A solution is 0.1 M in Cl^{-} and 0.001 M i | vedclass

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(A) $(i)$ $	ext{Concentration of } [Ag^{+}] 	ext{ required to precipitate } AgCl_{(s)}:$$K_{sp} = [Ag^{+}][Cl^{-}] = 1.7 	imes 10^{-10}$$[Ag^{+}] =

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$AgCl$ will precipitate first as the amount of $Ag^{+}$ needed to precipitate is low.

Vedclass · Answer

(A) $(i)$ $	ext{Concentration of } [Ag^{+}] 	ext{ required to precipitate } AgCl_{(s)}:$$K_{sp} = [Ag^{+}][Cl^{-}] = 1.7 	imes 10^{-10}$$[Ag^{+}] = \frac{1.7 	imes 10^{-10}}{0.1} = 1.7 	imes 10^{-9} \ M$$(ii)$ $	ext{Concentration of } [Ag^{+}] 	ext{ required to precipitate } Ag_{2}CrO_{4(s)}:$$K_{sp} = [Ag^{+}]^{2}[CrO_{4}^{2-}] = 1.9 	imes 10^{-12}$$[Ag^{+}]^{2} = \frac{1.9 	imes 10^{-12}}{0.001} = 1.9 	imes 10^{-9}$$[Ag^{+}] = \sqrt{1.9 	imes 10^{-9}} \approx 4.36 	imes 10^{-5} \ M$Since the $[Ag^{+}]$ required to precipitate $AgCl$ $(1.7 	imes 10^{-9} \ M)$ is lower than that required for $Ag_{2}CrO_{4}$ $(4.36 	imes 10^{-5} \ M)$,$AgCl$ will precipitate first.

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