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

Answer:

1) The equilibrium constant for the required reaction is $2.832\times 10^{-3}$.

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:

$K_{sp}=1.77\times 10^{-10}=[Ag^+][Cl^-]$..(1)

$Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^{+}(aq)$

Formation constant of $Ag(NH_3)_2^{+}$:

$K_f=1.6\times 10^7=\frac{[Ag(NH_3)_2^{+}]}{[Ag^+][NH_3]^2}$..(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:

$K=\frac{[Ag(NH_3)_2^{+}][Cl^-]}{[AgCl][NH_3]^2}$

[AgCl] = solid = 1

$K=\frac{[Ag(NH_3)_2^{+}][Cl^-]}{[1][NH_3]^2}\times \frac{[Ag^+]}{[Ag^+]}$

$K=K_f\times K_{sp}$ (from 1 and 2)

$K=1.6\times 10^7\times 1.77\times 10^{-10}=2.832\times 10^{-3}$

The equilibrium constant for the required reaction is $2.832\times 10^{-3}$.

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

$K=\frac{[Ag(NH_3)_2^{+}][Cl^-]}{[1][NH_3]^2}$

$2.832\times 10^{-3}=\frac{0.060 M\times 0.060 M}{(x-2(0.060))^2}$

x = 1.2474 M

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