Question

Consider the titration of butyric acid (HBut) with sodium hydroxide. In an experiment, \(50.00 \mathrm{~mL}\) of \(0.350 \mathrm{M}\) butyric acid is titrated with \(0.225 \mathrm{M} \mathrm{NaOH} . K_{\mathrm{a}}\) HBut \(=1.5 \times 10^{-5}\). (a) Write a balanced net ionic equation for the reaction that takes place during titration. (b) What are the species present at the equivalence point? (c) What volume of sodium hydroxide is required to reach the equivalence point? (d) What is the pH of the solution before any \(\mathrm{NaOH}\) is added? (e) What is the pH of the solution halfway to the equivalence point? (f) What is the pH of the solution at the equivalence point?

Short answer

Answer: The pH of the solution at the equivalence point is approximately 9.03.

Step-by-step solution

a) Balanced net ionic equation

The reaction between butyric acid and sodium hydroxide is an acid-base reaction, in which a proton (\(\mathrm{H}^{+}\)) will be transferred from the acid to the base. The balanced net ionic equation can be written as follows: \[\mathrm{HBut + OH}^{-} \rightarrow \mathrm{But}^{-} + \mathrm{H_2O}\]

b) Species at the equivalence point

At the equivalence point, the butyric acid (HBut) has been completely neutralized by sodium hydroxide (NaOH), so there are no free \(\mathrm{H}^{+}\) ions left. The remaining species are butyrate ion (\(\mathrm{But}^{-}\)) and water (\(\mathrm{H_2O}\)).

c) Volume of NaOH to reach the equivalence point

To reach the equivalence point, we need equal moles of butyric acid (HBut) and sodium hydroxide (NaOH). We can use the equation: Moles of HBut = Moles of NaOH \(50.00 \mathrm{~mL} \times 0.350\mathrm{~M} = V_\mathrm{NaOH} \times 0.225\mathrm{~M}\) Solving for \(V_\mathrm{NaOH}\), we get: \(V_\mathrm{NaOH} = \frac{50.00\mathrm{~mL} \times 0.350\mathrm{~M}}{ 0.225\mathrm{~M}}\) \(V_\mathrm{NaOH} = 77.78\mathrm{~mL}\)

d) pH of the solution before NaOH is added

Before any sodium hydroxide is added, the solution only contains butyric acid (HBut), which is a weak acid. We can use the Ka expression to find the concentration of \(\mathrm{H}^{+}\) ions and then calculate the pH using the formula: pH = -log10[H+] For HBut: \[K_{\mathrm{a}} = \frac{\mathrm{[H^{+}][But^{-}]}}{\mathrm{[HBut]}}\] Since the initial concentration of HBut is 0.35 M, we can write: \[\mathrm{1.5 \times 10^{-5} = \frac{[H^{+}][H^{+}]}{0.35 - [H^{+}]}}\] We can solve this quadratic equation for \(\mathrm{[H^{+}]}\) and then calculate the pH. Since the \(K_{a}\) value is very small, we can make an assumption that \([H^{+}] \ll 0.35\) and can approximate the equation as: \[\mathrm{1.5 \times 10^{-5} = \frac{[H^{+}^2]}{0.35}}\] Solving for \(\mathrm{[H^{+}]}\), we get: \([H^{+}] \approx \sqrt{1.5 \times 10^{-5} \times 0.35} \approx 0.00245\) The pH of the solution before any NaOH is added is: \(pH = -\log10(0.00245) \approx 2.61\)

e) pH of the solution halfway to the equivalence point

At halfway to the equivalence point, the concentration of HBut and butyrate ion (But\(^-\)) are equal. According to the Henderson-Hasselbalch equation: \(pH = pK_a + \log\frac{\mathrm{[But}^{-}]}{\mathrm{[HBut]}}\) Since at halfway point, \([But^{-}] = [HBut]\), the equation reduces to: \(pH = pK_a\) We can calculate the pKa value for HBut using: \(pK_a = -\log(1.5 \times 10^{-5}) \approx 4.82\) Therefore, the pH of the solution halfway to the equivalence point is approximately 4.82.

f) pH of the solution at the equivalence point

At the equivalence point, the solution contains only the conjugate base, butyrate ion (But\(^{-}\)), and water. Since the butyrate ion is the conjugate base of a weak acid, the resulting solution at the equivalence point will be slightly basic. We can calculate the concentration of \(\mathrm{OH^-}\) ions by using the Kb expression for the butyrate ion. \(K_{b}= \frac{K_{w}}{K_{a}}\) where \(K_w\) is the ion product of water (\(1\times 10^{-14}\)) \(K_{b} = \frac{1\times 10^{-14}}{1.5 \times 10^{-5}} = \mathrm{6.67 \times 10^{-10}}\) The Kb expression for the butyrate ion is: \[K_{\mathrm{b}} = \frac{\mathrm{[OH^{-}][HBut]}}{\mathrm{[But^{-}]}}\] The initial concentration of butyrate ion at the equivalence point can be calculated as: \([\mathrm{But}^-] = \frac{\mathrm{50.00\ mL \times 0.350\ \text{M}}}{50.00\ \mathrm{mL} \ +\ 77.78\ \mathrm{mL}} = \mathrm{0.140\ \text{M}}\) Since there is no HBut in the solution at the equivalence point, the concentration of \(\mathrm{OH^-}\) ions can be calculated as: \[6.67 \times 10^{-10} = \frac{\mathrm{[OH^{-}][0.140\ \text{M}]}}{\mathrm{[0.140\ \text{M} - OH^{-}]}}\] Assuming \(\mathrm{[OH^-]} \ll 0.140\ \text{M}\), \(\mathrm{[OH^{-}]} \approx \sqrt{6.67 \times 10^{-10} \times 0.140} \approx 1.08 \times 10^{-5}\) Now, we can calculate the pOH and then the pH: \(pOH = -\log(1.08 \times 10^{-5}) \approx 4.97\) \(pH = 14 - pOH = 14 - 4.97 \approx 9.03\) So, the pH of the solution at the equivalence point is approximately 9.03.