# H2O and Le Châtelier’s principle [duplicate]

According to Le Châtelier’s principle, the adding of a reactant causes a shift in the direction of products, to minimize the modification. But what about $$\ce{H2O}$$? When can it be considered as a reactant? Consider that it’s concentration doesn’t appear in the equilibrium constant when the solution is very diluted.

For example, in the following reaction:

$$\ce{CO2 (aq) + H2O (l) <=> H2CO3 (aq)}$$

does the adding of water cause the system to shift in the direction of products?

## marked as duplicate by Mithoron, Todd Minehardt, A.K., Tyberius, airhuffFeb 6 at 0:17

Equilibrium expressions in the thermodinamically strict way should be written with activities, not concentrations:

$$a_i = c_i \times \gamma_i$$

Where $$a_i$$ is activity, $$c_i$$ is the chosen concentration unit (molar fraction for water), and $$\gamma_i$$ is the activity coefficient, a number that measures how much the species i differs from ideal conditions in that particular reaction system. You can think of it as an "effective" concentration. In the ideal case, $$\gamma_i = 1$$ and $$a_i = c_i$$, so generally that is what we use to calculate, concentrations.

In a diluted aqueous solution, all that a water molecule "sees" is more water molecules, with an ocassional solute species passing by. Not only is $$c_{H_2O} \approxeq 1$$ (molar fraction), but also the interaction is practically zero, and each species behaves like it would by itself, so $$\gamma_{H_2O} \approxeq 1$$ as well. $$a_{H_2O} = 1$$, and adding more water only decreases interaction further; we can safely omit $$a_{H_2O}$$ from the equilibrium expression.

However, in a very concentrated solution water is closer to the solute, and there are significant interactions between them (e.g. attraction to anions by hydrogen bridging) so $$\gamma_i \neq 1$$ for all the species. If you add water (or remove it, in a mental experiment) in such a system, the interactions significantly change, and so does $$\gamma_{H_2O}$$ (and $$c_{H_2O}$$). The equilibrium shifts! We make big mistakes in calculations if we omit $$a_{H_2O}$$ from the equilibrium expression.

In the picture below (extracted from here) you can see a plot derived of experimental measures of water activity [$$a_{H_2O}$$] vs. molar fraction of the solute [x] in a perchloric acid solution.

It is up to you to decide when to start considering water in the equilibrium calculations, but normally we work with small concentrations so it's OK to ignore water activity (and the activity coefficients of the other species) altogether. Note that a molar fraction of 0.1 means a solution about 6 M in the acid, whereas a $$\mathbf{10^{-3} M}$$ solution represents a molar fraction of $$\mathbf{2 \times 10^{-5}}$$.

Additionally, as $$\gamma_i$$ for every species depends on the concentration of all the species in the system, the calculations can get really tricky if you switch to activities.