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

Answer

C

Explanation

Equilibrium expression and composition are
`{:(,HA(aq.)hArr,H^(+)(aq.)+,A^(-)(aq.)),("Initial moles",C,0,0),("Equilibrium moles",C-Calpha,Calpha,Calpha):}`
or `C(1-alpha)`
According to the law of chemical equilibrium,
`K_(a)= (C_(H^(+))C_(A^(-)))/(C_(HA))= ((Calpha)(Calpha))/(C(1-alpha))`
`= (Calpha^(2))/(1-alpha)`
Considering `alpha lt lt 1`, we have
`K_(a)=Calpha^(2)`
or `alpha=sqrt((K_(a))/(C ))=sqrt((1.0xx10^(-5))/(0.1))`
`=sqrt(10^(-4))=10^(-2)`
`:.` % Dissociation `= 10^(-2)xx100%`
`=1%`