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J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry 4th edition

Review Question 16.12

Explain why PbCl2(s) is less soluble in 1 M Pb(NO3)2 than in pure water, but somewhat more soluble in 1 M HCl(aq) than in pure water.




Answer:

Look at the different reactions:

In pure water,

PbCl2(s)Pb2+(aq) + 2 Cl(aq)

In 1 M Pb(NO3)2 both salts must be considered,

Pb(NO3)2(aq) Pb2+(aq) + 2 NO3(aq)

PbCl2(s)Pb2+(aq) + 2 Cl(aq)

The common lead ion pushes the equilibrium reaction towards reactants, i.e., less solubility.

In 1 M HCl three reactions must be considered,

HCl(aq) + H2O(l) H3O+(aq) + Cl(aq)

PbCl2(s)Pb2+(aq) + 2 Cl(aq)

Pb2+(aq) + 4 Cl(aq)[PbCl4]2–(aq)

The formation of the complex ion in the third reaction drives the solubilization equilibrium of the lead(II) chloride, thereby increasing the solubility somewhat.