a. Calculate the ratio of `pH` of a solution continaing `1mol`. Of `CH_(3)COONa +1 mol` of `HC1` per litre and of other solution containing `1mol` of `CH_(3)COONa + 1mol` of `CH_(3)COOH` per litre.
b. A `0.1M `solution of weak acid `HA` is `1%` dissociated at `298k`. what is its `K_(a)`? what will be the new degree of dissociation of `HA` and `pH` when `0.2M` of `NaA` is added to it.

Explanation

a. Case I: `pH` when `1mol CH_(3)COONa` and `1mol HCI` are present.
`{:(,CH_(3)COONa+,HCIrarr,CH_(3)COOH+,NACI),("Before reaction",1,1,0,0),("After reaction",0,0,1,1):}`
`:. [CH_(3)COOH] = 1M`
`:. [H^(o+)] = C.alpha = sqrt(((K_(a))/(c))) = sqrt(K_(a).C) = sqrt((K_(a)C))( because C=1)`
`:. pH_(1) =- (1)/(2) log K_(a)`
Case II: `pH` when `1mol CH_(3)COONa` and `1moles` of `CH_(3)COOH`, a buffer solution
`:. pH_(2) =- log K_(a) + "log" (["Salt"])/(["Acid"]) [[:.["Salt"]=,1M],[:.["Acid"]=,1M]]`
`pH =- log K_(a)`
`:. (pH_(1))/(pH_(2)) = (1)/(2) = 0.5`
b. For weak acid `HA : alpha_(HA) = (1)/(100) = 0.01, [HA] = 0.1M`
`:. K_(a) = C alpha^(2) = 0.1 xx (0.01)^(2) = 10^(-5)`
Now `0.2M NaA`, a salt of `HA`, is added to it resulting a buffer solution of `[HA]=0.1M` and `[NaA] = 0.2M`
`pH =- (log 10^(-5)) + "log" (0.2)/(0.1)`
`pH = 5.3010`
Also `{:(HA,hArr,H^(o+),+,A^(Theta)),(1,,0,,0),((1-alpha),,alpha,,alpha):}`
`because [A^(Theta)]` is procided by `NaA` since dissociation of `HA` in presure of `NaA` is suppressed due to common ion effect
`:. K_(a) = ([H^(o+)][A^(-1)])/([HA]) = ((Calpha)xx0.2)/(C(1-alpha))= 10^(-5)`
`:. alpha = 5 xx 10^(-5)`