If the pH of a saturated solution of is 12 at 25°C, the value of its solubility product is: — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

pH = 12 means pOH = 2, so $[OH^-] = 10^{-2} \text{ M}$. For $Ba(OH)_2 \rightleftharpoons Ba^{2+} + 2OH^-$, $[Ba^{2+}] = [OH^-]/2 = 0.5 \times 10^{-2} \text{ M}$. $K_{sp} = [Ba^{2+}][OH^-]^2 = (0.5 \times 10^{-2})(10^{-2})^2 = 5 \times 10^{-7}$.