1.

A monoprotic acid in a 0.1 M solution ionizes to 0.001%. Its ionisation constant is

Answer»

`1.0 xx 10^(-3)`
`1.0 xx 10^(-6)`
`1.0 xx 10^(-8)`
`1.0 xx 10^(-11)`

Solution :`:.` MONOPROTIC acid HA
`HA hArr H^(+) + A^(-)`
Ionisation constant = ?
`alpha= 0.001 % = (0.001)/(100) = 10^(-5) RARR K = (alpha^(2))/(V) = ([10^(-5)]^(2))/(10) = 10^(-11)`.


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