1.

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

Answer»

`6.0 `
`7.02`
`8.02`
`9.02`

SOLUTION :`[OvecH]_("total")=(10^(-8) + 10^(-7))M`
`thereforepOH=- log[10^(-8 ) + 10^(-7)]`
` ~~ 6.98 `
` THEREFORE pH=14 - 6.98= 7.02`


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