1.

Which one is correc for H_(2)O at 25^(0)C

Answer»

Ionic PRODUCT of water `K_(w) = 10^(-14)`
Equilibrium constant for dissociation of water `(Kc = 1.8 xx 10^(-16))`
Autoprotolysis constant of water `(k = 3.2 xx 10^(-) xx 10^(18))`
`K_(w)` increases with RISE in temperature

Solution :`H_(2)O hArr H^(+) + OH^(-)`
`K_(C) = ([H^(+)][OH^(-)])/([H_(2)O]) = (10^(-7) xx 10^(-7))/(55.55) = 1.8 xx 10^(-16)`


Discussion

No Comment Found