Question

A 0.140-mole quantity of NiCl2 is added to a liter of 1.20 M NH3 solution. What is the concentration of Ni2 ions at equilibrium? Assume the formation constant* of Ni(NH3)62 is 5.5 × 108.

please show all work and answer

Answer 1

Answer – Given, moles of NiCl₂ = 0.140 moles, [NH₃] = 1.20 M , volume = 1.0 L

Formation constant, Kf = 5.5*10⁸

We know, molarity = moles /L

So, [NiCl_­2] = 0.140 moles / 1 L = 0.140 M

We know the formation reaction of [Ni(NH₃)₆]²⁺

   NiCl₂ +   6 NH₃ ----> [Ni(NH₃)₆]²⁺ + 2Cl^-

I 0.140        1.20             0               0

C -x           -6x              +x            +2x

E 0.140-x 1.20-6x        +x            +2x

So,

K = [Ni(NH₃)₆]²⁺ [Cl^-]² / [NiCl₂] [NH₃]⁶

5.5*10⁸ = (x)*(2x)² / (0.140-x) (1.20-6x)⁶

5.5*10⁸[(0.140-x) (1.20-6x)⁶] = 4x³

By solving this one x = 0.1399

Means there is reaction gets completed

And the concentration of Ni²⁺ ions at equilibrium = 0.140 -x

                                                                                 = 0.140 – 0.1399

                                                                                = 1.0*10^-4 M

So, all the Ni²⁺ ions converted to its complex at the equilibrium