Verified Solution • 2025–26 Session moderate

QJEEM21Ch6fd1: JEE Main 2021 Chemistry - Class 11 JEEM21Ch6fd1 Solution: The solubility product of PbI 2 is 8.0 10 9 . The solubility of …

The solubility product of PbI2 is 8.0 ×\times 10 −- 9. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x ×\times 10 −- 6. mol/L. The value of x is __________. (Rounded off to the nearest integer) [Given 2\sqrt 2 = 1.41]

Step-by-Step Verified Solution

Given, [Ksp]Pbl2=8×10−9{[{K_{sp}}]_{Pb{l_2}}} = 8 \times {10^{ - 9}}

To calculate solubility of Pbl2 in 0.1 M solution of Pb(NO3)2,

(I) Pb(NO3)20.1M→Pb2+(aq)0.1M+2NO3−(aq)0.2M\mathop {Pb{{(N{O_3})}_2}}\limits_{0. 1 M} \to \mathop {P{b^{2 + }}(aq)}\limits_{0.1 M} + \mathop {2NO_3^ - (aq)}\limits_{0. 2 M}

(II) PbI2(s)Pb{I_2}(s) ⇌\rightleftharpoons Pb2+(aq)S+2I−(aq)2S\mathop {P{b^{2 + }}(aq)}\limits_S + \mathop {2{I^ - }(aq)}\limits_{2S}

∴\therefore [Pb2+] = S + 0.1 ≈\approx 0.1

∵\because S < < 0.1

Now, Ksp = 8 ×\times 10 −- 9

[Pb2] [I −- ]2 = 8 ×\times 10 −- 9

0.1 ×\times (2S)2 = 8 ×\times 10 −- 9

4S2 = 8 ×\times 10 −- 8 ⇒\Rightarrow S = 141 ×\times 10 −- 6 M

∴\therefore x = 141

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