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Question

The pH of a neutral solution at 50oC is:
(Kw=10−13.26 at 50oC)

A
7
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B
6.0
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C
7.23
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D
6.63
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Solution

The correct option is D 6.63
Ionic product of water is Kw=1013.26
Kw=[H+][OH]
Since [H+]=[OH]
[H+]2=Kw
[H+]2=Kw
[H+]=Kw=1013.26=2.344×107pH=log([H+])=log(2.344×107)=6.628
pH=6.63

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