The solubility product `(K_(sp) , mol^(3) dm^(-9)` ) of `MX_(2)` at 298 K based on the information available for the given concentration cell is
(take `2.303 xx R xx 298 // F = 0.059 `V)

Options

A. `1 xx 10^(-15)`
B. `4 xx 10^(-15)`
C. `1 xx 10^(-12)`
D. `4 xx 10^(-12)`

Answer

B

Explanation

The given concentration cell is `M | M^(2+) ` (saturated = `C_(1) ) || M^(2+) (0.001 M = C_(2))| M` EMF of concentration cell is :
`E_(cell) = (2.303 RT)/(nF) "log" (C_(2))/(C_(1)) = (0.059)/(n) "log" (C_(2))/(C_(1))`
`therefore 0.059 = (0.059)/(2) "log" (0.001)/(C_(1))`
`or 1 = (1)/(2) "log" (10^(-3))/(C_(1)) or log 10^(-3) - log C_(1) = 2`
`or -3-log C_(1) = 2 or log C_(1) = -5`
or `C_(1) = 10^(-5)` M
For the salt `MX_(2) , underset(S)(MX^(2)) to underset(S)(M^(2+)) + 2 underset(2 S)(X^(-))`
`K_(sp) = (S) (2S)^(2) = 4 S^(3) = 4 xx (10^(-5))^(3) = 4 xx 10^(-15)`