A solution is prepared by dissolving 0.56 g benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}, K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) in enough water to make \(1.0 \mathrm{L}\) of solution. Calculate \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right],\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\right],\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right]\) and the pH of this solution.

: We'll first calculate the initial concentration of benzoic acid \([\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}]\). To do this, we need to convert the grams of benzoic acid to moles and then divide by the volume of the solution. The molar mass of \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}\) is \(12 \times 7 + 6 \times 6 + 16 \times 2 = 122\) g/mol. Therefore, the initial concentration of \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}\) is: \[\frac{0.56 \, \mathrm{g}}{122\, \mathrm{g/mol}} \cdot \frac{1}{1.0\, \mathrm{L}} = 0.00459\, \mathrm{M}\]

: Next, we'll write the ionization equilibrium for benzoic acid as follows: \[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H} \longleftrightarrow \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-} + \mathrm{H}^{+}\] The equilibrium constant expression for this reaction is: \[K_a = \frac{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}][\mathrm{H}^{+}]}{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}]}\]

: We'll assume that the initial concentration of benzoic acid is approximately equal to the equilibrium concentration. With this assumption, we can now write the equation for the concentrations of the species in terms of \(x\), where \(x\) is the concentration of \(\mathrm{H}^{+}\): \([\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}] = 0.00459 - x\) \([\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}] = x\) \([\mathrm{H}^{+}] = x\) Substituting these expressions into the equilibrium constant equation and solving for \(x\): \(K_a = \frac{x^2}{0.00459 - x}\) \(6.4 \times 10^{-5} = \frac{x^2}{0.00459 - x}\)

: Now, we solve for \(x\) to get the concentration of \(\mathrm{H}^{+}\): \(x^2 + 6.4 \times 10^{-5}x - 0.00459 \times 6.4 \times 10^{-5} = 0\) We can disregard the negative root and solve for the positive root to get \(x \approx 7.4 \times 10^{-6}\): \([\mathrm{H}^{+}] \approx 7.4 \times 10^{-6}\, \mathrm{M}\) Now, we can find the other concentrations: \([\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}] \approx 0.00459 - 7.4\times10^{-6} \approx 0.00458\, \mathrm{M}\) \([\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}] \approx 7.4 \times 10^{-6}\, \mathrm{M}\) Lastly, we'll calculate the concentration of hydroxide ions, \([\mathrm{OH}^{-}]\), using the ion product of water: \(K_w = [\mathrm{H}^{+}][\mathrm{OH}^{-}]\) \(1.0 \times 10^{-14} = (7.4 \times 10^{-6})[\mathrm{OH}^{-}]\) \([\mathrm{OH}^{-}] \approx 1.35 \times 10^{-9}\, \mathrm{M}\)

: Finally, we'll calculate the pH of the solution using the equation pH = \(- \log [\mathrm{H}^{+}]\): pH \(\approx -\log(7.4 \times 10^{-6}) \approx 5.13\) The concentrations of the species in the solution are approximately as follows: \(\left[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}\right] \approx 0.00458\, \mathrm{M}\), \(\left[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}^{-}\right] \approx 7.4 \times 10^{-6}\, \mathrm{M}\), \(\left[\mathrm{H}^{+}\right] \approx 7.4 \times 10^{-6}\, \mathrm{M}\), and \(\left[\mathrm{OH}^{-}\right] \approx 1.35 \times 10^{-9}\, \mathrm{M}\). The pH of the solution is approximately 5.13.