The degree of dissociation of pure water at 25°C is found to be . The dissociation constant, of wate — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

The dissociation constant is given by $K_d = \frac{[\text{H}^+][\text{OH}^-]}{[\text{H}_2\text{O}]}$. We know $[\text{H}^+][\text{OH}^-] = K_w = 1.0 \times 10^{-14}$ at 25°C. The concentration of pure water is $[\text{H}_2\text{O}] = 1000\text{ g/L} / 18\text{ g/mol} = 55.55\text{ M}$. Thus, $K_d = \frac{10^{-14}}{55.55} = 1.8 \times 10^{-16}$.