7.68 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.

Let s be the solubility of Ag2CrO4.Then, AggCrO4Ag2++2CrO4-Ksp=(2s)2.s=4s31.1×10-10=4ss=6.5×10-5 MLet s' be the solubility of AgBr.AgBr(s)Ag++Br-Ksp=S'2=5.0×10-13s'=7.07×10-7 MTherefore, the ratio of the molarities of their saturated solution is2s'=6.5×10-5 M 7.07×10-7 M=91.9.