A solution is 0.1 M in $\mathrm{Cl}^{-}$ and $0.001 \mathrm{M}$ in $\mathrm{CrO}_{4}^{2-}$. Solid $\mathrm{AgNO}_{3}$ is gradually added to it. Assuming that the addition does not change in volume and $\mathrm{K}_{\mathrm{sp}}(\mathrm{AgCl})=1.7 \times 10^{-10} \mathrm{M}^{2}$ and $\mathrm{K}_{\mathrm{sp}}\left(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\right)=1.9 \times 10^{-12} \mathrm{M}^{3}$. Select correct statement from the following:

(C) AgCl will precipitate first as the amount of $\mathrm{Ag}^{+}$ needed to precipitate is low.

(A) $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$ precipitates first because the amount of $\mathrm{Ag}^{+}$ needed is low.

(B) $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$ precipitates first as its $\mathrm{K}_{\mathrm{sp}}$ is low.

(D) AgCl precipitates first because its $\mathrm{K}_{\mathrm{sp}}$ is high.

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Equilibrium

A solution is 0.1 M in

Cl^{-}

and

0.001 M

CrO_{4}^{2 -}

. Solid

AgNO_{3}

is gradually added to it. Assuming that the addition does not change in volume and

K_{sp} (AgCl) = 1.7 \times 1 0^{- 10} M^{2}

and

K_{sp} (Ag_{2} CrO_{4}) = 1.9 \times 1 0^{- 12} M^{3}

.
Select correct statement from the following:

(A) $Ag_{2} CrO_{4}$ precipitates first because the amount of $Ag^{+}$ needed is low.
(B) $Ag_{2} CrO_{4}$ precipitates first as its $K_{sp}$ is low.
(C)AgCl will precipitate first as the amount of $Ag^{+}$ needed to precipitate is low.✓ Correct
(D)AgCl precipitates first because its $K_{sp}$ is high.

Answer: (C)

Solution

A salt begins to separate when its ionic product reaches its solubility product, so the test is the silver ion concentration each salt needs. For silver chloride,

[A g^{+}] = \frac{1.7 \times 1 0 ^{- 10}}{0.1} = 1.7 \times 1 0^{- 9} M

For silver chromate,

[A g^{+}] = \frac{1.9 \times 1 0 ^{- 12}}{1.0 \times 1 0 ^{- 3}} = 4.4 \times 1 0^{- 5} M

The chloride needs about twenty-five thousand times less silver ion, so as silver nitrate is added slowly that threshold is reached long before the other. Silver chloride precipitates first because the amount of silver ion needed to precipitate it is low.

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Equilibrium

Solution

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