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

Explanation

Calculation `[H^+]` : There is following equilibrium in water.
`H_2O_((l))+ H_2O_((l)) =H_3O_((aq))^(+) + OH_((aq))^(-)`
`K_W=[H_3O^+][OH^-]= 2.7xx10^(-14)`
but `[H_3O^+]=[OH^-]`
`therefore [H_3O^+]^2=2.7xx10^(-14)`
`therefore [H_3O^+]= sqrt(2.7xx10^(-14)) =1.6432xx10^(-7)`
Calculation of pH :
pH=-log `[H^+]`
=-log `(1.6432xx10^(-7))`
=-(0.2157-7.0)=6.7843