Ka and PKa of Pure water

[SuperOrchestra]

Kw = 10 ^ -14
Kw = KaKb
Ka = 10 ^ -7
Kb = 10^ -7

From the EK 1001 chemistry questions,it states the PKA of pure water is 15.7 and the ka is 1.8 x 10 ^-16

Their explanation is that to get the Ka of pure water you take the Kw and divide it by the concentration of water which is 55.6 mol/L. However, I find that answer not satisfying because its edging on memorization not conceptualiztion, so here is how I interpret that, and please tell me if this is correct....

Ka = [H+][A-] / [HA]

in our situation of H2O, A- = OH- because its H2O minus the acidic proton, and HA = H2O because its the acid that dissociates. From the Kw of water Kw = [OH][H] = 10 ^ -14, so each equal 10 ^-7 in a neutral solution, and the numerator in will equal 10 ^ -14, which you then divide by the concentration of the undissociated acid which is H2O which is 55.6 mol/L

Having said that, I don't fully understand why its not 10^-7, if Kw = KaKb, can someone fill me in on what I understand correctly and incorrectly. Thank you.

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[BerkReviewTeach]

I don't fully understand why its not 10^-7, if Kw = KaKb, can someone fill me in on what I understand correctly and incorrectly. Thank you.

The short answer is that the relationship of Kw = Ka x Kb is for a conjugate pair, NOT the same compound. So, it should be true that:

• Kw = Ka H2O x Kb OH-
  
  OR
  
  Kw = Ka H3O+ x Kb H2O

But that is irrelevant, because Kw is based on autoionization, so you define it as [H3O+][OH-]/[H2O]. The reality is that some things you just need to know, such as the definition of Kw.