1.

K_(w) of H_(2)O at 373 K is 1 xx 10^(-12). Identify which of the following is/are correct

Answer»

`pK_(w)` of `H_(2)O` is 12
pH of `H_(2)O` is 6
`H_(2)O` is neutral
`H_(2)O` is acidic

Solution :`pK_(w) = -logK_(w) - log 1 xx 10^(-12) = 12`
`RARR K_(w) = [H^(+)][OH^(-)] = 10^(-12)`
`[H^(+)] = [OH^(-)] rArr`
`[H^(+)]^(2) = 10^(-12), [H^(+)] = 10^(-6), pH = -log [H^(+)] = -log 10^(-6) = 6`
`H_(2)O` is neutral because `[H^(+)] = [OH^(-)]` at 373 K even when pH = 6
(d) Is not correct at 373 K. Water cannot become acidic.


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