Calculate the pH of a solution prepared by mixing \(2.00 \mathrm{~g}\) of butyric acid \(\left(\mathrm{HC}_{4} \mathrm{H}_{7} \mathrm{O}_{2}\right)\) with \(0.50 \mathrm{~g}\) of \(\mathrm{NaOH}\) in water \(\left(K_{\mathrm{a}}\right.\) butyric acid \(\left.=1.5 \times 10^{-5}\right)\)

The butyric acid dissociation equation is as follows: \[\mathrm{HC}_4\mathrm{H}_7\mathrm{O}_2\leftrightharpoons\mathrm{H}^+ + \mathrm{C}_4\mathrm{H}_7\mathrm{O}_2^-\] And the reaction between butyric acid and sodium hydroxide is: \[\mathrm{HC}_4\mathrm{H}_7\mathrm{O}_2 + \mathrm{NaOH} \rightarrow \mathrm{NaC}_4\mathrm{H}_7\mathrm{O}_2 + \mathrm{H}_2\mathrm{O}\]

First, we need to find the molar mass of butyric acid and sodium hydroxide. Molar mass of butyric acid (\(\mathrm{HC}_4\mathrm{H}_7\mathrm{O}_2\)) = \(12.01 × 4 + 1.01 × 8 + 16 × 2 = 88.11\mathrm{~g/mol}\) Now, calculate the moles of butyric acid: \[\frac{2.00\mathrm{~g}}{88.11\mathrm{~g/mol}} = 0.02271\mathrm{~mol}\] Molar mass of sodium hydroxide (\(\mathrm{NaOH}\)) = \(22.99 + 16 + 1.01 = 40.00\mathrm{~g/mol}\) Now, calculate the moles of sodium hydroxide: \[\frac{0.50\mathrm{~g}}{40.00\mathrm{~g/mol}} = 0.01250\mathrm{~mol}\]

Since sodium hydroxide is a strong base, it will fully react with butyric acid. The limiting reagent is sodium hydroxide (0.01250 mol), so the reaction equation becomes: \(0.02271\mathrm{~mol}~\mathrm{HC}_4\mathrm{H}_7\mathrm{O}_2 - 0.01250\mathrm{~mol}~\mathrm{NaOH} = 0.01021\mathrm{~mol}~\mathrm{HC}_4\mathrm{H}_7\mathrm{O}_2\) (remaining) Now, we have 0.01021 mol of butyric acid and 0.01250 mol of butyrate ion (\(\mathrm{C}_4\mathrm{H}_7\mathrm{O}_2^-\)) in the solution.

Using the equilibrium expression, we can write: \[\frac{[\mathrm{H}^+][\mathrm{C}_4\mathrm{H}_7\mathrm{O}_2^-]}{[\mathrm{HC}_4\mathrm{H}_7\mathrm{O}_2]} = K_a = 1.5 \times 10^{-5}\] We can use the initial concentrations of the species and set up an equilibrium table to find the final concentrations: \[\frac{[\mathrm{H}^+][0.01250\mathrm{~mol}+x]}{[0.01021\mathrm{~mol}-x]} = 1.5 \times 10^{-5}\] Since \(K_a\) is very small, we can assume that \(x\) will be negligible compared to the initial concentrations: \[\frac{[\mathrm{H}^+][0.01250\mathrm{~mol}]}{[0.01021\mathrm{~mol}]} = 1.5 \times 10^{-5}\] Now, solve for \([\mathrm{H}^+]\): \[[\mathrm{H}^+] = \frac{1.5 \times 10^{-5} \times 0.01021\mathrm{~mol}}{0.01250\mathrm{~mol}} = 1.229 \times 10^{-6}\] Finally, calculate the pH – pH = -\(\log {[\mathrm{H}^+]}\): \[\mathrm{pH} = -\log{(1.229 \times 10^{-6})} = 5.91\] So, the pH of the solution is 5.91.