Question

A basic solution contains the iodide and phosphate ions that are to be separated via selective precipitation.

The I– concentration, which is 7.80×10-5 M, is 10,000 times less than that of the PO43– ion at 0.780 M .
A solution containing the silver(I) ion is slowly added. Answer the questions below.

Ksp of AgI is 8.30×10-17 and of Ag3PO4, 8.90×10-17.

Calculate the minimum Ag+ concentration required to cause precipitation of AgI.

Calculate the minimum Ag+ concentration required to cause precipitation of Ag3PO4.

Answer 1

For the precipitation of AgI remember the ksp definition

AgI ========= Ag⁺ + I^-

Ksp = [Ag⁺] [I^-] = 8.30×10^-17

8.30×10^-17 =  7.80×10^-5 * [I]

[I] = 8.30×10^-17 / 7.80×10^-5 = 1.064 x 10^-12

Ag₃PO₄ ---> 3 Ag⁺ + PO₄^3-, here we have a 3 stoichiometrical coefficient so we must consider it in the ksp equation

K_sp = [Ag⁺]³[PO₄^3-] = 8.9 x 10^-17

K_sp = [Ag⁺]³[0.78] = 8.9 x 10^-17

[Ag⁺]³ = 8.9 x 10^-17 / 0.78 = 1.141 x 10^-6

[Ag] = (1.141 x 10^-6)^1/3 = 4.85 x 10^-6