A weak monobasic acid is 1% ionized in 0.1 M solution at `25^(@)C`. The percentage of ionisation in its 0.025 M solution is

Options

A. 1
B. 2
C. 3
D. 4

Answer

B

Explanation

Ionisatin of weak acid at `C_(1)` = 0.1 M
`HA harr H^(+) (aq) + A^(-)(aq)`
`{:(C_(1),0,0,": Initial concentration"),(C_(1)(1 - alpha_(1)),C_(1)alpha_(1),C_(1)alpha_(1),": Conc. at equilibrium"):}`
`therefore" "K_(a) = (C_(1)alpha_(1).C_(1)alpha_(1))/(C_(1)(1-alpha_(2)))=(C_(1)alpha_(1))/((1-alpha_(1)))`
`= C_(1)alpha_(1)^(2)" "(because alpha_(1) lt lt lt 1)`
Ionisation of weak acid `C_(2)` = 0.025 M
`HA harr H^(+) (aq) + A^(-)(aq)`
`{:(C_(2),0,0,": Initial concentration"),(C_(2)(1-alpha_(2)),C_(2)alpha_(2),C_(2)alpha_(2),": Conc. at equilibrium"):}`
`therefore" "K_(a) = (C_(2)alpha_(2).C_(2)alpha_(2))/(C_(2)(1-alpha_(2)))= (C_(2)alpha_(2)^(2))/((1-alpha_(2)))`
`= C_(2)alpha_(2)^(2)" "(therefore alpha_(2) lt lt lt 1)`
`because` Ionisation constant of an acid is a constant and does not change with concentration.
`therefore" "C_(1)alpha_(1)^(2) = C_(2)alpha_(2)^(2)`
or `alpha_(2)^(2) = (C_(1)alpha_(1)^(2))/(C_(2)) = (0.1 xx (1)^(2))/(0.025)`
or `alpha_(2)^(2) = 4`
or `alpha_(2) = 2`
`therefore` The percentage of ionisation of weak acid in 0.025 M solution is 2%.