Question

A solution is $0.1M$ in $C{l}^{-}$and $0.001M$ in $Cr{O}_4{}^{2-}$.
Solid $AgN{O}_3$ 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 $K_{sp}\left(A{g}_{2}Cr{O}_4\right)=1.9\times 10^{-12}{M}^3$
Select correct statement from the following:

• A. $AgCl$ precipitates first because its $K_{sp}$ is high.
• B. $A{g}_{2}Cr{O}_4$ precipitates first as its $K_{sp}$ is low.
• C. $A{g}_{2}Cr{O}_4$ precipitates first because the amount of $A{g}^{+}$needed is low.
• D. $AgCl$ will precipitate first as the amount of $A{g}^{+}$ needed to precipitate is low.

Answer 1

Answer: (D)

(i) $\left[A{g}^{+}\right]$required to ppt $AgCl(s)$
$Ksp=IP=\left[A{g}^{+}\right]\left[C{l}^{-}\right]=1.7\times 10^{-10}$
$\left[A{g}^{+}\right]=1.7\times 10^{-9}$
(ii) $\left[A{g}^{+}\right]$required to $pptA{g}_{2}Cr{O}_4(s)$
$Ksp=IP=[A{g}^{+}{]}^2\left[Cr{O}_4^{-2}\right]=1.9\times 10^{-12}$
$\left[A{g}^{+}\right]=4.3\times 10^{-5}$
$\left[A{g}^{+}\right]$required to ppt $AgCl$ is low so $AgCl$ will ppt $1^{st}$