The solubility product constant of Ag_2CrO_4 and AgBr are 1.1xx10^(-12) and 5.0xx10^(-13) respectively. Calculate the ratio of the molarities of their saturated solutions.

The solubility product constant of Ag_2CrO_4 and AgBr are 1.1xx10^(-12

1.	The solubility product constant of Ag_2CrO_4 and AgBr are 1.1xx10^(-12) and 5.0xx10^(-13) respectively. Calculate the ratio of the molarities of their saturated solutions.
Answer» Solution :`{:(Ag_2CrO_(4(s))HARR, 2Ag_((aq))^(+)+, CrO_(4(aq))^(2-)),(S_1M,2S_1M +, S_1M):}` `therefore K_(SP)=[Ag^+]^2 [CrO_4^(2-)]=(2S_1)^2 (S_1)=(4S_1)^3` `therefore S_1=(K_(sp)/4)^(1/3) =root3(1.1/4 xx10^(-12))=0.6503xx10^(-4)` M `{:(AgBr_((s))hArr, Ag_((aq))^(+)+ , Br_((aq))^(-)),(S_2M,S_2M,S_2M):}` `K_(sp)=[Ag^+][Br^-]=(S_2)(S_2)=(S_2)^2` `therefore S_2=SQRT(K_(sp))=sqrt(5.0xx10^(-13))=sqrt(50xx10^(-14))` `=7.0711xx10^(-7)` M RATIO of solubilities = `S_1/S_2=(0.6503xx10^(-4))/(7.0711xx10^(-7))` `=0.091966xx10^(+3)M` =91.966