Question

Given:$Ag(N{H}_3{)}_2^{+}\rightleftharpoons A{g}^{+}+2N{H}_3,K_c=6.2\times {10}^{-8}$ & $K_{sp}$ of $AgCl=1.8\times {10}^{-10}$ at $298K$. Calculate the concetration of the complex in $1.0M$ aqueous ammoina.

Answer 1

$AgCl\rightleftharpoons A{g}^{+}+C{l}^{-}$

let at time of equilibrium $\left[Ag\right]=YM$

and $\left[Cl\right]=X$=solubility of Agcl

$XY=K_{SP}.............\left(1\right)$

now,

$A{g}^{+}+2N{H}_3⟷Ag\left(N{H}_3\right){}_2{}^{+}K=\frac{1}{K_C}$

now $\left[Ag\right(N{H}_3\left){}_2{}^{+}\right]=x-yM$

neglecting change in concentration of $N{H}_3$

$\Rightarrow \frac{X-Y}{Y}=\frac{1}{K_C}................\left(2\right)$

Solving 1 and 2 x=$\sqrt{\frac{K_{SP}}{K_C}}$