The ionization constnat for water is `2.9xx10^(-14)` at `40^(@)C`. Calculate `[H_(3)O^(+)],[OH],pH` and `pOH` for pure water at `40^(@)C`.

Explanation

`K_(w)=[H^(-)][OH^(-)]`
`2.9xx10^(-14)=[H^(+)][OH^(-)]`
For pure water `[H^(+)]=[OH^(-)]`
`:.[H^(+)]^(2)=2.9xx10^(-14)`
`[H^(+)]=1.4xx10^(-7)=[H_(3)O^(+)]`
`:. [OH^(-)]=1.7xx10^(-7)`
`pH=-log[H^(+)]`
`=-log[1.7xx10^(-7)]`
`=7-log1.7=6.7689`
`pOH=-log[OH^(-)]`
`=-log[1.7xx10^(-7)]`
`=7-log1.7=6.7689`