What is the activity of pure water?

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.