Equilibrium - Result Question 70
####74. 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 $Na_2 CrO_4$ ? [2015]
(a) $AgCl$
(b) $AgBr$
(c) $Ag_2 CrO_4$
(d) $AgI$
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Solution:
- (c) $Ag_2 CrO_4$
$K _{s p}=[Ag^{+}]^{2}[CrO_4^{-2}]=1.1 \times 10^{-12}$
$[Ag^{+}]=\sqrt{\frac{1.1 \times 10^{-12}}{[CrO_4^{-2}]}}$
$AgCl$
$K _{s p}=[Ag^{+}][Cl^{-}]=1.8 \times 10^{-10}$
$[Ag^{+}]=\frac{1.8 \times 10^{-10}}{[Cl^{-}]}$
$AgBr$
$K _{s p}=[Ag^{+}][Br^{-}]=5.0 \times 10^{-13}$
$[Ag^{+}]=\frac{5.3 \times 10^{-13}}{[Br^{-}]}$
$AgI$
$K _{s p}=[Ag^{+}][I^{-}]=8.3 \times 10^{-17}$
$[Ag^{+}]=\frac{8.3 \times 10^{-17}}{[I^{-}]}$
If we take $[CrO_4^{-2}]=[Cl^{-}]=[Br^{-}]=[I^{-}]=1$
then maximum $[Ag^{+}]$will be required in case of $Ag_2 CrO_4$.
