The `[H^(+)]` in `0.2M` solution of formic acid is `6.4xx10^(-3)` mol `litre^(-1)`. To this solution formate is added so as to adjusrt the conc.of sodium formate to one mol per litre. What will be pH of this solution ? `(K_(a) for HCOOH=2.4xx10^(-4))` and degree of dissociation of `HCOONa = 0.75`)

Explanation

In `0.2M HCOOH, [H^(+)]= 6.4xx10^(-3)`
`Calpha= 6.4xx10^(-3)`
`:. alpha= 3.2xx10^(-2)`
Now sodium formate is added and the dissociation will fiurther be suppressed and therefore, new degree of dissociation `(alpha_(1))` for HCOOH in presence of HCOONa is so small that it may be neglected.
`:. [HCOOH]` after dissociation
= [HCOOH] before dissociation
`[HCOOH]= 0.2`
`{:(,HCOONahArr,HCOO^(-)+,Na^(+)),("Conc. before dissociation", 1,0,0),("Conc. after dissociartion", (1-0.75),0.75,0.75):}`
So `[HCOOH]= 0.2`
`:. [HCOO^(-)]= 0.75`
`pH= -log 2.4xx10^(-4)+log((0.75)/(0.2))=4.19`