{"id":690,"date":"2016-01-11T20:01:52","date_gmt":"2016-01-11T20:01:52","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistryclone\/chapter\/autoionization-of-water-2\/"},"modified":"2020-04-20T16:38:20","modified_gmt":"2020-04-20T16:38:20","slug":"autoionization-of-water","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistryclone\/chapter\/autoionization-of-water\/","title":{"raw":"Autoionization of Water","rendered":"Autoionization of Water"},"content":{"raw":"
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Learning Objectives<\/h3>\r\n
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  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\n

    We have already seen that H2<\/sub>O can act as an acid or a base:<\/p>\r\nNH3<\/sub> +\u00a0H2<\/sub>O \u2192\u00a0NH4<\/sub>+<\/sup> +\u00a0OH\u2212<\/sup> (H2<\/sub>O acts as an acid)<\/span><\/span>\r\nHCl +\u00a0H2<\/sub>O \u2192\u00a0H3<\/sub>O+<\/sup> +\u00a0Cl\u2212<\/sup> (H2<\/sub>O acts as a base)<\/span><\/span>\r\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>\r\nH2<\/sub>O +\u00a0H2<\/sub>O \u2192\u00a0H3<\/sub>O+<\/sup> +\u00a0OH\u2212<\/sup><\/span><\/span>\r\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<\/a><\/span>. 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\u2014[ ]\u2014around a dissolved species to imply the molar concentration of that species, we have<\/p>\r\n[H+<\/sup>] = [OH\u2212<\/sup>] = 1.0 \u00d7 10\u22127<\/sup> M<\/span><\/span>\r\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>\r\n[H+<\/sup>] \u00d7 [OH\u2212<\/sup>] = (1.0 \u00d7 10\u22127<\/sup>)(1.0 \u00d7 10\u22127<\/sup>) = 1.0 \u00d7 10\u221214<\/sup><\/span><\/span>\r\n

    In acids, the concentration of H+<\/sup>(aq)\u2014[H+<\/sup>]\u2014is greater than 1.0 \u00d7 10\u22127<\/sup> M, while for bases the concentration of OH\u2212<\/sup>(aq)\u2014[OH\u2212<\/sup>]\u2014is greater than 1.0 \u00d7 10\u22127<\/sup> M. However, the product<\/em> of the two concentrations\u2014[H+<\/sup>][OH\u2212<\/sup>]\u2014is always<\/em> equal to 1.0 \u00d7 10\u221214<\/sup>, no matter whether the aqueous solution is an acid, a base, or neutral:<\/p>\r\n[H+<\/sup>][OH\u2212<\/sup>] = 1.0 \u00d7 10\u221214<\/sup><\/span><\/span>\r\n

    This value of the product of concentrations is so important for aqueous solutions that it is called the autoionization constant of water<\/a><\/span>\u00a0and is denoted K<\/em>w<\/sub>:<\/p>\r\nK<\/em>w<\/sub> = [H+<\/sup>][OH\u2212<\/sup>] = 1.0 \u00d7 10\u221214<\/sup><\/span><\/span>\r\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>\r\n\r\n

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    Example 9<\/h3>\r\n

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

    Solution<\/p>\r\n

    Using the expression and known value for K<\/em>w<\/sub>,<\/p>\r\nK<\/em>w<\/sub> = [H+<\/sup>][OH\u2212<\/sup>] = 1.0 \u00d7 10\u221214<\/sup> = (1.0 \u00d7 10\u22124<\/sup>)[OH\u2212<\/sup>]<\/span><\/span>\r\n

    We solve by dividing both sides of the equation by 1.0 \u00d7 10\u22124<\/sup>:<\/p>\r\n[OH\u2212<\/sup>] = (1.0\u00d710\u221214\u00a0<\/sup>\/ 1.0\u00d710\u22124)\u00a0<\/sup>= 1.0\u00d710\u221210<\/sup>\u2009M<\/span>\r\n

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

    Test Yourself<\/em><\/p>\r\n

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

    Answer<\/em><\/p>\r\n

    1.0 \u00d7 10\u22125<\/sup> M<\/p>\r\n\r\n<\/div>\r\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>\r\n\r\n

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    Example 10<\/h3>\r\n

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

    Solution<\/p>\r\n

    We begin by determining [OH\u2212<\/sup>]. 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. Now we can use the K<\/em>w<\/sub> expression:<\/p>\r\n[H+<\/sup>][OH\u2212<\/sup>] = 1.0 \u00d7 10\u221214<\/sup> = [H+<\/sup>](0.0088 M)<\/span><\/span>\r\n

    Dividing both sides by 0.0088:<\/p>\r\n[H+]= 1.0\u00d710\u221214<\/sup> \/ (0.0088) = 1.1\u00d710\u221212<\/sup>\u00a0M<\/span>\r\n

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

    Test Yourself<\/em><\/p>\r\n

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

    Answer<\/em><\/p>\r\n

    1.6 \u00d7 10\u221211<\/sup> M<\/p>\r\n\r\n<\/div>\r\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>\r\n\r\n

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    Example 11<\/h3>\r\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>\r\n

    Solution<\/p>\r\n

    Because 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>\r\n[H+<\/sup>] = 0.030 \u00d7 0.0788 = 0.00236 M<\/span><\/span>\r\n

    With this [H+<\/sup>], then [OH\u2212<\/sup>] can be calculated as follows:<\/p>\r\n[OH\u2212] = 1.0\u00d710\u221214 <\/sup>\/\u00a0<\/sup>0.00236 = 4.2\u00d710\u221212<\/sup>\u00a0M<\/span>\r\n

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

    Test Yourself<\/em><\/p>\r\n

    A 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>\r\n

    Answer<\/em><\/p>\r\n

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

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    Key Takeaways<\/h3>\r\n