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`        Percentage ionization of water at a certain temperature is 3.6×10^-7 . Calculate Kw and ph`
one year ago

Sahani Kumar
99 Points
```							Percentage ionization = 3.6 * 10^-7Actual ionization = (3.6 * 10^-7)/(100) = 3.6 * 10^-9                                                                   2(H2O)--------------->OH-  +  H3O+initial number.            1.                            0.             0of 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^-18Now we know that - pH = log[H3O+]-pH= log[3.6 * 10^-9]-pH =( log3.6) +(log 10^-9)-pH = 0.556 - 9pH = 9 - 0.556 = 8.4Regards Askiitian member If you like this answer please encourage us by approving thus answer.
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one year ago
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