JEE Main 24 Feb 2021 Shift 1 Solved Paper

Given, stability constant value, k1=104
k2=1.58×103
k3=5×102
k4=102
Cu2++NH3

k1

⇌

[Cu(NH3)]2+ . . . (i)
[Cu(NH3)]2++NH3

k2

⇌

[Cu(NH3)2]2+ . . . (ii)
[Cu(NH3)2]2++NH3

k3

⇌

[Cu(NH3)3]2+ . . . (iii)
[Cu(NH3)3]2++NH3

k4

⇌

[Cu(NH3)4]2+. . . (iv)
On adding Eqs. (i), (ii), (iii) and (iv), we get
Cu2++4NH3‌

stackrel‌k

⇌⇌

[Cu(NH3)4]2+
∴ The overall reaction constant (k) or equilibrium constant for formation of [Cu(NH3)4]2+ is
‌k=k1×k2×k3×k4
‌k=104×1.58×103×5×102×102
‌k=7.9×1011
where, k= equilibrium constant for formation of [Cu(NH3)4]2+
So, equilibrium constant k′ for dissociation of [Cu(NH3)4]2+ is ‌

1

k

.
k′=‌

1

k

=‌

1

7.9×1011

=1.26×10−12
Hence, k′‌‌=x×10−12
x‌‌=1.26