What is the pH of a 0.1 M solution of acetic acid (CH3COOH) given that its acid dissociation constant (Ka) is 1.8 x 10^-5?

• A. 2.87
• B. 5.25
• C. 3.45
• D. 4.75

Answer: (A)

Verified step by step guidance

1. Start by writing the chemical equation for the dissociation of acetic acid: \( \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \).
2. Use the expression for the acid dissociation constant \( K_a \), which is given by \( K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \). Substitute \( K_a = 1.8 \times 10^{-5} \) and the initial concentration of acetic acid \( [\text{CH}_3\text{COOH}] = 0.1 \text{ M} \).
3. Assume that the concentration of \( \text{CH}_3\text{COO}^- \) and \( \text{H}^+ \) ions formed is \( x \). Therefore, \( [\text{CH}_3\text{COO}^-] = x \) and \( [\text{H}^+] = x \), while \( [\text{CH}_3\text{COOH}] = 0.1 - x \).
4. Substitute these values into the \( K_a \) expression: \( 1.8 \times 10^{-5} = \frac{x^2}{0.1 - x} \). Since \( x \) is small compared to 0.1, you can approximate \( 0.1 - x \approx 0.1 \).
5. Solve for \( x \) to find \( [\text{H}^+] \), and then calculate the pH using the formula \( \text{pH} = -\log[\text{H}^+] \).