Conductivity of a saturated solution of a sparingly soluble salt AB at 298 K is `1.85 xx 10^(-5) S m^(-1)`. Solubility product of the salt AB at 298 K is
Given `^^_m^(0)(AB) = 140 xx 10^(-4)S m^(2) " mol"^(-1)`

Options

A. `5.7xx10^(-12)`
B. `1.32xx10^(-12)`
C. `7.5xx10^(-12)`
D. `1.74xx10^(-12)`

Answer

D

Explanation

For a sparingly soluble salt,
solubility `=(k xx 1000)/(Lambda_(m)^(@))`
`k=1.85xx10^(-5) Sm^(-1)`
` = 1.85 xx 10^(-7) S cm^(-1)`
`Lambda_(m)^(@)=140xx10^(4) S m^(2) "mol"^(-1)`
`= 140 S cm^(2) "mol"^(-1)`
Solubility `=(1.85xx10^(-7)xx1000)/(140)`
`=1.321xx10^(-6) ` mol/L
For an electrolyte, AB,
`K_(sp)=S^(2)`
where is the solubility of AB
`:.K_(sp)=(1.321xx10^(-6))^(2)`
`=1.745xx10^(-12) "mol"^(2) L^(-2)`