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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
$\mathrm{AgBr} ;\left[\mathrm{Ag}^{+}\right]=\frac{5 \times 10^{-13}}{0.1}=5 \times 10^{-12}\, \mathrm{M}$
$\mathrm{AgCl} ;\left[\mathrm{Ag}^{+}\right]=\frac{1.8 \times 10^{-10}}{0.1}=1.8 \times 10^{-9}\, \mathrm{M}$
$\mathrm{Ag}_{2} \mathrm{CO}_{3}:\left[\mathrm{Ag}^{+}\right]^{2}=\frac{8.1 \times 10^{-12}}{0.1} ;\left[\mathrm{Ag}^{+}\right]=9 \times 10^{-6}\, \mathrm{M}$
$\mathrm{Ag}_{3} \mathrm{AsO}_{4} ;\left[\mathrm{Ag}^{+}\right]^{3}=\frac{1 \times 10^{-22}}{0.1}:\left[\mathrm{Ag}^{+}\right]=10^{-7} \,\mathrm{M}$
