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. `5.7 xx 10^(-12)`
B. `1.32 xx 10^(-12)`
C. `7.5xx10^(-12)`
D. `1.74xx10^(-12)`

Answer

D

Explanation

Using the relationship between solubility of sparingly soluble salt and `Lambda_(m)^(@)`,
`k=1.85xx10^(-5)S m^(-1)= 1.85xx10^(-7)S cm^(-1)`
`Lambda_(m)^(@)=140xx10^(-4)S m^(2)"mol"^(-1)=140S cm^(2) "mol"^(-1)`
`s=(k xx 1000)/(Lambda_(m)^(@))=(1.85xx10^(-7)xx1000)/(140)`
`=1.32xx10^(-6)`
`K_(sp)=(s)^(2)" "` [For AB type of salt]
`=(1.32xx10^(-6))^(2)=1.74xx10^(-12)`