1.

A weak acid, HA has a K_(a) of 1.00xx10^(-5). If 0.100 mol of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closest to :

Answer»

`1.00%`
`99.9%`
`0.100%`
`99.0%`

Solution :`alpha = sqrt((K_(a))/(c ))=sqrt((1.0xx10^(-5))/(10^(-1)))=10^(-2)`.
`therefore` PERCENTAGE of acid dissociated `=10^(-2)xx100`
`=1%`


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