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At `25^(@)C`, pH of a `10^(-8)M` aqueous KOH solution will beA. `6.0`B. 7.01C. 8.02D. 9.02 |
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Answer» 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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