Exercise 7 | Solubility and Precipitation Reactions

Solubility reaction: AgCl (s) $\rightleftarrows$ Ag⁺ (aq) + Cl^- (aq)

Complexation reaction: Ag⁺ (aq) + 2 NH₃ (aq) $\rightleftharpoons$ [Ag(NH₃)₂]⁺ (aq)

General reaction: AgCl (s) + 2 NH₃ (aq) $\rightleftarrows$ Cl^- (aq) + [Ag(NH₃)₂]⁺ (aq)

K = $\frac{\left[\mathrm{Ag}{{\left({\mathrm{NH}}_3\right)}_2}^{+}\right]\left[{\mathrm{Cl}}^{-}\right]}{\left[{\mathrm{NH}}_3\right]}$ = K_sp x K_f

⇒ $\frac{{\mathrm{x}}^2}{{\left(0.50 - 2\mathrm{x}\right)}^2}$ = K_sp x K_f

⇒ $\frac{\mathrm{x}}{0.50 - 2\mathrm{x}}$ = $\sqrt{{\mathrm{K}}_{\mathrm{sp}} {\mathrm{K}}_{\mathrm{f}}}$

⇒ x = 2.7 x 10^-2 M  (x must be positive)

At t = t_equilibrium:

[NH₃] = 4.5 x 10^-1 M

[Cl^-] = 2.7 x 10^-2 M 

[Ag(NH₃)₂⁺] = 2.7 x 10^-2 M 

The total solubility s of AgCl (s) in M is: s = [Ag⁺] + [Ag(NH₃)₂⁺]

K_f = $\frac{\left[\mathrm{Ag}{{\left({\mathrm{NH}}_3\right)}_2}^{+}\right]}{\left[{\mathrm{Ag}}^{+}\right]{\left[{\mathrm{NH}}_3\right]}^2}$

⇒ [Ag⁺] = $\frac{\left[\mathrm{Ag}{{\left({\mathrm{NH}}_3\right)}_2}^{+}\right]}{{\mathrm{K}}_{\mathrm{f}} {\left[{\mathrm{NH}}_3\right]}^2}$

⇒ [Ag⁺] = 6.7 x 10^-9 M

s = 2.7 x 10^-2 M

S = solubility of AgCl (s) in g.L^-1 = s x M_AgCl

⇒ S = 3.8 g.L^-1