Ionic product of water at 310 K is 2.7 times 10^-14. What is the pH of – InterviewSolution

1.	Ionic product of water at 310 K is 2.7 times 10^-14. What is the pH of neutral water at thin temperature?
Answer» SOLUTION :`[H^+]=sqrtK_w=sqrt(2.7 TIMES 10^-14)=1.643 times 10^-7 M` `pH=-LOG[H^+]=-log(1.643 times 10^-7)=7-0.2156=6.78`

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