Ionic product of water at 310 K is `2.7 × 10^(-14)`. What is the pH of neutral water at this temperature?

Explanation

Ionic product of water `=2.7xx10-14`
`H_(2)O rarr underset(x)(H^(o+))+underset(x)(overset(Θ)(O)H`
`[H^(o+)][overset(Θ)(O)H]=2.7xx10^(-14)`
`x^(2)=2.7xx10^(-14)`
`x=sqrt(2.7xx10^(-14))=1.643xx10^(-7)`
`pH=-log[H^(o+)]`
`=-log[1.643xx10^(-7)]=[0.2156-7]`
`=6.78`