Calculate the pH of a 0.0095 M solution of acetic acid, CH3COOH, given that the acid dissociation constant (Ka) is 1.8 x 10^-5.

• A. pH = 5.25
• B. pH = 3.02
• C. pH = 4.75
• D. pH = 2.74

Answer: (B)

Verified step by step guidance

1. Start by writing the chemical equation for the dissociation of acetic acid in water: \( \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \).
2. Set up the expression for the acid dissociation constant \( K_a \) using the formula: \( K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \).
3. Assume that the initial concentration of acetic acid is 0.0095 M and that the change in concentration due to dissociation is \( x \). Therefore, at equilibrium, \([\text{CH}_3\text{COO}^-] = x\), \([\text{H}^+] = x\), and \([\text{CH}_3\text{COOH}] = 0.0095 - x\).
4. Substitute these equilibrium concentrations into the \( K_a \) expression: \( 1.8 \times 10^{-5} = \frac{x^2}{0.0095 - x} \).
5. Assume \( x \) is small compared to 0.0095, so \( 0.0095 - x \approx 0.0095 \). Solve the simplified equation \( 1.8 \times 10^{-5} = \frac{x^2}{0.0095} \) for \( x \), and then calculate the pH using \( \text{pH} = -\log_{10}(x) \).