1.

A monoprotic acid in 1.00 M solution is 0.01% ionized. The dissociation constant of this acid is

Answer»

`1 xx 10^(-8)`
`1 xx 10^(-4)`
`1 xx 10^(-6)`
`10^(-5)`

Solution :`K = (alpha^(2)C)/(1- alpha), alpha = (0.01)/(100) LT lt lt 1 :. K = alpha^(2)C = [(0.01)/(100)]^(2) xx 1`
`= 1 xx 10^(-8)`.


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