Question

When 0.001 M Na2SO4 solution is saturated with CaSO4, its conductivity increases from 2.6 x 10-4 S cm-1 to 7.0 x 10-4 S cm-1. If the molar conductivities of Na+ and Ca2+ respectively are 50 S cm2 mol-1 and 120 S cm2 mol-1, then the solubility product of CaSO4 is:[Assume that conductivity of water used is negligible.]

• A. (A) 7.0 x 10-6 M2
• B. (B) 4.0 x 10-6 M2
• C. (C) 3.5 x 10-6 M2
• D. (D) 2.0 x 10-6 M2

Answer 1

Answer: (B)

(B) 4.0 x 10-6 M2

Answer 2

Explanation:
Conductivity of Na2SO4=2.6×10−4Scm−1Λm(N2SO4)=1000×26×10−40.001=260cm2mof−1Λm(SO42−)=Λm(Na2SO4)−2Λm(Na+)=260−2×50=160Scm2mol−1Conductivity of CaSO 4 solution=7×10−4−2.6×10−4=4.4×10−4Scm−1Λm(CaSO4)=Λm(Ca2+)+Λm(SO42−)Λm=120+160=280Scm2mol−1M=1000×KΛm=1000×4.4×10−4280M=1.57×10−3MKSP=[Ca2+][SO42−]total =(0.00157)(0.00157+0.001)=4.0×10−6M2