At 298 K, The specific conductivity of a saturated solution of silver chloride in water is `2.30xx10^(-5)Scm^(-1)`. Calculate its solubiility in `gL^(-1)` at 298 K. Given `lambda_(m)^(@)(Ag^(+))"and"lambda_(m)^(@)(CI^(-))`are61.9 and 76.3 S `cm^(2)mol^(-1)` respectively.

Options

A. `5.7 xx 10^(-12)`
B. `1.32 xx 10^(-12)`
C. `7.5 xx 10^(-12)`
D. `1.74 xx 10^(-12)`

Answer

D

Explanation

For a sparingly soluble salt
Solubility = `(kappa xx 1000)/(Lambda_(m)^(@))`
`kappa = 1.85 xx 10^(-5) S m^(-1) = 1.85 xx 10^(-7) S cm^(-1)`
`Lambda_(m)^(@) = 140 xx 10^(-4) S m^(2) mol^(-1) = 140 S cm^(2) mol^(-1)`
Solubility = `(1.85 xx 10^(-7) xx 1000)/(140)`
`= 1.321 xx 10^(-6) mol L^(-1)`
For an electrolyte AB ,
`AB hArr A^+ + B^-`
If s is the solubility , then
`K_(sp) = s^(2) = (1.321 xx 10^(-6))^(2)`
`= 1.745 xx 10^(-12) mol^(2) L^(-2)`