Skip to main content
Chemistry LibreTexts

10.15: Ion-Product of Water

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    At one time, you could take the little caps off the top of a car battery and check the condition of the sulfuric acid inside. If it got low, you could add more acid. But, sulfuric acid is hazardous stuff, so the batteries are now sealed to protect people. Because of the acid's dangerous nature, it is not a good idea to cut into a battery to see what it looks like—you could get acid burns.

    The Ion-Product of Water

    The self-ionization of water (the process in which water ionizes to hydronium ions and hydroxide ions) occurs to a very limited extent. When two molecules of water collide, there can be a transfer of a hydrogen ion from one molecule to the other. The products are a positively charged hydronium ion and a negatively charged hydroxide ion.

    \[\ce{H_2O} \left( l \right) + \ce{H_2O} \left( l \right) \rightleftharpoons \ce{H_3O^+} \left( aq \right) + \ce{OH^-} \left( aq \right)\nonumber \]

    We often use the simplified form of the reaction:

    \[\ce{H_2O} \left( l \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{OH^-} \left( aq \right)\nonumber \]

    The equilibrium constant for the self-ionization of water is referred to as the ion-product for water and is given the symbol \(K_\text{w}\).

    \[K_\text{w} = \left[ \ce{H^+} \right] \left[ \ce{OH^-} \right]\nonumber \]

    The ion-product of water \(\left( K_\text{w} \right)\) is the mathematical product of the concentration of hydrogen ions and hydroxide ions. Note that \(\ce{H_2O}\) is not included in the ion-product expression because it is a pure liquid. The value of \(K_\text{w}\) is very small, in accordance with a reaction that favors the reactants. At \(25^\text{o} \text{C}\), the experimentally determined value of \(K_\text{w}\) in pure water is \(1.0 \times 10^{-14}\).

    \[K_\text{w} = \left[ \ce{H^+} \right] \left[ \ce{OH^-} \right] = 1.0 \times 10^{-14}\nonumber \]

    In pure water, the concentrations of hydrogen and hydroxide ions are equal to one another. Pure water or any other aqueous solution in which this ratio holds is said to be neutral. To find the molarity of each ion, the square root of \(K_\text{w}\) is taken.

    \[\left[ \ce{H^_+} \right] = \left[ \ce{OH^-} \right] = 1.0 \times 10^{-7}\nonumber \]

    An acidic solution is a solution in which the concentration of hydrogen ions is greater than the concentration of hydroxide ions. For example, hydrogen chloride ionizes to produce \(\ce{H^+}\) and \(\ce{Cl^-}\) ions upon dissolving in water.

    \[\ce{HCl} \left( g \right) \rightarrow \ce{H^+} \left( aq \right) + \ce{Cl^-} \left( aq \right)\nonumber \]

    This increases the concentration of \(\ce{H^+}\) ions in the solution. According to Le Chatelier's principle, the equilibrium represented by \(\ce{H_2O} \left( l \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{OH^-} \left( aq \right)\) is forced to the left, towards the reactant. As a result, the concentration of the hydroxide ion decreases.

    A basic solution is a solution in which the concentration of hydroxide ions is greater than the concentration of hydrogen ions. Solid potassium hydroxide dissociates in water to yield potassium ions and hydroxide ions.

    \[\ce{KOH} \left( s \right) \rightarrow \ce{K^+} \left( aq \right) + \ce{OH^-} \left( aq \right)\nonumber \]

    The increase in concentration of the \(\ce{OH^-}\) ions causes a decrease in the concentration of the \(\ce{H^+}\) ions and the ion-product of \(\left[ \ce{H^+} \right] \left[ \ce{OH^-} \right]\) remains constant.

    Example \(\PageIndex{1}\)

    Hydrochloric acid \(\left( \ce{HCl} \right)\) is a strong acid, meaning it is \(100\%\) ionized in solution. What is the \(\left[ \ce{H^+} \right]\) and the \(\left[ \ce{OH^-} \right]\) in a solution of \(2.0 \times 10^{-3} \: \text{M} \: \ce{HCl}\)?

    Step 1: List the known values and plan the problem.
    • \(\left[ \ce{HCl} \right] = 2.0 \times 10^{-3} \: \text{M}\)
    • \(K_\text{w} = 1.0 \times 10^{-14}\)
    • \(\left[ \ce{H^+} \right] = ? \: \text{M}\)
    • \(\left[ \ce{OH^-} \right] = ? \: \text{M}\)

    Because \(\ce{HCl}\) is \(100\%\) ionized, the concentration of \(\ce{H^+}\) ions in solution will be equal to the original concentration of \(\ce{HCl}\). Each \(\ce{HCl}\) molecule that was originally present ionizes into one \(\ce{H^+}\) ion and one \(\ce{Cl^-}\) ion. The concentration of \(\ce{OH^-}\) can then be determined from the \(\left[ \ce{H^+} \right]\) and \(K_\text{w}\).

    Step 2: Solve.

    \[\begin{align*} \left[ \ce{H^+} \right] &= 2.0 \times 10^{-3} \: \text{M} \\ K_\text{w} &= \left[ \ce{H^+} \right] \left[ \ce{OH^-} \right] = 1.0 \times 10^{-14} \\ \left[ \ce{OH^-} \right] &= \frac{K_\text{w}}{\left[ \ce{H^+} \right]} = \frac{1.0 \times 10^{-14}}{2.0 \times 10^{-3}} = 5.0 \times 10^{-12} \: \text{M} \end{align*}\nonumber \]

    Step 3: Think about your result.

    The \(\left[ \ce{H^+} \right]\) is much higher than the \(\left[ \ce{OH^-} \right]\) because the solution is acidic. As with other equilibrium constants, the unit for \(K_\text{w}\) is customarily omitted.


    • The self-ionization of water is described and an ionization constant for the process is stated.
    • Acidic and basic solutions are defined.
    • Calculations using \(K_\text{w}\) are illustrated.

    10.15: Ion-Product of Water is shared under a CC BY-NC license and was authored, remixed, and/or curated by LibreTexts.