Question

One millimole of Ni(NO3)2 dissolves in 210.0 mL of a solution that is 0.500 M in ammonia.
The formation constant of Ni(NH3)62+ is 5.5×108.

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

Answer 1

Sol:-

Given

Number of moles of Ni(NO₃)₂ or Ni²⁺ = 1 millimole = 0.001 mol

Molarity of NH₃ = 0.500 M

Volume = 210.0 mL = 0.210 L

We know

Molarity = number of moles / volume of solution in litre

so

Number of moles of NH₃ = Molarity x Volume = 0.500 M x 0.210 L = 0.105 mol

ICF table for the reaction between Ni(NO₃)₂ or Ni²⁺ and NH₃, we have

...........................Ni²⁺............+..........6NH₃ ------------> [Ni(NH₃)₆]²⁺

Initial (I).............0.001 mol..................0.105 mol..............0.0 mol

Change (C).......-0.001........................-6(0.001)...............+0.001

Final (F)..............0.0 mol.....................0.099 mol...............0.001 mol

Molarity of NH₃ i.e. [NH₃] = 0.099 mol / 0.210 L = 0.471 M and

Molarity of [Ni(NH₃)₆]²⁺ = 0.001 mol / 0.210 L = 0.00476 M

Now the ICE table between Ni²⁺ and NH₃ is :

...........................Ni2+............+..........6NH3 ------------> [Ni(NH₃)₆]²⁺

Initial (I).............0.0 M......................0.471 M..................0.00476 M mol

Change (C).......+X...........................+ 6X..........................- X

Final (F)..............X M.....................(0.471 + 6X)M............(0.00476 - X) M

Now expression of K_f is :

K_f = [Ni(NH₃)₆]²⁺ / [Ni²⁺] [NH₃]⁶

K_f = (0.00476 - X) / (X) (0.471 + 6X)

As X will be very very small therefore neglect X as compare to 0.00476 and 0.471 i.e.

put (0.00476 - X) = 0.00476 and (0.471 + 6X )= 0.471

we have

K_f = 0.00476 / (X) 0.471

5.5 x 10⁸ = 0.00476 / (X) (0.471)

X = 0.00476 / (5.5 x 10⁸) (0.471)

X = 1.84 x 10^-11 M

Hence equilibrium concentration of Ni²⁺ i.e. [Ni]²⁺= X = 1.84 x 10^-11 M