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Deffense [45]
4 years ago
7

A solution containing a mixture of 0.0379 M potassium chromate ( K2CrO4 ) and 0.0739 M sodium oxalate ( Na2C2O4 ) was titrated w

ith a solution of barium chloride ( BaCl2 ) for the purpose of separating CrO2−4 and C2O2−4 by precipitation with the Ba2+ cation. The solubility product constants ( Ksp ) for BaCrO4 and BaC2O4 are 2.10×10−10 and 1.30×10−6 , respectively. Which barium salt will precipitate first?
Chemistry
1 answer:
motikmotik4 years ago
5 0

Answer:

BaCrO₄ will precipitate first.

Explanation:

For a generic salt (AB) that is soluble in some solvent, the solubility product constant (Kps) is:

AB(s) ⇄ A⁺ + B⁻

Kps = [A⁺]*[B⁻]

So, as greater is Kps, as higher is the solubility, and more difficult will be to form a precipitated (the solid form of the salt).

Kps of BaCrO₄ < Kps of BaC₂O₄, so BaCrO₄ will precipitate first.

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Answer:

NaCl.

Explanation:

In the solution, ZnSe ionizes to Zn^2^+ and Se^2^- . Following reaction represents the ionization of ZnSe in solution -

ZnSe ⇄ Zn^2^+ + Se^2^-

As we want to increase the solubility of ZnSe, we must decrease the concentration of dissociated ions so that the reaction continues to forward direction.

If we add NaCl to this solution, then we have Na^+ and Cl^- in the solution which will be formed by the ionization of NaCl.

Now, Zn^2^+ in the solution will react with two Cl^- ions to form ZnCl_2 as follows -

Zn^2^++2Cl^- ⇄ ZnCl_2

Due to this reaction the concentration of Zn^2^+ will decrease in the solution and more ZnSe can be soluble in the solution.

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3 years ago
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4 years ago
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