JEE Main 2021 — Chemical Equilibrium Question with Solution

${\mathrm{Cu}}^{2+}+{\mathrm{NH}}_3\stackrel{{\mathrm{K}}_1}{\rightleftharpoons }{\left[\mathrm{Cu}\left({\mathrm{NH}}_3\right)\right]}^{2+}$

${\left[\mathrm{Cu}\left({\mathrm{NH}}_3\right)\right]}^{2+}+{\mathrm{NH}}_3\stackrel{{\mathrm{K}}_2}{\rightleftharpoons }{\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_2\right]}^{2+}$

${\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_2\right]}^{2+}+{\mathrm{NH}}_3\stackrel{{\mathrm{K}}_3}{\rightleftharpoons }{\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_3\right]}^{2+}$

${\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_3\right]}^{2+}+{\mathrm{NH}}_3\stackrel{{\mathrm{K}}_4}{\rightleftharpoons }{\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]}^{2+}$

${\mathrm{Cu}}^{2+}+4{\mathrm{NH}}_3\stackrel{\mathrm{K}}{\rightleftharpoons }{\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]}^{2+}$

So,

$\mathrm{K}={\mathrm{K}}_1\times {\mathrm{K}}_2\times {\mathrm{K}}_3\times {\mathrm{K}}_4$

$={10}^4\times 1.58\times {10}^3\times 5\times {10}^2\times {10}^2$

$\mathrm{K}=7.9\times {10}^{11}$

Where $\mathrm{K}\to$ Equilibrium constant for formation of ${\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]}^{2+}$
So equilibrium constant $\left({\mathrm{K}}^{\text{'}}\right)$ for dissociation
$\text{ of }{\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]}^{2+}\text{ is }\frac{{\displaystyle 1}}{{\displaystyle \mathrm{K}}}$

${\mathrm{K}}^{\text{'}}=\frac{1}{ \mathrm{K}}$

${\mathrm{K}}^{\text{'}}=\frac{1}{7.9\times {10}^{11}}$

$=1.26\times {10}^{-12}=\left(\mathrm{x}\times {10}^{-12}\right)$

So the value of $\mathrm{x}=1.26$

$\mathrm{OMR}$ Ans $=1$ (After rounded off to the nearest integer)