The dissociation constant of a weak monoprotic acid is numerically equal to the dissociation constant of its conjugate base. What is pH of 0.1 M solution of this acid?

Question

Prashant Kotian · Accepted Answer

Correct answer: D. We are given Ka = Kb. Since Ka × Kb = Kw = 10^-14, we have Ka^2 = 10^-14 => Ka = 10^-7. For a 0.1 M solution of this weak monoprotic acid: [H+] = sqrt(Ka × C) = sqrt(10^-7 × 0.1) = sqrt(10^-8) = 10^-4 M. pH = -log[H+] = -log(10^-4) = 4.0.

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