An aqueous solution of a metal bromide is saturated with . What is the minimum pH at which MS will p — Ionic Equilibrium Chemistry Question
Question
An aqueous solution of a metal bromide $\text{MBr}_2 \ (0.04\text{ M})$ is saturated with $\text{H}_2\text{S}$. What is the minimum pH at which MS will precipitate? $K_{sp} \text{ for MS} = 6.0 \times 10^{-21}$; concentration of saturated $\text{H}_2\text{S} = 0.1\text{ M}$, $K_1 = 10^{-7}$ and $K_2 = 1.5 \times 10^{-13} \text{ for H}_2\text{S}$.
💡 Solution & Explanation
$[\text{M}^{2+}] = 0.04\text{ M}$. For MS to precipitate, $[\text{S}^{2-}] = \frac{K_{sp}}{[\text{M}^{2+}]} = \frac{6.0 \times 10^{-21}}{0.04} = 1.5 \times 10^{-19}\text{ M}$. For $\text{H}_2\text{S}$, the overall dissociation constant is $K_a = K_1 K_2 = 1.5 \times 10^{-20}$. Using $K_a = \frac{[\text{H}^+]^2[\text{S}^{2-}]}{[\text{H}_2\text{S}]} \implies 1.5 \times 10^{-20} = \frac{[\text{H}^+]^2 (1.5 \times 10^{-19})}{0.1}$. Solving for $[\text{H}^+]^2 = 10^{-2} \implies [\text{H}^+] = 10^{-1}\text{ M}$. The minimum pH required is 1.0.