The value of `K_(w)` is `9.55xx10^(-14)` at certain temperature. Calculate the pH of water at this temperature.

Explanation

Since `K_(W)=[H^(+)][OH^(-)]`
`K_(W)=9.5xx10^(-14) "mol"//"lit"^(2)`
`[OH^(+)][OH^(-)]=9.88xx10^(-14) "mol"//"lit"^(2)`
`:. [H^(+)]= sqrt(9.55xx10^(-14))`
`=3.09xx10^(-7)` mol/lit
`pH=-log[H^(+)]`
`=-log[3.09xx10^(-7)]`
`=-[log3.09+log10^(-7)]`
`=-[0.49-7]`
`pH=7-0.49=6.51`