JEE Main 2021 — Ionic Equilibrium Question with Solution
From: JEE Main 2021 (Online) 24th February Evening Shift
Question
The solubility product of PbI2 is 8.0 109. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x 106. mol/L. The value of x is __________. (Rounded off to the nearest integer) [Given = 1.41]
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Show full solutionCorrect answer: 141
Correct answer
141
Step-by-step explanation
Given,
To calculate solubility of Pbl2 in 0.1 M solution of Pb(NO3)2,
(I)
(II)
[Pb2+] = S + 0.1 0.1
S < < 0.1
Now, Ksp = 8 109
[Pb2] [I]2 = 8 109
0.1 (2S)2 = 8 109
4S2 = 8 108 S = 141 106 M
x = 141
To calculate solubility of Pbl2 in 0.1 M solution of Pb(NO3)2,
(I)
(II)
[Pb2+] = S + 0.1 0.1
S < < 0.1
Now, Ksp = 8 109
[Pb2] [I]2 = 8 109
0.1 (2S)2 = 8 109
4S2 = 8 108 S = 141 106 M
x = 141
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