Skip to main content
Chemistry LibreTexts

14.4: The pH of Solutions of Weak Bases

  • Page ID
    49687
  • [ "article:topic", "ChemPrime", "pH", "pH of Solutions of Weak Bases", "weak base", "authorname:chemprime", "showtoc:no" ]

    The pH of a solution of a weak base can be calculated in a way which is very similar to that used for a weak acid. Instead of an acid constant Ka, a base constant Kb must be used. If a weak base B accepts protons from water according to the equation

    \[\text{B} + \text{ H}_{\text{2}}\text{O}\rightleftharpoons\text{BH}^{+} + \text{OH}^{-}  \label{1}\]

    then the base constant is defined by the expression

    \[K_{b}=\dfrac{\text{ }\!\![\!\!\text{ BH}^{\text{+}}\text{ }\!\!]\!\!\text{ }\!\![\!\!\text{ OH}^{-}\text{ }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ B }\!\!]\!\!\text{ }} \label{2}\]

    A list of Kb values for selected bases arranged in order of strength is given in the table below. This table is part of our larger collection of acid-base resources.

    Table \(\PageIndex{1}\): The Base Constants for Some Bases at 25°C.
    Base Formula and Ionization Equation Kb Molecular Shape
    Ammonia \(NH_3 + H_2O \rightleftharpoons NH^+_4 + OH^–\) 1.77 × 10–5
    Aniline \(C_6H_5NH_2 + H_2O \rightleftharpoons C_6H_5NH^+_3 + OH^–\) 3.9 × 10–10
    Carbonate ion \(CO_3^{2–} + H_2O \rightleftharpoons HCO^-_3 + OH^–\) 2.1 × 10–4
    Hydrazine \(N_2H_4 + H_2O \rightleftharpoons N_2H^+_5 + OH^–\)
    \(N_2H^+_5 + H_2O \rightleftharpoons N_2H_6^{2+} + OH^–\)
    K1 = 1.2 × 10–6
    K2 = 1.3 × 10–15
    Hydride ion \(H^– + H_2O \rightarrow H_2 + OH^–\) large
    Phosphate ion \(PO_4^{3–} + H_2O \rightleftharpoons HPO^{2-}_4 + OH^–\) 5.9 × 10–3
    Pyridine \(C_5H_5N + H_2O \rightleftharpoons C_5H_5NH^+ + OH^–\) 1.6 × 10–9
    • Taken from Hogfelt, E. Perrin, D. D. Stability Constants of Metal Ion Complexes, 1st ed. Oxford; New Pergamon, 1979-1982. International Union of Pure and Applied Chemistry, Commission on Equilibrium. ISBN: 0080209580

    To find the pH we follow the same general procedure as in the case of a weak acid. If the stoichiometric concentration of the base is indicated by cb, the result is entirely analogous to equation 4 in the section on the pH of weak acids; namely,

    \[K_{b}=\dfrac{\text{ }\!\![\!\!\text{ OH}^{-}\text{ }\!\!]\!\!\text{ }^{\text{2}}}{c_{b}-\text{ }\!\![\!\!\text{ OH}^{-}\text{ }\!\!]\!\!\text{ }} \label{3}\]

    Under most circumstances we can make the approximation

    \[c_b – [OH^–] \approx c_b\]

    in which case Equation \(\ref{3}\) reduces to the approximation

    \[[OH^–] ≈ \sqrt{K_{b}c_{b}} \label{4}\]

    which is identical to the expression obtained in the acid case (approximation shown in equation 6 in the section on the pH of weak acids) except that OH replaces H3O+ and b replaces a. Once we have found the hydroxide-ion concentration from this approximation, we can then easily find the pOH, and from it the pH.

    Example \(\PageIndex{1}\): pH using Kb

    Using the value for Kb listed in the table, find the pH of 0.100 M NH3.

    Solution

    It is not a bad idea to guess an approximate pH before embarking on the calculation. Since we have a dilute solution of a weak base, we expect the solution to be only mildly basic. A pH of 13 or 14 would be too basic, while a pH of 8 or 9 is too close to neutral. A pH of 10 or 11 seems reasonable. Using Eq. (4) we have

    \(\begin{align}\text{ }\!\![\!\!\text{ OH}^{-}\text{ }\!\!]\!\!\text{ }=\sqrt{K_{b}c_{b}} \\ \text{ }=\sqrt{\text{1}\text{.8 }\times \text{ 10}^{-\text{5}}\text{ mol L}^{-\text{1}}\text{ }\times \text{ 0}\text{.100 mol L}^{-\text{1}}} \\\text{ }=\sqrt{\text{1}\text{.8 }\times \text{ 10}^{-\text{6}}\text{ mol}^{\text{2}}\text{ L}^{-2}}=\text{1}\text{.34 }\times \text{ 10}^{-\text{3}}\text{ mol L}^{-\text{1}} \\\end{align}\)

    Checking the accuracy of the approximation, we find

    \(\dfrac{\text{ }\!\![\!\!\text{ OH}^{-}\text{ }\!\!]\!\!\text{ }}{c_{\text{b}}}=\dfrac{\text{1}\text{.34 }\times \text{ 10}^{-\text{3}}}{\text{0}\text{.1}}\approx \text{1 percent}\)

    The approximation is valid, and we thus proceed to find the pOH.

    \(\text{pOH}=-\text{log}\dfrac{\text{ }\!\![\!\!\text{ OH}^{-}\text{ }\!\!]\!\!\text{ }}{\text{mol L}^{-\text{1}}}=-\text{log(1}\text{.34 }\times \text{ 10}^{-\text{3}}\text{)}=\text{2}\text{.87}\)

    From which

    pH = 14.00 – pOH = 14.00 – 2.87 = 11.13

    This calculated value checks well with our initial guess.

    Occasionally we will find that the approximation

    cb – [OH] ≈ cb

    is not valid, in which case we must use a series of successive approximations similar to that outlined above for acids. The appropriate formula can be derived from Eq. (3) and reads

    [OH] ≈ \(\sqrt{K_{b}\text{(}c_{b}-\text{ }\!\![\!\!\text{ OH}^{-}\text{ }\!\!]\!\!\text{ )}}\)      (5)

    Contributors