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Question

Percentage ionization of H2O at certain temperature is 3.6×107%. Calculate Kw at this temperature.

A
Kw=4×1014
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B
Kw=3×1014
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C
Kw=1×1014
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D
none of these
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Solution

The correct option is D none of these
Actual ionisation of water at certain temperature=%ofionization100=3.6×107100=3.6×109
2H2OOH+H3O+
initial 1 0 0
Eqb. (13.6×109) (3.6×109) (3.6×109)
Kw=[OH][H3O+]
=(3.2×109)2
Kw=12.96×1018

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