InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Two buffer, (X) and (Y) of pH `4.0` and `6.0` respectively are prepared from acid HA and the salt NaA. Both the buffers are `0.50` M in HA. What would be the pH of the solution obtained by mixing equal volumes of the two buffers ? `(K_(HA)=1.0xx10^(-5))` |
|
Answer» pH of buffer is given by: `pH= -log K_(a)+log ((["Salt"])/(["Acid"]))` For I: `4= - log 1.0xx10^(-5)+log ((["Salt"])/([0.5]))` `:. Log ((["Salt"])/(0.5))= -1` or `["Salt"]= 0.1xx0.5=0.05M` For II: `6= - log 1.0xx10^(-5)+log ((["Salt"])/(0.5))` `:. log ((["Salt"])/(0.5))=1` `:. ["Salt"]= 10xx0.5=5M` Now the two buffer `[(I NaA = 0.05M and HA = 0.5M)` and II `(NaA= 5M and HA= 0.5M)]` are mixed in equal proportion. Thus, new conc. of NaA is mixed buffer `=(0.05xxV+5xxV)/(2V)=(5.05)/(2)` New conc. of HA in mixed buffer `=(0.5xxN+0.5xxV)/(2V)=0.5M` Thus, `pH= -log 1.0xx10^(-5)+log (([5.05//2])/([0.5]))` `pH=5+0.7033= 5.7033` |
|