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14.2: Ionization of Water

  • Page ID
    49684
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    In the section on amphiprotic species, we saw that water can act as a very weak acid and a very weak base, donating protons to itself to a limited extent:

    \[\text{2H}_{2}\text{O}\text{ }({l}) \rightleftharpoons \text{H}_{3}\text{O}^{+} ({aq}) + \text{OH}^{-} ({aq}) \nonumber \]

    Ball and stick models of 2 water molecules becoming a hydronium ion and hydroxyl group.

    The equilibrium constant \(K\) for this reaction can be written as follows:

    \[K_{a}=\dfrac{a_{H_3O^+}·a_{OH^-}}{a_{H_2O}^2} \approx \frac{[H_{3}O^{+}][HO^{-}]}{(1)^{2}}=[H_{3}O^{+}][HO^{-}] \label{16.3.4} \]

    where \(a\) is the activity of a species. Because water is the solvent, and the solution is assumed to be dilute, the activity of the water is approximated by the activity of pure water, which is defined as having a value of 1. The activity of each solute is approximated by the molarity of the solute.

    Note

    It is a common error to claim that the molar concentration of the solvent is in some way involved in the equilibrium law. This error is a result of a misunderstanding of solution thermodynamics. For example, it is often claimed that Ka = Keq[H2O] for aqueous solutions. This equation is incorrect because it is an erroneous interpretation of the correct equation Ka = Keq(\(\textit{a}_{H_2O}\)). Because \(\textit{a}_{H_2O}\) = 1 for a dilute solution, Ka = Keq(1), or Ka = Keq.

    In this reaction, one water molecule acts as an acid and one water molecule acts as a base. Thus, this reaction actually can be designated as the \(K_a\) of water and as the \(K_b\) of water. It is most common, however, to designate this reaction and the associated law of mass action as the \(K_w\) of water:

    \[K_{w}=[H_{3}O^{+}][HO^{-}] \label{16.3.5} \]

    Measurements of the electrical conductivity of carefully purified water indicate that at 25°C [H3O+] = [OH] = 1.00 × 10–7 mol/L, so that

    \[\begin{align}K_{w}={1.00}\times{10}^{-7}\text{ mol L}^{-1}\times{1.00}\times{10}^{-7}\text{ mol L}^{-1} \nonumber \\\text{ }\\\text{ }={1.00}\times{10}^{-14}\text{ mol}^2\text{L}^{-2} \nonumber \end{align} \nonumber \]

    (Since the equilibrium law is not obeyed exactly, even in dilute solutions, results of most equilibrium calculations are rounded to three significant figures. Hence the value of Kw = 1.00 × 10–14 mol2/L2 is sufficiently accurate for all such calculations.)

    The equilibrium constant Kw applies not only to pure water but to any aqueous solution at 25°C. Thus, for example, if we add 1.00 mol of the strong acid HNO3 to H2O to make a total volume of 1 L, essentially all the HNO3 molecules donate their protons to H2O:

    \[\text{HNO}_{3} + \text{H}_{2}\text{O} \rightarrow \text{NO}_{3}^{-} + \text{H}_{3}\text{O}^{+} \nonumber \]

    Ball and stick models of a water molecule and nitric acid molecule reacting to form a hydronium ion and a nitrate molecule.

    and a solution in which [H3O+] = 1.00 mol/L is obtained. Although this solution is very acidic, there are still hydroxide ions present. We can calculate their concentration by rearranging Eq. \(\ref{3}\):

    \[\begin{align}\text{ }[\text{OH}^{-}]=\dfrac{K_{w}}{[\text{ H}_{3}\text{O}^{+}]}=\dfrac{\text{1.00 }\times \text{ 10}^{-14}\text{ mol}^{2}\text{ L}^{-2}}{\text{1.00 mol L}^{-1}}\\\text{ }\\\text{ }=\text{1.00 }\times \text{ 10}^{-14}\text{ mol L}^{-1}\end{align} \nonumber \]

    The addition of the HNO3 to H2O not only increases the hydronium-ion concentration but also reduces the hydroxide-ion concentration from an initially minute 10–7 mol/L to an even more minute 10–14 mol/L.

    Example \(\PageIndex{2}\): Ion Concentration

    Calculate the hydronium-ion concentration in a solution of 0.306 M Ba(OH)2.

    Solution

    Since 1 mol Ba(OH)2 produces 2 mol OH in solution, we have

    \[[OH^-] = 2 \times 0.306 \dfrac{mol}{L} = 0.612 \dfrac{mol}{L} \nonumber \]

    Then

    \[\begin{align}\text{ }[\text{ H}_{3}\text{O}^{+}]=\dfrac{K_{w}}{[\text{OH}^{-}]}=\dfrac{\text{1.00 }\times \text{ 10}^{-14}\text{ mol}^{2}\text{ L}^{-2}}{\text{0.612 mol L}^{-1}} \nonumber \\\text{ }\\\text{ }=\text{1.63 }\times \text{ 10}^{-14}\text{ mol L}^{-1} \nonumber \end{align} \nonumber \]

    Note

    Note that since strong acids like HNO3 are completely converted to H3O+ in aqueous solution, it is a simple matter to determine [H3O+], and from it, [OH]. Similarly, when a strong base dissolves in H2O it is entirely converted to OH, so that [OH], and from it [H3O+] are easily obtained.


    This page titled 14.2: Ionization of Water is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Ed Vitz, John W. Moore, Justin Shorb, Xavier Prat-Resina, Tim Wendorff, & Adam Hahn.

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