JEE Main 2018 — Ionic Equilibrium Question with Solution
From: JEE Main 2018 (Online) 15th April Morning Slot
Question
The minimum volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution (Ksp of PbCl2 = 3.2 10-8 atomic mass of Pb = 207 u ) is :
Choose an option
Show full solutionCorrect option: C
Correct answer
C0.18 L
Step-by-step explanation
Given,
Ksp of PbCl2 = 3.2 108
Ksp = [s] [2s]2
Ksp = 4s3
4s3 = 3.2 108
s3 = 8 109
s = 2 10 3 M
Solubility =
2 103 =
V = = 0.18 L
Ksp of PbCl2 = 3.2 108
| PbCl2 | Pb2+ | 2Cl | |
|---|---|---|---|
| Initially | 1 | 0 | 0 |
| At equilibrium | 1-s | s | 2s |
Ksp = [s] [2s]2
Ksp = 4s3
4s3 = 3.2 108
s3 = 8 109
s = 2 10 3 M
Solubility =
2 103 =
V = = 0.18 L
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