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Silver ions are added to a solution with
$[Br^-] = [Cl^-] = [CO_3^{2-} ] = [AsO_4^{3-}] = 0.1\,M$.
Which compound will precipitate with lowest $[Ag^+]$ ?
$AgBr\,(K_{sp} = 5 \times 10^{-13})$
$AgCl\,(K_{sp} = 1.8 \times 10^{-10})$
$Ag_2CO_3\,(K_{sp} = 8.1 \times 10^{-12})$
$Ag_3AsO_4(K_{sp} = 1 \times 10^{-22})$
Solution
$(1)$ $\left[\mathrm{Ag}^{+}\right]_{\text {required }}=\frac{\mathrm{K}_{\mathrm{sp}}}{\left[\mathrm{Br}^{-}\right]}=5 \times 10^{-12}$
$(2)$ $\left[\mathrm{Ag}^{+}\right]_{\text {required }}=1.8 \times 10^{-9}$
$(3)$ $\left[\mathrm{Ag}^{+}\right]_{\text {required }}=\sqrt{\frac{\mathrm{K}_{\mathrm{r}}}{\left[\mathrm{CO}_{3}^{2-}\right]}}=\sqrt{\frac{8.1 \times 10^{-12}}{0.1}}=9 \times 10^{-6}$
$(4)$ $\left[\mathrm{Ag}^{+}\right]_{\text {required }}=\sqrt{\frac{\mathrm{K}_{\mathrm{r}_{2}}}{\left[\mathrm{AsO}_{4}^{3-}\right]}}=3 \sqrt{\frac{1 \times 10^{-12}}{0.1}}=10^{-7}$
