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 $K_w$** The ionic product of water is: $$K_w = [H_3O^+][OH^-]$$ **Step 2: Identify the given information** At 90°C in pure water: - $[H_3O^+] = 10^{-6}$ mol/L **Step 3: Use the property of pure water** In pure water, water molecules ionize equally to produce $H_3O^+$ and $OH^-$ ions: $$H_2O \rightleftharpoons H_3O^+ + OH^-$$ Therefore, in pure water: $$[H_3O^+] = [OH^-]$$ **Step 4: Calculate $[OH^-]$** $$[OH^-] = [H_3O^+] = 10^{-6} \text{ mol/L}$$ **Step 5: Calculate $K_w$** $$K_w = [H_3O^+][OH^-] = (10^{-6})(10^{-6}) = 10^{-12}$$ **Answer: $K_w = 10^{-12}$ mol²/L² at 90°C** This makes physical sense: as temperature increases, water ionization increases, so $K_w$ increases from $10^{-14}$ at 25°C to $10^{-12}$ at 90°C.