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For this reaction, at equilibrium,

$\ce{[Co(H_2O)_6]^2+ + 4Cl^- <=> [CoCl_4]^2- + 6H_2O}$

$\ce{[Co(H_2O)_6]^2+}$ and $\ce{[CoCl_4]^2-}$ are red and blue in aqueous solution, respectively.

When acetone is then added, the solution changes from red to blue. What causes this?

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I haven't tried this before but it sounds interesting. As the blue chloro-complex is in an equilibrium with the red aqua-complex, I'd suggest the equilibrium is shifted as the water is absorbed by the acetone. As water and acetone are miscible this may take some water from the equilibrium, causing the shift towards the chloro side.

If you use ethanol to dissolve the anhydrous $\ce{CoCl2}$ it will turn blue as well. Yutaka Fukuda mentioned a chloro-ethanol-complex in his book 'Inorganic Chromotropism' for acetone I doubt it will coordinate.

Therefore I'd rather tend to the first one.

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    $\begingroup$ Half of this answer is ‘I think’, ‘I assume’ or irrelevant to the question. I was indifferent of it at first but seeing how it gained so many upvotes I have to counter that. Please attempt to find sources and references to back your assumptions. $\endgroup$ – Jan Oct 14 '17 at 2:09
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    $\begingroup$ I think an answer should also include similar cases. Here two possibilities can arise and I know of only this one book mentioned above that considered a coordination of the solvent as well (for this particular equilibrium) so why not give this as an option for someone having a similar question? $\endgroup$ – Justanotherchemist Oct 14 '17 at 13:28
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    $\begingroup$ Please don’t misunderstand me: I’m all for noting similar or seemingly similar cases once the key case in point has been explained. However, your answer explains the key case with words that suggest you’re just guessing. I could have written a guesswork answer like that but I didn’t because as a scientist I prefer the experimental evidence where I am unsure. $\endgroup$ – Jan Oct 15 '17 at 14:28
  • $\begingroup$ I would rather you consider the problem using concentrations and or activities. I think we need to consider the change in the activity of the water and the chloride in the system. $\endgroup$ – Nuclear Chemist Jun 21 '18 at 14:04
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The great problem is that the ability of a chemical substance to act in some way, is controlled by the activity of the substance and not by the concentration of it. Oftein people consider the concnetration in a first approximation where the activity coefficent of everything is assumed to be one. But for a good prediction or good undertsanding we need the activity coefficents.

for a species X

activity (a) = concentration x activity coefficent (gamma)

We need to understand that as the composition of the solvent changes or the temperture changes the activity coefficents will change.

I think that the answer relates to the relative dielectric constant of the medium, as you replace water with ethanol (or acetone) or increase the temperature then the relative dielectric constant decreases.

By lowering the relative dielectric constant you will make the coordination of the anions to the cations more favourable. I think that this is the reason why a solution of cobalt chloride in ethanol/water is thermochromic. It is pink when cold and blue when hot.

The replacement of the water with a less coordinating solvent will also help to shift the equilibrium in favour of the blue tetrachloro anionic complex.

If you look at the Debye-Huckel equation for the activity coefficient and make sure that you look at what is inside the A term, you will find that when the relative dielectric constant changes that the activity coefficients of ions will change. I think that the change of colour that you see is due to a change in activity coefficient.

The extended Debye-Huckel equation in its full glory is

$$\log (\gamma i) = \left(\frac{-Z_i^2 e^3}{4 \pi \left({\epsilon_0 \epsilon_r k_\mathrm B T}^{1.5}\right)}\right)\frac 1{1 + r_i(8 \pi n e^2/\epsilon_0 \epsilon_r k_\mathrm B T)^{0.5}}(n/2)^{0.5}$$

  • $Z_i$ is the charge on the ion of interest
  • $e$ is the electronic charge
  • $\epsilon_0$ is the vacuum permittivity
  • $\epsilon_r$ is the relative dielectric constant
  • $k_\mathrm B$ is Boltzmann’s constant
  • $T$ is the temperature in kelvin
  • $n$ is the number of ions in one cubic meter.
  • $r_i$ is the sum of the radii of the cation and anion which are in contact.

I have done the calculations for the activity coefficient of chloride in mixtures of water and acetone for a 500 μM solution of cobalt(II) chloride at 298 K in mixtures of water and acetone. As the acetone replaces the water the activity coefficient of the chloride goes down, I think that this will be due to the electrostatic interactions of the chloride anions with the cobalt cations (ion pairing):

enter image description here

I repeated the calculation for the activity coefficients of the cobalt ion and the $\ce{CoCl4^2-}$ ion. Bear in mind that the sum of the radii for $\ce{Co^2+}$ and $\ce{CoCl4^2-}$ will be greater than the sum of the radii for $\ce{Co^2+}$ and $\ce{Cl-}$.

