1.

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

Answer»

`6.0`
7.01
8.02
9.02

Solution :`[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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