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Question

Percentage ionization of H2O at certain temperature is 3.6×107. Calculate Kw and pH of water.
(Kw=4×1014, ph=6.7)

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Solution

Degree of ionization= α=3.6×107×102=3.6×109
H2OH++OH
1α α α
KW=[OH][H+]=α2=(3.6×109)2=12.96×1018
Also, pH=log[H+]=log(3.6×1019)=8.44

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