Question

A 0.130-mole quantity of CoCl² is added to a liter of 1.20 M NH₃ solution. What is the concentration of Co² ions at equilibrium? Assume the formation constant* of Co(NH₃)₆²⁺ is 5.0 × 10³¹ M^–6.

The balanced equation is:

Co²⁺ + 6NH₃ <==> Co(NH₃)₆²⁺

The ICE table is:

I:    0.13,      1.2,    0

C:      -x,       -6x,   +x

E: 0.13-x, 1.2-6x,    x

k=[Co(NH₃)₆²⁺]/[(Co²⁺)(NH₃)⁶]

5.0x10³¹=(x)/[(0.13-x)(1.2-6x)⁶]

Answer 1

SInce the amount of Co²⁺ in this mixture is much lesser than the 6 times the amount of NH₃ present and the formation constant of Co(NH₃)₆²⁺ is very high all of the Co is in complex form which is 0.13 mole so the equilibrium needs to be

Amount of NH₃ consumed is 0.13 x 6 = 0.78 moles

Amount of free NH₃ remaining = 1.2 - 0.78 = 0.42 moles

Co(NH₃)₆²⁺ Co²⁺ + 6NH₃ to bring some Co²⁺ into solution

I: 0.13, 0 0.42

C: -x, +x +6x

E: 0.13-x, x 0.42+6x,

k=1/5.0x10³¹ = [Co²⁺][NH₃]_⁶/[Co(NH₃)₆²⁺]

2 x 10^-32 = x*(0.42 +6x)⁶/(0.13-x) the x will be very small in this case and can be ignore in the summations and differences

2 x 10^-32 = x*(0.42)⁶/(0.13)

2 x 10^-32 = x*0.0054/(0.13)

2.6 x 10^-33 = x*0.0054

x = 2.6 x 10^-33 /0.0054

x = 4.81 x 10^-31

The concentration of Co²⁺ is 4.81 x 10^-31M