Question

# If Kw for pure H2O at 45∘C is 4×10−14 M2. The pH of water at this temperature is :

A
5.7
B
3.7
C
4.7
D
6.7

Solution

## The correct option is C 6.7We know that, Kw=[H+] [OH−] As water is neutral and for a neutral solution: [H+]=[OH−]∴Kw=[H+]2∵Kw=4×10−14   (Given)⇒[H+]2=4×10−14     [H+] =√4×10−14=2×10−7pH=−log10[H+]∴pH=−log10(2×10−7)pH=−(0.3−7)pH=6.7Chemistry

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