The ionization constant of `HF` is `3.2xx10^(-4)`. Calculate the degree of dissociation of `HF` in its `0.02M` solution. Calculate the concentration of all species present (`H_(3)O^(+)`, `F^(-)` and `HF`) in the solution and its `pH`.

Explanation

The following proton, transfer reactions are possible :
`(1) HF+H_(2)O hArr H_(3)O^(+) FK_(a)=3.2xx10^(-4)`
(2) `H_(2)O+H_(2)O H_(3)O^(+)+OH^(-) K_(W)=1.0xx10^(-14)`
As `K_(a gt gt K_(w)` [1] is the principle reaction.
Initial concentration (M)
`-0.02 alpha " "+0.02 alpha " " +0.02 alpha`
Equilibrium (M)
`0.02-0.02alpha " " 0.02 alpha " " 0.02 alpha`
substituting equilibrium concentrations in the equailibrium reaction for principle reaction gives.
`K_(a)=(0.02 alpha)^(2)//(0.02-0.2 alpha)`
`=0.02alpha^(2)//(1-alpha)=3.2xx10^(-4)`
We obtain the following quadratic equation:
`alphaA^(2)+1.6+10^(-2)alpha-1.6xx10^(-2)=0`
The quadratic equation in `alpha` can be solved and the two values of the roots are :
`alpha=0.12 and 0.12`
The negative root is not acceptable and hence.
a=0.12
This means that the degree. of ionizaton `alpha 0.12` then, equilibrium concentrations of other speceis viz. `HF,F^(-) and H_(3)O^(+)` are given by :
`[H_(3)O^(+)]=[F]= calpha=0.2xx0.12`
`=2.4xx10^(-3)M`
`[HF]=c(1-alpha)=0.02(1-0.12)`
`=17.6xx10^(-3)M`
`pH=-log[H^(+)]=-log(2.4xx10^(-3))=2.62`