JEE MAIN - Chemistry (2021 - 24th February Evening Shift - No. 16)
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 $$\sqrt 2 $$ = 1.41]
Answer
141
Explanation
Given, $${[{K_{sp}}]_{Pb{l_2}}} = 8 \times {10^{ - 9}}$$
To calculate solubility of Pbl2 in 0.1 M solution of Pb(NO3)2,
(I) $$\mathop {Pb{{(N{O_3})}_2}}\limits_{0. 1 M} \to \mathop {P{b^{2 + }}(aq)}\limits_{0.1 M} + \mathop {2NO_3^ - (aq)}\limits_{0. 2 M} $$
(II) $$Pb{I_2}(s)$$ $$\rightleftharpoons$$ $$\mathop {P{b^{2 + }}(aq)}\limits_S + \mathop {2{I^ - }(aq)}\limits_{2S} $$
$$\therefore$$ [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
To calculate solubility of Pbl2 in 0.1 M solution of Pb(NO3)2,
(I) $$\mathop {Pb{{(N{O_3})}_2}}\limits_{0. 1 M} \to \mathop {P{b^{2 + }}(aq)}\limits_{0.1 M} + \mathop {2NO_3^ - (aq)}\limits_{0. 2 M} $$
(II) $$Pb{I_2}(s)$$ $$\rightleftharpoons$$ $$\mathop {P{b^{2 + }}(aq)}\limits_S + \mathop {2{I^ - }(aq)}\limits_{2S} $$
$$\therefore$$ [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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