Exercise 7 | Solubility and Precipitation Reactions

Calculate the solubility in g.L-1 of AgCl (s) in 0.50 M NH3 (aq).


Data: Ksp (AgCl) = 1.8 x 10-10 2; Kf ( [Ag(NH3)2]) = 2.0 x 107 -2

Solubility reaction: AgCl (s)  Ag+ (aq) + Cl- (aq)

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

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


 

At t = t0:

[NH3] = 0.50 M

[Cl-] = 0 M

[Ag(NH3)2+] = 0 M

At t = tequilibrium:

[NH3] = 0.50 – 2x

[Cl-] = x

[Ag(NH3)2+] = x

 

K = Ag(NH3)2+Cl-NH3 = Ksp x Kf

⇒ x20.50 - 2x2 = Ksp x Kf

x0.50 - 2x = Ksp Kf

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


At t = tequilibrium:

[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+]

Kf = Ag(NH3)2+Ag+NH32

⇒ [Ag+] = Ag(NH3)2+Kf NH32

⇒ [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 MAgCl

⇒ S = 3.8 g.L-1