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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 68
Total: 73
The solubility product constant of Ag2CrO4 and AgBr are 1.1 × 10–12 and 5.0 × 10–13 respectively. Calculate the ratio of the molarities of their saturated solutions.
Solution:  
Ksp(Ag2CrO4) > Ksp(AgBr)
∴ Ag2CrO4 is more soluble.
Ag2CrO4(s) ⇌ 2Ag(aq)++CrO2–4(aq);
Ksp = 4S3
∴ S = (
1.1×10−12
4
)
1
3
 = 6.50 × 10−5M
AgBr(s) ⇌ Ag(aq)++Br(aq)− ; Ksp = S′2
∴ S' = (5×10−13)
1
2
 = 7.07 × 10−7M
The ratio of the molarities =
S
S
 =
6.50×10−5
7.07×10−7
 = 91.94
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