If a solution containing 0.10 M Cl-, Br-, I-, and LaTeX: CrO_4^2-C r O 4 2 − is treated with Ag+, in what order will the anions precipitate? From first to last: (Ksp of AgCl: 1.8e-10, Ksp of AgBr: 5.0e-13, Ksp of AgI: 8.3e-17, Ksp of LaTeX: Ag_2CrO_4A g 2 C r O 4: 1.2e-12)

The first thing to have in mind is that the Ksp (constant of solubility) gives you an idea of how much soluble a salt is. If the concentration of ions of that salt exceeds the kps, the salt will precipitate.

The Kps of all these salts are very small compared to the 0.1 M concentration of the anions so all of them will precipitate.

The smaller the constant, the faster the salt precipitates

Having that concept in mind, the order of precipitation from first to last would be:

[tex]AgI \longrightarrow AgBr \longrightarrow Ag_2CrO_4 \longrightarrow AgCl[/tex]

ajeigbeibraheem ajeigbeibraheem · Answer 2 · 2021-11-13T08:27:56+01:00

The order in which the anions will precipitate is: I⁻ > Br⁻ > Cl⁻ > CrO₄²⁻

Precipitation in an aqueous solution is the method of changing a dissolved substance to an insoluble solid via a supersaturated solution.

However, since the solubility of a substance is based on its Ksp Values, Then, the Ksp Value and solubility vary directly proportional to each other.

Ksp of AgCl = 1.8 × 10⁻¹⁰
Ksp of AgBr = 5.0 × 10⁻¹³
Ksp of AgI = 8.3 × 10⁻¹⁷
Ksp of AgCrO₄ = 1.2 × 10⁻¹²

The higher the Ksp value, the more the solubility and, the lesser the precipitation. Thus, the order is as follows: I⁻ > Br⁻ > Cl⁻ > CrO₄²⁻

Therefore, in a solution of 0.10 M Cl-, Br-, I-, and [tex]\mathbf{CrO_4^2-}[/tex], the order in which the anions will precipitate is: I⁻ > Br⁻ > Cl⁻ > CrO₄²⁻

Learn more about Ksp value here:

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