Question

The maximum amount of silver phosphate that will dissolve in a 0.119 M silver acetate solution is? _______M

Answer 1

Maximum amount soluble in water = Solubility of that compound in water.

The solubility product of Ag₃PO₄ = 2.6 X 10^-18

Ag₃PO₄ 3Ag⁺ + PO₄^3-

Ksp = [Ag⁺]³ X [PO₄^3-] = (3S)³S = 27S⁴

or, 27S⁴ = 2.6 X 10^-18

or, S = 1.762 X 10^-5

Hence the solubility of Ag₃PO₄ in aqueous solution = 1.762 X 10^-5

Silver acetate is solublein water. Hence, in 0.119 M silver acetate solution, [Ag⁺] = 0.119 M

Now,

Ksp = [Ag⁺]³ X [PO₄^3-] Here, [Ag⁺] = 0.119 and [PO₄^3-] = S

or, 0.119 S= 2.6 X 10^-18

or, S = 1.762 X 10^-5/0.119 = 1.48 X 10^-4.

Hence the maximum amount of silver phosphate that will dissolve in a 0.119 M silver acetate solution is

1.48 X 10^-4 M