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 K a , a base constant K b 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 K b 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. Base Formula and Ionization Equation Kb Molecular Shape Ammonia \(NH_3 + H_2O \rightleftharpoons NH^+_4 + OH^–\) 1.77 × 10–5 NaN Aniline \(C_6H_5NH_2 + H_2O \rightleftharpoons C_6H_5NH^+_3 + OH^–\) 3.9 × 10–10 NaN Carbonate ion \(CO_3^{2–} + H_2O \rightleftharpoons HCO^-_3 + OH^–\) 2.1 × 10–4 NaN 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 NaN Hydride ion \(H^– + H_2O \rightarrow H_2 + OH^–\) large NaN Phosphate ion \(PO_4^{3–} + H_2O \rightleftharpoons HPO^{2-}_4 + OH^–\) 5.9 × 10–3 NaN Pyridine \(C_5H_5N + H_2O \rightleftharpoons C_5H_5NH^+ + OH^–\) 1.6 × 10–9 NaN Taken from Hogfelt, E. Perrin, D. D. Stability Constants of Metal Ion Complexes, 1 st 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 c b , 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 H 3 O + 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 K b Using the value for K b listed in the table, find the pH of 0.100 M NH 3 . 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 c b – [OH – ] ≈ c b 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 Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.