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. (Answer in mol/L)

Calculate the minimum Ag+ concentration required to cause precipitation of Ag3PO4. (Answer in mol/L)

Answer 1

given

[I^-]=9.6*10-5M

[PO4^3-]=0.960M

Ksp(AgI)= 8.3*10-17

Ksp(Ag₃PO₄)=8.9*10-17

substituting I-concentrtion and PO4^3- concntration in their respective solubility product expressions we can ge t minimum Ag+concentration requird for precipitation. because for a salt to precipitate ionic product has to be more than solubility product.

8.3*10-17=[Ag+]*(0.960*10^-5)

[Ag+]= 8.64*10^-12

Ksp= [Ag+]3*[PO4^3-]

8.9*10^-17= [Ag+]3* 0.960

[Ag+]3=9.27*10-17

[Ag+]=(9.27*10-17)^1/3

[Ag+]= 4.525*10^-6