InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
At `25^(@)C`, the dissociation constant of a base. BOH is `1.0xx10^(-12)`. The concentration of hydroxyl ions in `0.01`M aqueous solution of the base would beA. `2.0xx10^(-6)mol L^(-1)`B. `1.0xx10^(-5)mol L^(-1)`C. `1.0xx10^(-6)mol L^(-1)`D. `1.0xx10^(-7)mol L^(-1)` |
|
Answer» Correct Answer - D Base, BOH is dissociated as follows `BOH hArr B^(+)+OH^(-)` So, the dissociation constant of BOH base `K_(b)=([B^(+)][OH^(-)])/([BOH])` ….(i) At equilibrium `[B^(+)]=[OH^(-)]` `therefore K_(b)=([OH^(-)]^(2))/([BOH])` Given that `K_(b)=1.0xx10^(-12)` and `[BOH]=0.01 M` Thus, `1.0xx10^(-12)=([OH^(-)]^(2))/(0.01)` `[OH^(-)]^(2)=1xx10^(-14)` `[OH^(-)]=1.0xx10^(-7)"mol L"^(-1)` |
|