# 21.15: Calculating pH of Weak Acid and Base Solutions

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

Bees are beautiful creatures that help plants flourish. They carry pollen from one plant to another to facilitate plant growth and development. However, they can be troublesome when they sting! For those who are allergic to bee venom, this can be a serious, life-threatening problem. For all other humans, it can be a painful experience. When stung by a bee, one first-aid treatment is to apply a paste of baking soda (sodium bicarbonate) to the stung area. This weak base helps with the itching and swelling that accompanies the bee sting.

## Calculating pH of Weak Acid and Base Solutions

The $$K_\text{a}$$ and $$K_\text{b}$$ values have been determined for a great many acids and bases, as shown in Tables 21.12.2 and 21.13.1. These can be used to calculate the pH of any solution of a weak acid or base whose ionization constant is known.

##### Example $$\PageIndex{1}$$

Calculate the pH of a $$2.00 \: \text{M}$$ solution of nitrous acid $$\left( \ce{HNO_2} \right)$$. The $$K_\text{a}$$ for nitrous acid is $$4.5 \times 10^{-4}$$.

###### Known
• Initial $$\left[ \ce{HNO_2} \right] = 2.00 \: \text{M}$$
• $$K_\text{a} = 4.5 \times 10^{-4}$$
###### Unknown

First, an ICE table is set up with the variable $$x$$ used to signify the change in concentration of the substance due to ionization of the acid. Then the $$K_\text{a}$$ expression is used to solve for $$x$$ and calculate the pH.

###### Step 2: Solve.

$\begin{array}{l|ccc} & \ce{HNO_2} & \ce{H^+} & \ce{NO_2^-} \\ \hline \text{Initial} & 2.00 & 0 & 0 \\ \text{Change} & -x & +x & +x \\ \text{Equilibrium} & 2.00 - x & x & x \end{array}\nonumber$

The $$K_\text{a}$$ expression and value are used to set up an equation to solve for $$x$$.

$K_\text{a} = 4.5 \times 10^{-4} = \frac{\left( x \right) \left( x \right)}{2.00 - x} = \frac{x^2}{2.00 - x}\nonumber$

The quadratic equation is required to solve this equation for $$x$$. However, a simplification can be made of the fact that the extent of ionization of weak acids is small. The value of $$x$$ will be significantly less than 2.00, so the "$$-x$$" in the denominator can be dropped.

\begin{align*} 4.5 \times 10^{-4} &= \frac{x^2}{2.00 - x} \approx \frac{x^2}{2.00} \\ x &= \sqrt{ 4.5 \times 10^{-4} \left( 2.00 \right)} = 2.9 \times 10^{-2} \: \text{M} = \left[ \ce{H^+} \right] \end{align*}\nonumber

Since the variable $$x$$ represents the hydrogen-ion concentration, the pH of the solution can now be calculated.

$\text{pH} = -\text{log} \left[ \ce{H^+} \right] = -\text{log} \left[ 2.9 \times 10^{-2} \right] = 1.54\nonumber$

The pH of a $$2.00 \: \text{M}$$ solution of a strong acid would be equal to $$-\text{log} \left( 2.00 \right) = -0.30$$. The higher pH of the $$2.00 \: \text{M}$$ nitrous acid is consistent with it being a weak acid and therefore not as acidic as a strong acid would be.
The procedure for calculating the pH of a solution of a weak base is similar to that of the weak acid in the example. However, the variable $$x$$ will represent the concentration of the hydroxide ion. The pH is found by taking the negative logarithm to get the pOH, followed by subtracting from 14 to get the pH.