To 0.35 l of is added 0.15 L of . What mass of should be added to cause the to re-dissolve? for , fo — Ionic Equilibrium Chemistry Question
Question
To 0.35 l of $0.1 \text{ M} - \text{NH}_3$ is added 0.15 L of $0.1 \text{ M} - \text{MgCl}_2$. What mass of $\text{(NH}_4\text{)}_2\text{SO}_4$ should be added to cause the $\text{Mg(OH)}_2$ to re-dissolve? $K_{sp}$ for $\text{Mg(OH)}_2 = 1.2 \times 10^{-11}$, $K_b$ for $\text{NH}_3 = 2.0 \times 10^{-5}$.
💡 Solution & Explanation
Total volume = 0.50 L. $[\text{Mg}^{2+}] = \frac{0.15 \times 0.1}{0.50} = 0.03 \text{ M}$. $[\text{NH}_3] = \frac{0.35 \times 0.1}{0.50} = 0.07 \text{ M}$. To prevent precipitation, $[\text{OH}^-] \le \sqrt{\frac{1.2 \times 10^{-11}}{0.03}} = 2.0 \times 10^{-5} \text{ M}$. From the buffer equation, $[\text{OH}^-] = K_b \frac{[\text{NH}_3]}{[\text{NH}_4^+]} \implies 2.0 \times 10^{-5} = 2.0 \times 10^{-5} \frac{0.07}{[\text{NH}_4^+]} \implies [\text{NH}_4^+] = 0.07 \text{ M}$. Total moles of $\text{NH}_4^+$ required in 0.50 L is $0.07 \times 0.50 = 0.035 \text{ moles}$. Since each mole of $\text{(NH}_4\text{)}_2\text{SO}_4$ provides two moles of $\text{NH}_4^+$, we need $\frac{0.035}{2} = 0.0175 \text{ moles}$. Molar mass of $\text{(NH}_4\text{)}_2\text{SO}_4 = 132 \text{ g/mol}$. Mass = $0.0175 \times 132 = 2.31 \text{ g}$. Therefore, correct answer is B.