Question

What is the concentration of $A{g}^{+}$ ion in $0.01MAgN{O}_3$ that is also $1.0MN{H}_3$ ?
Will $AgCl$ precipitate from a solution that is $0.01MAgN{O}_3$ , $0.01MNaCl$ and $1MN{H}_3$?
Given : Formation constant,$K_{f}of\left[Ag\right(N{H}_3{)}_2{]}^{+}=1.70\times {10}^{7}and{K}_{sp}AgCl=1.8\times {10}^{-10}$

• A. $3.5\times {10}^{-9}M$, precipitate of $AgCl$ forms
• B. $6.1\times {10}^{-10}M$, precipitate of $AgCl$ does not form.
• C. $5.3\times {10}^{-8}M$, precipitate of $AgCl$ forms
• D. $2.1\times {10}^{-4}M$, precipitate of $AgCl$ does not form.

Answer 1

The correct option is B $6.1\times {10}^{-10}M$, precipitate of $AgCl$ does not form.

The reaction for the formation of the complex is given where $0.01MAgN{O}_3$ shall combines with $0.02MN{H}_3$ to form $Ag(N{H}_3{)}_2^{+}$

$A{g}^{+}+2N{H}_3\to \left[Ag\right(N{H}_3{)}_2{]}^{+}Initially:0.0110Equilibrium:0(1-0.02)=0.980.01$

Now consider the dissociation equilibrium of the complex :
$\left[Ag\right(N{H}_3{)}_2{]}^{+}\rightleftharpoons A{g}^{+}+2N{H}_{3}Initially:0.0100.98Equilibrium:0.01-x\approx 0.01x0.98+2x\approx 0.98$

Since,the above forward reaction is the dissociation of a complex, it will hardly occur
$0.98+2x\approx 0.980.01-x\approx 0.01$
Thus the formation constant$\left(K_f\right)$ is the inverse of dissociation constant $\left(K_d\right)$:
$K_d=\frac{1}{K_f}=\frac{\left[A{g}^{+}\right][N{H}_3{]}^2}{\left[Ag\right(N{H}_3{)}_2{]}^{+}]}\left[A{g}^{+}\right]=\frac{\left[Ag\right(N{H}_3{)}_2{]}^{+}}{K_f\times [N{H}_3{]}^2}\left[A{g}^{+}\right]=\frac{0.01}{(1.70\times {10}^7)\times (0.98{)}^2}\left[A{g}^{+}\right]=6.1\times {10}^{-10}M$

Ionic product of $AgCl=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$
$Q_{sp}=(6.1\times {10}^{-10})\times \left(0.01\right)Q_{sp}=6.1\times {10}^{-12}$
Since $Q_{sp}<K_{sp}$, precipitate does not form.