Question

For the aqueous [CO(NH₃)₆]³⁺ complex K_f = 4.6 x 10³³ at 25 degrees celcius.

Suppose equal volumes of 0.0038 M Co(NO₃)₃ solution and 0.24 M NH₃ solution are mixed. Calculate the equilibrium molarity of aqueous Co³⁺ ion.

Round your answer to 2 significant digits.

Answer 1

Given,

The concentration of Co(NO₃)₃ solution = 0.0038 M

The concentration of NH₃ solution = 0.24 M

Also given,

K_f of [Co(NH₃)₆]³⁺ = 4.6 x 10³³

Suppose, the volume of each solution is 1 L.since given equal volumes of solutions are mixed.

Calculating the number of moles of Co³⁺ and NH₃ from the given concentrations and volumes.

We know, the formula of molarity,

Molarity = Number of moles / Volume of solution(L)

Rearranging the formula,

Number of moles = Molarity x Volume of solution(L)

For Co(NO₃)₃ Or Co³⁺,

= 0.0038 M x 1 L

= 0.0038 mol Co³⁺

Similarly, for NH₃,

= 0.24 M x 1 L

= 0.24 mol NH₃

Now, the formation reaction of [Co(NH₃)₆]³⁺ is,

Co³⁺(aq) + 6NH₃(aq) [Co(NH₃)₆]³⁺(aq)

Drawing an ICE chart,

	Co3+(aq)	6NH3(aq)	[Co(NH3)6]3+(aq)
I(moles)	0.0038	0.24	0
C(moles)	-0.0038	-6x0.0038	+0.0038
E(moles)	0	0.2172	0.0038

Total volume of solution = 1L +1L = 2L

Now, the new concentrations of NH₃ and [Co(NH₃)₆]³⁺ is,

[NH₃] = 0.2172 mol / 2L = 0.1086 M

[[Co(NH₃)₆]³⁺] = 0.0038 mol / 2L = 0.0019 M

Now, the decomposition equilibrium reaction for [Co(NH₃)₆]³⁺ is,

[Co(NH₃)₆]³⁺(aq) Co³⁺(aq) + 6NH₃(aq)

Drawing an ICE chart,

	[Co(NH3)6]3+(aq)	Co3+(aq)	6NH3(aq)
I(M)	0.0019	0	0.1086
C(M)	-x	+x	+6x
E(M)	0.0019-x	x	0.1086+6x

Now, the K_d expression is,

K_d = [Co³⁺][NH₃]⁶ / [[Co(NH₃)₆]³⁺]

Calculating the K_d value from the given K_f value,

K_d = 1/K_f -------since we have reversed the formation reaction.

K_d = 1/4.6 x 10³³

K_d = 2.17 x 10^-34

Now,

2.17 x 10^-34 = [x][0.1086+6x]⁶ / [0.0019-x]

2.17 x 10^-34 = [x][0.1086]⁶ / [0.0019]-------Here,[0.1086+6x]0.1086 and [0.0019-x]0.0019, since,x<<<0.1086 and 0.0019

x = 2.5 x 10^-31

Now, From the ICE chart,

[Co³⁺] = x = 2.5 x 10^-31 M [2 S.F]