Chemistry - Chemical Equilibrium Question with Solution | TestHub

Solubility of$\text{AgBr}$in$\text{0}\text{.01}\text{M}{\text{CaBr}}_{\text{2}}\left({\text{S}}_{\text{2}}\right)$......(i)

${\text{CaBr}}_{\text{2}}$is a strong electrolyte and dissociates completely to give common ion${\text{Br}}^{-}$

$\underset{0.01\text{M}}{{\text{CaBr}}_{\text{2}}}\to \underset{\text{0}\text{.01}\text{M}}{{\text{Ca}}^{\text{2+}}}+\underset{\text{0}\text{.02}\text{M}}{{\text{2Br}}^{-}}$

$\text{AgBr(s)}\rightleftharpoons {\text{Ag}}^{\text{+}}{\text{(aq)+Br}}^{-}\text{(aq);}{\text{K}}_{\text{sp}}$

${\text{S}}_{\text{2}}\text{0}{\text{.02+S}}_{\text{2}}\Rightarrow {\text{S}}_{\text{2}}\times 0.02={\text{K}}_{\text{sp}}$

$~0.02$or${\text{S}}_{\text{2}}=50{\text{K}}_{\text{sp}}$

Similarly,$\text{KBr}$gives common ion${\text{Br}}^{-}$

And${\text{AgNO}}_{\text{3}}$gives common ion${\text{Ag}}^{\text{+}}$and we can calculate

${\text{S}}_{\text{3}}\text{=}{\text{100 K}}_{\text{sp}}$......(iii)

And${\text{S}}_{\text{4}}=20{\text{K}}_{\text{sp}}$......(iv) respectively

From (i), (ii), (iii) and (iv)

${\text{S}}_{\text{1}}\text{>}{\text{S}}_{\text{3}}\text{>}{\text{S}}_{\text{2}}\text{>}{\text{S}}_{\text{4}}$

(Note:$\sqrt{{\text{K}}_{\text{sp}}}\text{>}\text{100}{\text{K}}_{\text{sp}}\because {\text{K}}_{\text{sp}}$order$-{10}^{-12}$)