Ionic EquilibriumHard

Question

The solubility of PbCl2 when it is 80% ionized is

Options

A.25% less than the solubility of PbCl2 when it is 100% ionized.
B.50% less than the solubility of PbCl2 when it is 100% ionized.
C.More than the solubility of PbCl2 when it is 100% ionized.
D.is equal to the solubility of PbCl2 when it is 100% ionized.

Solution

PbCl2(s) $\rightleftharpoons$ Pb2+ + 2Cl–

S1M 2S1M

When it is 80% ionized, then

PbCl2(s) $\rightleftharpoons$Pb2+ + 2Cl–

0.8 S2 M 1.6 S2M

Now, Ksp = [Pb2+][Cl–] = constant

or, S1 × (2S1)2 = (0.8S2) × (1.6S2)2 ⇒ S1 = 0.512S2 ⇒ S2 > S1

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