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.

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: \[ pH = 14 - pOH = 14 - 5.1 \approx 8.9 \]