1.

2H2O ⇄ H3O+ + OH-,KW=1x10-14 at 25°C hence Ka is:

Answer» Molar mass of water = 18 gmol^-1

Mass of 1L water is 1000g

Hence molarity of water

 [H2O]= 1000/18 molL^-1

Now dissociation constant of water

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

= Kw/[H2O]

= 10^-14×18/1000=18×10^-17


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