If pH of 0.001 M potassium propionate solution be 8.0, then the dissociation constant of propionic a — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

$\text{pH} = 8.0 \implies \text{pOH} = 6.0 \implies [\text{OH}^-] = 10^{-6}\text{ M}$. For the anionic hydrolysis of potassium propionate, $[\text{OH}^-] = \sqrt{\frac{K_w \cdot C}{K_a}}$. Substituting the values: $10^{-6} = \sqrt{\frac{10^{-14} \times 10^{-3}}{K_a}} \implies 10^{-12} = \frac{10^{-17}}{K_a}$. Solving for $K_a$ gives $K_a = \frac{10^{-17}}{10^{-12}} = 10^{-5}$.