Question

One millimole of Ni(NO3)2 dissolves in 210mL of a solution that is .500M in NH3.

(a) What is the initial concentration of Ni(NO3)2 in the solution?

------> The answer to (a) is .00476 M. I just can't figure out part (b)

(b) What is the equilibrium concentration of Ni^2+(aq) in the solution?

HINT: The Ni^2+ forms a hexammonia complex: Ni(NH3)6^2+. Kf = 5.5 x 10^8.

Answer 1

Given,

millimoles of Ni(NO₃)₂ = 1

Volume of solution = 210 mL x ( 1L /1000 mL) = 0.210 L

Concentration of NH₃ solution = 0.500 M

K_f of [Ni(NH₃)₆]²⁺ = 5.5 x 10⁸

a) calculating the initial concentration of Ni(NO₃)₂ in the solution,

= 1 millimole Ni(NO₃)₂ x ( 1 mole / 1000 millimole)

= 0.001 mol Ni(NO₃)₂

Now, we know the formula to calculate molarity,

Molarity = Number of moles / L of solution

Molarity = 0.001 mol / 0.210 L

Molarity = 0.00476 M

b) Given,

Moles of Ni(NO₃)₂ Or Ni²⁺ = 0.001 mol

Calculating the number of moles of NH₃ from the given volume of solution and concentration,

= 0.500 M x 0.210 L

= 0.105 mol NH₃

Now, the reaction between Ni²⁺ and NH₃ is,

Ni²⁺(aq) + 6NH₃(aq) [Ni(NH₃)₆]²⁺(aq)

Drawing an ICE chart,

	Ni2+(aq)	6NH3(aq)	[Ni(NH3)6]2+(aq)
I(moles)	0.001	0.105	0
C(moles)	-0.001	-6x0.001	+0.001
E(moles)	0	0.099	0.001

Now, the new concentrations of NH₃ and [Ni(NH₃)₆]²⁺ are,

[NH₃] = 0.099 mol / 0.210 L = 0.4714 M

[Ni(NH₃)₆]²⁺] = 0.001 mol / 0.210 L = 0.004762 M

Now, the decomposition reaction of [Ni(NH₃)₆]²⁺]

Ni(NH₃)₆]²⁺(aq) Ni²⁺(aq) + 6NH₃(aq)

Drawing an ICE chart,

	Ni(NH3)6]2+(aq)	Ni2+(aq)	6NH3(aq)
I(M)	0.004762	0	0.4714
C(M)	-x	+x	+6x
E(M)	0.004762-x	x	0.4714+6x

Now, the K_d expression is,

K_d = [Ni²⁺][NH₃]⁶ / [Ni(NH₃)₆]²⁺]

Now, calculating the K_d value from the given K_f value,

K_d = 1/ K_f

K_d = 1/ 5.5 x 10⁸

K_d = 1.818 x 10^-9

1.818 x 10^-9 = [x][0.4714+6x]⁶ / [0.004762-x]

1.818 x 10^-9 = [x][0.4714]⁶ / [0.004762] ------Here, [0.4714+6x]0.4714 and [0.004762-x]0.004762, since, x <<0.471 and 0.004762

x = 7.89 x 10^-10

From the ICE table,

x = [Ni²⁺]_eq = 7.89 x 10^-10 M