A solution contains and . The concentration of in solution is adjusted to 0.05 M. Determine the pH r — Ionic Equilibrium Chemistry Question
Question
A solution contains $0.1 \text{ M} - \text{Mg}^{2+}$ and $0.1 \text{ M} - \text{Sr}^{2+}$. The concentration of $\text{H}_2\text{CO}_3$ in solution is adjusted to 0.05 M. Determine the pH range which would permit the precipitation of $\text{SrCO}_3$ without any precipitation of $\text{MgCO}_3$. $\text{H}^+$ ion concentration is controlled by external factors. Given: $K_{sp}(\text{MgCO}_3) = 4 \times 10^{-8} \text{ M}^2$; $K_{sp}(\text{SrCO}_3) = 9 \times 10^{-10} \text{ M}^2$; $K_{a, \text{overall}}(\text{H}_2\text{CO}_3) = 5 \times 10^{-17}$; $\log 2 = 0.3$; $\log 3 = 0.48$.
💡 Solution & Explanation
$\text{SrCO}_3$ requires $[\text{CO}_3^{2-}] > \frac{9 \times 10^{-10}}{0.1} = 9 \times 10^{-9} \text{ M}$. $\text{MgCO}_3$ requires $[\text{CO}_3^{2-}] > \frac{4 \times 10^{-8}}{0.1} = 4 \times 10^{-7} \text{ M}$. We need $[\text{CO}_3^{2-}]$ to be strictly between $9 \times 10^{-9} \text{ M}$ and $4 \times 10^{-7} \text{ M}$. From the acid dissociation: $[\text{H}^+]^2 = \frac{K_{a, \text{overall}}[\text{H}_2\text{CO}_3]}{[\text{CO}_3^{2-}]} = \frac{2.5 \times 10^{-18}}{[\text{CO}_3^{2-}]}$. For max $[\text{CO}_3^{2-}] = 4 \times 10^{-7} \text{ M}$: $[\text{H}^+]^2 = 6.25 \times 10^{-12} \implies [\text{H}^+] = 2.5 \times 10^{-6} \text{ M} \implies \text{pH} = 5.60$. For min $[\text{CO}_3^{2-}] = 9 \times 10^{-9} \text{ M}$: $[\text{H}^+]^2 = \frac{25}{9} \times 10^{-10} \implies [\text{H}^+] = \frac{5}{3} \times 10^{-5} \text{ M} \approx 1.67 \times 10^{-5} \text{ M} \implies \text{pH} = 4.78$. Thus, the pH range is 4.78 to 5.60. Therefore, correct answer is A.