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Consider the insoluble compound silver bromide , AgBr . The silver ion also forms a complex with ammonia . Write a balanced net ionic equation to show why the solubility of AgBr (s) increases in the presence of ammonia and calculate the equilibrium constant for this reaction. For Ag(NH3)2+ , Kf = 1.6×107 . Use the pull-down boxes to specify states such as (aq) or (s).

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Answer:

- [tex]AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)[/tex]

- [tex]K=1.2x10^{-5}[/tex]

Explanation:

Hello,

In this case, by considering the dissolution of silver bromide:

[tex]AgBr(s)\rightleftharpoons Ag^+(aq)+Br^-(aq) \ \ \ Ksp=[Ag^+][Br^-]=7.7x10^{-13}[/tex]

And the formation of the complex:

[tex]Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)\ \ \ Kf=\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}=1.6x10^7[/tex]

We obtain the balanced net ionic equation by adding the aforementioned equations:

[tex]AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)+Ag^+(aq)\\\\AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)[/tex]

Now, the equilibrium constant is obtained by writing the law of mass action for the non-simplified net ionic equation:

[tex]AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-+Ag^+\\\\K=[Ag^+][Br^-]*\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}[/tex]

So we notice that the equilibrium constant contains the solubility constant and formation constant for the initial reactions:

[tex]K=Ksp*Kf=7.7x10^{-13}*1.6x10^{7}\\\\K=1.2x10^{-5}[/tex]

Best regards.

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