InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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`) |
|
Answer» 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` |
|