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)OhArrH_(3)O^(+)+F^(-)` `K_(a)=3.2xx10^(-4)`
`(2)` `H_(2)O+H_(2)OhArrH_(3)O^(+)+OH^(-)` `K_(w)=1.0xx10^(-14)`
As `K_(a) gt gt K_(w)`, `[1]` is the principle reaction.
`{:(,,HF,+,H_(2)O,hArr,H_(3)O^(+),+,F^(-)),("Initial",,,,,,,,),("concentration(M)",,,,,,,,),(,,0.02,,,,0,,0),("Change(M)",,,,,,,,),(,,-0.02alpha,,,,+0.02alpha,,+0.02alpha),("Equilibrium",,,,,,,,),("concentration(M)",,,,,,,,),(,,0.02-0.02alpha,,,,0.02alpha,,0.02alpha):}`
Substituting equilibrium concentrations in the equilibrium reaction for principal reaction gives :
`K_(a)=(0.02alpha)^(2)//(0.02-0.02alpha)`
`=0.02alpha^(2)//(1-alpha)=3.2xx10^(-4)`
We obtain the following quadratic equation :
`alpha^(2)+1.6xx10^(-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,
`alpha=0.12`
This means that the degree of ionization, `alpha=0.12`, then equilibrium concentrations of other species viz., `HF,F^(-)` and `H_(2)O^(+)` are given by :
`[H_(3)O^(+)]=[F^(-)]=calpha=0.02xx0.12`
`=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`