Question
A solution contains a mixture of Ag+ (0.10 M) and Hg22+ (0.10 M), which are to be separated by selective precipitation. Calculate the maximum concentration of iodide ion at which one of them gets precipitated almost completely. What per cent of that metal ion is precipitated before the start of precipitation of second metal ion? Ksp(AgI) = 8.5 × 10−17 and Ksp(Hg2I2) = 2.5 × 10−26.
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Solution
[I– ]min for ppt of AgI $= \frac{K_{sp}}{\left\lbrack Ag^{+} \right\rbrack} = \frac{8.5 \times 10^{- 17}}{0.1} = 8.5 \times 10^{- 16}\text{ M}$
[I– ]min for ppt of Hg2I2 = $\sqrt{\frac{K_{sp}}{\left\lbrack Hg_{2}^{2 +} \right\rbrack}} = \sqrt{\frac{2.5 \times 10^{- 26}}{0.1}} = 5 \times 10^{- 13}\text{ M}$
Hence, AgI will precipitate first. The [I–] to start precipitation of other metal ion = 5 × 10–13 M
Now, [Ag+]left when Hg2I2 start precipitating $= \frac{K_{sp}}{\left\lbrack I^{-} \right\rbrack} = \frac{8.5 \times 10^{- 17}}{5 \times 10^{- 13}} = 1.7 \times 10^{- 4}\text{ M}$
Hence, % of Ag+ precipitate $= \frac{0.1 - 1.7 \times 10^{- 4}}{0.1} \times 100 = 99.83\%$
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