# 0.010 m sltn AgNO3 is made 0.50 M in NH3, forming  [Ag(HN3)2]+ .Will AgCl precipitate if enough NaCl is added making the sltn 0.010 Min Cl-?

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• So the first equilibrium is:

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

Kf = [Ag(NH3)2+] / [Ag+][NH3]^2

I found that Kf for this reaction is 1.7X10^7 (Check that value with the information you have.)

So, because Kf is so large, it is easiest to assume that the reaction goes completely to the right, and then some of the complex dissociates. So, once you form the solution,

[Ag(NH3)2+] = 0.010 M

[NH3] = 0.50 M - 0.020 M = 0.48 M and

[Ag+] = 0.

Now assume that x M of the complex dissociates to establish the equilibrium. Then, [Ag(NH3)2+] = 0.010 - x

[NH3+] = 0.48 + 2x

[Ag+] = x

Because Kf is so large, x will be very small, and we can assume that x will be insignificant compared to both 0.010 and to 0.48. So,

Kf = 0.01 / (x)(0.48) = 1.7X10^7

x = [Ag+] = 1.2X10^-9 M

Now, the question of whether AgCl will precipitate...

Ksp = [Ag+][Cl-] = 1.7X10^-10

Now, using the concentrations of Ag+ and Cl-, calculate the reaction quotient Q for the precipitation of AgCl:

Q = [Ag+][Cl-]

Q = 1.2X10^-9 (0.010) = 1.2X10^-11

Because Q is less than Ksp, AgCl WILL NOT precipitate from this solution.

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