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

# If the solubility product of a sparingly soluble salt Li3PO4 is 2.7×10−9. The concentration (mol L−1) of Li+ ions in the saturated solution of the salt is: Given : (100)14=3.16

A
3.78×109 molL1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2.7×109 molL1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3.16×103 molL1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
9.48×103 mol L1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is D 9.48×10−3 mol L−1The equilibrium between the undissolved solid Li3PO4 and the ions in a saturated solution can be represented by the solution. Li3PO4(s)⇌3Li+(aq)+PO3−4(aq) c 0 0 c−s 3s s "s" is solubility. The solubility product is given by Ksp=[Li+]3[PO3−4] Ksp=(3s)3×s=27s4 Ksp=27s4=2.7×10−9 s4=10−10=100×10−12 s=3.16×10−3[Li+]=3×(3.16×10−3)=9.48×10−3 mol L−1

Suggest Corrections
0
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program