The value of `k_(w) " is " 9.55 xx 10^(-14)` at a certain temperature . Calculate the pH of water at this temperature .

Answer

`6.51`

Explanation

`K_(w) =9.55 xx10^(-14) , " For water " [H_(3)O^(+)] =[OH^(-)]`
`K_(w) [H_(3)O^(+)][OH^(-)] =[H_(3)O^(+)]^(2)`
`:." "[H_(3)O^(+)] =(K_(w))^(1//2) =(9.55 xx 10^(-14))^(1//2) =3.09 xx 10^(-7) M`
`pH =-log [H_(3)O^(+)] =-(log 3.09 xx 10^(-7))`
=-(log 3.09 -7 log 10)= (7-log 3.09) =(7-0.49) =6.51`