Question

Calculate the solubility (a) in mol/L and (b) in g/L of AgCl in 2.0 M NH3(aq). What is the concentration of Ag+ in this solution? (Ksp = 1.6 x 10-10; Kf = 1.7 x 107 for Ag(NH3)2+

Answer 1

we have two reactions

AgCl (s) <--> Ag+ (aq) + Cl- (aq)   , Ksp = 1.6 x 10^ -10

Ag+ (aq) + 2NH3 (aq) <--> Ag(NH3)2+ (aq) , Kf = 1.7 x 10^7

combining both we get

AgCl (s) + 2NH3 (aq) <---> Ag(NH3)2+(aq) + Cl- (aq)   K = Ksp x Kf = 0.00272

Initially [NH3] = 2M , [AG(NH3)2=] =[Cl-] = 0

at equilibrium [NH3] = 2-2X , [Ag(NH3)2+] = [Cl-] = X

K = [Ag(NH3)2+] [Cl-] /[NH3]^2

0.00272 = (X^2) / ( 2-2X)^2

0.00272 = [( X/2-2X) ]^2

0.052154= X / ( 2-2X)

0.1043 = 1.1043X

X = 0.09445 = [Cl-]

hence

(a) solubility in moles/Liter = 0.09445 mols/liter

b) solubility in g/L = Solubility in mols/liter x molar mas of AgCl

                 = 0.09445 mol/L x 143.32 g/mol

                = 13.54 g

Based on Ksp [Ag+] formed = 1.265 x 10^ -5       ( Ksp = S^2 , where S = [Ag+] )

Ag+ formed reacts with NH3 to form complex , hence [Ag(NH3)2+] = 1.265 x 10^ -5 M

now we have reverse reaction

Ag(NH3)2+ (aq) <--> Ag+ (aq) + 2NH3 , K = 1/Kf = 5.88 x 10^ -8

K = [Ag+][NH3]^2 /[ Ag(NH3)2+]     where [NH3] = 2-2X = 2-( 2x 0.09445) = 1.81 ( from previous calclations)

5.88 x 10^ -8 = (X) ( 1.81) / ( 1.265 x 10^ -5)

[Ag+] = 4.1 x 10^ -13 M