Question

The $K_{sp}$ of $A{g}_{2}Cr{O}_4$, 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 $AgN{O}_3$ solution is added to the solution containing equal moles of NaCl, NaBr, NaI and $N{a}_{2}Cr{O}_4$?

• A. AgCl
• B. AgBr
• C. $A{g}_{2}Cr{O}_4$
• D. AgI

Answer 1

The correct option is C $A{g}_{2}Cr{O}_4$

$A{g}_2{C}_r\to 2A_g^{+}+C_r{O}_4^2$

1 0 0

1-s 2s s

$K_{sp}=\frac{(2s{)}^{2}s}{1-s}$, s< < 1

$=1.1\times {10}^{-12}=4s^3$

$=s=6.5\times {10}^{-5}$

$A_g{C}_l\to A{g}^{+}+C{l}^{-}$

s s

$1.8\times {10}^{-10}=s^2$

$=s=1.34\times {10}^{-5}$

silmilarly s for $A_g{B}_r$ and $A_{g}Cl$ is $7.1\times {10}^{-7}$ and $9\times {10}^{-9}$ respectively since solubility of $A{g}_2{C}_r{O}_4$ is lighest, $s_a$ it will perticipate lost.

Option C) $A{g}_2{C}_r{O}_4$ is correct option