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Question

The Kw for water for the following equillibrium.
2H2OH3O+OH
changes from 1014 at 25 to 9.62×1014 at 60C. Calculate the pH and nature at 60C


A
pH=5.51, solution is basic
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B
pH=6.51, solution is neutral
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C
pH=5.51, solution is acidic
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D
pH=6.51, solution is basic
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Solution

The correct option is C pH=6.51, solution is neutral
pH=log[H+], Kw=[H+] [OH]9.62×1014=x2
x=9.62×1014 x=3.18×107pH=log[3.18×107]pH=6.51
pH value: Negative logarithem if H+ ion concentration.
pH=log[H+]
pH=log1[H+]
Kw=[H+] [OH]
log[H+]log[OH]=logKw
pH+pOH=pKw
at 25C,Kw=1014
At  65C ,solution is neutral.

Chemistry

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