A weak monoprotic acid ionizes as . The exact pH of a solution of this acid is measured to be 5. Wha — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Given $pH = 5$, the $[H^+]$ is $10^{-5} \text{ M}$. From the equilibrium $HA \rightleftharpoons H^+ + A^-$, $[H^+] = [A^-] = 10^{-5} \text{ M}$. The equilibrium concentration of $HA$ is $C - [H^+] = 1.0 - 10^{-5} \approx 1.0 \text{ M}$. The dissociation constant $K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(10^{-5})^2}{1} = 10^{-10}$. Therefore, $K_a \times 10^{10} = 10^{-10} \times 10^{10} = 1$.