Let's go step by step: --- ### **a) The equation used to calculate the concentration of [OH⁻] ions added to the buffer solution is:** \[ [\text{OH}^-] = \frac{n \text{ (mol)}}{V \text{ (volume)}} \] where: - \( n \) is the moles of LiOH added \( = 0.01 \) mol - \( V \) is the volume of solution \( = 0.5 \) L So, the answer is: - \( 0.02 \) M (since \( 0.01 \) mol / \( 0.5 \) L = \( 0.02 \) M) --- ### **b) Reaction Table (concentration):** The reaction is: \[ \mathrm{RCOOH} + \mathrm{OH}^- \rightarrow \mathrm{RCOO}^- + \mathrm{H_2O} \] - **Initial:** - RCOOH: 0.35 M - OH⁻: 0.02 M - RCOO⁻: 0.65 M - **Changes:** - RCOOH: -0.02 (since OH⁻ is the limiting reagent, all 0.02 M will react) - OH⁻: -0.02 - RCOO⁻: +0.02 - **Final:** - RCOOH: 0.35 - 0.02 = **0.33 M** - OH⁻: 0 - RCOO⁻: 0.65 + 0.02 = **0.67 M** --- ### **c) What is the pH of this solution?** This is a buffer solution (weak acid + its conjugate base), so we use the **Henderson-Hasselbalch equation**: \[ \text{pH} = \text{p}K_a + \log \left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \] Given: - \( K_a = 5.8 \times 10^{-5} \implies \text{p}K_a = -\log(5.8 \times 10^{-5}) = 4.24 \) - \( [\text{A}^-] = [\text{RCOO}^-] = 0.67 \) - \( [\text{HA}] = [\text{RCOOH}] = 0.33 \) Plug in the values: \[ \text{pH} = 4.24 + \log \left(\frac{0.67}{0.33}\right) \] \[ \frac{0.67}{0.33} \approx 2.03 \] \[ \log(2.03) \approx 0.308 \] \[ \text{pH} = 4.24 + 0.308 = 4.55 \] --- ### **Final Answers:** - **a)** [OH⁻] = **0.02 M** - **b)** (see table filled above) - **c)** pH = **4.55**

a) \( 0.02\, \text{M} \) b) Final concentrations: - \([\text{RCOOH}] = 0.33\, \text{M}\) - \([\text{RCOO}^-] = 0.67\, \text{M}\) c) pH ≈ **4.55**