Question

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

Answer 1

Here, we are given : `K_(w)= 9.55xx10^(-14)`
Now, as for water, `[H_(3)O^(+)]=[OH^(-)] :. K_(w)=[H_(3)O^(+)][OH^(-)]=[H_(3)O^(+)][H_(3)O^(+)]=[H_(3)O^(+)]^(2)`,
i.e., `[H_(3)O^(+)]^(2) = 9.55xx10^(-14) or [H_(3)O^(+)]=sqrt(9.55xx10^(-14))=3.09xx10^(-7)M`
`:. pH = - log [H_(3)O^(+)]=-log (3.09xx10^(-7))=-[log 3.09+ log 10^(-7)]=- [ 0.49-7]=6.51`.