# 7.15: Calculating pH of Weak Acid and Base Solutions

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## 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 $$4.5 \times 10^{-4}$$.

Solution

Step 1: List the known values and plan the problem.

Known

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

Unknown

• pH $$= ?$$

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.

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

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

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

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}

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

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

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.

## Summary

• The procedure for calculating the pH of a weak acid or base is illustrated.