# Le Chatelier's Principle - Increasing concentration of coloured ion

Take the following equilibrium:

$$\ce{[CoCl4]^2- + 6H2O <=> [Co(H2O)6]^2+ + 4 Cl^-}$$

Where $$\ce{CoCl4^2-}$$ is blue and $$\ce{[Co(H2O)6]^2+}$$ is pink. My understanding is that Le Chatelier's Principle states that the system would act to partially offset any changes. K value is ~ 588, thus heavily favouring products.

Thus, if I add more $$\ce{CoCl4^2-}$$ and thus increase its concentration within the system, the equilibrium would initially turn much more blue, before the partial offset of this change which would lead to a new equilibrium being re-established which, whilst less blue than the solution immediately after the $$\ce{CoCl4^2-}$$ was added, is more blue than the original equilibrium since the system can only partially offset this change.

However, on a few sites on the internet, I have read that the solution will actually turn pink after equilibrium is re-established as the equilibrium is pushed to the right hand side as a result of Le Chatelier's Principle, due to a greater concentration of $$\ce{[Co(H2O)6]^2+}$$. Whilst I understand that $$\ce{[Co(H2O)6]^2+}$$ concentration will be greater at the re-established equilibrium, I believed that its increase in pink colour would not be enough to fully counteract the increase in blue colour as a result of the added $$\ce{CoCl4^2-}$$.

Can someone clarify on whether my existing beliefs are correct or whether the solution does actually turn pink?

Thanks

• It depends on the equilibrium constant. Does it favor products or reactants? Sep 10, 2022 at 3:28
• @BuckThorn I was not aware it made a difference. Could you please elaborate?
– bio
Sep 10, 2022 at 3:30
• Le Chateliers principle simply says that the system tends in the direction that takes it back to a state of equilibrium. It doesn't make much sense to discuss changes to a system if you don't know where this equilibrium lies in the first place. Sep 10, 2022 at 3:32
• @BuckThorn Edited. Pink side heavily favoured according to chemdemos.uoregon.edu/demos/…
– bio
Sep 10, 2022 at 3:36
• You forget that to get the pink, blue is sacrificed. So at that point there is no blue left Sep 10, 2022 at 4:06

If $$\ce{CoCl2·6H2O}$$ is dissolved in water, the solution is pink, because of the pink ion $$\ce{[Co(H2O)6]^2+}$$. Matter of fact, the amount of blue $$\ce{CoCl4^2-}$$ ion produced by the following equilibrium (as proposed in the original post) is negligible $$\ce{CoCl4^2- + 6 H2O <=> [Co(H2O)6]^2+ + 4 Cl-}$$ because $$6$$ $$\ce{H2O}$$ will push the equilibrium to the right-hand side.

If the same substance $$\ce{CoCl2·6H2O}$$ is dissolved in ethanol, the reaction is $$\ce{2 CoCl2·6H2O -> "2 Co^2+ + 4 Cl^- + 6 H2O" -> CoCl4^2- + [Co(H2O)6]^2+}$$ This reaction produces the same amount of both cobaltic ions. But as the blue color of $$\ce{CoCl4^2-}$$ ion is much more intense than the pale pink color of $$\ce{[Co(H2O)6]^2+}$$, the mixture of both ions appears blue. In this system, water is lacking. So the mixture turns deep blue, for entropic reasons : there are more independent species ($$1 + 6 = 7$$) on the left-hand side than on the right-hand side ($$1 + 4 = 5$$).

Now, adding a few drops water to this blue ethanol solution makes it pale pink, because the equilibrium is pushed to the right-hand side by the excess of water.

Now it you heat a bit the obtained pink solution in aqueous ethanol, it turns blue again. This is because heating a system where an equilibrium exists favors the endothermic direction of the reaction, which is the reaction right to left. And as soon as the hot solution is cooled down, the solution turns pink again.

• I'm not exactly sure I follow on the comments about addition of water - I believed that water did not have any effect on LCP when it was a liquid as it did not affect concentration? By adding more water, you don't get more concentrated water, you're just increasing volume
– bio
Sep 10, 2022 at 15:00
• No ! By adding water to the blue ethanol solution, containing practically no water, you change the position of the equilibrium, as written by you. When adding water, the system reacts by consuming a part of this water, according Le Chatelier's principle applied to the equilibrium equation. So the system consumes some $\ce{H2O}$ and also some $\ce{CoCl4^2-}$. And it produces more $\ce{[Co(H2O)6]^2+}$ plus of course some $\ce{Cl-}$ Sep 10, 2022 at 16:36
• @bio If you formulate the equation for the reaction constant, you will realize changing concentration of all by water dilution strongly disbalances the reaction quotient from equilibrium, which has to be re-established. Sep 11, 2022 at 5:08