Question

The units of solubility product of silver chromate `(AgCrO_(4))` will be
A. `mol^(2)L^(-2)`
B. `mol^(3)L^(-3)`
C. `mol L^(-1)`
D. `mol^(-1)L`

Answer 1

Correct Answer - B
The solubility product is an equilibrium constant.
`Ag_(2)CrO_(4)(s)hArr2Ag^(+)(aq.)+CrO_(4)^(2-)(aq.)`
`K_(sp) = C_(Ag^(+))^(2)C_(CrO_(4)^(2-))`
`= (mol L^(-1))^(2)(mol L^(-1))= mol^(3)L^(-3)`