KwforH2Ois9.62×10−14at60∘C The pH of water at this temperature is :
A
6.5
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B
7.5
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C
7
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D
6
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Solution
The correct option is A 6.5 We know that, Kw=[H+][OH−] As water is neutral and for a neutral solution: [H+]=[OH−]∴Kw=[H+]2∵Kw=9.62×10−14(Given)⇒[H+]2=9.62×10−14[H+]=√9.62×10−14=31×10−8pH=−log10[H+]∴pH=−log10(31×10−8)pH=6.5