Ionic product of water at `310 K` is `2.7xx10^(-14)`. What is the `pH` of netural water at this temperature?

Answer

`pH = 6.78`

Explanation

Ionic product, `K_(w)=[H^(+)][OH^(-)]`
Let `[H^(+)]=x` ltBrgt Since `[H^(+)]=[OH^(-)], K_(w)=x^(2)`
`rArrK_(w)"at310K is "2.7xx10^(-14)`
`therefore 2.7xx10^(-14)=x^(2)`
`rArrx=1.64xx10^(-7)`
`rArr[H^(+)]=1.64xx10^(-7)`
`rArrpH=-log[H^(+)]`
`=-log[1.64xx10^(-7)]` = 6.78
Hence, the pH of neutral water is 6.78.