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

For ` Ag_2CrO_4` (a ternary salt) ,
` s= ((K_(sp))/(4))^(1//3) = ((1.1 xx 10 ^(-12))/(4))^(1//3) = 6.5 xx 10 ^(-5) M`
For AgBr (a binary salt )
` s. = sqrt(K_(sp) ) = sqrt(5.0 xx10 ^(-13) )= 7.1 xx 10 ^(-7)`
` therefore ` Ratio of molarities of their saturated solutions
` " " = ( s)/(s.) = (6.5 xx 10 ^(-5))/(7.1 xx 10 ^(-7)) = 91.5`