A weak acid, `HA`, has a `K_(a)` of `1.00xx10^(-5)`. If `0.100` mol of the acid is dissolved in 1 L of water, the percentage of the acid dissociated at equilibrium is the closed to

Options

A. `99.0%`
B. `1.00%`
C. `99.9%`
D. `0.100%`

Answer

B

Explanation

`HA hArr H^(+)+A^(-)`
At equilibrium `[H^(+)=A^(-)]`
`K_(a)=([H^(+)][A^(-)])/([HA])=([H^(+)]^(2))/([HA])`
`[H^(+)]=sqrt(K_(a)[HA])=sqrt(1xx10^(-5)xx0.1)`
`=sqrt(1xx10^(-6))=1xx10^(-3)`
`alpha = ("Actual ionisation")/("Molar concentration")=(10^(-3))/(0.1)=10^(-2)`
% of acid dissociated `=10^(-2)xx1.00=1%=100%`