{"id":604,"date":"2017-12-14T21:42:06","date_gmt":"2017-12-14T21:42:06","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/autoionization-of-water\/"},"modified":"2018-09-02T01:43:24","modified_gmt":"2018-09-02T01:43:24","slug":"autoionization-of-water","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/autoionization-of-water\/","title":{"raw":"10.4 Autoionization of Water","rendered":"10.4 Autoionization of Water"},"content":{"raw":"
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Learning Objectives<\/h3>\r\nBy the end of this module, you will be able to:\r\n
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  • Describe the autoionization of water.<\/li>\r\n \t
  • Calculate the concentrations of H+<\/sup> and OH\u2212<\/sup> in solutions, knowing the other concentration.<\/li>\r\n<\/ul>\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\n

    Water acting as a acid,<\/p>\r\n

    [latex]\\large{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow {\\text{NH}}_{4}^{\\text{+}}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]<\/span><\/span><\/p>\r\n

    Water acting as a base,<\/span><\/span><\/p>\r\n

    [latex]\\large{\\text{HCl}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{Cl}}^{-}\\left(aq\\right)[\/latex]<\/p>\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\n

    [latex]\\large{\\text{H}}_{2}\\text{O}\\left(l\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]<\/span><\/span><\/p>\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 autoionizaton of water. \u00a0<\/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\u2014[ ]\u2014around a dissolved species to imply the molar concentration of that species, we have<\/p>\r\n

    [latex]\\large{[\\text{H}}_{3}{\\text{O}}^{\\text{+}}]={[\\text{OH}}^{-}]=1.0\\times {10}^{-7}M[\/latex]<\/span><\/span><\/p>\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

    [latex]\\large{[\\text{H}}_{3}{\\text{O}}^{\\text{+}}]\\times{{[\\text{OH}}^{-}]}=(2.0\\times {10}^{-6})\\times(5.0\\times {10}^{-9})=1.0\\times {10}^{-14}[\/latex]<\/span><\/span><\/p>\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

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

    [latex]\\large \\text{K}_w={[\\text{H}}_{3}{\\text{O}}^{\\text{+}}]{{[\\text{OH}}^{-}]}=1.0\\times {10}^{-14}[\/latex]<\/span><\/span><\/p>\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<\/div>\r\n

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    Example 1: Ion Concentrations in Pure Water<\/h3>\r\nWhat are the hydronium ion concentration and the hydroxide ion concentration in pure water at 25 \u00b0C?\r\n[reveal-answer q=\"160586\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"160586\"]\r\n\r\nThe autoionization of water yields the same number of hydronium and hydroxide ions. Therefore, in pure water, [latex]\\large\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=\\left[\\text{OH}^{-}\\right][\/latex]. At 25 \u00b0C: [latex]{K}_{\\text{w}}=\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{OH}}^{-}\\right]={\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}^{\\text{2+}}={\\left[{\\text{OH}}^{-}\\right]}^{\\text{2+}}=1.0\\times {10}^{-14}[\/latex]\r\n\r\nSo:\r\n\r\n[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=\\left[{\\text{OH}}^{-}\\right]=\\sqrt{1.0\\times {10}^{-14}}=1.0\\times {10}^{-7}M[\/latex]\r\n\r\nThe hydronium ion concentration and the hydroxide ion concentration are the same, and we find that both equal 1.0 \u00d7 10\u22127<\/sup>M<\/em>.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nExample 2 demonstrates the quantitative aspects of this relation between hydronium and hydroxide ion concentrations.\r\n
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    Example 2: The Inverse Proportionality of [H3<\/sub>O+<\/sup>] and [OH\u2212<\/sup>]<\/h3>\r\nA solution of an acid in water has a hydronium ion concentration of 2.0 [latex]\\times [\/latex] 10\u22126<\/sup>M<\/em>. What is the concentration of hydroxide ion at 25 \u00b0C?\r\n\r\n[reveal-answer q=\"647487\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"647487\"]\r\n\r\nWe know the value of the ion-product constant for water at 25 \u00b0C:\r\n\r\n[latex]\\large 2{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right){K}_{\\text{w}}=\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{OH}}^{-}\\right]=1.0\\times {10}^{-14}[\/latex]\r\n\r\nThus, we can calculate the missing equilibrium concentration.\r\n\r\nRearrangement of the K<\/em>w<\/sub> expression yields that [OH\u2212<\/sup>] is directly proportional to the inverse of [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right][\/latex]:\r\n\r\n[latex]\\large\\left[{\\text{OH}}^{-}\\right]=\\frac{{K}_{\\text{w}}}{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}=\\frac{1.0\\times {10}^{-14}}{2.0\\times {10}^{-6}}=5.0\\times {10}^{-9}[\/latex]\r\n\r\n[\/hidden-answer]\r\n

    Check Your Learning<\/h4>\r\nWhat is the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 M<\/em> at 25 \u00b0C?\r\n[reveal-answer q=\"366443\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"366443\"][latex]\\large\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=1\\times 10^{-11}M[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n
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    Key concepts and summary<\/h3>\r\nIn any aqueous solution, the product of [H+<\/sup>] and [OH\u2212<\/sup>] equals 1.0 \u00d7 10\u221214<\/sup>.\r\n

    Key Equations<\/h4>\r\n