CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The conductivity of 0.001M Na2SO4 solution is 2.6×104 Scm1 and increases to 7.0×104 Scm1, when the solution is saturated with CaSO4. The molar conductivities of Na+ and Ca2+ are 50 and 120 Scm2mol1, respectively. Neglect conductivity of used water. What is the solubility product of CaSO4?

A
4×106
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
1.57×103
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
4×104
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2.46×106
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 4×106
Conductivity of Na2SO4(KNa2SO4)=2.6×104 Scm1

Molar conductivity of Na2SO4(m(Na2SO4)=conductivity (K)concentration (C)

=2.6×104×10000.001
=260 Scm2 mol1
m(SO24)=m(Na2SO4)2m(Na+)
m(SO24)=260(2×50)
=160Scm2 mol1
Conductivity of CaSO4 solution =7×1042.6×104=4.4×104Scm1
m(CaSO4)=m(Ca2+)+m(SO24)
=120+160
=280Scm2 mol1
Solubility of Ca2+=1000×Km=1000×4.4×104280=1.57×103M
Solubility product (KSP)ofCaSO4=[Ca2+][SO24]total
=(0.00157) (0.00157+0.001)
=4.0×106M2

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Introduction
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon