Question

Sodium chloride is added slowly to a solution that is 0.010 M in Cu+, Ag+, and Au+. The Ksp values for the chloride salts are 1.9× 10−7, 1.6× 10−10, and 2.0× 10−13, respectively. What will be the order of the precipitation of the metals, first to last? Explain or prove your reasoning.

Answer 1

Expression for solubility product of CuCl is given as follows:

Ksp(CuCl) = [Cu+][Cl-]

Substitute, 1.9 x 10^-7 for Ksp(CuCl), 0.010 M for [Cu+]

(1.9 x 10^-7) = 0.010 M x [Cl-]

[Cl-] = (1.9 x 10^-7) / (0.010 M)

[Cl-] = 1.9 x 10^-5 M

Thus, CuCl will start precipitating when chloride ion concentration reaches 1.9 x 10^-5 M

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Expression for solubility product of AgCl is given as follows:

Ksp(AgCl) = [Ag+][Cl-]

Substitute, 1.6 x 10^-10 for Ksp(AgCl), 0.010 M for [Ag+]

(1.6 x 10^-10) = 0.010 M x [Cl-]

[Cl-] = (1.6 x 10^-10) / (0.010 M)

[Cl-] = 1.6 x 10^-8 M

Thus, AgCl will start precipitating when chloride ion concentration reaches 1.6 x 10^-8 M

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Expression for solubility product of AuCl is given as follows:

Ksp(AuCl) = [Au+][Cl-]

Substitute, 2.0 x 10^-13 for Ksp(AuCl), 0.010 M for [Au+]

(2.0 x 10^-13) = 0.010 M x [Cl-]

[Cl-] = (2.0 x 10^-13) / (0.010 M)

[Cl-] = 2.0 x 10^-11 M

Thus, AuCl will start precipitating when chloride ion concentration reaches 2.0 x 10^-11 M

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Therefore, concentration of chloride ions required to precipitate Au+ is lowest (2.0 x 10^-11 M) and that for Cu+ is highest (1.9 x 10^-5 M).

Thus, AuCl will precipitate first, followed by AgCl and CuCl will precipitate last.