What is the pH of a 0.050 M solution of HNO2 (Ka = 4.6 × 10^-4)?

• A. 3.12
• B. 2.89
• C. 2.34
• D. 3.45

Answer: (B)

Verified step by step guidance

1. Identify that HNO2 is a weak acid and will partially dissociate in water. The dissociation can be represented as: HNO2 ⇌ H+ + NO2−.
2. Write the expression for the acid dissociation constant (Ka) for HNO2: \( K_a = \frac{[H^+][NO_2^-]}{[HNO_2]} \).
3. Set up an ICE (Initial, Change, Equilibrium) table to determine the concentrations of the species at equilibrium. Initially, [HNO2] = 0.050 M, [H+] = 0, and [NO2−] = 0.
4. Assume that the change in concentration of HNO2 is 'x', so at equilibrium, [HNO2] = 0.050 - x, [H+] = x, and [NO2−] = x. Substitute these into the Ka expression: \( 4.6 \times 10^{-4} = \frac{x^2}{0.050 - x} \).
5. Assume x is small compared to 0.050 M, simplifying the equation to \( 4.6 \times 10^{-4} = \frac{x^2}{0.050} \). Solve for x to find [H+], then calculate pH using the formula \( \text{pH} = -\log[H^+] \).