Question

Quest. 6 (20 pts). Nickel (II) ions forms a strong complex in the presence of ammonia (NH). Calculate the equilibrium concentrations of NP and [Ni(NHLP complex after 0.120 mol of Ni(NO3)2 has been dissolved in 1.00 L of 2.30 M NH, solution. The Kiof (Ni(NHSP is 2.0 x 10. N" (aq) +6 NHỊ (aq) # [NINH " (aq)

Answer 1

Consider a complex formation reaction, Ni ²⁺ (aq) + 6 NH ₃ (aq) [ Ni (NH ₃) ₆ ] ²⁺(aq)

Formation constant for above reaction is K f = [ Ni (NH ₃) ₆ ] ²⁺ / [ Ni ²⁺] [ NH ₃ ] ⁶ = 2.0 10 ⁸

Value of formation constant ( 2.0 10 ⁸ ) is very large, hence we can assume that all metal ion is converted into complex. Therefore, initially there is no metal ions present in the solution.

Now find out initial concentrations of metal ion , ligand and complex.

[ Ni ²⁺] initial = 0.00 M

[ Ni (NH ₃) ₆ ] ²⁺ initial = 0.120 M

[NH ₃ ] initial =2.30 M - 6 (0.120 ) = 1.58 M

At equilibrium concentration of metal ion will be due to dissociation of complex ion. Let's use ICE table to find out equilibrium concentrations of metal ion, ligand and complex ion.

Concentration (M)	Ni 2+	6 NH 3	[ Ni (NH 3) 6 ] 2+
Initial		1.58	0.120
Change	+X	+6 X	- X
Equilibrium	X	1.58 + 6 X	0.120 - X

$\therefore$ K f = ( 0.120 - X) / X ( 1.58 + 6 X) ⁶ = 2.0 10 ⁸

Reaction is at equilibrium and value of formation constant is very high. Hence, X will be very small.

Therefore, we can write 1.58 + 6 X $\approx$ 1.58 and 0.120 - X $\approx$ 0.120.

$\therefore$ K f = 0.120 / X ( 1.58 ) ⁶ = 2.0 10 ⁸

$\therefore$ X = 0.120 / 2.0 10 ⁸ ( 1.58 ) ⁶

X = 3.86 10 ^-11 M = [ Ni ²⁺] _eq

[ Ni (NH ₃) ₆ ] ²⁺ = 0.120 - X = 0.120 - 3.86 10 ^-11 = 0.120 M

ANSWER : Equilibrium concentration of  Ni ²⁺ = 3.86 10 ^-11 M

Equilibrium concentration of [ Ni (NH ₃) ₆ ] ²⁺ = 0.120 M