InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The ionization constant of chloroacetic acid is `1.35xx10^(-3)`. What will be the `pH` of `0.1 M` acid and its `0.1M` sodium salt solution? |
|
Answer» Correct Answer - (a) `pH` of acid solution `= 1.9` , (b) `pH ` of its salt solution `= 7.9` It is given that `K_(a)` for `CICH_(2)COOH" is "1.35xx10^(-3)`. `rArr K_(a)=calpha^(2)` `thereforealpha=sqrt((K_(a))/(c))` `=sqrt((1.35xx10^(-3))/(0.1))" "(therefore"concentration of acid = 0.1 m")` `alpha=sqrt(1.35xx10^(-2))=0.116` `therefore[H^(+)]=calpha=0.1xx0.116` = .0116 `rArrpH=-log[H^(+)]=1.94` `ClCH_(2)COONa` is the salt of weak acid i.e., `ClCH_(2)COOH` and a strong base i.e., NaOH. `ClCH_(2)COO^(-)+H_(2)OleftrightarrowCICH_(2)COOH+OH^(-)` `K_(h)=([ClCH_(2)COOH][OH^(-)])/([ClCH_(2)COO^(-)])` `K_(h)=(K_(v))/(K_(a))` `K_(h)=(10^(-14))/(1.35xx10^(-3))` `=0.740xx10^(-11)` Also, `K_(h)=(x^(2))/(0.1)` (Where x is the concentration of `OH^(-)` and `ClCH_(2)COOH`) `0.740xx10^(-11)=(x^(2))/(0.1)` `0.074xx10^(-11)=x^(2)` `rArr x^(2)=0.74xx10^(-12)` `x=0.86xx10^(-6)` `[OH^(-)]=0.86xx10^(-6)` `therefore [H^(+)]=(K_(w))/(0.86xx10^(-6))` `=(10^(-14))/(0.86xx10^(-6))` `[H^(+)]=1.162xx10^(-8)` `pH=-log[H^(+)]` = 7.94. |
|