Question

6. What is the silver ion concentration in a solution prepared by mixing 389 mL of 0.391 M silver nitrate with 441 mL of 0.487 M sodium phosphate? The Ksp of silver phosphate is 2.8

Answer 1

volume of AgNO3 , V1= 389 mL = 0.389 L

molarity of AgNO3 , M1 = 0.391 M

Moles of AgNO3 = M1xV1 = 0.389 L x 0.391 M = 0.152 mol

volume of Na3PO4 , V2= 441 mL = 0.441 L

molarity of Na3PO4 , M2 = 0.487 M

Moles of Na3PO4 = M2xV2 = 0.441 L x 0.487 M = 0.215 mol

Concentration of AgNO3 after they are added = 0.152 mol / (0.389+0.441) L = 0.183 M

Concentration of Na3PO4 after they are added = 0.215 mol / (0.389+0.441) L = 0.178 M

Ksp of Ag3PO4 = 2.8x10^-18

Ionic product (IP) of Ag3PO4 = [Ag⁺ ]³x[PO₄^3- ] = (0.183 M)³ x (0.178 M) = 0.0011 which is greater than Ksp of Ag3PO4. Hence precipitation of Ag3PO4 will occur and the net ionic equation is

3 Ag+ (aq)+ PO4^3- (aq)-------------> Ag3PO4 (s)

3 moles of AgNO3 reacts with 1 mole of Na3PO4.

Hence 0.152 mol of AgNO3 that will react with the moles of Na3PO4 = 0.152/3 = 0.051 mol to form 0.051 mole of Ag3PO4

Hence moles of Na3PO4 remained in the solution = 0.215 mol - 0.051 mol = 0.164 mol

Hence molarity of Na3PO4 ( or PO43-) after reaction = 0.164 mol / (0.389+0.441) L = 0.198 M

For the dissociation of Ag3PO4

Ag3PO4 (s)-------------> 3 Ag+ (aq)+ PO4^3- (aq)

Ksp = 2.8x10^-18   = [Ag⁺ ]³x[PO₄^3- ] = [Ag⁺ ] x 0.198 M

=> [Ag⁺ ] = cube root ( 2.8x10^-18 /0.198) = 2.42x10^-6 M   (answer)