What is the approximate ionization constant () of pure water at 25°C? (Given density of water is ) — Ionic Equilibrium Chemistry Question
Question
What is the approximate ionization constant ($K$) of pure water at 25°C? (Given density of water is $1.0 \text{ g/cm}^3$)
Answer: B
💡 Solution & Explanation
The ionization constant is $K = [H^+][OH^-] / [H_2O]$. The ionic product $K_w = 10^{-14}$. $[H_2O] = 1000/18 = 55.5 \text{ M}$. Thus, $K = 10^{-14} / 55.5 = 1.8 \times 10^{-16}$.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes