A. You mix a 125.5 mL sample of a solution that is 0.0111 M in NiCl2 with a 183.

Question

A. You mix a 125.5 mL sample of a solution that is 0.0111 M in NiCl2 with a 183.0 mL sample of a solution that is 0.225 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.

B. A 120.0 mL sample of a solution that is 2.7×103 M in AgNO3 is mixed with a 230.0 mL sample of a solution that is 0.10 M in NaCN.

After the solution reaches equilibrium, what concentration of Ag+(aq) remains?

Express your answer using two significant figures.

Explanation / Answer

Answer for question A:

Total volume of the solution = 125.5 + 183.0 = 308.5 mL

The initial concentrations of ions in solution are as follows:

[Ni²?]? = 0.0111 × (0.1255/0.3085) = 0.00451 M

[NH?]? = 0.225 × (0.2000/0.3085) = 0.14587 M

The Ni2+ ion forms a complex with ammonia as follows :

Ni²?(aq) + 6NH?(aq) ? Ni(NH?)?²?(aq)

And whose complex Formation constant, Kf = 2.0 × 10?

According to balanced equation 1 mol of Ni²? reacts with 6 moles of NH?.

Therefore

[Ni²?]? × 6 = (0.00451 M) × 6 = 0.02706 M =

Since NH? is in large excess and Kf is very large, the equilibrium is almost completely to the right.

In other words, Ni²? is almost completely reacted.

Hence decrease in [Ni²?]o = 0.00451 M

At equilibrium :

[NH?] = 0.290 – (0.00451 × 6)

[NH?] = 0.26294 M

[Ni(NH?)?²?] = 0.00451 M

Kf = [Ni(NH?)?²?] / ([Ni²?] [NH?]?)

0.00451 / ([Ni²?] × 0.26294?) = 2.0 × 10?

[Ni²?] = 0.00451 / (0.26294? × 2.0 × 10?)

[Ni²?] = 0.00451 / (0.000330 x 2.0 × 10?)

[Ni²?] = 0.00451 / 66000

[Ni²?] = 6.833 × 10?? M

At equilibrium, the concentration of Ni2+(aq) remaining = 6.833 × 10?? M

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Answer for question B:

Total volume of the solution = 120 + 230 = 350 mL

The initial concentrations of ions in solution are as follows:

[Ag?]? = 0.0027 × (0.120/0.350) = 0.000926 M

[CN-]? = 0.100 × (0.230/0.350) = 0.065714 M

The Ag+ ion forms a complex with CN-

Ag+ + 2 CN- <========> Ag(CN)2-

And whose complex Formation constant, Kf = 1 x 1021

According to balanced equation 1 mol of Ag? reacts with 2 moles of CN-.

Since CN- is in large excess and Kf is very large, the equilibrium is almost completely to the right.

In other words, Ag? is almost completely reacted.

Hence decrease in [Ag?]o = 0.000926 M

At equilibrium :

[CN-] = 0.065714 – (0.000926 × 2)

[CN-] = 0.063862 M

Ag(CN)2- = 0.000926 M

Kf = [Ag(CN)2-]/ [Ag?][CN-]2

1 x 1021 = 0.000926 / [Ag?][0.063862]2

[Ag?] = 0.000926 / 1 x 1021 [0.063862]2

[Ag?] = 0.000926 / 0.004078 x 1021

[Ag?] = 2.27 x 10-22 M

After the solution reaches equilibrium, the concentration of Ag+(aq) remaining = 2.27 x 10-22 M

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