The exact pH of a saturated aqueous solution of is found to be at a certain temperature. What is the — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

In a saturated solution with $pH = 12.0$, the $pOH = 14 - 12 = 2.0$. Thus, the equilibrium $[OH^-] = 10^{-2} \text{ M}$. The dissolution reaction is $Ba(OH)_2 \rightleftharpoons Ba^{2+} + 2OH^-$. From stoichiometry, the $[Ba^{2+}]$ must be exactly half of $[OH^-]$, so $[Ba^{2+}] = 0.5 \times 10^{-2} \text{ M}$. The solubility product is $K_{sp} = [Ba^{2+}][OH^-]^2 = (0.5 \times 10^{-2})(10^{-2})^2 = 0.5 \times 10^{-6} = 5 \times 10^{-7}$. Multiplying by $10^7$ yields exactly 5.