Find the pH of a 0.031 M H2SO4 (aq) solution at 25 degrees Celsius. Ka for HSO4- is 1.2 × 10^-2.

• A. 1.20
• B. 1.50
• C. 2.00
• D. 0.85

Answer: (A)

Verified step by step guidance

1. Start by recognizing that sulfuric acid (H2SO4) is a strong acid, which means it dissociates completely in its first ionization step: H2SO4 → H+ + HSO4-. Therefore, the initial concentration of H+ ions from the first dissociation is equal to the concentration of H2SO4, which is 0.031 M.
2. Next, consider the second ionization step of HSO4-, which is a weak acid: HSO4- ⇌ H+ + SO4^2-. Use the given Ka value for HSO4- (1.2 × 10^-2) to determine the additional concentration of H+ ions produced from this equilibrium.
3. Set up an equilibrium expression for the second ionization: Ka = [H+][SO4^2-] / [HSO4-]. Assume that the change in concentration of HSO4- is 'x', which is also the change in concentration of H+ and SO4^2-.
4. Substitute the known values into the equilibrium expression: 1.2 × 10^-2 = (0.031 + x)(x) / (0.031 - x). Solve for 'x' to find the additional concentration of H+ ions from the second ionization.
5. Finally, calculate the total concentration of H+ ions by adding the concentration from the first ionization (0.031 M) to the additional concentration 'x' from the second ionization. Use the formula for pH: pH = -log[H+], to find the pH of the solution.