Abasic 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.

i think it's letter a.