InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The ionization constant of ammonium hydroxide is `1.77xx10^(-5)` at `298 K`. Hydrolysis constant of ammonium chloride isA. `5.65xx10^(-10)`B. `6.50xx10^(-12)`C. `5.65xx10^(-13)`D. `5.65xx10^(-12)` |
|
Answer» Correct Answer - A Given, `K_(a)(NH_(4)OH)=1.77xx10^(-6)` `NH_(4)OH hArr NH_(4)^(+)+OH^(-)` `K_(a)=([NH_(4)^(+)][OH^(-)])/([NH_(4)OH])=1.77xx10^(-5)` ..(i) Hydrolysis of `NH_(4)Cl` takes place as, `NH_(4)Cl+H_(2)O rarr NH_(4)OH+HCl` or `NH_(4)^(+)+H_(2)O rarr NH_(4)OH+H^(+)` Hydrolysis constant, `K_(h)=([NH_(4)OH][H^(+)])/([NH_(4)^(+)])` ....(ii) or `K_(h)=([NH_(4)OH][H^(+)][OH^(-)])/([NH_(4)^(+)][OH^(-)])` ....(iii) From Eqs. (i), (ii) and (iii) `K_(h)=(K_(w))/(K_(a)) " " [because [H^(+)][OH^(-)]=K_(w)]` `=(10^(-14))/(1.77xx10^(-5))=5.65xx10^(-10)` |
|