The ionisation constant of `HF` is `3.2 xx 10^(-4)`.
a. Calculate the dergee of dissociation of aall species present `M` solution.
b. Calculate the concentration of all species present `(H_(3)O^(o+),F^(Θ)` and `HF`) in the solution.
c. Calculate method:

Explanation

a. The following proton transfter reactions are possible,
`HF +H_(2)O hArr H_(3)O^(oplus)+F^(Theta)(K_(a)=3.2 xx10^(-4))` ....(i)
`H_(2)O+H_(2)O hArr H_(3)O^(oplus)+overset(Theta)(O)H (Kw=10^(14))`....(ii)
Since `K_(a)gt gt K_(w)`, reaction (i) is the main reaction.
`{:(,HF+, H_(2)O hArr,H_(3)O^(oplus),F^(Theta)),("Initial conc",rArr0.02,-,0,0),("Change in conc",rArr-0.02alpha,-,+0.02alpha,+0.02alpha),("Eq conc",rArr0.02-0.02alpha,-,,),(,=0.021(1-alpha),,0.02alpha,0.02alpha):}`
`:.K_(a) = ((0.02alpha)^(2))/(0.02(1-alpha)) = (0.02alpha^(2))/(1-alpha) = 3.2 xx 10^(-4) ..(i)`
Solving the equaiton (i), the following quadratic equaiton is obained,
`alpha^(2) +1.6 xx 10^(-2)alpha - 1.6 xx 10^(-2) = 0`.
`:. alpha = +0.12` and `-0.12`. (-ve value is not taken)
Thus, `alpha = 0.12`
b. `[H_(3)O^(o+)] = [F^(Θ)] = C alpha = 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`
c. `pH = - log [H^(o+)] =- log (2.4 xx 10^(-3)) = 2.62`
Note: `alpha - 0.12` which is greater than `0.05, (alpha ge 0.05` or `alpha le 5%)` So the term `(1-alpha) != 1`, is taken.
Second method: By using formula of `alpha`, and `ph`.
a. `alpha = sqrt((K_(a))/(C)) = sqrt((3.2 xx 10^(-4))/(0.02)) = sqrt(16xx10^(-4)xx10)`
`= 4 xx 10^(-2) xx sqrt(10)`
`(sqrt(10) = 3.162)`
`= 4xx3.162 xx 10^(-2)`
`= 0.126`
`:. alpha= 0.12`
b. `[H_(3)O^(o+)] = [F^(Θ)] = Calpha = 0.02 xx 0.12 = 2.4 xx 10^(-3)M`
`[HF] = C(1-alpha) = 0.02 (1-0.12) = 1.76 xx 10^(-3)M`
c. `pK_(a) =- log (3.2 xx 10^(-4)) =- log (32 xx 10^(-5))`
`=- log (25 xx 10^(-5))`
`=- 5 log 2+5`
`=- 5 xx 0.3 +5 = 3.5`
`pK_(a) ~~ 3.5`
`pH = (1)/(2) (pK_(a) - logC) = (1)/(2) (3.5-log0.02)`
`= (1)/(3) (3.5 - log 2 xx 10^(-2))`
`= (1)/(2) (3.5 - 0.3 +2)`
`= (1)/(2) (5.2) = 2.6`
`:. pH ~~ 2.6`.