Question

You mix a 120.5 mL sample of a solution that is 0.0123 M in NiCl2 with a 183.0 mL sample of a solution that is 0.270 M in NH3. After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? (Formation constant is Kf=2.0×108.)

Answer 1

Volume of the solution = 120.5 + 183.0 = 303.5 mL

Consider the dilute solution of solution just before reaction occurs :
[Ni²⁺]ₒ = 0.0123 × (120.5/303.5) = 0.00488 M
[NH₃]ₒ = 0.270 × (183.0/303.5) = 0.162 M

Consider the formation of complex :
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 = (0.00488 M) × 6 = 0.0292 M < 0.162 M = [NH₃]ₒ
NH₃ is in large excess.

As Kf is very large, the equilibrium is almost completely to the right.
In other words, Ni²⁺ is almost completely reacted.
Decrease in [Ni²⁺] = 0.00488 M

At equilibrium :
[NH₃] = 0.162 - (0.00488 × 6) = 0.133 M
[Ni(NH₃)₆²⁺] = 0.00488 M

Kf = [Ni(NH₃)₆²⁺] / ([Ni²⁺] [NH₃]⁶)
0.00488 / ([Ni²⁺] × 0.133⁶) = 2.0 × 10⁸
[Ni²⁺] = 4.40 × 10⁻6 M