Question

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.

Short answer

Question: Calculate the pH and percentage ionization of a benzoic acid solution with a volume of 726 mL and containing 0.288 mol of benzoic acid, given a Ka value of 6.6 x 10^-5. Answer: The pH of the benzoic acid solution is approximately 2.301, and the percentage ionization is approximately 1.26%.

Step-by-step solution

Calculate the initial concentration of benzoic acid

Given the amount of benzoic acid (0.288 mol) and the volume of the solution (726 mL), we can calculate the initial concentration of benzoic acid ([C7H5O2]): Initial concentration (\(C_{C7H5O2}\)) = \(\frac{0.288~\text{mol}}{0.726~\text{L}} = 0.397~\text{M}\)

Write the Ka expression

For benzoic acid in equilibrium with its ions, the reaction is: \(C7H5O2 (aq) \rightleftharpoons C7H5O2^- (aq) + H^+ (aq)\). The equilibrium constant \(K_a\) for benzoic acid is given. Write the equilibrium expression for the reaction: \(K_a = \frac{[C7H5O2^-][H^+]}{[C7H5O2]}\)

Calculate the concentration of H+ ions

Let the concentration of H+ ions (\(x\)) increase at equilibrium. Then, \(K_a = \frac{x^2}{0.397 - x}\). Solve for the concentration of H+ ions (\(x\)) using the given value of \(K_a = 6.6 \times 10^{-5}\): \(6.6 \times 10^{-5}= \frac{x^2}{0.397 - x}\). To simplify calculations, we can assume \(x << 0.397\), which gives: \(x^2 = 6.6 \times 10^{-5} \times 0.397\). Solving for \(x\), we get \(x = 4.999 \times 10^{-3}~\text{M}\) representing the concentration of H+ ions.

Calculate the pH of the solution

Now that we have the concentration of H+ ions, we can calculate the pH using the formula: \(pH = -\log_{10}[H^+]\). \(pH = -\log_{10}(4.999 \times 10^{-3}) \approx 2.301\). The pH of the solution is approximately 2.301.

Calculate percentage ionization

To find the percentage ionization of benzoic acid (BA), we first need to calculate the ratio of ionized BA (H+ ions) to the initial BA concentration: \(\%~\text{ionization} = \frac{[H^+]}{C_{C7H5O2}} \times 100\%\). \(\%~\text{ionization} = \frac{4.999 \times 10^{-3}}{0.397} \times 100\% \approx 1.26\%\). The percentage ionization of the benzoic acid solution is approximately 1.26%.