Question

You mix a 116.5 mL sample of a solution that is 0.0127 M in NiCl2 with a 174.5 mL sample of a solution that is 0.220 M in NH3. After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? (Formation constant is Kf=2.0×108.) Express your answer with the appropriate units.

Answer 1

The reaction will be

Ni⁺² + 6NH₃ --> [Ni(NH₃)₆]⁺²

Kf =   [Ni(NH₃)₆]⁺² / [Ni⁺² ][NH₃]^6 = 2.0×10⁸

Initial moles of ni+2 = Molarity of NiCl2 X volume = 0.0127 X 16.5 = 1.48 millimoles

Initial moles of NH3 = 174.5 x 0.22 = 38.39 millimoles

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

Initial               1.48            38.39                     0

Change             -x              -6x                        x

Equilibrium      1.48-x         38.39-6x                 x

however we can assume here that as the value of Kf is very high the reaction almost goes to completion

as per stoichiometry , 1 moles of Ni+2 will react with 6 moles of NH3 to give 1 mole of complex

so for 1.48 millimoles we will need = 6X1.48 moles of NH3 = 8.88 millimoles

so complex formed = 1.48 millmoles

[NH3] = 38.39- 8.88 / Total volume = 29.51 millimoles / 291 mL = 0.101 M

[Complex] = 8.88 / 291 = 0.031

Kf =   [Ni(NH₃)₆]⁺² / [Ni⁺² ][NH₃]^6 = 2.0×10⁸

2.0×10⁸ = 0.031 / [Ni⁺² ][0.101]^6

[Ni⁺² ] = 29203.4 / 2.0×10⁸ = 14601.7 X 10^-8 = 1.46 X 10^-5 M