1.

A weak acid HA has a K_(a) of 1.00 xx 10^(-5). If 0.100 mol of this acid is dissolved in one litre of water the percentage of acid dissociated at equilibrium is closed to

Answer»

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

Solution :`K_(a) = 1 xx 10^(-5)`
`{:(HA,HARR,H^(+),+,A^(-)),(.1m//l,,0,,0),(.1-x,,x,,x):}`
`10^(-5) = (x^(2))/(1-x)`
Assume `x lt lt.1`
`x^(2) = 10^(-6)` or `x = 10^(-3)`, % DISSOCIATION `= (10^(-3))/(.1) xx 100 = 1%`


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