Question

Commercial silver-plating operations frequently use a solution containing the complex Ag+ ion. Because the formation constant (Kf) is quite large, this procedure ensures that the free Ag+ concentration in solution is low for uniform electrodeposition. In one process, a chemist added 9.0 L of 5.0 M NaCN to 90.0 L of 0.21 M AgNO3.

Calculate the concentration of free Ag+ ions at equilibrium.

See your textbook for Kf values.

Answer 1

Answer:

[Ag⁺] = 1.10x10⁻²³ M

Explanation:

The equilibrium that takes place is:

Ag⁺ + 2CN⁻  ↔ [Ag(CN)₂]⁻

Kf = 1.0x10²¹ = $\frac{[Ag(CN)_{2}]}{[Ag^{+}][CN^{-}]^2}$

We calculate the moles of each reagent:

Ag⁺ ⇒ 90.0 L * 0.21 M = 18.9 moles Ag⁺

CN⁻ ⇒ 90.0 L * 5.0 M = 450 moles CN⁻

Because Ag⁺ is the limiting reactant, we use that value to calculate the moles of [Ag(CN)₂]⁻ formed.

We assume 18.9 moles of [Ag(CN)₂]⁻ are formed, because Kf is quite large.

The moles of CN⁻ that reacted are:

18.9 molAg⁺ * 2molCN⁻/1molAg⁺ = 37.8 moles CN⁻

So the remaining CN⁻ moles in solution are 450-37.8 = 412.2 mol CN⁻

The final volume is 99.0 L, so now we recalculate the molar concentrations of CN⁻ and the complex:

412.2 mol CN⁻ / 99L = 4.16 M

18.9 mol [Ag(CN)₂]⁻ / 99L = 0.19 M

Finally we put the data for the complex and [CN⁻] in the expression of Kf and solve for [Ag⁺]:

$\frac{[Ag(CN)_{2}]}{[Ag^{+}][CN^{-}]^2}$ = 1.0x10²¹

$\frac{0.19}{[Ag^{+}]*4.16^2}$ = 1.0x10²¹

[Ag⁺] = 1.10x10⁻²³ M