Question

Consider $aM$ of $N{H}_3\left(aq\right)$ is present in a solution containing $AgCl$ and $x$ is the solublity of $AgCl$ in this solution.
Given: $K_f$ is the formation constant of $\left[Ag\right(N{H}_3{)}_2{]}^{+}$,
$K_{sp}$ is the solubility product of $AgCl$.
Which of the following relation is correct.

• A. $\frac{K_{sp}}{K_f}=\frac{x}{(a+2x)}$
• B. $K_{sp}\times K_f=\frac{x}{(a-2x)}$
• C. $\frac{K_{sp}}{K_f}=\frac{x^2}{(a+2x{)}^2}$
• D. $K_{sp}\times K_f=\frac{x^2}{(a-2x{)}^2}$

Answer 1

The correct option is D $K_{sp}\times K_f=\frac{x^2}{(a-2x{)}^2}$

Theory:
Complex salt :
Formed due to the combination of simple salts or molecular compounds and they retain their identity in the solution as well as the solid state.
Example - $\left[Ag\right(N{H}_3{)}_2]Cl,[Co\left(N{H}_3{)}_{4}C{l}_2\right]Cl$
For $\left[Ag\right(N{H}_3{)}_2]Cl$

$\left[Ag\right(N{H}_3{)}_2\left]Cl\right(aq)\rightleftharpoons [Ag\left(N{H}_3{)}_2{]}^{+}\right(aq)+C{l}^{-}(aq)$
Formation of a complex ion occurs in steps
Example - Formation of $\left[Ag\right(N{H}_3{)}_2{]}^{+}$ complex ion,
$A{g}^{+}\left(aq\right)+N{H}_3\left(aq\right)\stackrel{K_1}{\rightleftharpoons }\left[Ag\right(N{H}_3\left){]}^{+}\right(aq)$
Where $K_1$ is formation constant or stability constant.
$\left[Ag\right(N{H}_3\left){]}^{+}\right(aq)+N{H}_3(aq\left)\stackrel{K_2}{\rightleftharpoons }\right[Ag\left(N{H}_3{)}_2{]}^{+}\right(aq)$
Where $K_2$ is formation constant or stability constant.
Adding above equations we get :
$A{g}^{+}\left(aq\right)+2N{H}_3\left(aq\right)\stackrel{K_{form}}{\rightleftharpoons }\left[Ag\right(N{H}_3{)}_2{]}^{+}\left(aq\right)$
$K_{form}=K_1\times K_2$
where, $K_{form}$ is the formation constant or the stability constant for $\left[Ag\right(N{H}_3{)}_2{]}^{+}$ .
Higher the value of the stability constant,more stable will be the complex.
Instability constant $K_{inst}$ is numerically equal to the inverse of stability constant $K_{form}$
$K_{inst}=\frac{1}{K_{form}}$

let $x$ be the solubility of $AgCl$, so
$AgCl\left(s\right)\rightleftharpoons A{g}^{+}\left(aq\right)+C{l}^{-}\left(aq\right)$
$x-yx$
$A{g}^{+}\left(aq\right)+2N{H}_3\left(aq\right)\rightleftharpoons \left[Ag\right(N{H}_3{)}_2{]}^{+}\left(aq\right)$
$xa0$
$x-ya-2yy$

$K_{sp}=(x-y)x$

$K_f=\frac{y}{(x-y)(a-2y{)}^2}$
$K_f$ is the formation constant of $\left[Ag\right(N{H}_3{)}_2{]}^{+}$
If the complex is stable, then $K_f$ will be high (in most of cases $K_f$ is high like in this case) .
$\therefore x-y\approx 0\to x\approx y$
Adding both the equations, we get :
$AgCl\left(s\right)+2N{H}_3\left(aq\right)\rightleftharpoons \left[Ag\right(N{H}_3{)}_2{]}^{+}\left(aq\right)+Cl-\left(aq\right)$
$a-2xxx$
Since the equations are added, the equilibrium constant gets multiplied and the new equilibrium constant for this reaction will be $(K_{sp}\times K_f)$:

$K_{sp}\times K_f=\frac{x^2}{(a-2x{)}^2}$