What is the Kw of pure water at 51.4ºC if the pH is 6.78? Enter your answer in scientific notation using "e" instead of "×10^" (1.23×10-7 = 1.23e-7) and round to three sig figs.
Kw is called as Ionic product of water
The product of Molar Concentration of H+ ion and Molar concentration of OH- is called as ionic product of water.
Kw = [H+] [OH-]
The given PH = 6.78
according to the definiation of PH
PH = -log[H+]
6.78 = -log[H+]
[H+] = 10^-6.78
[H+] = 1.66 x10^-7
Concentration of H+ = 1.66 x10^-7M
but , In a pure water
Concentration of H+ = concentration of OH-
i,e [H+] = [OH-]
SO, the concentration of OH- = 1.66 x10^-7M
Kw = [H+][OH-]
Kw =[ 1.66 x10^-7 ] [ 1.66 x10^-7]
Kw = 2.76 x10^-14
Kw = 2.76 e^-14
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