Question

Silver ions are added to the solution with:
$\left[{Br}^{-}\right]=\left[{Cl}^{-}\right]=\left[{{CO}_3}^{2-}\right]=\left[A{{sO}_3}^{2-}\right]=0.1M$
Which compound will precipitate at the lowest $\left[{Ag}^{+}\right]$?

• A. $AgBr(K_{sp}=5\times {10}^{-13})$
• B. $AgCl(K_{sp}=1.8\times {10}^{-10})$
• C. ${Ag}_2{CO}_3(K_{sp}=8.1\times {10}^{-12})$
• D. ${Ag}_2{AsO}_4(K_{sp}={10}^{-22})$

Answer 1

The correct option is A $AgBr(K_{sp}=5\times {10}^{-13})$

Since here the anions are ${Br}^{-},{Cl}^{-},{CO}_3^{2-}$ and ${AsO}_3^{2-}$, So the probable precipitates will be $AgBr,AgCl$, ${Ag}_2{CO}_3$ only $AgBr$ has lowest value of solubility product i.e. $K_{sp}$ of $AgBr=5\times {10}^{-13}$. So when ${Ag}^{+}$ ions are added to the solution, $AgBr$ will precipitate at the lowest concentration of ${Ag}^{+}$.