Question

You mix a 120.0 mL sample of a solution that is 0.0130 M in NiCl2 with a 178.5 mL sample of a solution that is 0.223 M in NH3. After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? (Formation constant is Kf=2.0×108.)

Answer 1

Solution-

Volume of the solution = 120.0 + 178.5 = 298.5 mL

Now let us consider the dilute solution of solution just before reaction occurs :
[Ni²⁺]ₒ = 0.0103 × (120.0/298.5) = 0.0041 M
[NH₃]ₒ = 0.223 × (178.5/298.5) = 0.133 M

Now consider complex formation:
Ni²⁺(aq) + 6NH₃(aq) ⇌ Ni(NH₃)₆²⁺(aq)[ Kf = 2.0 × 10⁸ ]

1 mol of Ni²⁺ reacts with 6 moles of NH₃.
[Ni²⁺]ₒ × 6 = (0.0041 M) × 6 = 0.0246 M < 0.133 M = [NH₃]ₒ
So, NH₃ is in large excess.
Here, Ni²⁺ is almost completely reacted.
Decrease in [Ni²⁺] = 0.0041 M

At equilibrium :
[NH₃] = 0.133 - 0.0041 × 6 = 0.108 M
[Ni(NH₃)₆²⁺] = 0.0041 M

Kf = [Ni(NH₃)₆²⁺] / ([Ni²⁺] [NH₃]⁶)
0.0041 / ([Ni²⁺] × 0.108⁶) = 2.0 × 10⁸
[Ni²⁺] = 0.0041 / (0.108⁶ × 2.0 × 10⁸) = 0.0000129 M