Chemistry - Ionic Equilibrium Question with Solution | TestHub

Initially $\left[{\mathrm{Ag}}^{+}\right]=0.2\mathrm{M}$ and $[\mathrm{N}{\mathrm{H}}_3]=1.2\mathrm{M}$. Almost all silver forms $\mathrm{Ag}({\mathrm{NH}}_3{)}_2^{+}=0.2\mathrm{M}$, using 0.4 M ${\mathrm{NH}}_3$, leaving $[\mathrm{N}{\mathrm{H}}_3]=0.8\mathrm{M}$.

$K_{f2}=\frac{[\mathrm{Ag}(\mathrm{N}{\mathrm{H}}_3{)}_2^{+}]}{[\mathrm{Ag}(\mathrm{N}{\mathrm{H}}_3{)}^{+}][\mathrm{N}{\mathrm{H}}_3]}$.

$[\mathrm{Ag}(\mathrm{N}{\mathrm{H}}_3{)}^{+}]=\frac{0.2}{2.5\times 10^4\times 0.8}=1.0\times 10^{-5}\mathrm{M}$.

(Using 1.2 M for free ${\mathrm{NH}}_3$ would give $6.7\times 10^{-6}\mathrm{M}$, the trap option.)

Answer: (A).