At 90oC, pure water has + ion concentration of 10-6 mol/L. The Kw at 90oC is — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

# Solution: Finding $K_w$ at 90°C **Step 1: Recall the definition of $K_w$** $$K_w = [H_3O^+][OH^-]$$ **Step 2: Use the property of pure water** In pure water, the concentration of $H_3O^+$ equals the concentration of $OH^-$ due to auto-ionization: $$[H_3O^+] = [OH^-]$$ **Step 3: Substitute the given concentration** Given: $[H_3O^+] = 10^{-6}$ mol/L at 90°C Therefore: $[OH^-] = 10^{-6}$ mol/L **Step 4: Calculate $K_w$** $$K_w = (10^{-6})(10^{-6}) = 10^{-12}$$ **Answer: $K_w = 10^{-12}$ at 90°C** This makes physical sense: at higher temperatures (90°C vs. standard 25°C), water auto-ionization increases significantly, so $K_w$ is larger than the standard value of $10^{-14}$ at 25°C.