How to Find the Ph of a Solution Weak Base

14.5: The pH of Solutions of Weak Bases

Last updated
Save as PDF

Page ID: 49687

ChemPRIME at Chemical Education Digital Library (ChemEd DL)

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.

Table \(\PageIndex{1}\): The Base Constants for Some Bases at 25°C.
Base	Formula and Ionization Equation	K_b
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, 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 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 K_b

Using the value for K_b 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

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)

How to Find the Ph of a Solution Weak Base

Source: https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/14%3A_Ionic_Equilibria_in_Aqueous_Solutions/14.05%3A_The_pH_of_Solutions_of_Weak_Bases