>] in a 0.344 M solution of HNO3<\/sub>?<\/li>\n
                                16. What are [H+<\/sup>] and [OH\u2212<\/sup>] in a 2.86 M solution of HBr?<\/li>\n
                                17. What are [H+<\/sup>] and [OH\u2212<\/sup>] in a 0.00338 M solution of KOH?<\/li>\n
                                18. What are [H+<\/sup>] and [OH\u2212<\/sup>] in a 6.02 \u00d7 10\u22124<\/sup> M solution of Ca(OH)2<\/sub>?<\/li>\n
                                19. If HNO2<\/sub> is dissociated only to an extent of 0.445%, what are [H+<\/sup>] and [OH\u2212<\/sup>] in a 0.307 M solution of HNO2<\/sub>?<\/li>\n
                                20. If (C2<\/sub>H5<\/sub>)2<\/sub>NH is dissociated only to an extent of 0.077%, what are [H+<\/sup>] and [OH\u2212<\/sup>] in a 0.0955 M solution of (C2<\/sub>H5<\/sub>)2<\/sub>NH?<\/li>\n<\/ol>\n

                                  Answers<\/h1>\n
                                    \n
                                  1. [H+<\/sup>] varies with the amount of acid or base in a solution.<\/li>\n<\/ol>\n
                                      \n
                                    1. [H+<\/sup>] = K<\/em>w<\/sub>[OH\u2212<\/sup>]<\/li>\n<\/ol>\n
                                        \n
                                      1. H2<\/sub>O +\u00a0H2<\/sub>O \u2192\u00a0H3<\/sub>O+<\/sup> +\u00a0OH\u2212<\/sup>; H2<\/sub>O\/H3<\/sub>O+<\/sup> and H2<\/sub>O\/OH\u2212<\/sup><\/li>\n<\/ol>\n
                                          \n
                                        1. 1.0 \u00d7 10\u221211<\/sup> M<\/li>\n<\/ol>\n
                                            \n
                                          1. 1.26 \u00d7 10\u221210<\/sup> M<\/li>\n<\/ol>\n
                                              \n
                                            1. 1.0 \u00d7 10\u22129<\/sup> M<\/li>\n<\/ol>\n
                                                \n
                                              1. 2.65 \u00d7 10\u221211<\/sup> M<\/li>\n<\/ol>\n
                                                  \n
                                                1. [H+<\/sup>] = 0.344 M; [OH\u2212<\/sup>] = 2.91 \u00d7 10\u221214<\/sup> M<\/li>\n<\/ol>\n
                                                    \n
                                                  1. [OH\u2212<\/sup>] = 0.00338 M; [H+<\/sup>] = 2.96 \u00d7 10\u221212<\/sup> M<\/li>\n<\/ol>\n
                                                      \n
                                                    1. [H+<\/sup>] = 0.00137 M; [OH\u2212<\/sup>] = 7.32 \u00d7 10\u221212<\/sup> M<\/li>\n<\/ol>\n<\/div>\n<\/div>\n","protected":false},"author":90,"menu_order":5,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-7631","chapter","type-chapter","status-publish","hentry"],"part":7615,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7631","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/users\/90"}],"version-history":[{"count":7,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7631\/revisions"}],"predecessor-version":[{"id":8926,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7631\/revisions\/8926"}],"part":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/parts\/7615"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/7631\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/media?parent=7631"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=7631"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/contributor?post=7631"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/introductorychemistry\/wp-json\/wp\/v2\/license?post=7631"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}