A solution is 0.1 M in $C l^{-}$ and 0.001 M in $C r O_{4}^{2 -}$ . Solid $A g N O_{3}$ is gradually added to it.
Assuming that the addition does not change in volume and
$K_{s p} (A g C l) = 1.7 \times 10^{- 10} M^{2}$ and
$K_{s p} (A g_{2} C r O_{4}) = 1.9 \times 10^{- 12} M^{3}$
Select correct statement from the following

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$A g C l$ will precipitate first as the amount of $A g^{+}$ needed to precipitate is low

$A g_{2} C r O_{4}$ precipitates first as its $K_{s p}$ is low.

$A g_{2} C r O_{4}$ precipitates first because the amount of $A g^{+}$ needed is low.

$A g C l$ precipitates first because its $K_{s p}$ is high,

answer is D.

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Detailed Solution

Solution: Given data, $K_{s p} = 1.7 \times 10^{- 10} M^{2}$
$K_{s p} A g_{2} C r O_{4} = 1.9 \times 10^{- 12} M^{3}$
$[C l] = 0.1 M, [C r O_{4}^{2 -}] = 0.001 M$
i) $[A g^{+}]$ required to precipitate $A g C l (s)$
$∴ K_{s p} = I . P = [A g^{+}] [C l^{-}] = 1.7 \times 10^{- 10}$
$[A g^{+}] = \frac{1.7 \times 10^{- 10}}{0.1} = 1.7 \times 10^{- 9}$
ii) $[A g^{+}]$ required to precipitate $A g_{2} C r O_{4} (s)$
$K_{s p} = I . P . [A g^{+}] [C r O_{4}^{- 2}] = 1.9 \times 10^{- 12}$
$[A g^{+}] = 4.3 \times 10^{- 5}$
$[A g^{+}]$ required to precipitate $A g C l$ is low. So $A g C l$ will precipitate $1^{st}$