Grade 12Physical Chemistry
Percentage ionization of water at a certain temperature is 3.6×10^-7 . Calculate Kw and ph
Percentage ionization of water at a certain temperature is 3.6×10^-7 . Calculate Kw and ph
1 Answer
Sahani Kumar
Percentage ionization = 3.6 * 10^-7
Actual ionization = (3.6 * 10^-7)/(100) = 3.6 * 10^-9
2(H2O)--------------->OH- + H3O+
initial number. 1. 0. 0
of moles.
Number of moles (1 - (3.6 * 10^-9)) (3.6 * 10^-9
at equilibrium: of OH-and H3O+)
So now Kw = [OH-] [H3O+]
= [3.6 * 10^-9] [3.6 * 10^-9]
Kw=12.96 * 10^-18
Now we know that
- pH = log[H3O+]
-pH= log[3.6 * 10^-9]
-pH =( log3.6) +(log 10^-9)
-pH = 0.556 - 9
pH = 9 - 0.556 = 8.4
Regards
Askiitian member
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