Question

In an excess of NH3(aq), Cu2+ ion forms a deep blue complex ion, Cu(NH3)42+, which has a formation constant Kf=5.6×1011. Calculate the concentration of Cu2+ in a solution prepared by adding 4.6×10−3mol of CuSO4 to 0.550 L of 0.37 M NH3.

Express your answer using two significant figures.

Answer 1

Given,

Moles of CuSO₄ or Cu²⁺ = 4.6 x 10^-3 mol

Volume of NH₃ solution = 0.550 L

Concentration of NH₃ solution = 0.37 M

Also given,

K_f of [Cu(NH₃)₄]²⁺ = 5.6 x 10¹¹

Firstly calculating the number of moles of NH₃ from the given concentration and volume,

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)

Substituting the known values,

Number of moles = 0.37 M x 0.550 L

Number of moles = 0.2035 mol NH₃

Now, the reaction between Cu²⁺ and NH₃ is,

Cu²⁺(aq) + 4NH₃(aq) [Cu(NH₃)₄]²⁺(aq)

Drawing an ICE chart,

	Cu2+(aq)	4NH3(aq)	[Cu(NH3)4]2+(aq)
I(moles)	0.0046	0.2035	0
C(moles)	-0.0046	-4x0.0046	+0.0046
E(moles)	0	0.1851	0.0046

Now, the new concentrations of NH₃ and [Cu(NH₃)₄]²⁺ are,

[NH₃] = 0.1851 mol / 0.550 L = 0.3365 M

[[Cu(NH₃)₄]²⁺] = 0.0046 mol / 0.550 L = 0.008364 M

Now, the dissociation equilibrium reaction for [Cu(NH₃)₄]²⁺ is,

[Cu(NH₃)₄]²⁺(aq) Cu²⁺(aq) + 4NH₃(aq)

Drawing an ICE chart,

	[Cu(NH3)4]2+(aq)	Cu2+(aq)	4NH3(aq)
I(M)	0.008364	0	0.3365
C(M)	-x	+x	+4x
E(M)	0.008364-x	x	0.3365+4x

Now, the K_d expression is,

K_d = [Cu²⁺] [NH₃]⁴ / [[Cu(NH₃)₄]²⁺]

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 = 1/5.6 x 10¹¹

K_d = 1.786 x 10^-12

Now,

1.786 x 10^-12= [x] [0.3365+4x]⁴ / [0.008364-x]

1.786 x 10^-12= [x] [0.3365]⁴ / [0.008364] --------Here, [0.3365+4x]0.3365 and [0.008364-x]0.008364, since x <<<< 0.3365,0.008364

x = 1.2 x 10^-12

Thus, from the ICE chart,

[Cu²⁺] = x = 1.2 x 10^-12 M [2S.F]