Equivalent conductance of saturated `BaSO_4` is `400 ohm^(-1) cm^(2) "equiv"^(-1)` and specific conductance is ` 8 xx 10^(-5) ohm^(-1), cm^(-1)` . Hence `K_(sp)` of `BaSO_4 ` is

Options

A. `4 xx 10^(-8) M^(2) `
B. `1 xx 10^(-8) M^(2)`
C. ` 2 xx 10^(-4) M^(2)`
D. `1 xx 10^(-4) M^(2)`

Answer

B

Explanation

`^^_(eq) = K xx (1000)/(N), N = K xx (1000)/(^^_(eq)) = 8 xx 10^(-5) xx (1000)/(400)= 2 xx 10^(-4)`
N = valency factor ` xx ` M , `Ba^(+2) SO_4^(-2) v.f = 2 :. M = (N)/(v.f) = (2 xx 10^(-4))/(2) = 10^(-5)`
` BaSO_4 iff Ba^(+2) + So_4^(-2) : K_(gp) = [Ba^(+2)][So_4^(-2)] = 10 -4 xx 10-4 = 10^(-8) M^(2)`