JEE Main 2021 — Chemical Equilibrium Question with Solution

Given, stability constant value,

K₁ = 10⁴

K₂ = 1.58 $$\times$$ 10³

K₃ = 5 $$\times$$ 10²

K₄ = 10²

Cu²⁺ + NH₃ $$\frac{\text{buildrel}{K}_1}{\rightleftharpoons }$$ [Cu(NH₃)]²⁺ .... (i)

[Cu(NH₃)]²⁺ + NH₃ $$\frac{\text{buildrel}{K}_2}{\rightleftharpoons }$$ [Cu(NH₃)₂]²⁺.... (ii)

[Cu(NH₃)₂]²⁺ + NH₃ $$\frac{\text{buildrel}{K}_3}{\rightleftharpoons }$$ [Cu(NH₃)₃]²⁺..... (iii)

[Cu(NH₃)₃]²⁺ + NH₃ $$\frac{\text{buildrel}{K}_4}{\rightleftharpoons }$$ [Cu(NH₃)₄]²⁺ ..... (iv)

On adding Eqs. (i), (ii), (iii) and (iv), we get

Cu²⁺ + 4NH₃ $$\frac{\text{buildrel}K}{\rightleftharpoons }$$ [Cu(NH₃)₄]²⁺

$$\therefore$$ The overall reaction constant (k) or equilibrium constant for formation of [Cu(NH₃)₄]²⁺ is

K = K₁ $$\times$$ K₂ $$\times$$ K₃ $$\times$$ K₄

$$\Rightarrow$$ K = 10⁴ $$\times$$ 1.58 $$\times$$ 10³ $$\times$$ 5 $$\times$$ 10² $$\times$$ 10²

$$\Rightarrow$$ K = 7.9 $$\times$$ 10¹¹

where, K = equilibrium constant for formation of [Cu(NH₃)₄]²⁺

So, equilibrium constant 'K' for dissociation of [Cu(NH₃)₄]²⁺ is $$\frac{1}{K}$$.

$$K^{\prime }=\frac{1}{K}=\frac{1}{7.9\times {10}^{11}}=1.26\times 10^{-12}$$

Hence, K' = x $$\times$$ 10^$$-$$12

x = 1.26