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# A solution is 0.1 M in

Question:

A solution is $0.1 \mathrm{M}$ 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 :

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

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

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

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

Correct Option: , 4

Solution:

(i) $\left[\mathrm{Ag}^{+}\right]$required to ppt $\mathrm{AgCl}(\mathrm{s})$

$\mathrm{Ksp}=\mathrm{IP}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=1.7 \times 10^{-10}$

$\left[\mathrm{Ag}^{+}\right]=1.7 \times 10^{-9}$

(ii) $\left[\mathrm{Ag}^{+}\right]$required to $\mathrm{ppt} \mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{~s})$

$\mathrm{Ksp}=\mathrm{IP}=[\mathrm{Ag}+]^{2}\left[\mathrm{CrO}_{4}^{-2}\right]=1.9 \times 10^{-12}$

$\left[\mathrm{Ag}^{+}\right]=4.3 \times 10^{-5}$

$\left[\mathrm{Ag}^{+}\right]$required to $\mathrm{ppt} \mathrm{AgCl}$ is low so $\mathrm{AgCl}$ will $\mathrm{ppt}$ $1^{\mathrm{st}}$.