Question

What is the activity of pure water?

Answer 1

I assume you mean the activity `a` of liquid water... which is just the "real life" version of concentration.

In general, we define activity as:

`color(blue)(a_A = chi_Agamma_A) = (chi_Agamma_AP_A^@)/(P_A^@) = color(blue)(P_A/P_A^@)`,

where:

• `a_A` is the activity of substance `A`.
• `gamma_A` is the activity coefficient of substance `A`.
• `chi_A = n_A/(sum_(i=1)^(N) n_i)` is the mol fraction of substance `A` and `n_i` is the mols of substance `i`.
• `P_A` is the partial vapor pressure of substance `A`.
• `P_A^@` is the vapor pressure of pure `A` under the same conditions.
• `P_A = chi_Agamma_AP_A^@` is the real-life version of Raoult's law (i.e. for nonideal solutions).

You can find a more tailored definition here, but we can cover this in general using water as an example.

Let's say we had a pure water solution of `"pH"` `7` at `25^@ "C"` and `"1 atm"`, with concentrations `["H"^(+)] = 10^(-7) "M"` and `["OH"^(-)] = 10^(-7) "M"`. In a `"1 L"` solution, we thus have:

`n_(H^(+)) = 10^(-7) "mols"`
`n_(OH^(-)) = 10^(-7) "mols"`
`n_(H_2O) = cancel"1 L" xx (997.0749 cancel"g")/cancel"L" xx "1 mol water"/(18.015 cancel"g") = "55.34 mols"`

As a result, the mol fraction of water in water is:

`chi_(H_2O) = n_(H_2O)/(n_(H^(+)) + n_(OH^(-)) + n_(H_2O))`

`= "55.34 mols"/(10^(-7) "mols" + 10^(-7) "mols" + "55.34 mols")`

`= 0.9999999964 cdots ~~ 1`

It is known that as `chi_A -> 1`, `gamma_A -> 1`. Since `chi_A ~~ 1`, it follows that `a_A ~~ 1`, and the activity of water in water is `bb1`.

Another way to recognize this is to realize that, and this is a redundant description, but water by itself can be treated as "water in water", so:

`P_A/P_A^@ = 1`

since the vapor pressure of water in this pure water "solution", `P_A`, is (effectively) not reduced by any solutes relative to the vapor pressure of pure water, `P_A^@`, both at the same temperature and pressure.