Question

Please answer both questions.

One millimole of Ni(NO₃)₂ dissolves in 270.0 mL of a solution that is 0.400 M in ammonia.
The formation constant of Ni(NH₃)₆²⁺ is 5.5×10⁸.

1. What is the initial concentration of Ni(NO₃)₂ in the solution?

2. What is the equilibrium concentration of Ni²⁺(aq ) in the solution?

Answer 1

1)

Initial concentration of Ni(NO3)2

= (0.001mol/270.0ml)× 1000ml

= 0.003704M

2)

First consider complete formation of Ni(NH3)6²⁺

Ni²⁺(aq) + 6NH3 -------> Ni(NH3)6²⁺(aq)

stoichiometrically, 1mol of Ni²⁺react with 6moles of NH3

0.003704M of Ni²⁺ react with 0.02222M of NH3 to give 0.003704M of Ni(NH3)6²⁺

After completion of reaction

remaining concentration of NH3 = 0.400M - 0.02222M = 0.3778M

Now , consider dissociation equilibrium of Ni(NH3)6²⁺

Ni(NH3)6²⁺(aq) <-------> Ni²⁺(aq) + 6NH3(aq)

K_d = [Ni²⁺][NH3]⁶/[Ni(NH3)6²⁺

  K_d = 1/K_f = 1/5.5 ×10⁸ = 1.82 ×10^-9

Initial concentration

[Ni(NH3)6²⁺] = 0.003704

[Ni²⁺] = 0

[NH3] = 0.3778

Change in concentration

[Ni(NH3)6²⁺] = - x

[Ni²⁺] = +x

[NH3] = 0.3778 +6x

Equilibrium concentration

[Ni(NH3)6²⁺] = 0.003704 - x

[Ni²⁺] = x

[NH3] =0.3778 + 6x

so,

x( 0.3778 + 6x)⁶/(0.003704 -x) = 1.82 ×10^-9

we can assume 0.3778 + 6x = 0.3778 and 0.003704 -x = 0.003704 because x is small value

x( 0.3778)⁶ / 0.003704 = 1.82 × 10^-9

x 0.7851 = 1.82 ×10^-9

x = 2.32 × 10^-9

Therefore,

  Equilibrium concentration of Ni²⁺ = 2.32 ×10^-9M