A saturated solution of `Ag_(2)SO_(4) is 2.5 xx 10^(-2)M`. The value of its solubility product is

Options

A. `62.5xx10^(-6)`
B. `6.25xx10^(-4)`
C. `15.625xx10^(-6)`
D. `3.125xx10^(-6)`

Answer

A

Explanation

`Ag_(2)SO_(4)tounderset(2x)(2Ag^(+))+underset(x)(SO_(4)^(2-))` (moles after diss.)
`K_(sp)=[Ag^(+)]^(2)[SO_(4)^(2-)]=(2x)^(2)xxx=4x^(3)`
Given solubility `=x=2.5xx10^(-2)` mole/litre
`:.K_(sp)=4xx(2.5xx10^(-2))^(3)`
`=4xx15.6xx10^(-6)=62.5xx10^(-6)`