14.5: The pH of Solutions of Weak Bases
 Page ID
 49687
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 acidbase resources.
Base  Formula and Ionization Equation  K_{b}  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  \(N_2H_4 + H_2O \rightleftharpoons N_2H^+_5 + OH^–\) \(N_2H^+_5 + H_2O \rightleftharpoons N_2H_6^{2+} + OH^–\) 
K_{1} = 1.2 × 10^{–6} K_{2} = 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, 1^{st} ed. Oxford; New Pergamon, 19791982. 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 hydroxideion 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 (UWMadison), Justin Shorb (Hope College), Xavier PratResina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.