Calculating the pH of a salt solution Calculate the pH at $$ 25^{\circ} \mathrm{C} $$ of a 0.34 M solution of sodium benzoate $$ \left(\mathrm{NaC}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}\right) $$. Note that benzoic acid $$ \left(\mathrm{HC}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}\right) $$ is a weak acid with a $$ p K_{a} $$ of 4.20 . Round your answer to 1 decimal place.

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We start by writing the hydrolysis reaction for the benzoate ion in water:

$$
\mathrm{C}_6\mathrm{H}_5\mathrm{CO}_2^- + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{HC}_6\mathrm{H}_5\mathrm{CO}_2 + \mathrm{OH}^-
$$

Because benzoic acid is weak, the benzoate ion acts as a weak base with a base ionization constant $$ K_b $$. The relationship between the acid and base constants is:

$$
K_a \cdot K_b = K_w
$$

Given $$ K_a = 10^{-4.20} $$ and $$ K_w = 1.0 \times 10^{-14} $$, we calculate $$ K_b $$:

$$
K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{10^{-4.20}} = \frac{1.0 \times 10^{-14}}{6.31 \times 10^{-5}} \approx 1.58 \times 10^{-10}
$$

Next, let $$ x $$ be the concentration of $$\mathrm{OH}^-$$ produced (in M). We assume the initial concentration of benzoate is 0.34 M and that the change $$ x $$ is small relative to 0.34 M. The equilibrium expression becomes:

$$
K_b = \frac{x^2}{0.34}
$$

Solving for $$ x $$:

$$
x^2 = K_b \times 0.34 \quad \Longrightarrow \quad x = \sqrt{1.58 \times 10^{-10} \times 0.34}
$$

$$
x = \sqrt{5.37 \times 10^{-11}} \approx 7.33 \times 10^{-6}
$$

This $$ x $$ represents the concentration of $$\mathrm{OH}^-$$ in the solution. The $$ pOH $$ is calculated as:

$$
pOH = -\log (7.33 \times 10^{-6}) \approx 5.1
$$

Finally, the $$ pH $$ of the solution is:

$$
pH = 14 - pOH = 14 - 5.1 \approx 8.9
$$
The pH of the 0.34 M sodium benzoate solution at $$ 25^{\circ} \mathrm{C} $$ is approximately 8.9.

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