A 0.140-mole quantity of CoCl2 is added to a liter of 1.20 M NH3 solution. What is the concentration of Co^2+ ions at equilibrium? Assume the formation constant of Co(NH3)6^2+ is 5.0× 10^31.

The way these complex ion problems are solved is almost like working it backwards but that is what we do. You see such a large K (5E31 is huge) that it makes it almost impossible to solve the complex equations. So we FORCE the equation to go to the right ASSUMING that all of it goes to completion. Then we make it go backwards (because that's what our calculations can do) and solve for the small amount that IT DID NOT GO. The idea here is that it's easier to solve for a small amount than a large amount so we make it work as we can calculate.

.........Co^2+ + 6NH3 ==> Co(NH3)6^2+
I.......0.140M...1.20M......0
change.-0.140...-0.84......+0.140
equil....0......0.36.......0.140
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That's the forcing part. We just acted as if all if it reacted completely because K is so large. Now we make it go backwards and calculate the x part.
I.........0........0.36......0.140
C.........x........6x...........-x
E.........x........6x........0.140-x

K = [Co(NH3)6]^2+/(Co^2+)(NH3)^6
Substitute from the ICE chart (the lower one) and solve for x = (Co^2+).

To find the concentration of Co^2+ ions at equilibrium, we need to determine the equilibrium constant for the formation of Co(NH3)6^2+ complex and use it in the equilibrium expression.

The formation constant (Kf) of Co(NH3)6^2+ is given as 5.0×10^31.

The equilibrium reaction is:
Co^2+ + 6NH3 ⇌ Co(NH3)6^2+

Let's assume "x" to be the concentration of Co^2+ ions that reacts with NH3 to form the complex.

The initial concentration of Co^2+ ions is determined by the moles of the substance and the volume of the solution. In this case, we have 0.140 moles of CoCl2 in a 1 L solution. So the initial concentration of Co^2+ ions is 0.140 M.

The initial concentration of NH3 is 1.20 M.

After the reaction reaches equilibrium, the concentration of Co^2+ ions is (0.140 - x) M, and the concentration of NH3 is (1.20 - 6x) M (since 6 moles of NH3 react with 1 mole of Co^2+ ion to form the complex).

Using the formation constant, we can write the equilibrium expression:
Kf = [Co(NH3)6^2+] / [Co^2+] [NH3]^6

Since we know the initial concentrations and the equilibrium concentrations of the Co^2+ ions and NH3, we can plug these values into the equilibrium expression and solve for "x".

Kf = (0.140 - x) / (x * (1.20 - 6x)^6)

Now, we can solve this equation for "x" using numerical methods such as trial and error or computational software. The value of "x" represents the concentration of Co^2+ ions at equilibrium.

Please note that due to the extremely large magnitude of the formation constant, the concentration of Co^2+ ions will be very low because most of the Co^2+ ions will form the Co(NH3)6^2+ complex.