7.65 Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at this temperature?

Ionic product, 
Kw=[H+][OH-]Let [H+]=x.Since [H+]=[OH-], Kw=x2.Kw at 310K is 2.7×10-14.2.7×10-14=x2x=1.64×10-7[H+]=1.64×10-7 pH=-log[H+]=- log [1.64×10-7]=6.78
Hence, the pH of neutral water is 6.78.