Question

What is the pKa of water and how do you solve for it?

Answer 1

H2O   <------------------> H+ +   OH-

The [H+] and [OH-] = 1 x 10^-7 M in pure water

1 L water = 1000 g water

moles of water = 55.5 moles H2O

concentration = 55.5 moles H2O / 1 L H2O = 55.5 M H2O

Pure water has a molarity of 55.5.

H2O <---> H+ + OH-

Ka = [H+][OH-] / [H2O]

Ka = (1 x 10^-7)^2 / 55.5 = 1.8 x 10^-16

pKa = -log[Ka] = 15.7

pKa = 15.7

in another way of calculate :

H2O(l) -------------> H+ (aq) + OH−(aq)

The Gibbs free energy is given by:

Δ Gf = Gproducts - Greactants

Δ Gf = (GH+ + GOH− ) - GH2O

Δ Gf = ( 0 + -157.2 ) - ( -237.14 )

ΔG o = 79.94 kJ mol−1

  ΔG o = -RT ln(K)

79.94 × 10^3 = - 8.314 × 298.15 × ln(K)

ln(K) = - 32.25

K = 9.863×10^−15

pKa = -log (Ka)

pKa = -log (9.863×10^−15)

pKa = 14.0