A basic solution contains the iodide and phosphate ions that are to be separated via selective precipitation. The I - concentration, which is 7.60*10-5 M, is 10,000 times less than that of the PO43 - ion at 0.760 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*10-17.1.) Calculate the minimum Ag + concentration required to cause precipitation of AgI. 2.) Calculate the minimum Ag + concentration required to cause precipitation of Ag3PO4.3.) Which salt will precipitate first?4.) Is the separation of the I - and PO43 - ion complete (is the percentage of "remaining" ion less than 0.10% or its original value) ?

1) 1.092x10⁻¹² M

2) 4.89x10⁻⁶ M

3) Ag₃PO₄

4) Yes, it is.

Explanation:

1) The dissolution reaction of AgI is:

AgI (s) ⇄ Ag⁺ (aq) + I⁻ (aq)

The constant of equilibrium Kps is:

Kps = [Ag⁺]*[I⁻]

So, the minimum concentration of Ag⁺ is that one in equilibrium

8.30x10⁻¹⁷ = [Ag⁺]*7.60x10⁻⁵

[Ag+] = 1.092x10⁻¹² M

2) The dissolution reaction of Ag₃PO₄ is:

Ag₃PO₄ (s) ⇄ 3Ag⁺ (aq) + PO₄³⁻ (aq)

The constant of equilibrium, Kps is:

Kps = [Ag⁺]³*[PO₄³⁻]

So, the minimum concentration of Ag⁺ is that one in equilibrium

8.90x10⁻¹⁷ = [Ag⁺]³*0.760

[Ag⁺]³ = 1.171x10⁻¹⁶

[Ag⁺] = 4.89x10⁻⁶ M

3) Because Ag₃PO₄ needs less Ag⁺ to precipitated, it will precipitate first.

4) To see if the separation will occur, let's verify is the ratio of the molar concentration of Ag⁺ for the substances is less then 0.1%:

(1.092x10⁻¹²) / (4.89x10⁻⁶) * 100% = 2.23x10⁻⁵ %

So, the separation is complete.