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

Silver chromate is more soluble and the ratio of their molarities `= 91.9`

Explanation

Let s be the solubility of `Ag_(2)CrCO_(4)`
Then, `Ag_(2)CrCO_(4)leftrightarrowAg^(2+)+2CrO_(4)^(-)`
`K_(sp)=(2s)^(2).s=4s^(3)`
`1.1xx10^(-12)=4s^(3)`
`s=6.5xx10^(-5)` M
Let s´ be the solubility of AgBr.
`AgBr_((s))leftrightarrowAg^(+)+Br^(-)`
`K_(sp)=s^('2)=5.0xx10^(-13)`
`therefores^(')=7.07xx10^(-7)` M
Therefore, the ratio of the molarities of their saturated solution is
`(s)/(s^('))=(6.5xx10^(-5)"M")/(7.07xx10^(-7)"M")=91.9.`