A solution contains $0.05 m o l . {l i t r e}^{- 1}$ of ${B a}^{2 +}$ ions and $0.002 m o l . l i t r e$ of ${A g}^{+}$ ions. The metals are to be precipitated by addition of chromate ions, $C r O_{4}^{2 -}$ . Which ion precipitates first? What percentage of this ion remain in the solution when the second ion begins to precipitate?
$K_{s p} ({A g}_{2} C r O_{4}) = 3 \times 10^{12}; K_{s p} (B a C r O_{4}) = 1 \times 10^{- 10}$

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Solution

$C r O_{4}^{2 -}$ ion concentration required for precipitation:
For $B a C r O_{4}$ :
$[C r O_{4}^{2 -}] = \frac{K_{s p} B a C r O_{4}}{[{B a}^{2 +}]} = \frac{10^{- 10}}{0.05} = 20 \times 10^{- 10} M$
For ${A g}_{2} C r O_{4}$
$[C r O_{4}^{2 -}] = \frac{K_{s p} B a C r O_{4}}{[{B a}^{2 +}]} = \frac{10^{- 10}}{0.05} = 20 \times 10^{- 10} M$
${B a}^{2 +}$ will be precipitated first because it requires low concentration of $C r O_{4}^{2 -}$ . Remaining conc. of ${B a}^{2 +}$ when ${A g}^{+}$ will start precipitation.
${[{B a}^{2 +}]}_{R e m a i n i n g} = \frac{K_{s p} B a C r O_{4}}{[C r O_{4}^{2 -}]} = 1.33 \times 10^{- 4} M$
% of remaining ${B a}^{2 +} = \frac{1.33 \times 10^{- 4}}{0.05} \times 100 = 0.26$ %

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