Question

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

• 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. $A{g}_{3}As{O}_4(K_{sp}=1\times {10}^{-22})$

Answer 1

Answer: (B)

$AgCl(K_{sp}=1.8\times {10}^{-10})$

The extent of precipitation is directly proportional to the solubility of the product $\left(K_{sp}\right)$.

Here, the solubility product $\left(K_{sp}\right)$ follows the order:

$\left(D\right)<\left(A\right)<\left(C\right)<\left(B\right)$

Therefore, option (B) is the answer.