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At `25^(@)C`, pH of a `10^(-8)M` aqueous KOH solution will be

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At `25^(@)C`, pH of a `10^(-8)M` aqueous KOH solution will be
A. `6.0`
B. 7.01
C. 8.02
D. 9.02
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Correct Answer - B
`[Obar(H)]_("Total") = (10^(-8) + 10^(-7)) M :. pOH = -log [10^(-8) + 10^(-7)]`
`= ~ 6.98`
`:. pH = 14 - 6.98 = 7.02`
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