The degree of dissociation of pure water at 25°C is found to be 1.8 * 10^-9. The dissociation consta — Ionic Equilibrium Chemistry Question
Question
The degree of dissociation of pure water at 25°C is found to be 1.8 * 10^-9. The dissociation constant, Kd, of water, at 25°C is
Answer: B
💡 Solution & Explanation
The dissociation constant of water Kd is defined as: Kd = [H+][OH-] / [$H_2O$]. At 25°C, [H+][OH-] = Kw = 1.0 * 10^-14 and [$H_2O$] = 55.56 M. Therefore, Kd = 1.0 * 10^-14 / 55.56 = 1.8 * 10^-16.
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