Question

7. A 25 ml aliquot of 0.200 M AgNO3 is mixed with 50 ml of 0.076 M K2CrO4 solution. Find the composition of the solution at equilibrium. Is the precipitation of silver quantitative? 8. How many grams of K2Cr04 should be added to a 1 liter saturated solution of Ag2Cr04 so that the concentration of chromate is exactly twice the concentration of silver? 9. Calculate the molar solubilities of silver acetate and silver chromate in a solution saturated by both salts. By what percentage is the solubility of Ag2CrO4 decreased by the presence of AgoAc?

Answer 1

7)

25 ml of 0.2M AgNO₃ is mixed with 50 ml of 0.076 K₂CrO₄.

Solubility product of Silver chromate (Ag2CrO4), k_sp = [Ag⁺]² x [CrO₄^2-] = 1.2 x 10^-12

Concentration of CrO₄^2- in final solution = 50*0.076/75   [since total volume of final solution is 75 ml]

= 0.05066 M

Solubility product is the ionic product of respective ions in solution. i.e. at equilibrium, the product of square of silver ion concentration times the chromate concentration is equal to solubility product.

Therefore, silver ion concentration at equilibrium, [Ag⁺] = (k_sp/[CrO₄^2-])^0.5

Substituting the value of [CrO₄^2-] = 0.05066M and k_sp = 1.2 x 10^-12, we get

[Ag⁺] = 4.866 x 10^-6 M

Atomic weight of silver = 107.8682 g

Therefore, amount of silver present in solution at equilibrium = 4.866 x 10^-6 x 107.8682 = 5.25 x 10^-4 g/l = 0.525 ppm of silver

This is quantitative precipitation of silver

8)

A 1 liter saturated solution of Ag₂CrO₄ will have  [Ag⁺]² x [CrO₄^2-] = 1.2 x 10^-12

Ag₂CrO₄ + H₂O -----> 2Ag⁺ + CrO₄^2-

so for, every x moles of Ag₂CrO₄ dissolving, there will be 2x moles of Ag⁺ and x moles of CrO₄^2-

=> (2x)²_*x = 1.2 x 10^-12

=> x = 6.789 x 10^-5

Solubility of silver chromate = 6.789 x 10^-5 M

K₂CrO₄ is added such that final concentration of chromate ions, [CrO₄^2-] = 2 x  [Ag⁺]

Since x moles of Ag₂CrO₄, gives x moles of CrO₄^2-, we need additional 3x moles of K₂CrO₄ to be added to get 4x moles of K₂CrO₄ (since each mole of K₂CrO₄ gives one mole of CrO₄^2-)

Therefore additional moles of K2CrO4 = 3 x 6.789 x 10^-5 moles = 2.03 x 10^-4 moles

Molecular weight of K2CrO4 = 194.1896 g

Therefore additional amount of K2CrO4 that needs to be added to 1 liter of saturated Ag₂CrO₄ solution = 2.03 x 10^-4 x 194.1896 = 3.94 x 10^-2 g

9)

Solubility of Silver acetate = 10.2 g/l

Molecular weight of Silver acetate = 166.912 g/mol

Concentration of silver acetate in its saturated solution = 10.2/166.912 M =  0.06111 M

Therefore concentration of silver ions = concentration of acetate ions = 0.06111 M

Solubility product of silver chromate, k_sp = [Ag⁺]² x [CrO₄^2-] = 1.2 x 10^-12

[Ag⁺] = 0.06111 M

Therefore, [CrO₄^2-] = k_sp/[Ag⁺]² = 3.213 x 10^-10 M

Since 1 mole of Ag₂CrO₄ gives 1 mole of CrO₄^2-, solubility of Ag₂CrO₄ in presence of saturated solution of silver acetate = 3.213 x 10^-10 M

In the absence of silver acetate, Solubility of Ag2CrO4 = 6.7589 x 10^-5 M (from problem 8)

Thus in presence of silver acetate, solubility of silver chromate decreases by 2.1 x 10⁵ times