Question

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

• A. $AgCl$ precipitates first because its $K_{s p}$ is high.
• B. $Ag _{2} CrO _{4}$ precipitates first as its $K_{s p}$ is low.
• C. $Ag _{2} CrO _{4}$ precipitates first because the amount of $Ag ^{+}$needed is low.
• D. $AgCl$ will precipitate first as the amount of $Ag ^{+}$ needed to precipitate is low.

Answer 1

Answer: (D)

(i) $\left[ Ag ^{+}\right]$required to ppt $AgCl (s)$
$K s p=I P=\left[ Ag ^{+}\right]\left[ Cl ^{-}\right]=1.7 \times 10^{-10}$
$\left[A g^{+}\right]=1.7 \times 10^{-9}$
(ii) $\left[ Ag ^{+}\right]$required to $ppt Ag _{2} CrO _{4}(s)$
$K s p=I P=[ Ag^ +]^{2}\left[ CrO _{4}^{-2}\right]=1.9 \times 10^{-12}$
$\left[A g^{+}\right]=4.3 \times 10^{-5}$
$\left[ Ag ^{+}\right]$required to ppt $AgCl$ is low so $AgCl$ will ppt $1^{s t}$