Question

The conductivity of 0.001M `Na_2SO_4` solution is `2.6xx10^(-4)Scm^(-1)` and increases to `7.0xx10^(-4)Scm^(-1)`, When the solution is saturated with `CaSO_4`. The molar conductivities of `Na^+` and `Ca^(2+)` are 50 and 120`Scm^2mol^(-1)`, respectively. Neglect conductivity of used water. What is the solubility product fo `CaSO_4`?
A. `4 xx 10^(-6)`
B. `1.57 xx 10^(-3)`
C. `4 xx 10^(-4)`
D. `2.46 xx 10^(-6)`

Answer 1

Correct Answer - A
Conductivity of `Na_(2)SO_(4)=2.6xx10^(-4)`
`Lambda_(m)(Na_(2)SO_(4))=(1000xx2.6xx10^(-4))/(0.001)`
`=260"S cm"^(2)mol^(-1)`
`lambda_(m)(SO_(4)^(2-))=Lambda_(m)`
`(Na_(2)SO_(4))-2lambda_(m)(Na^(+))`
`=260-2xx50=160Scm^(2)mol^(-1)`
Conductivity of `CaSO_(4)` solution
`=7xx10^(-4)-2.6xx10^(-4)=4.4xx10^(-4)Scm^(-1)`
`^^_(m)(CaSO_(4))=lambda_(m)(Ca^(+2))+lambda_(m)(SO_(4)^(2-))`
`=120+160=280Scm^(2)mol^(-1)`
`S=(1000xxK)/(^^_(m))=(1000xx4.4xx10^(-4))/(280)=1.57xx10^(-3)m`
`K_(sp)=[Ca^(2+)][SO_(4)^(2-)]"total"`
`K_(sp)=(1.57xx10^(-3))(1.57xx10^(-3)+0.001)`
`K_(sp)=4xx10^(-6)M^(2)`