Question
A solution contains 0.1 M – Mg2+ and 0.1 M – Sr2+. The concentration of H2CO3 in solution is adjusted to 0.05 M. Determine the pH range which would permit the precipitation of SrCO3 without any precipitation of MgCO3. The H+ ion concentration is controlled by external factors. Given: Ksp(MgCO3) = 4 × 10−8 M2; Ksp(SrCO3) = 9 × 10−10 M2; Ka,overall(H2CO3) = 5 × 10−17; log 2 = 0.3; log 3 = 0.48.
Options
Solution
$\left\lbrack CO_{3}^{2 -} \right\rbrack = K_{a}\left( \text{overall} \right).\frac{\left\lbrack H_{2}CO_{3} \right\rbrack}{\left\lbrack H^{+} \right\rbrack^{2}}$
To prevent precipitation of MCO3,
Or, $\left\lbrack M^{2 +} \right\rbrack\left\lbrack CO_{3}^{2 -} \right\rbrack \leq K_{sp}$
Or, $\left\lbrack M^{2 +} \right\rbrack.\frac{K_{a}\left\lbrack H_{2}CO_{3} \right\rbrack}{\left\lbrack H^{+} \right\rbrack} \leq K_{sp}$
$\therefore\left\lbrack H^{+} \right\rbrack \geq \sqrt{\frac{\left\lbrack M^{2 +} \right\rbrack.K_{a}\left\lbrack H_{2}CO_{3} \right\rbrack}{K_{sp}}}$
For MgCO3: $\left\lbrack H^{+} \right\rbrack \geq \sqrt{\frac{0.1 \times 5 \times 10^{- 17} \times 0.05}{9 \times 10^{- 8}}} = 2.5 \times 10^{- 6}\text{ M}$
For SrCO3: $\left\lbrack H^{+} \right\rbrack \geq \sqrt{\frac{0.1 \times 5 \times 10^{- 17} \times 0.05}{9 \times 10^{- 10}}} = \frac{5 \times 10^{- 5}}{3}\text{ M}$
∴ PH ≤ 4.78
For precipitation of SrCO3 without any precipitation of MgCO3, the PH range should be 4.78 to 5.6
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