The solubility product constant of `Ag_(2)CrO_(4)` and `AgBr` are `1.1 xx 10^(-12)` and `5.0 xx 10^(-13)` respectively. Calculate the ratio of the molarities of their saturated solutions.

Explanation

`Ag_(2)CrO_(4)` dissolves in water and the equilibrium in the saturated solution is
`Ag_(2)CrO_(4)(s)= 2Ag^(o+)(aq)+CrO_(4)^(2-)(aq)`
Let the solubolty of `AgBr` be `S_(1)`
`[Ag^(o+)(aq)]=2S_(1)`
and `[CrO_(4)^(2-)(aq)=S_(1)]`
`K_(sp)=[Ag^(o+)(aq)]^(2)[CrO_(4)^(2-)(aq)]=(2S_(1))^(2)xxS=4S_(1)^(3)`
or `S_(1)=(K_(sp)/4)^(1//3)=((1.1xx10^(-12))/4)^(1//3)=0.65xx10^(-14) M`
Similarly `AgBr=Ag^(o+)(aq)+Br^(Θ)(aq)`
Let the solubility of `AgBr be S_(2)`
`[Ag^(o+)(aq)]=S_(2)` and `Br^(Θ)(aq)]=S_(2)`
`K_(sp)=[Ag^(o+)(aq)][Br^(Θ)(aq)]`
`S_(2)xxS_(2)=S_(2)^(2)`
or `S_(2)=(K_(sp))^(1//2)=(5.0xx10^(-13))^(1//2)`
`=(0.5xx10^(-12))^(1//2)=0.707xx10^(-6) mol L^(-1)`
Ratio of the molarities of silver chromate to silver bromide comes to `=(0.65xx10^(-4))/(0.707xx10^(-6))=91.9` silver chromate is more soluble.