The stepwise formation of $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$ is given b

Question

The stepwise formation of $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}ight)_{4}ight]^{2+}$ is given below \The value of stability constants $K_{1}, K_{2}, K_{3}$ and $K_{4}$ are $10^{4}, 1.58 	imes 10^{3}, 5 	imes 10^{2}$ and $10^{2}$ respectively. The overall equilibrium constant for dissociation of $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}ight)_{4}ight]^{2+}$ is $x 	imes 10^{-12}$. The value of $x$ is $\_\_\_\_$ . (Rounded off to the nearest integer)\

Professor PC Thomas Classes · Answer

(1) : Equilibrium constant for the overall reaction\ $\mathrm{Cu}^{2+}+4 \mathrm{NH}_{3} ightleftharpoons\left[\mathrm{Cu}\left(\mathrm{NH}_{3}ight)_{4}ight]^{2+}$\ can be given as $K=K_{1} 	imes K_{2} 	imes K_{3} 	imes K_{4}$\ $K=10^{4} 	imes 1.58 	imes 10^{3} 	imes 5 	imes 10^{2} 	imes 10^{2}$ $$ =7.9 	imes 10^{11} $$ For dissociation of $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}ight)_{4}ight]^{2+}$,\ Equilibrium constant $K^{\prime}=\frac{1}{K}=\frac{1}{7.9 	imes 10^{11}}$\ $=1.26 	imes 10^{-12}$\ Hence, $x=1$.\