Question

A basic solution contains the iodide and phosphate ions that are to be separated via selective precipitation. The I– concentration, which is 9.60×10-5 M, is 10,000 times less than that of the PO43– ion at 0.960 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

¡) Dissolution equillibrium of AgI

AgI(s) <--------> Ag+(aq) + I-(aq)

Ksp = [Ag+] [I-]= 8.30×10^-17

given concentration of I- =9.60×10^-5M

Therefore,

[Ag+]×9.60×10^-5M = 8.30×10^-17M^2

[ Ag+ ] = 8.65×10^-13M

Therefore,

Minimum concentration of Ag+ required to cause precipitation of AgI is 8.65×10^-13M

ii) Dissociation equillibrium of Ag3PO4 is

Ag3PO4(s) --------> 3Ag+(aq) + PO4^3-

Ksp = [ Ag+ ]^3 [ PO4^3- ] = 8.90×10^-17

given concentration of PO4^3- = 0.960M

So,

[ Ag+ ]^3 × 0.960M = 8.90×10^-17M^4

[ Ag+ ]^3 = 9.27×10^-17M^3

[ Ag+ ] = 4.53 × 10^-6M

Therefore,

the minimum Ag+ concentration to cause precipitation of Ag3PO4 is 4.53×10^-6M