Question

Silver plating processes use Ag(CN)₂^- as the source of Ag⁺ in solution. To make the solution: 4.0 L of 3.00 M NaCN and 50.0 L of 0.200 M AgNO₃ are combined.

(a) What is the concentration of free Ag⁺ ions in solution?

(b) Why is AgCl not used in this process?

NOTE: We have not yet introduced the equilibrium constant K_f. An answer is needed that gets around K_f or slowly introduces the concept. Thanks!

(Jespersen 7th edition 21.98)

Answer 1

a) The reaction taking place is

Ag⁺ (aq) + 2 CN^- (aq) -------> [Ag(CN)₂]^- (aq)

K_f = [Ag(CN)₂^-]/[Ag⁺][CN^-]² = 5.6*10¹⁸ where K_f = formation constant of [Ag(CN)₂]^-

K_f is extremely high; hence we can assume that almost all the Ag⁺ added as AgNO₃ is used up in forming [Ag(CN)₂]^-.

Find out the initial concentration of Ag⁺ in the final solution using the dilution equation. The dilution equation is

M₁*V₁ = M₂*V₂ where M₁ = concentration of the original solution; M₂ = concentration of the dilute solution; V₁ = volume of the original solution and V₂ = volume of the dilute solution.

The total volume of the solution is (4.0 + 50.0) L = 54.0 L.

Ag⁺: (50.0 L)*(0.200 M) = (54.0 L)*M₂

===> M₂ = (50.0*0.200)/(54.0) M = 0.185 M

Since the entire amount of Ag⁺ is converted to [Ag(CN)₂^-], hence the equilibrium concentration of [Ag(CN)₂^-] formed is 0.185 M.

Again, the molar ratio of Ag⁺ and CN^- is 1:2, hence, the equilibrium concentration of CN^- is [(3.0 M) – 2*(0.185 M)] = 2.63 M.

Let the equilibrium concentration of free Ag⁺ in the solution be x M; therefore,

5.6*10¹⁸ = (0.185)/(x)(2.63 – 2x)²

Assume that x is much smaller than 2.63 M, hence (2.63 – 2x) ≈ 2.63 M; hence,

5.6*10¹⁸ = (0.185)/(x.)(2.63)²

===> 5.6*10¹⁸ = (0.185)/(x).(6.9169)

===> x = (0.185)/(6.9169*5.6*10¹⁸) = 4.776*10^-21 ≈ 4.78*10^-21

The concentration of free Ag⁺ in the solution is 4.78*10^-21 M (ans).

b) We cannot use AgCl as the source of Ag⁺ since AgCl is insoluble in water and will not furnish the requisite Ag⁺ for complex formation (ans).