Question

$K_{sp}$ of $AgBr,AgCl,{Ag}_2{CO}_3$ and ${Ag}_{3}As{O}_4$ are $5\times {10}^{-13},1.8\times {10}^{-10},8.1\times {10}^{-12}$ and ${10}^{-22}$ respectively. Silver nitrate is added to the solution each having $0.1M$ concentration of ${Br}^{-},{Cl}^{-},{CO}_3^{2-}$ and $As{O}_4^{3-}$. Select the correct answer.

• A. $AgBr$ will be precipitated before $AgCl$
• B. ${Ag}_2{CO}_3$ will be precipitated after ${Ag}_{3}As{O}_4$
• C. ${Ag}_{3}As{O}_4$ will be precipitated after $AgCl$
• D. ${Ag}_{3}As{O}_4$ will be precipitated after ${Ag}_2{CO}_3$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $AgBr$ will be precipitated before $AgCl$
B ${Ag}_2{CO}_3$ will be precipitated after ${Ag}_{3}As{O}_4$
C ${Ag}_{3}As{O}_4$ will be precipitated after $AgCl$

According to given data,

$\left[A{g}^{+}\right]$ needed for precipitation of

$AgCl=\frac{1.8\times {10}^{-10}}{{10}^{-1}}=1.8\times {10}^{-9}M$

$AgBr=\frac{5\times {10}^{-13}}{{10}^{-1}}=5\times {10}^{-12}M$

$A{g}_{2}C{O}_3=\sqrt{\frac{8.1\times {10}^{-12}}{{10}^{-1}}}=9\times {10}^{-6}M$

$A{g}_{3}As{O}_4=(\frac{{10}^{-22}}{{10}^{-1}}{)}^{1/3}={10}^{-7}M$

Thus, minimum $\left[A{g}^{+}\right]$ is for $AgBr$ and so it will be precipitated first.

The order is $AgBr>AgCl>A{g}_{3}As{O}_4>A{g}_{2}C{O}_3$