Question

Calculate the equilibrium concentration of Ag⁺ (aq) in a solution that is initially 0.130 M AgNO₃ and 0.610 M NH₃ and in which the following reaction takes place:

Ag⁺ (aq) + 2NH₃(aq) <--> Ag(NH₃)₂⁺ (aq) K_f = 1.7 x 10⁷

Answer 1

The formation constant for the given reaction is K_f = 1.7*10⁷. Since K_f is high, it is assumed that all the Ag⁺ (added as AgNO₃) has coupled with NH₃ to form the complex, [Ag(NH₃)₂]⁺ and the equilibrium concentration of [Ag(NH₃)₂]⁺ is equal to the initial concentration of Ag⁺, i.e, 0.130 M.

Again, as per the stoichiometric equation,

1 M [Ag(NH₃)₂]⁺ = 2 M NH₃.

Therefore,

0.130 M [Ag(NH₃)₂]⁺ = (2 M NH₃)*(0.130 M [Ag(NH₃)₂]⁺)/(1 M [Ag(NH₃)₂]⁺)

= 0.260 M NH₃.

Therefore, equilibrium concentration of NH₃ = (0.610 – 0260) M = 0.350 M.

Write down the expression for the formation constant of [Ag(NH₃)₂]⁺ as

K_f = [Ag(NH₃)₂⁺]/[Ag⁺][NH₃]²

where [..] denote concentrations in mol/L or M.

Plug in values and get

1.7*10⁷ = (0.130 M)/[Ag⁺](0.350 M)²

======> [Ag⁺] = (0.130 M)/(1.7*10⁷)(0.350 M)²

Ignore units and obtain

[Ag⁺] = (0.130)/(1.7*10⁷)(0.1225)

======> [Ag⁺] = 6.242*10^-8 ≈ 6.24*10^-8 (ans, correct to 3 sig. figs).

The equilibrium concentration of Ag⁺ is 6.24*10^-8 M (ans).