Question

A basic solution contains the iodide and phosphate ions that are to be separated via selective precipitation. The I concentration, which is 8.80x10 5M, is 10,000 times less than that of the PO4on at 0.880 M.A solution containing the silver() ion is slowly added. Answer the questions below.Ksp of Agl is 8.30x1017 and of Ag3PO4, 8.90x1017 1st attempt Part 1 (1 point) See Periodic Table ? See Hint Calculate the minimum Ag concentration required to cause precipitation of Agl mol/L Part 2 (1 point) Calculate the minimum Ag concentration required to cause precipitation of Ag3PO mol/L Part 3 (1 point) Which salt will precipitate first? Choose one: AgsPO4 Cannot be determined. Part4 (1 point) is the separation of the l-and PO43-ion complete (is the percentage of "remaining" ion less than 0.10% or its original value)? Choose one: o Yes Cannot be determined

Answer 1

a)

min. Ag+ required for AgI

Ksp = [Ag+][I-]

(8.3*10^-17) = [Ag+](8.8*10^-5)

[Ag+] = (8.3*10^-17)/(8.8*10^-5)

[Ag+] = 9.4318*10^-13 M required

b)

find Ag+ for Ag3PO4

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

8.9*10^-17 =  [Ag+]^3 * (0.88)

[Ag+] = ((8.9*10^-17)/0.88)^(1/3)

[Ag+] = 0.0000046591 = 4.65*10^-6 M

c)

clearly,

[Ag+] for AgI requires much less than Ag3PO4

AgI should form first

D)

apply

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

[Ag+] = (8.9*10^-17) = (s + 9.4318*10^-13)^3 * (s)

since [PO43-] >> [I-]

then choose "yes" it can be