InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
How many moles of AgBr `(K_(sp)=5xx10^(-13)"mol"^(2)L^(-2))` will dissolve in 0.01 M NaBr solution ? |
|
Answer» Correct Answer - `5xx10^(-11) "mol" L^(-1)` Suppose solubility of AgBr in 0.01 M NaBr = s mol `L^(-1)`. Then as `AgBr rarr Ag^(+) + Br^(-)`, `[Ag^(+)]=s "mol " L^(-1) and "Tota" [Br^(-)]=0.01 + s ~~ 0.01 M` `K_(sp) = [ Ag^(+)][Br^(-)], i.e., 5xx10^(-13) = s xx0.01 or s = 5 xx 10^(-11) "mol" L^(-1)` |
|