[lokifenrir96]

Hi, I've read in other posts that Ka of water is 1.8x10^-16 mol dm^-3. The explanation given was:

H2O -> H+ + OH-

Ka = [H+][OH-]/[H2O]
Since [H2O] is a constant of 55 mol dm^-3, and [H+][OH-] is equivalent to Kw, which is 10^-14, then Ka = 1.8x10^-16 mol dm^-3 at 25 degrees celcius.

But I thought the equation should really be H2O + H2O -> H3O+ + OH-, so shouldn't Ka = [H+][OH-]/[H2O]^2? So shouldn't Ka be 3.31 x 10^-18? Also, why is the concentration of water even included? I thought it's usually excluded since it's a constant? In which case, shouldn't Ka of water = Kw = [H=][OH-] = 10^-14?

[ramboacid]

Equilibrium constants depend more on the activities of the reactants than the concentrations, but the concentration of an aqueous solute is its activity so we think of substituting concentrations into the equilibrium calculations. Remember that water is in the liquid state, not aqueous, and so for our purposes it's activity is 1. Because of the activities of solids and liquids is taken to be 1, we can safely ignore them in routine calculations.

The real reaction is H₂O + H₂O -> H₃O⁺ + OH^- as you have said, and so the equilibrium constant is supposed to be Ka = [H⁺][OH^-]/[H₂O]². But when the activity of water is 1, we can safely ignore it.

The [H⁺] and [OH^-] concentrations in water have been determined by experiment, according to Chemical Principles 4th Ed. by Atkins and Jones.

[lokifenrir96]

Ah so if we ignore the [H2O]^2, then should the Ka of water just be [H+][OH-] which is equivalent to the Kw of water? Sorry I have no idea what the 'activity' refers to >.<

[prankster1590]

H₂O + H₂O -> H₃O⁺ + OH^-

K_eq = [H₃O⁺][OH^-]/[H₂O]²  K_w = K_eq[H₂O]² = [H₃O⁺][OH^-] = 10^-14

K_w/[H₂O] = k_a = K_eq[H₂O] = 10^-14/[H₂O]= 10^-14/[55,51] = 1,8*10^-16 = 10^-15,74

[Aldebaran]

The link here: https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/What_is_the_pKa_of_water
gives an extensive explanation on this topic.