What will be the effect of adding 100 ml of 0.001 M – HCl solution to 100 ml of a solution having 0. — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Initially, 0.1 M HA has $[H^+] = \sqrt{10^{-5} \times 0.1} = 10^{-3} \text{ M}$, and degree of dissociation $\alpha = 10^{-3}/0.1 = 10^{-2}$. Upon adding 100 ml of 0.001 M HCl, the volume doubles. The new concentrations are $C_{HA} = 0.05 \text{ M}$ and $C_{HCl} = 5 \times 10^{-4} \text{ M}$. Let $[H^+]_{final} = h$. We use the exact equation $h^2 - C_{HCl}h - K_a C_{HA} = 0 \implies h^2 - 5 \times 10^{-4}h - 5 \times 10^{-7} = 0$. Plugging in $h = 10^{-3} \text{ M}$ perfectly satisfies this equation ($(10^{-3})^2 - 5 \times 10^{-7} - 5 \times 10^{-7} = 0$). Thus, final $[H^+] = 10^{-3} \text{ M}$, meaning pH remains unchanged. Final $\alpha = [\text{A}^-] / C_{HA} = (K_a / h) = 10^{-5} / 10^{-3} = 10^{-2}$. Both remain unchanged. Therefore, correct answer is C.