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

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

A

`6.0 `

B

`7.02`

C

`8.02`

D

`9.02`

Text Solution

Verified by Experts

The correct Answer is:
B
`[OvecH]_("total")=(10^(-8) + 10^(-7))M`
`therefore pOH =- log [10^(-8 ) + 10^(-7)]`
` ~~ 6.98 `
` therefore pH =14 - 6.98 = 7.02`
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