How much AgBr could dissolve in 1.0 L of 0.4 M NH3? Assume that [Ag(NH3)2]+ is the only complex formed given, Kf [Ag(NH3)2 +]= 1 × 108 , Ksp(AgBr)= 5.0 × 10-13

Question Collected from Telegram Group.

How much AgBr could dissolve in 1.0 L of 0.4 M NH₃? Assume that [Ag(NH₃)₂]⁺ is the only complex formed given, K_f [Ag(NH₃)₂ ⁺]= 1 × 10⁸ , K_sp(AgBr)= 5.0 × 10^-13? 

Solution:

AgBr⇌Ag⁺+Br⁻ ---------------(1)
Ag⁺+2NH₃⇌[Ag(NH₃)₂]⁺ ---------------(2)
Let x = solubility
Then x=[Br−]=[Ag+] initially

However due to very high K_f and presence of sufficient NH₃, almost all of it will converted to [Ag(NH₃)₂]⁺ causing more AgBr to dissolve.

∴ $\frac{{{{[Ag{{\left( {N{H_3}} \right)}_2}]}^ + }}}{{[A{g^ + }].{{[{{\left( {N{H_3}} \right)}_2}]}^2}}} = 1 \times {10^8}$

$or,{\rm{ }}\frac{x}{{[A{g^ + }].{{[0.4 - 2x]}^2}}} = 1 \times {10^8}$

Also, $[A{g^ + }] = \frac{{{K_{sp}}}}{x} = \frac{{5 \times {{10}^{ - 13}}}}{x}$

So, ${\rm{ }}\frac{{{x^2}}}{{5 \times {{10}^{ - 13}}.{{[0.4 - 2x]}^2}}} = 1 \times {10^8}$

$ \Rightarrow x = 2.8 \times {10^{ - 3}}mols/lit$