Question

Calculate the concentration of Ag+ present in solution at equilibrium when concentrated ammonia is added to a 0.013 M solution of AgNO3 to give an equilibrium concentration of [NH3] = 0.17 M. The formation constant for Ag(NH3)2+ at this temperature is 1.7 × 107. How do you solve a problem like this? Thanks!

Answer 1

We assume that the AgNO3 is completely dissociated, giving 0.10 M Ag+. Because Kf for the formation of Ag(NH3)2 + is large, essentially all the Ag+ is then converted to Ag(NH3)2+. But the complex ion will dissociate so we will look at that dissociation rather than the formation.

We will need to reverse the equation to represent the formation of Ag+ and NH3 from Ag(NH3)2 + and also make the corresponding change to the Kf.

Ag(NH3)2+ <--------> Ag+ + 2NH3

0.013____________ 0

0.013 -x __________ x_____0.17

Kf-1 = 1/Kf= 1/1.7x10⁷= 5.9x10^-8= [Ag+][NH3]²/[Ag(NH3)2+]= x (0.17)²/ 0.013-x

Now, the Kf-1 is really small so we can assume that x<<0.013M, so, 0.013 -x is almost the same as 0.013.

x= 2.65x10^-8M= [Ag+]