1.

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

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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