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

Explanation

If c is the initial concentration and `alpha` is the degree of ionization of HF then
`{:(,HF+H_2O hArr , H_3O^(+)+, F^(-)),("Initial conc.",c,0,0),("at equi,",c(1-alpha),c alpha, calpha):}`
`K_a=([H_3O^+][F^-])/([HF])`
`=((c alpha)xx(c alpha))/(c(1-alpha))=(calpha^2)/(1-alpha)`
Since `alpha` is very small, `1-alpha approx 1`
`K_a=calpha^2`
`alpha=sqrt(K_a /c)=sqrt((3.2xx10^(-4))/0.02)=0.126`
Degree of ionization of HF =0.126
Equilibrium concentrations can be calculated as :
`[H_3O^+]= calpha =0.02xx0.126=2.52xx10^(-3)` M
`[F^-]=c alpha =0.02xx0.126=2.52xx10^(-3)` M
`[HF]=c (1-alpha)=0.02(1-0.126)=1.748xx10^(-2)` M
Now pH =-log `[H_3O^+]`
=-log `(2.52xx10^(-3))` =2.60