Question

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.

Answer 1

(I).Calculation of molar solubility (molarity) of `Ag_(2)CrO_(4)` solution
`Ag_(2)CrO_(4) (s) overset((aq))(hArr) 2Ag^(+) (aq) CrO_(4)^(2-)(aq)`
Let the solubility of the salt in water be S
`[Ag^(+)(aq)] =2S " and "[CrO_(4)^(2-) (aq)]=S`
`K_(sp) =[Ag^(+)(aq)]^(2)[CrO_(4)^(2-)(aq)](2S)^(2)xxS =4S^(3)`
`S=(k_(sp)/(4))^(1//3)=((1.1 xx 10^(-12))/(4))^(1//3) = 0.65 xx10^(-4) " mol " L^(-1)`
(II). Calculation of molar solubility (molarity) of AgBr solution.
`AgBr(s) overset((aq))(hArr) Ag^(+) (aq) +Br^(-) (aq)`
`[Ag^(+) (aq)] =S " and " [Br^(-) (aq)] =S`
`K_(sp) =[Ag^(+) (aq)][Br^(-)(aq)]=Sxx S=S^(2)`
`S=(K_(sp))^(1//2) =(5.0 xx 10^(-13))^(1//2) =(0.5 xx 10^(-12))^(1//2)`
`=0.707 xx 10^(-6) " mol " L^(-1)`
(III). Calculation of ratio of the molarities of the saturated solutions.
`(M_((Ag2CrO_(4))))/(M_((AgBr)))=((0.65xx10^(-4) " mol "L^(-1)))/((0.707 xx 10^(-6) " mol " L^(-1))) =91.9.`