I have the following equilibrium given:

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

(I imagine the (l) for water is wrong, as the water is a part of the solution)

The $\ce{[Co(H2O)6]^2+}$ complex is pink, and the $\ce{[CoCl4]^2-}$ complex is blue.

I had to more or less answer the following question: "Would adding water shift the equilibrium?"

How would you describe the shift in the equilibrium by adding more water?

As I understand it, Le chatelier's principle would certainly shift the equilibrium towards causing a pink coloration. Two chemistry teachers I have spoken to have disregarded this and tell me that I am wrong, whilst a higher-level chemistry teacher, and multiple videos demonstrating the exercise agrees with it "causing a shift". Since a "dilution" often does not shift an equilibrium, they claim that only a little water could cause a shift. I do, however, think that even at a microscale, the water molecules would still cause a shift, even if we used an absurdly large amount of water. Realistically, you would use an amount equivalent to a beaker for example, which is used in the video below.

For those curious about empirical evidence: Cobalt Chloride Equilibrium on Youtube video, RamZland channel

This question is grade deciding, as I got a split between two grades, so it would be nice to hear some professional input. The question was specifically "What will happen to the equilibrium and what will we observe when we add water?" I specifically answered that it would shift towards the side that caused a pink coloration, as I think the equilibrium will always change at a microscale, although be it less at larger volumes of water. Using an infinite amount of water is not a realistic consideration, just like asking if $\pu{0.000001 g}$ of chloride would. Keep also in mind I got zero points for this answer, and that the solution just says the equilibrium does not shift at all. No water or solution concentration is given in the question.

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    – andselisk
    May 3, 2023 at 19:52

1 Answer 1


I have the following equilibrium given: $$\ce{[Co(H2O)6]^2+(aq) + 4 Cl-(aq) <=> [CoCl4]^2-(aq) + 6 H2O(l)}$$

This above equation may rather confuse you, for your purpose is better to write it just as:

$$\ce{Co^2+ + 4 Cl- <=> CoCl4^2-} \tag{1}$$

with the equilibrium constant:

$$K = \frac{a_{\ce{CoCl4^2-}}}{a_{\ce{Co^2+}} \cdot \left(a_{\ce{Cl-}}\right)^4} \tag{2}$$

Activities of ions in concentrated solutions ( as implied from scenario) is very difficult to estimate or to obtain as experimental data. I will therefore use for this educational illustrative purpose concentrations instead of activities, being fully aware of significant errors:

$$K_c = \frac{[\ce{CoCl4^2-}]}{[\ce{Co^2+}] \cdot [\ce{Cl-}]^4}\label{3}\tag{3}$$

Dilution of the solution by water has similar effect as decreasing pressure ( and therefore partial component pressures) for this synthesis:

$$\ce{3 H2 + N2 <=> 2 NH3}\tag{4}$$

with the equilibrium constant

$$K_p = \frac{p^2_{\ce{NH3}}}{p_{\ce{N2}} \cdot p^3_{\ce{H2}}}\label{5}\tag{5}$$

In both cases, $\eqref{3}$ and $\eqref{5}$, dilution, respectively pressure decrease, causes increase of the value of the respective fraction, as denominators decrease in both cases faster then numerators.

From the equation below, dilution to twice as much volume leads to 16 times lower ratio of concentrations of chloro and aqua complexes.

$$\frac{[\ce{CoCl4^2-}]}{[\ce{Co^2+}]}\label{6}\tag{6} = K_c \cdot [\ce{Cl-}]^4$$

Therefore, dilution causes $\ce{CoCl4^2-}$ to gradually converted back to $\ce{Co^2+}$. In reality, the effect is even stronger, as activity coefficients start to steeply raise for concentrated solutions.


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