For Practice 18.15 You mix a 129.0 mL sample of a solution that is 0.0129 M in N

Question

For Practice 18.15

You mix a 129.0 mL sample of a solution that is 0.0129 M in NiCl2 with a 181.5 mL sample of a solution that is 0.228 M in NH3.

Part A

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.

For Practice 18.15

You mix a 129.0 mL sample of a solution that is 0.0129 M in NiCl2 with a 181.5 mL sample of a solution that is 0.228 M in NH3.

Part A

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.

Explanation / Answer

Volume of the solution = 129.0 + 181.5 = 310.5 mL

Consider the dilute solution of solution just before reaction occurs :
[Ni²] = 0.0129 × (129.0/310.5.0) = 0.00535 M
[NH] = 0.228 × (181.5/310.5) = 0.133 M

Consider the formation of complex :
Ni²(aq) + 6NH(aq) Ni(NH)²(aq) ... Kf = 2.0 × 10

1 mol of Ni² reacts with 6 moles of NH.
[Ni²] × 6 = (0.00535 M) × 6 = 0.0321 M < 0.133 M = [NH]
NH 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] = 0.00535 M

At equilibrium :
[NH] = 0.133 - 0.00535 × 6 = 0.101 M
[Ni(NH)²] = 0.00535 M

Kf = [Ni(NH)²] / ([Ni²] [NH])
0.00535 / ([Ni²] × 0.101) = 2.0 × 10
[Ni²] = 0.00535 / (0.101 × 2.0 × 10) = 2.519 × 105 M

The concentration of [Ni²] remains at equilibrium is 2.519 × 105 M

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