1.

Calculate the pOH of a solution at 25^(@)C that contains 1xx10^(-10)M of hydronium ions i.e H_(3)O^(+) :

Answer»

`4.00`
`9.00`
`1.00`
`7.00`

Solution :`[OH^(-)]=(10^(-14))/([H_(3)O^(+)])=(10^(-14))/(10^(-10))=10^(-4)MOL L^(-1)`
`POH = log [OH^(-)]=-log (1xx10^(-4))=4`.


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