Benzoic acid \(\left(K_{\mathrm{a}}=6.6 \times 10^{-5}\right)\) is present in many berries. Calculate the \(\mathrm{pH}\) and \(\%\) ionization of a \(726-\mathrm{mL}\) solution that contains \(0.288 \mathrm{~mol}\) of benzoic acid.

We have 0.288 moles of benzoic acid in a 726 mL solution, so we will convert the volume to liters and then calculate the molarity: volume in liters = \(\frac{726 \mathrm{~mL}}{1000 \mathrm{~mL/L}} = 0.726 \mathrm{~L}\) molarity (M) = \(\frac{0.288 \mathrm{~mol}}{0.726 \mathrm{~L}} = 0.397 \mathrm{~M}\)

Write the equation for benzoic acid ionization: \(\mathrm{HC_7H_5O_2} \longleftrightarrow \mathrm{H^+} + \mathrm{C_7H_5O_2^-}\) Then, write the equilibrium expression using the Ka for benzoic acid: \(K_\mathrm{a} = \frac{[\mathrm{H^+}][\mathrm{C_7H_5O_2^-}]}{[\mathrm{HC_7H_5O_2}]}\)

We initially have 0.397 M of benzoic acid and 0 M of \(\mathrm{H^+}\) and \(\mathrm{C_7H_5O_2^-}\), so let's set up a table for the equilibrium concentrations: HC7H5O2 H+ C7H5O2^- Initial (M) 0.397 0 0 Change (M) -x +x +x Equilibrium (M) 0.397-x x x Now we can plug the equilibrium values into the Ka expression: \(6.6 \times 10^{-5} = \frac{x^2}{0.397-x}\) We can assume that x is small compared to 0.397 and make this simplification to solve for x: \(6.6 \times 10^{-5} \approx \frac{x^2}{0.397}\) \(x \approx \sqrt{6.6 \times 10^{-5} \times 0.397} = 0.00437 \mathrm{~M}\) x represents the concentration of \(\mathrm{H^+}\) ions at equilibrium, so we can use this to determine the pH: pH \(= -\log_{10}(0.00437) = 2.36\)

Percent ionization is given by the following formula: \(\% \mathrm{ionization} = \frac{\text{concentration of ionized acid}}{\text{initial concentration of acid}} \times 100\) In this case, the concentration of ionized acid is the concentration of \(\mathrm{H^+}\) ions and the initial concentration of acid is the initial concentration of benzoic acid: \(\% \mathrm{ionization} = \frac{0.00437}{0.397} \times 100 = 1.10\%\) The pH of the benzoic acid solution is 2.36 and the percent ionization is 1.10%.