JEE MAIN - Chemistry (2021 - 24th February Morning Shift - No. 15)
The stepwise formation of $${\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}$$ is given below:
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The value of stability constants K1, K2, K3 and K4 are 104, 1.58 x 103, 5 x 102 and 102 respectively.
The overall equilibrium constants for dissociation of $${\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}$$ is x $$ \times $$ 10-12.
The value of x is ________. (Rounded off to the nearest integer)
The value of stability constants K1, K2, K3 and K4 are 104, 1.58 x 103, 5 x 102 and 102 respectively.
The overall equilibrium constants for dissociation of $${\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}$$ is x $$ \times $$ 10-12.
The value of x is ________. (Rounded off to the nearest integer)
Answer
1
Explanation
Given, stability constant value,
K1 = 104
K2 = 1.58 $$\times$$ 103
K3 = 5 $$\times$$ 102
K4 = 102
Cu2+ + NH3 $$\buildrel {K_1} \over \rightleftharpoons $$ [Cu(NH3)]2+ .... (i)
[Cu(NH3)]2+ + NH3 $$\buildrel {K_2} \over \rightleftharpoons $$ [Cu(NH3)2]2+.... (ii)
[Cu(NH3)2]2+ + NH3 $$\buildrel {K_3} \over \rightleftharpoons $$ [Cu(NH3)3]2+..... (iii)
[Cu(NH3)3]2+ + NH3 $$\buildrel {K_4} \over \rightleftharpoons $$ [Cu(NH3)4]2+ ..... (iv)
On adding Eqs. (i), (ii), (iii) and (iv), we get
Cu2+ + 4NH3 $$\buildrel {K} \over \rightleftharpoons $$ [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
K1 = 104
K2 = 1.58 $$\times$$ 103
K3 = 5 $$\times$$ 102
K4 = 102
Cu2+ + NH3 $$\buildrel {K_1} \over \rightleftharpoons $$ [Cu(NH3)]2+ .... (i)
[Cu(NH3)]2+ + NH3 $$\buildrel {K_2} \over \rightleftharpoons $$ [Cu(NH3)2]2+.... (ii)
[Cu(NH3)2]2+ + NH3 $$\buildrel {K_3} \over \rightleftharpoons $$ [Cu(NH3)3]2+..... (iii)
[Cu(NH3)3]2+ + NH3 $$\buildrel {K_4} \over \rightleftharpoons $$ [Cu(NH3)4]2+ ..... (iv)
On adding Eqs. (i), (ii), (iii) and (iv), we get
Cu2+ + 4NH3 $$\buildrel {K} \over \rightleftharpoons $$ [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
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