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Question
The degree of dissociation of pure water at 25 oC is found to be 1.8×10−9. Find Kw and Kd at 25 oC ?
The degree of dissociation of pure water at 25 oC is found to be 1.8×10−9. Find Kw and Kd at 25 oC ?
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Solution
The correct option is B 1.00×10−14 ; 1.80×10−16
Since α=1.8×10−9
and for water concentration , C=100018=55.56 M
[H+]=[OH−]=Cα=55.56×1.8×10−9=1×10−7 M
Kw=[H+][OH−]=(1×10−7)2=1×10−14
and Kd=[H+][OH]−[H2O]Kd=Kw[H2O]Kd=10−1455.56=1.8×10−16
Hence, Kw and Kd are 1.00×10−14 1.8×10−16 respectively.
Since α=1.8×10−9
and for water concentration , C=100018=55.56 M
[H+]=[OH−]=Cα=55.56×1.8×10−9=1×10−7 M
Kw=[H+][OH−]=(1×10−7)2=1×10−14
and Kd=[H+][OH]−[H2O]Kd=Kw[H2O]Kd=10−1455.56=1.8×10−16
Hence, Kw and Kd are 1.00×10−14 1.8×10−16 respectively.
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