Inorganic and Applied Chemistry

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Inorganic and Applied Chemistry

[Pb2+] 0 = 0 M

[I-] 0 = 3.33 · 10-2 M

The end concentrations is found on the basis of the reaction scheme:

PbI 2 (s)  Pb2+(aq) + 2 I-(aq)

[Pb2+] = x

[I-] = 3.33 · 10-2 M + 2x

And by insertion into the expression for Ksp in the equation we achieve:

Ksp = 1.4 · 10-8 M^3 = [Pb2+] · [I-] = x · (3.33 · 10-2 M + 2x)^2 

x = 1.3 · 10-5 M

Thus, the equilibrium concentrations become:

[Pb2+]equilibrium = 1.3 · 10-5 M

[I-] equilibrium = 3.33 · 10-2 M + 2x  3.33 · 10-2 M

Thereby we have determined the equilibrium concentrations of lead and iodine ions in a solution that is

prepared by mixing solutions of sodium iodine and lead.

Equilibrium