# 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 = $\frac{\left[\mathrm{Ag}{{\left({\mathrm{NH}}_{3}\right)}_{2}}^{+}\right]\left[{\mathrm{Cl}}^{-}\right]}{\left[{\mathrm{NH}}_{3}\right]}$ = Ksp x Kf

⇒  = Ksp x 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 = $\frac{\left[\mathrm{Ag}{{\left({\mathrm{NH}}_{3}\right)}_{2}}^{+}\right]}{\left[{\mathrm{Ag}}^{+}\right]{\left[{\mathrm{NH}}_{3}\right]}^{2}}$

⇒ [Ag+] =

⇒ [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