A solution is 0.1M in Cl^- and 0.001M in $C r O_{4}^{2 -}$ . Solid AgNo₃ is gradually added to it. Assuming that the addition does not change in volume and K_sp(AgCl) = 1.7 * 10^-10M² and K_sp (Ag₂CrO₄) = 1.9 * 10^-12M³

Option 1 - Ag2CroO4 precipitates first because the amount of Ag+ of needed is low.

Option 2 - Ag2CrO4 precipitates first as its Ksp is low.

Option 3 - AgCl will precipitate first as the amount of Ag+ needed to precipitate is low.

Option 4 - AgCl precipitates first because its Ksp is high.

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Vishal Baghel | Contributor-Level 10

6 months ago

Correct Option - 3

Detailed Solution:

$[C l^{-}] = 1 0^{- 1} M$

$[C r {O_{4}}^{- -}] = 1 0^{- 3} M$

for AgCl ppt¹, ${[A g^{+}]}_{r e q} = \frac{1.7 * 1 0^{- 10}}{1 0^{- 1}} M$

For Ag₂CrO₄ppt², ${[A g^{+}]}_{r e q} = \sqrt[]{\frac{1.9 * 1 0^{- 12}}{1 0^{- 3}}} M$

${[A g^{+}]}_{r e q} = 4.3 * 1 0^{- 5} M$

Being lower concentration of [Ag⁺] in case of AgCl, it will precipitate first.

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A solution is 0.1M in Cl- and 0.001M in C r O 4 2 − . Solid AgNo3 is gradually added to it. Assuming that the addition does not change in volume and Ksp(AgCl) = 1.7 * 10-10M2 and Ksp (Ag2CrO4) = 1.9 * 10-12M3

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