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The ionic product of water at 310 K is 2...

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

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`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`
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