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 ion product of water is: $$K_w = [H_3O^+][OH^-]$$ **Step 2: Use the property of pure water** In pure water, the autoionization produces equal moles of $H_3O^+$ and $OH^-$: $$[H_3O^+] = [OH^-]$$ **Step 3: Substitute given values** Given: $[H_3O^+] = 10^{-6}$ mol/L at 90°C Therefore: $[OH^-] = 10^{-6}$ mol/L **Step 4: Calculate $K_w$** $$K_w = (10^{-6})(10^{-6}) = 10^{-12} \text{ mol}^2/\text{L}^2$$ **Key insight:** $K_w$ increases with temperature because water's autoionization is endothermic. At 90°C, $K_w = 10^{-12}$ (compared to $10^{-14}$ at 25°C), meaning water is more ionized at higher temperatures. **Answer: C** (which should be $10^{-12}$ or equivalent form like $1 \times 10^{-12}$)