The conductivity of saturated solution of `BaSO_(4)` is `3.06xx10^(-6) ohm^(-1) cm^(-1)` and its equivalent conductance is 1.53 `ohm^(-1) cm^(2) "equiv"^(-1)`. The `K_(sp)` for `BaSO_(4)` will be:

Options

A. `4 xx 10^(-12)`
B. `2.5 xx 10^(-9)`
C. `2.5xx10^(-13)`
D. `4 xx 10^(-6)`

Answer

D

Explanation

Solubility , `s=(kxx1000)/(^^_(eq))`
`=(3.06xx10^(-6)xx1000)/(1.53)=2xx10^(-3)`
`K_(sp)=s^(2) = 4xx10^(-6)`