7.65. Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at this temperature?
7.65. Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at this temperature?
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1 Answer
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Ionic product,
Kw= [H+] [OH−]
Let [H+]= x
Since [H+]= [OH−], Kw=x2.
⇒Kw at 310K is 2.7×10−14
∴2.7×10−14=x2
⇒x=1.64×10−7
⇒ [H+]=1.64×10−7
⇒ pH= −log [H+]=−log [1.64×10−7]=6.78
Hence, the pH of neutral water is 6.78.
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