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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 68 of 73
Marks: +1, -0
The solubility product constant of Ag2CrO4\mathrm{Ag_2CrO_4} and AgBr are 1.1 × 10−1210^{-12} and 5.0 × 10−1310^{-13} respectively. Calculate the ratio of the molarities of their saturated solutions.
Solution:  
Ksp(Ag2CrO4)K_{sp}(\mathrm{Ag_2CrO_4}) > Ksp(AgBr)K_{sp}(\mathrm{AgBr})
∴ Ag2CrO4\mathrm{Ag_2CrO_4} is more soluble.
Ag2CrO4(s)\mathrm{Ag_2CrO_{4(s)}} ⇌ 2Ag(aq)++CrO2(aq)−42\mathrm{Ag}^{+}_{(aq)} + \mathrm{CrO_2}^{-4}_{(aq)};
KspK_{sp} = 4S34S^3
∴ S = (1.1×10−124)1/3\left(\frac{1.1\times10^{-12}}{4}\right)^{1/3} = 6.50 × 10−5 M10^{-5}\,\text{M}
AgBr(s)\mathrm{AgBr}_{(s)} ⇌ Ag(aq)++Br(aq)−\mathrm{Ag}^{+}_{(aq)} + \mathrm{Br}^{-}_{(aq)} ; KspK_{sp} = S′2S'^2
∴ S' = (5×10−13)1/2\left(5\times10^{-13}\right)^{1/2} = 7.07 × 10−7 M10^{-7}\,\text{M}
The ratio of the molarities = SS\frac{S}{S} = 6.50×10−57.07×10−7\frac{6.50\times10^{-5}}{7.07\times10^{-7}} = 91.94
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