The solubility product constant of `Ag_(2)CrO_(4)` and `AgBr` are `1.1xx10^(-12)` and `5.0xx10^(-13)` respectively. Calculate the ratio of the molarities of their saturated solutions.

Explanation

For `Ag_2CrO_4 " " K_(sp)=1.1xx10^(-12)`
If s is the solubility
`Ag_2CrO_4 hArr 2Ag^(+) + CrO_4^(2-)`
`[Ag^+]=2s, CrO_4^(2-) =s`
`K_(sp)=(2s)^2 (s)=4s^3`
`4s^3 =K_(sp) " " therefore s=(K_"sp"/4)^(1//3)`
`s=((1.1xx10^(-12))/4)^(1/3)=6.50xx10^(-5) "mol L"^(-1)`
For AgBr, `K_(sp)=5.0xx10^(-13)`
If s is the solubility ,
`[Ag^+]=s, [Br^-]=s`
`K_(sp)=(s)xx(s)=s^2`
`s^2=K_(sp)`
`s=(5.0xx10^(-13))^(1//2) =7.07xx10^(-7) " mol L"^(-1)`
Ratio of the molarities of the saturated solutions
`=(6.50xx10^(-5))/(7.07xx10^(-7))=91.9` times