After solid was equilibrated with a buffer at pH 8.6, the solution was found to have , what is of ? — Ionic Equilibrium Chemistry Question
Question
After solid $\text{SrCO}_3$ was equilibrated with a buffer at pH 8.6, the solution was found to have $[\text{Sr}^{2+}] = 2.0 \times 10^{-4}\text{ M}$, what is $K_{sp}$ of $\text{SrCO}_3$? ($K_{a2}$ for $\text{H}_2\text{CO}_3 = 5.0 \times 10^{-11}$, $\log 2 = 0.3, 5.1 \times 0.196 = 1.0$)
💡 Solution & Explanation
At $\text{pH} = 8.6$, $[\text{H}^+] = 10^{-8.6} \approx 2.5 \times 10^{-9}\text{ M}$ (since $\log 2.5 \approx 0.4$). The total dissolved carbonate $C_{\text{carb}} = [\text{Sr}^{2+}] = 2.0 \times 10^{-4}\text{ M}$. At this pH, carbonate exists mostly as $\text{CO}_3^{2-}$ and $\text{HCO}_3^-$. The ratio $\frac{[\text{HCO}_3^-]}{[\text{CO}_3^{2-}]} = \frac{[\text{H}^+]}{K_{a2}} = \frac{2.5 \times 10^{-9}}{5.0 \times 10^{-11}} = 50$. This means $[\text{HCO}_3^-] = 50[\text{CO}_3^{2-}]$, making total $C_{\text{carb}} = 51[\text{CO}_3^{2-}]$. Given the hint $5.1 \times 0.196 = 1.0$, we know $1/51 \approx 0.0196$. Thus, $[\text{CO}_3^{2-}] = C_{\text{carb}} / 51 = 2.0 \times 10^{-4} \times 0.0196 = 3.92 \times 10^{-6}\text{ M}$. $K_{sp} = [\text{Sr}^{2+}][\text{CO}_3^{2-}] = (2.0 \times 10^{-4})(3.92 \times 10^{-6}) \approx 7.84 \times 10^{-10} \approx 8.0 \times 10^{-10}$. Therefore, correct answer is C.