QJEEM21Chb4a2: JEE Main 2021 Chemistry - Class 11 JEEM21Chb4a2 Solution: The stepwise formation of is given below: The value of…

Given, stability constant value,

K₁ = 10⁴

K₂ = 1.58 $\times$ 10³

K₃ = 5 $\times$ 10²

K₄ = 10²

Cu²⁺ + NH₃ $$\stackrel{K_1}{\rightleftharpoons }$$ [Cu(NH₃)]²⁺ .... (i)

[Cu(NH₃)]²⁺ + NH₃ $$\stackrel{K_2}{\rightleftharpoons }$$ [Cu(NH₃)₂]²⁺.... (ii)

[Cu(NH₃)₂]²⁺ + NH₃ $$\stackrel{K_3}{\rightleftharpoons }$$ [Cu(NH₃)₃]²⁺..... (iii)

[Cu(NH₃)₃]²⁺ + NH₃ $$\stackrel{K_4}{\rightleftharpoons }$$ [Cu(NH₃)₄]²⁺ ..... (iv)

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

Cu²⁺ + 4NH₃ $$\stackrel{K}{\rightleftharpoons }$$ [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 ${1 \over K}$ .

$K' = {1 \over K} = {1 \over {7.9 \times {{10}^{11}}}} = 1.26 \times {10^{ - 12}}$

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

x = 1.26