What is the pH at the equivalence point for a titration involving 25 mL of 0.1 M acetic acid and 0.2 M KOH, given that the acid dissociation constant (Ka) for acetic acid is 1.8 × 10^-5?

• A. 5.28
• B. 7.00
• C. 3.75
• D. 8.72

Answer: (D)

Verified step by step guidance

1. Determine the moles of acetic acid initially present using the formula: \( \text{moles} = \text{concentration} \times \text{volume} \). For acetic acid, \( \text{moles} = 0.1 \text{ M} \times 0.025 \text{ L} \).
2. Calculate the volume of KOH needed to reach the equivalence point. Since the reaction is a 1:1 molar ratio, the moles of KOH needed are equal to the moles of acetic acid. Use the formula: \( \text{volume} = \frac{\text{moles}}{\text{concentration}} \) for KOH.
3. At the equivalence point, all acetic acid is converted to acetate ion. The solution now contains the acetate ion, which is the conjugate base of acetic acid. Calculate the concentration of acetate ion by dividing the moles of acetate by the total volume of the solution at the equivalence point.
4. Use the hydrolysis of the acetate ion to find the \( \text{pH} \). The reaction is: \( \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \). Use the expression for the base dissociation constant \( K_b \), which is related to \( K_a \) by \( K_w = K_a \times K_b \), where \( K_w = 1.0 \times 10^{-14} \).
5. Calculate the \( \text{pOH} \) from the concentration of \( \text{OH}^- \) ions using the formula \( \text{pOH} = -\log[\text{OH}^-] \), and then find the \( \text{pH} \) using \( \text{pH} = 14 - \text{pOH} \).