The freezing point of `0.20M` solution of weak acid `HA` is `272.5K`. The molality of the solution is `0.263mol Kg^(-1)`.
Find the `pH` of the solution on adding `0.20M` solution of `NaOH`. Given: `K_(f)` of water `= 1.86 Km^(-1)`

Explanation

a. `Delta_(f)T = (272- 272.5) = 0.5K, Delta_(f)T = iK_(f)m`
`:. I = Delta_(f)T//K_(f)m = 0.5//1.86 xx 0.263 = 1.022`
`{:(,HArarr,H^(o+)+,A^( Θ),),("Initial",1,0,0,),("Final",1-alpha,alpha,alpha,):}`
`I = 1 + alpha :. 1.022 = 1 + alpha :. alpha = 0.022`
`K_(a) = C alpha^(2) = 0.2 xx (0.022)^(2) = 9.6 xx 10^(-5)`
`pK_(a) = 4.0177`
On adding `0.25M NaA`, buffer is formed.
`:. ["Salt"] = 0.25 M, ["Acid"] = 0.20M`
`ph = pK_(a) + log ["salt/Acid"] = 4.0177 + log (0.25//0.20)`
`= 4.0177 + 0.0969 = 4.1146`
b. On adding `0.2M NaOH`, salt of `W_(A)` and `S_(B)` is formed
`:. ["Salt"] = 0.2M`
`pH = (1)/(2) (pK_(w) + pK_(a) + logC)`
`= (1)/(2) (14 + 4.0177 + log 2 xx 10^(-1))`
`[log 2 xx 10^(-1) =- 1 + 0.3 =- 0.7]`
`= (1)/(2) (18. 0177 - 0.7) = 8.65`