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