Question
To 0.35 L of 0.1 M – NH3 0.15 L of 0.1 M-MgCl2 is added. What minimum mass of (NH4)2SO4 should be added to cause the Mg(OH)2 to re-dissolve? The value of Ksp for Mg(OH)2 = 1.2 × 10–11, Kb for NH3 = 2.0 × 10–5.
Options
Solution
$Mg(OH)_{2}(s) + 2NH_{4}^{+} \rightleftharpoons Mg^{2 +} + 2NH_{4}OH$
To re-dissolve Mg(OH)2, $Q \leq K_{eq}$
Or, $\frac{\left\lbrack Mg^{2 +} \right\rbrack\left\lbrack NH_{4}OH \right\rbrack^{2}}{\left\lbrack NH_{4}^{+} \right\rbrack^{2}} \leq \frac{K_{sp}}{K_{b}^{2}}$
Or, $\frac{\left( \frac{0.15 \times 0.1}{0.5} \right)\left( \frac{0.35 \times 0.1}{0.5} \right)^{2}}{\left( \frac{n}{0.5} \right)^{2}} \leq \frac{1.2 \times 10^{- 11}}{\left( 2.0 \times 10^{- 5} \right)^{2}}$
∴ n ≥ 0.035
Hence, minimum mass of (NH4)2SO4 needed
$= \frac{0.035}{2} \times 132 = 2.31\text{ gm}$
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