Now if we consider the equation

$$K_\mathrm{eq} = a\frac{\ce{CoCl4^2-}}{a(\ce{Co^2+}){a\ce{(Cl-)^4}}}$$

Now if $K_\mathrm{eq}$ is a constant regardless of the solvent system, then as

$$a(\ce{Cl-}) = [\ce{Cl-}] \times \text{activity coefficient of } \ce{Cl-}$$

$$a(\ce{CoCl4^2-}) = [\ce{CoCl4^2-}] \times \text{activity coefficient of } \ce{CoCl4^2-}$$

$$a(\ce{Co^2+}) = [\ce{Co^2+}] \times \text{activity coefficient of } \ce{Co^2+}$$

Then if we have a new equation which is

$$\text{Random greek symbol} = \text{activity coefficient } \ce{CoCl4^2-} / {\text{activity coefficient } \ce{Co^2+} \times (\text{activity coefficient } \ce{Cl-})^4}$$

If we multiply "random greek symbol" with $K_\mathrm{eq}$' we will get a new equilibrium constant for the system.

($K_\mathrm{eq}$' is the $K$ value determined in water where it is an apparent constant - one calculated making the assumption that all activity coefficients are one)

We can get a new equilibrium constant $K$ (solvent composition) which we can plot against the fraction of the solvent mixture which is acetone. Here we have the graph.

Here K' is the true equilibrium constant which would be for a system in water with a zero concentration of cobalt chloride.

We can see as the acetone content increases that the cobalt cations are more likely to react with the chloride anions to form $\ce{CoCl4}$ dianions than they are in the aqueous system.

enter image description here

If you read the papers by people like Bromley[3]The Journal of Chemical Thermodynamics 1972, 4 (5), 669–673. who clearly expresses the ideas of Guggenheim you will find a specific ion interaction term in the activity coefficient equation (Bromley equation). I am sure that the value of the specific ion interaction constant for cobalt and chloride will increase as the water is replaced with acetone.

Sadly I very much doubt if the constants for water/acetone mixtures exist, most of the research effort has gone into aqueous only systems. You could make a lot of the measurements by doing UV/vis spectroscopy on solutions of chloride salts in acetone/water mixtures. But it will be a complex problem, I know that while sodium chloride is a strong electrolyte in water it is a weak one in acetone. I know a classic reaction in organic chemistry for the formation of alkyl iodides is to treat an alkyl bromide or alkyl chloride with sodium iodide in acetone. The larger anion size makes sodium iodide more soluble (and a stronger electroyte) in acetone than sodium chloride/sodium bromide.

If you look at the supporting information for [1], you will see at the end an example of thermochromism. Here mixtures of water, cobalt chloride and the lactic acid / choline chloride eutectic are shown both hot and cold.

This effect has also been reported with nickel chloride in deep eutectic solvents made from choline chloride by some chemists in China. Similar effects were seen for nickel by [2]. They have observed thermochromism which I think is due to a change in the coordination chemistry which in turn is induced by the temperature induced change in relative dielectric constant.


References:

  1. Foreman, M. R. S. J.; Holgersson, S.; Mcphee, C.; Tyumentsev, M. S. Activity coefficients in deep eutectic solvents: implications for the solvent extraction of metals. New J. Chem. 2018, 42 (3), 2006–2012. DOI: 10.1039/c7nj03736h.
  2. Gu, C.; Tu, J. Thermochromic behavior of chloro-nickel(II) in deep eutectic solvents and their application in thermochromic composite films. RSC Adv. 2011, 1 (7), 1220. DOI: 10.1039/C1RA00345C.

  3. Bromley, L. A. Approximate individual ion values of β (or B) in extended Debye-Hückel theory for uni-univalent aqueous solutions at 298.15 K. The Journal of Chemical Thermodynamics 1972, 4 (5), 669–673. DOI: 10.1016/0021-9614(72)90038-9.

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    $\begingroup$ I also tried it experimentally now. I prepared solutions of cobalt chloride in the solvents, added enough water to discolor them and then added more solvent until it turned blue again. For acetone this shift does not take much solvent and the color is quite intense. For THF however, which has a lower dielectricity constant it takes a lot of solvent and only gives a pale blue. DMF on the other hand has no sudden jump from pink to blue but a huge gradient where it turns purple inbetween and the color is intense, too. I don't really understand what happens with the THF though. $\endgroup$ – Justanotherchemist Jun 26 '18 at 13:52
  • $\begingroup$ THF is a better donor than acetone, I think that in some ways THF is a bit like water. It would be nice if you could post some pictures. Depending on your reputation you could edit my answer to add the pictures. Failing that we will find a way of doing it. $\endgroup$ – Nuclear Chemist Jun 26 '18 at 18:25
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    $\begingroup$ I can try that tomorrow when I'm back at the lab. I know that THF can act as a donor here but it was blue as well when I 'dissolved' the cobalt chloride in pure THF, I only added enough of the salt to fully dissolve it in THF so therefore it should turn blue again at some point. $\endgroup$ – Justanotherchemist Jun 26 '18 at 18:49

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