Chemistry Why is the activity of water as a solvent very nearly 1?

[zenterix]

Homework Statement

Consider the proton transfer reaction between water molecules

Relevant Equations

$$\mathrm{H_2O(l)+H_2O(l)\leftrightharpoons H_3O^+(aq)+OH^-(aq)}$$

The equilibrium constant is

$$K=\frac{a_{\mathrm{H_3O^+}}a_{\mathrm{OH^-}}}{(a_{\mathrm{H_2O}})^2}$$

where $a_J$ denotes the activity of a solute $\text{J}$ in a dilute solution, $\mathrm{[J]}/c^\circ$ with $c^\circ=1\mathrm{mol\cdot L^{-1}}$.

The book I am following says that the solvent, water, is very nearly pure, "and so its activity may be taken to be 1".

Where does this 1 come from?

It certainly is not $$\mathrm{[H_2O]/c^\circ}$$.

 

[zenterix]

Okay, so it seems that "activity" is not as simple as it was presented to me so far.

Apparently, if a substance $J$ is a pure solid or liquid then whatever concentration it has is the "standard concentration" that you divide by when you compute activity. Thus, pure water as a solvent in a dilute solution has approximately this standard concentration and so when we calculation activity we have the same concentration in the numerator and denominator.

I don't know much more that this very vague explanation but I can see that there is a relatively simple reason for the 1.

 

[Borek]

There are probably historical reasons why we ended with 1, but to some extent it is an arbitrary number. We could choose any other number for the activity in the standard state, as activity in every other state is in a relation to that basic one. So if you choose 1 you just make the reference easier to remember. Sure, choosing any other number would require rescaling all values of experimentally determined thermodynamic data - but if you were to use these new numbers in practical calculations changes would cancel out and final results would be exactly the same they are now.

So let's stick with 1

 

[DrDu]

The point is, that if all components are in their standard states, $\Delta G=\Delta G^o$ by definition.
As $\Delta G=\Delta G^o+\sum_i \nu_i RT \ln a_i$, the $a_i$ have to be 0 in the standard state as $\ln 1=0$. For solvents, the standard state is the pure solvent, for solutes, the solute at infinite dilution extrapolated to a concentration of 1 mol/L (or some molality condition). For gasses, a pressure of 1 atm, strictly speaking also extrapolated from infinitely low pressures.