Question

Write the equilibrium reactions on a scratch paper, calculate K from Ksp and Kf and determine the concentration of NH3 needed to form 0.060 M Ag(NH3)2+, Kf = 1.6 x 107; Ksp AgCl = 1.77 x10-10

Answer 1

1) The equilibrium constant for the required reaction is
`2.832* 10^(-3)`.

2) 1.2474 M the concentration of ammonia needed to form 0.060 M of complex.Explanation:

`AgCl(s)\rightarrow Ag^+(aq)+Cl^-(aq)`

Solubility product of silver chloride:

`K_(sp)=1.77* 10^(-10)=[Ag^+][Cl^-]`..(1)

`Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^(+)(aq)`

Formation constant of
`Ag(NH_3)_2^(+)`:

`K_f=1.6* 10^7=([Ag(NH_3)_2^(+)])/([Ag^+][NH_3]^2)`..(2)

Reactions solid silver chloride and liquid ammonia:

`AgCl(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^(+)(aq)+Cl^-(aq)`

Expression of an equilibrium constant of the above reaction can be written as:

`K=([Ag(NH_3)_2^(+)][Cl^-])/([AgCl][NH_3]^2)`

[AgCl] = solid = 1

`K=([Ag(NH_3)_2^(+)][Cl^-])/([1][NH_3]^2)* ([Ag^+])/([Ag^+])`

`K=K_f* K_(sp)` (from 1 and 2)

`K=1.6* 10^7* 1.77* 10^(-10)=2.832* 10^(-3)`

The equilibrium constant for the required reaction is
`2.832* 10^(-3)`.

2)

Concentration of complex at equilibrium :
`[Ag(NH_3)_2^(+)]`= 0.060 M

`AgCl(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^(+)(aq)+Cl^-(aq)`

Initaly

x 0 0

At equilibrium

x- 2(0.060) 0.060 0.060

`K=([Ag(NH_3)_2^(+)][Cl^-])/([1][NH_3]^2)`

`2.832* 10^(-3)=(0.060 M* 0.060 M)/((x-2(0.060))^2)`

x = 1.2474 M

1.2474 M the concentration of ammonia needed to form 0.060 M of complex.