A solution contains 0.1 M of ${Mg}^{+ 2}$ and 0.1 M of ${Sr}^{+ 2}$ . The concentration of $H_{2} C O_{3}$ is adjusted to 0.05 M. Determine the pH range which would permit the precipitation of $S r C O_{3}$ without any precipitation of $M g C O_{3}$ . The $H^{+}$ ion concentration is controlled by external factors. Given that $K_{s p}$ of $S r C O_{3}$ and $M g C O_{3}$ are $9 \times 10^{- 10}$ and $4 \times 10^{- 8}$ respectively. $K_{a}$ (overall) of $H_{2} C O_{3} = 5 \times 10^{- 17}$

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4.78 to 5.6

4.6 to 5.78

5.78 to 6.4

5.22 to 5.4

answer is A.

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Detailed Solution

(a) $[C O_{3}^{2 -}] = K_{a} \frac{[H_{2} C O_{3}]}{{[H^{+}]}^{2}}$
To prevent metal carbonate’s precipitation
$[M^{+ 2}] [C O_{3}^{2 -}] \leq K_{S P}$
$[M^{+ 2}] K_{a} [H_{2} C O_{3}] / {[H^{+}]}^{2} \leq K_{S P}$
$[H^{+}] \geq \sqrt{\frac{[M^{+ 2}] . K_{a} [H_{2} C O_{3}]}{K_{S P}}}$
For $M g C O_{3}$ ,
$[H^{+}] \geq \sqrt{\frac{0.1 \times 5 \times 10^{- 17} \times 0.005}{9 \times 10^{- 8}}}$
$= 2.5 \times 10^{- 6} M$
Hence, $p H \leq 5.6$
Similarly, for SrCO3, we get
$[H^{+}] \geq 1.67 \times 10^{- 5} M : p H \leq 4.78$ .