The dissociation constant of a weak monoprotic acid is numerically equal to the dissociation constan — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Let the acid be HA. Given $K_a = K_b$. For a conjugate pair, $K_a \times K_b = K_w = 10^{-14}$. Thus, $K_a^2 = 10^{-14} \implies K_a = 10^{-7}$. For a $0.1 \text{ M}$ solution of this weak acid, $[H^+] = \sqrt{K_a \cdot C} = \sqrt{10^{-7} \times 0.1} = \sqrt{10^{-8}} = 10^{-4} \text{ M}$. The pH is $-\log(10^{-4}) = 4.0$. Therefore, correct answer is D.