`K_(w)` of `H_(2)O` at `373` K is `1xx10^(-12)` Identify which of the following is/are correct?

Options

A. `pH+pOH=12` for every aqueous solutions.
B. `pH` of `H_(2)O` is `6`.
C. `alpha_(H_(2)O)` has increased from its value at `298`K
D. `H_(2)O` is acidic.

Explanation

`pK_(w)=-logK_(w)=-log1xx10^(-12)=12`
`K_(w)=[H^(+)][OH^(-)]=10^(-12)`
`[H^(+)]=[OH^(-)]`
`rArr [H^(+)]^(2)=10^(-12),[H^(+)]=10^(-6),pH=-log[H^(+)]=-log10^(-6)=6`
`H_(2)O` is neutral because `[H^(+)]=[OH^(-)]` at `373` K even when `pH=6`
`(D)` is not correct at `373K` .Water cannot become acidic.