Question

can you teach me how to solve this problem?

Calculate the molar solubility of silver chloride in 0.10 M NH3(aq), given that K=1.6x10-10 for silver chloride and K=1.6x107 for the ammonia complex of Ag* ions, Ag(NH3)2.

Answer 1

The formation of silver ammonium ion can be illustrated by the following chemical reaction

Ag⁺ (aq) + 2NH₃ (aq) ---------> [Ag(NH₃)₂]⁺ (aq)

the formation constant can be given as follows :

K_f =  ([Ag(NH₃)₂⁺]) / ([Ag]⁺ [NH₃]² )

= 1.6 X 10⁷

the chemical reaction for the dissociation of AgCl at equilibrium is shown below:

AgCl(s) -------> Ag⁺(aq) + Cl^-(aq)

substitute the given value of Ksp in above equation and the solubility product of AgCl at equilibrium can be determined as follows :

K_sp = [Ag⁺] [Cl^- ]

= 1.6 X 10^-10

the overall chemical reaction is shown below:

AgCl(s) + 2NH₃ (aq) --------->   [Ag(NH₃)₂]⁺(aq) + Cl^-(aq)

therefore the overall formation constant for the above reaction can be given as follows

K_eq = ([Ag(NH₃)₂⁺])[Cl^-] / [NH₃]²

multiply and divide the above equation by [Ag⁺]

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

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

= K_f X K_sp

substitute the given value of K_f and K_sp in the above equation

K_eq = (1.6) X 10⁷ X 1.6 X 10^-10

= 2.56 X 10^-3  

therefore the overall equilibrium is 2.56 X 10^-3

An ICE table for the overall chemical reaction is tabulated as shown below

AgCl(s) + 2NH₃ (aq) --------->   [Ag(NH₃)₂]⁺(aq) + Cl^-(aq)

I (M) 0.10 0 0

C(M) -2x x x

E(M) 0.10 - 2x x x

here the concentration of products so formed is x

the equilibrium constant can also be given in terms of x which is shown below:

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

2.56 X 10^-3 = x X x / (0.10 - 2x)²

2.56 X 10^-3 = ( x / (0.10 - 2x))²

x / (0.10 - 2x) = square root of (2.56 X 10^-3)

x / (0.10 - 2x) = 0.0506

(0.10 - 2x) / x =`19.76

0.10/x -2 = 19.76

0.10 / x = 21.76

x = 4.59 X 10^-3

therefore the solubility product of AgCl is 4.59 X 10^-3 M

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