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

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

1.	The ionic product of water at 310 K is 2.7xx10^(-14). What is the pH of neutral water at this temperature ?
Answer» SOLUTION :`[H^(+)]=SQRT(K_(w))=sqrt(2.7xx10^(-14))=1.643xx10^(-7)M` `pH= - LOG [H^(+)]=-log (1.643xx10^(-7))=7-0.2156=6.78`