A solution is prepared in which 0.1 mole each of HCl, and is present in a litre. If the ionization c — Ionic Equilibrium Chemistry Question
Question
A solution is prepared in which 0.1 mole each of HCl, $\text{CH}_3\text{COOH}$ and $\text{CHCl}_2\text{COOH}$ is present in a litre. If the ionization constant of $\text{CH}_3\text{COOH}$ is $10^{-5}$ and that of $\text{Cl}_2\text{CHCOOH}$ is 0.15, the pH of solution is (log 2 = 0.3, log 3 = 0.48)
💡 Solution & Explanation
HCl is a strong acid, providing an initial $[H^+] = 0.1 \text{ M}$. $\text{CHCl}_2\text{COOH}$ is a moderately strong weak acid with $K_a = 0.15$. Let it dissociate to give $y \text{ M}$ of $H^+$. $K_a = \frac{y(0.1 + y)}{0.1 - y} = 0.15 \implies y^2 + 0.1y = 0.015 - 0.15y \implies y^2 + 0.25y - 0.015 = 0$. Solving the quadratic gives $y = 0.05 \text{ M}$. Total $[H^+] = 0.1 (\text{from HCl}) + 0.05 (\text{from CHCl}_2\text{COOH}) = 0.15 \text{ M}$. The $pH = -\log(0.15) = -\log(15 \times 10^{-2}) = 2 - (\log 3 + \log 5) = 2 - (0.48 + 0.70) = 0.82$. Therefore, correct answer is B.