Answer:
1.1 M
Explanation:
The dissociation of [tex]AgCl_{(s)}[/tex] is as follows:
[tex]AgCl \leftrightharpoons Ag^+_{(aq)}+Cl^-_{(aq)}[/tex]
Given Value for [tex]K_{sp} = 1.77*10^{-10}[/tex]
The equation for the reaction for the formation of complex ion [tex]Ag(NH_3)^+_2[/tex] is :
[tex]Ag^+_{(aq)} + 2NH_3_{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)}[/tex]
The value of [tex]K_f = 1.6*10^7[/tex]
If we combine both equation and find the overall equilibrium constant will be:
[tex]AgCl \leftrightharpoons Ag^+_{(aq)}+Cl^-_{(aq)}[/tex]
[tex]Ag^+_{(aq)} + 2NH_3_{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)}[/tex]
[tex]AgCl_{(s)}+2NH_3_{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)} + Cl^-_{(aq)}[/tex]
[tex]K = (1.77*10^{-10})(1.6*10^7)[/tex]
[tex]K = 0.00283[/tex]
If [tex][NH_3][/tex] = x M
The solubility of [tex]AgCl_{(s)}[/tex] in the [tex]NH_3[/tex] solution will be:
[tex]x = 743*10^{-3}g * \frac{mol AgCl}{143.32g}*\frac{1}{0.1000L}[/tex]
[tex]x = 0.0518 M[/tex]
Constructing an ICE Table; we have :
[tex]AgCl_{(s)} + 2 NH_3{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)} + Cl^-_{(aq)}[/tex]
Initial (M) x 0 0
Change (M) -2 (0.0518) + 0.0518 + 0.0518
Equilibrium (M) x - 0.1156 0.0518 0.0518
Equilibrium constant;
(K) = [tex]\frac{[Ag(NH_3)_2^+][Cl^-]}{[NH_3]^2}[/tex]
[tex]0.00283 = \frac{(0.0518)^2}{(x-0.1156)^2}[/tex]
[tex]0.00283 = (\frac{0.0518}{x-0.1156})^2[/tex]
[tex]x = 0.1156 + \sqrt{\frac{0.0518^2}{0.00283} }[/tex]
x = [NH₃] = 1.089 M
[NH₃] ≅ 1.1 M
