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12.2: Autoionization of Water

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    170063

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    Learning Objectives

    • Use the ion-product constant for water to calculate hydronium and hydroxide ion concentrations

    In the preceding section we saw that water is an amphiprotic substance. It can function as either an acid or a base, depending on the nature of the solute dissolved in it. In fact, in pure water or in any aqueous solution, water acts both as an acid and a base. A very small fraction of water molecules donate protons to other water molecules to form hydronium ions and hydroxide ions:

    This figure has two rows. In both rows, a chemical reaction is shown. In the first, structural formulas are provided. In this model, in purple, O atom which has H atoms singly bonded above and to the right. The O atom has pairs of electron dots on its left and lower sides. This is followed by a plus sign, which is followed in green by an O atom which has H atoms singly bonded above and to the right. The O atom has pairs of electron dots on its left and lower sides. A double arrow follows. To the right, in brackets is a structure with a central O atom in green, with green H atoms singly bonded above and to the right. A pair of green electron dots is on the lower side of the O atom. To the left of the green O atom, a purple H atom is singly bonded. Outside the brackets to the right is a superscript plus. This is followed by a plus sign and an O atom in purple with pairs of electron dots above, left, and below. An H atom is singly bonded to the right. This atom has a superscript negative sign. The reaction is written in symbolic form below. H subscript 2 O is labeled in purple below as “Acid subscript 1.” This is followed by plus H subscript 2 O, which is labeled in green below as “Base subscript 2.” A double sided arrow follows. To the right is H subscript 3 O superscript plus, which is labeled in green as below in as “Acid subscript 2.” This is followed by plus and O with pairs of dots above, below, and to the left with a singly bonded H on the right with a superscript negative. The label below in purple reads, “ Base subscript 1.”

    This type of reaction, in which a substance ionizes when one molecule of the substance reacts with another molecule of the same substance, is referred to as autoionization. Pure water undergoes autoionization to a very slight extent. Only about two out of every \(10^9\) molecules in a sample of pure water are ionized at 25 °C. The equilibrium constant for the ionization of water is called the ion-product constant for water (Kw):

    \[\ce{H_2O(l) + H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)}\;\;\; K_w = \ce{[H_3O^+][OH^- ]}\]

    The slight ionization of pure water is reflected in the small value of the equilibrium constant; at 25 °C, Kw has a value of \(1.0 \times 10^{−14}\). The process is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion increase with temperature. For example, at 100 °C, the value for \(K_w\) is approximately \(5.1 \times 10^{−13}\), roughly 100-times larger than the value at 25 °C.

    Example \(\PageIndex{1}\): Ion Concentrations in Pure Water

    What are the hydronium ion concentration and the hydroxide ion concentration in pure water at 25 °C?

    Solution

    The autoionization of water yields the same number of hydronium and hydroxide ions. Therefore, in pure water, \(\ce{[H_3O^+]} = \ce{[OH^- ]}\). At 25 °C:

    \[K_\ce{w}=\ce{[H_3O^+][OH^- ]}=\ce{[H_3O^+]^2+}=\ce{[OH^- ]^2+}=1.0 \times 10^{−14}\]

    So:

    \[\ce{[H_3O^+]}=\ce{[OH^- ]}=\sqrt{1.0 \times 10^{−14}} =1.0 \times 10^{−7}\; M\]

    The hydronium ion concentration and the hydroxide ion concentration are the same, and we find that both equal \(1.0 \times 10^{−7}\; M\).

    Exercise \(\PageIndex{1}\)

    The ion product of water at 80 °C is \(2.4 \times 10^{−13}\). What are the concentrations of hydronium and hydroxide ions in pure water at 80 °C?

    Answer

    \(\ce{[H_3O^+]} = \ce{[OH^- ]} = 4.9 \times 10^{−7}\; M\)

    It is important to realize that the autoionization equilibrium for water is established in all aqueous solutions. Adding an acid or base to water will not change the value of the equilibrium constant. The following example demonstrates the quantitative aspects of this relation between hydronium and hydroxide ion concentrations in an aqueous solution.

    Example \(\PageIndex{2}\): The Inverse Proportionality of \(\ce{[H_3O^+]}\) and \(\ce{[OH^- ]}\)

    A solution of carbon dioxide in water has a hydronium ion concentration of \(2.0 \times 10^{−6}\; M\). What is the concentration of hydroxide ion at 25 °C?

    Solution

    We know the value of the ion-product constant for water at 25 °C:

    \[\ce{2 H_2O}_{(l)} \rightleftharpoons \ce{H_3O^+}_{(aq)} + \ce{OH^-}_{(aq)}\]

    \[K_\ce{w}=\ce{[H3O+][OH^- ]}=1.0 \times 10^{−14}\]

    Thus, we can calculate the missing equilibrium concentration.

    Rearrangement of the Kw expression yields that \([\ce{OH^- }]\) is directly proportional to the inverse of [H3O+]:

    \[[\ce{OH^- }]=\dfrac{K_{\ce w}}{[\ce{H_3O^+}]}=\dfrac{1.0 \times 10^{−14}}{2.0 \times 10^{−6}}=5.0 \times 10^{−9}\]

    The hydroxide ion concentration in water is reduced to \(5.0 \times 10^{−9}\: M\) as the hydrogen ion concentration increases to \(2.0 \times 10^{−6}\; M\). This is expected from Le Chatelier’s principle; the autoionization reaction shifts to the left to reduce the stress of the increased hydronium ion concentration and the \(\ce{[OH^- ]}\) is reduced relative to that in pure water.

    A check of these concentrations confirms that our arithmetic is correct:

    \[K_\ce{w}=\ce{[H_3O^+][OH^- ]}=(2.0 \times 10^{−6})(5.0 \times 10^{−9})=1.0 \times 10^{−14}\]

    Exercise \(\PageIndex{2}\)

    What is the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 M at 25 °C?

    Answer

    \[\ce{[H3O+]} = 1 \times 10^{−11} M\]

    Summary

    Water is an important amphiprotic species. It can form both the hydronium ion, H3O+, and the hydroxide ion, \(\ce{OH^-}\) when it undergoes autoionization:

    \[\ce{2 H_2O}_{(l)} \rightleftharpoons H_3O^+_{(aq)} + OH^-_{(aq)}\]

    The ion product of water, Kw is the equilibrium constant for the autoionization reaction:

    \[K_\ce{w}=\mathrm{[H_2O^+][OH^- ]=1.0 \times 10^{−14} \; at\; 25°C}\]

    {Key Equations

    • \[K_{\ce w} = \ce{[H3O+][OH^- ]} = 1.0 \times 10^{−14}\textrm{ (at 25 °C)}\]

    Glossary

    autoionization
    reaction between identical species yielding ionic products; for water, this reaction involves transfer of protons to yield hydronium and hydroxide ions
    ion-product constant for water (Kw)
    equilibrium constant for the autoionization of water

    Contributors and Attributions

    12.2: Autoionization of Water is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.