InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the `pK_(a)` bromoacetic acid. |
|
Answer» `{:(,CH_(2)(Br)CO OH ,hArr,CH_(2)(Br)CO O^(-) ,+,H^(+),,,),("Initial conc.",C,,0,,0,,,),("Conc. at eqm.",C-C alpha ,,C alpha ,,C alpha ,,,),(,,,,,,,,):}` `K_(a) = (C alpha . C alpha)/(C ( 1-alpha))=(C alpha^(2))/(1-alpha)~=C alpha^(2) = 0.1 xx (0.132)^(2) = 1.74xx10^(-3)` `pK_(a) = - log (1.74xx10^(-3))=3-0.2405=2.76` `[H^(+)]=C alpha = 0.1 xx 0.32 = 1.32 xx 10^(-2) M` `pH = - log (1.32xx10^(-2))=2-0.1206=1.88` |
|