Question

What is the silver ion concentration [Ag+] in a solution prepared by mixing 395 mL 0.360 M395 mL 0.360 M silver nitrate with 463 mL 0.492 M463 mL 0.492 M sodium phosphate?

The Ksp of silver phosphate is 2.8×10−18.

Answer 1

The amount of excess reagent Na₃PO₄ remains unreacted = total number of moles of sodium phosphate - reacted number of moles of sodium phosphate with silver nitrate

reacted number of moles of sodium phosphate with silver nitrate can be obtained by using the balanced chemical equation.

3 mol of AgNO₃ reacts with 1 mol of sodium phosphate

0.142 mol of AgNO3 requires how many moles of sodium phosphate?

=> (0.142 mol AgNO₃ /3 mol AgNO3 ) X 1 mol sodium phosphate

= 0.0473mol of sodium phosphate

The amount of excess reagent Na₃PO₄ remains unreacted = 0.227 mol - 0.0472mol

= 0.1797 mol = 0.180mol

[PO₄^3-] = 0.180 mol / total volume in litres

= 0.180 mol / [(395 +463)/1000]L

= 0.210M

The dissociation of saturated solution of silver phosphate is as shown below:

Ag₃PO_4(S) <-> 3 Ag⁺_(aq) + PO₄^3-_(aq)

_{x 3x x+0.210M}

K_SP = ( 3X)³ . (X+0.210M)

Given Ksp = 2.8 x 10^-18

therefore :

2.8 x 10^-18 = ( 3X)³ . (X+0.210M)

On simplification we get the value of x. The concentration of Ag⁺ is 3x, so we should multiply the value of x with 3.

2.8 x 10^-18 = 27X³ . (X+0.210M)

as x<< 0.210

so , (X+0.210M) = 0.210M

2.8 x 10^-18 = 27X³ . (0.210M)

=> X³ = 2.8 x 10^-18 /27 x (0.210M)

=> X = CUBE ROOT OF [(2.8 x 10^-18 /27 x (0.210M)]

=> X = 7.89 x 10^-7 M

3 X is equal to  3 x 7.89 x 10^-7 M = 2.36 x 10^-6M

Hence,

[Ag⁺] = 2.36 x 10^-6M