The pH of a 0.22 M solution of a weak monoprotic acid, HA, is 2.74. Calculate K_a for this acid.

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Quizplus · Accepted Answer

The pH of the solution is 2.74, which means the concentration of H⁺ ions is 10^(-2.74) = 1.82 x 10^(-3) M. For a weak acid HA dissociating as HA ⇌ H⁺ + A⁻, the equilibrium concentration of H⁺ is equal to the concentration of A⁻, and the initial concentration of HA is 0.22 M. The equilibrium concentration of HA is approximately 0.22 - 1.82 x 10^(-3) ≈ 0.218 M. The expression for Kₐ is [H⁺][A⁻]/[HA] = (1.82 x 10^(-3))^2 / 0.218 = 1.5 x 10^(-5).

The PH of a 0