Question

A 0.160-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)6^2 is 5.5 × 10^8

Answer 1

Initial Moles of NH₃ = 1.20 mol/L x 1 L = 1.20 Moles

Initial Moles of NiCl2 = 0.160 Moles

The problem reaction can be restated as : Ni²⁺ + 6NH₃ --> [Ni(NH₃)₆]²⁺

Since the formation Constant is very large( of te order 10⁸), it can be assumed that the entire Ni²⁺ forms the complex, without any residual Ni²⁺ in the product. Thus , it can be asserted that the amount of [Ni(NH₃)₆]²⁺ formed is equal to the Initial amount of NiCl2,i.e. 0.160 moles.

By stoichiometry, the amount of NH₃ unconsumed = 1.20 mol - (6x 0.160) = 1.20 - (0.96) = 0.24 moles

Since volume of the solution is constant at 1 L , Concentration will be numerically equal to the number of moles.

K_f =  [ [Ni(NH₃)₆]²⁺ ]/ [ Ni²⁺][NH₃]⁶

Using values from above,

5.5x10⁸= 0.160/ [ Ni²⁺](0.24⁶)

Solving the Above, we get :

[ Ni²⁺] = [ 5.5x10⁸ x (0.24⁶) / 0.160 ] ^-1

[ Ni²⁺] = 1.522263 x 10^-6 M at Equillibrium.