A solution is 0.1 Min each of ions ${Cl}^{-}, I^{-}, {CO}_{3}^{2 -}$ and ${AsO}_{4}^{3 -}$ . To this solution, ${Ag}^{+}$ ions is gradually added, the salt which starts precipitating first is

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$AgCl (K_{s} = 1.8 \times 10^{- 10} M^{2})$

${Ag}_{3} {AsO}_{4} (K_{s} = 1.0 \times 10^{- 22} M^{4})$

${Ag}_{2} {CO}_{3} (K_{s} = 8.0 \times 10^{- 12} M^{3})$

$AgI (K_{s} = 8.0 \times 10^{- 17} M^{2})$

answer is B.

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Detailed Solution

The concentration of ${Ag}^{+}$ ions needed to precipitate given ions are as follows.

$AgCl : [{Ag}^{+}] = \frac{K_{s} (AgCl)}{[{Cl}^{-}]} = \frac{1.8 \times 10^{- 10} M^{2}}{0.1 M} = 1.8 \times 10^{- 9} M$

$AgI : [{Ag}^{+}] = \frac{K_{s} (AgI)}{[I^{-}]} = \frac{8.0 \times 10^{- 17} M^{2}}{0.1 M} = 8.0 \times 10^{- 17} M$

${Ag}_{2} {CO}_{3} : [{Ag}^{+}] = {(\frac{K_{s} ({Ag}_{2} {CO}_{3})}{[{CO}_{3}^{2 -}]})}^{1 / 2} = {(\frac{8.0 \times 10^{- 12} M^{3}}{0.1 M})}^{1 / 2} = 8.94 \times 10^{- 6} M$

${Ag}_{3} {AsO}_{4} : [{Ag}^{+}] = {(\frac{K_{s} ({Ag}_{3} {AsO}_{4})}{[{AsO}_{4}^{3 -}]})}^{1 / 3} = {(\frac{1.0 \times 10^{- 22} M^{4}}{0.1 M})}^{1 / 3} = 1.0 \times 10^{- 7} M$ ${Ag}^{+}$ ions needed to precipitated out $AgI$ is minimum. This salt precipitates first.