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Caculate the pH of 10^(-8) mNaOH...

Caculate the pH of `10^(-8) mNaOH`

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The given box solution is very dilute. Hence `OH^(-)` obtained from the box and water must be taken into consideration.
`:. [OH^(-)]=10^(-8)+10^(-7)=1.1xx10^(-7)`
`pOH=-log1.1xx10^(-7)`
`pOH=6.995`
`:. pOH=14-6.995=7.005`
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