Question

The $K_{s p}$ of $Ag_{2}CrO_{4}$ , AgCl, AgBr and AgI are respectively, $1.1\times 10^{- 12}$ , $1.8\times 10^{- 10}$ , $5.0\times 10^{- 13}$ , $8.3\times 10^{- 17}$ . Which one of the following salts will precipitate last if $AgNO_{3}$ solution is added to the solution containing equal moles of NaCl, NaBr, NaI and $Na_{2}CrO_{4}$ ?

• A. $AgI$
• B. $AgCl$
• C. $Ag_{2}CrO_{4}$
• D. $AgBr$

Answer 1

Answer: (C)

$\underset{\underset{1 - s}{1}}{A g_{2} C r O_{4}} \rightarrow \underset{\underset{2 s}{0}}{2 A g^{+}}+\underset{\underset{s}{0}}{C r O_{4}^{- 2}}$
$K_{s p}=\frac{\left(\right. 2 s \left(\left.\right)^{2} s}{1 - s}$
$s < < 1$
As 1 >> s so 1 – s = 1
$K_{s p}=4s^{3}$
$=1.1\times 10^{- 12}=4s^{3}$
$=s=6.5\times 10^{- 5}$
$AgCl \rightarrow \underset{s}{A g^{+}}+\underset{s}{C l^{-}}$
$1.8\times 10^{- 10}=s^{2}$
or $s=1.34\times 10^{- 5}$
Similarly 's' for AgBr and AgCl is found to be $7.1\times 10^{- 7}$ and $9\times 10^{- 9}$ respectively since solubility of $Ag_{2}CrO_{4}$ is highest, so it will participate last.