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: Identify the ion concentration in pure water** In pure water, the $H_3O^+$ and $OH^-$ ions are produced in equal amounts from water's self-ionization: $$2H_2O \rightleftharpoons H_3O^+ + OH^-$$ Therefore: $[H_3O^+] = [OH^-] = 10^{-6} \text{ mol/L}$ **Step 3: Calculate $K_w$** $$K_w = (10^{-6})(10^{-6}) = 10^{-12}$$ **Answer: $K_w = 10^{-12}$ at 90°C** (This is option C) **Key insight:** At higher temperatures (90°C vs standard 25°C where $K_w = 10^{-14}$), water's self-ionization increases, so $K_w$ becomes larger. The value $10^{-12}$ correctly reflects this temperature dependence—water is more ionized at elevated temperature.