What is the pH of 1 M solution of acetic acid ? To what volume one litre of this solution be diluted so that pH of the resulting solution will be twice of the original value ? `(K_(a)=1.8xx10^(-5))`

Explanation

Case I `{:(,CH_(3)COOH,hArr,CH_(3)COO^(Theta)+,H^(oplus)),("Concentration before dissociation",1,,0,0),("Concentration after dissociation",1-alpha,,alpha,alpha):}`
`[H^(o+)] = Calpha = C sqrt((K_(a)//C))`
`=sqrt((K_(a)C)) = sqrt((1.8 xx 10^(-5)xx1))`
`= sqrt((18 xx 10^(-6)) = 4.24 xx 10^(-3) M`
`:. pH =- log H^(o+) =- log (4.24 xx 10^(-3)) = 2.3724`or
Use formula for the `pH` of weak acid.
`pH = (1)/(2) (pK_(a) - log C) = (1)/(2) (4.7447 - 0) = 2.3724`
Case II: New `pH` is given as `= 2.3724 xx 2 = 4.7488`
Let new concentration be `C_(1)` and dergee of dissociation be `alpha_(1)`
`:. -log [H^(o+)] = 4.7448`
`:. [H^(o+)] = 1.8 xx 10^(-5)`or `C_(1) alpha_(1) = 1.8 xx 10^(-5)`
Now again `K_(a) = ([H^(o+)][CH_(3)COO^(Θ)])/([CH_(3)COOH])`
`K_(a) = (C_(1)alpha_(1) xx C_(1)alpha_(1))/(C_(1)(1-alpha_(1))) = (C_(1)alpha_(1)alpha_(1))/((1-alpha_(1)))`
`:. 1.8 xx 10^(-5) = (1.8 xx 10^(-5) xx alpha_(1))/((1-alpha_(1))) :. alpha_(1) = 0.5`
Now `C_(1)alpha_(1) = 1.8 xx 10^(-5)`
`:. C_(1) = (1.8 xx 10^(-5))/(alpha_(1)) = (1.8 xx 10^(-5))/(0.5) = 3.6 xx 10^(-5)M`
Let `1L` of concentrated solution be diluted to `VL` Equivalent of dilute solution `=` Equivalent of concentrated solution
`3.6 xx 10^(-5) xxV = 1xx1 [M_(1)V_(1) = M_(2)V_(2)]`
`:. V = (1)/(3.6 xx 10^(-5)) = 2.77 xx 10^(4)L`
Note: iIn `II` case `alpha` comes `0.5` by `K_(a) = (C_(1)alpha_(1)^(2))/((1-alpha_(1)))`, and thus, it is not advisible to assume `(1-alpha_(1)) ~~1`.