Question

Silver ions are added to the solution with
$\left[B{r}^{-}\right]=\left[C{l}^{-}\right]=\left[C{O}_3^{2-}\right]=\left[As{O}_4^{3-}\right]=0.1M$
Which compound will precipitate at the lowest [$A{g}^{+}$] ?

• A. $AgBr(K_{sp}=5\times {10}^{-13})$
• B. $AgCl(K_{sp}=1.8\times {10}^{-10})$
• C. $A{g}_{2}C{O}_3(K_{sp}=8.1\times {10}^{-12})$
• D. $A{g}_{3}As{O}_4(K_{sp}={10}^{-22})$

Answer 1

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

Let S moles per litre be the solubility of the salt $A_x{B}_y$.
The expression for the solubility product will be $K_{sp}=\left[A^{y+}{]}^x\right[B^{x-}{]}^y=(x\times S{)}^x(y\times y{)}^y$.
Also $\left[A^{y+}\right]=(x\times S$ and $B^{x-})=y\times y$.
When values are substituted in the above expressions for $AgBr,AgCl$ , $A{g}_{2}C{O}_3$ and $A{g}_{3}As{O}_4$, lowest $\left[A{g}^{+}\right]$ (that will precipitate the compound) is obtained for $AgBr.$