The ionic product of water at 310 K is 2.7 xx10^-14. What is the pH of neutral water at this temperature.

The ionic product of water at 310 K is 2.7 xx10^-14. What is the pH of

1.	The ionic product of water at 310 K is 2.7 xx10^-14. What is the pH of neutral water at this temperature.
Answer» SOLUTION :`K_w=[H^+][OH^-]=2.7xx10^-14` `THEREFORE[H^+]=SQRT(2.7xx10^-14)=1.643xx10^-7` `therefore pH=-log(1.643xx10^-7)=7-0.2156=6.7844=6.78`