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

The following proton transfer reactions are possible:
1) `HF + H_(2)O hArr H_(3)O^(+) + F^(-)`
`K_(a) = 3.2 xx 10^(-4)`
2) `H_(2)O+ H_(2)O hArr H_(3)O^(+) + OH^(-)`
`K_(w) = 1.0 xx 10^(-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),("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)`
`= 0.2alpha^(2)//(1-alpha)=3.2 xx 10^(-4)`
We obtain the following quadratic equation :
`alpha^(2) + 1.6 xx 10^(-2)alpha- 1.6 xx 10^(-2) = 0`
The quadratic equation in `alpha` can be solved and the two valus 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_(3)O^(+)` are given by:
`[H_(3)O^(+)] = [F^(-)] = calpha = 0.02 xx 0.12`
`= 2.4 xx 10^(-3) M`
`[HF] = c(1-alpha) = 0.02 (1-0.12)`
`= 17.6 xx 10^(-3) M`
`pH = - "log" [H^(+)] = -"log"[2.4 xx 10^(-3)] = 2.62`