07 Question (4 points) a See page 816 A basic solution contains the iodide and p

Question

07 Question (4 points) a See page 816 A basic solution contains the iodide and phosphate ions that are to be separated via selective precipitation. The I concentration, which is 9.20x10 M, is 10,000 times less than that of the PO43 ion at 0.920 M.A solution containing the silverl)ion is slowly added.Answer the questions below. Kp of Agl is 8.30x10-17 and of Ag PO4 8.90x10-17 2nd attempt d See Periodic Table See Hint Part 1 (1point) Feedback Calculate the minimum Ag concentration required to cause precipitation of Agl. 1.02 x 101mol/L Part2 (1 point)Feedback Calculate the minimum Ag concentration required to cause precipitation of AggPO4 4.8 x 106mol/L

Explanation / Answer

Dissociation is as
AgI Ag+ + I¯

Ksp of AgI is 8.3 x 10^-17

Let the solubility of AgI be x. Then, Ksp = [Ag+][I-] = x.x = x2

solubility = x = = 9.11 x 10-9 mol/L
Dissociation is as

Ag3PO4 ---> 3 Ag+ + PO43-

Ksp = [Ag+]3[PO43-] = 8.9 x 10-17

Let solubility of Ag3PO4 be x. Then, Ksp = x3.x = 8.9 x 10-17

= 9.71 x 10-5 mol/L

So the solubility of Ag3PO4 is larger than AgI, so Ag3PO4 is more soluble. Therefore, AgI would precipitate first.

(a) Minimum Ag+ required to precipitate AgI is:

Ksp = [Ag+][I-]

8.3 x 10-17 = [Ag+](9.20*10^-5)

[Ag+] = 9.02 x 10-13

b) minimum Ag+ required to precipitate phosphate ions is:

Ksp = [Ag+]3[PO43-]

8.9 x 10-17 = [Ag+]^3(0.920)

[Ag+]= 4.59 x 10-6 mol/L

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