InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The solubility product for silver choride is `1.2xx10^(-10)` at 298 K. Calculate the solubility of silver chloride at 298 K. |
|
Answer» Silver chloride dissociates according to the equation : `AgCl(s) hArr AgCl(aq) hArr AG^(+) (aq) + Cl^(-)(aq)` Let s be the solubility of AgCl in moles per litre. Consequently, the molar concentration of `Ag^(+) and Cl^(-)` will also be s each. Substituting in the expression for solubility product of AgCl, `K_(sp)=[Ag^(+)][Cl^(-)]=s xx s =s^(2)` But `K_(sp)= 1.2xx10^(-10) ` (Given) `:. s^(2)=1.2xx10^(-10) or s = sqrt(1.2xx10^(-19))=1.1xx10^(-5) "mol " L^(-1)` |
|