Question

What is the silver ion concentration in a solution prepared by mixing 337 mL of 0.376 M silver nitrate with 475 mL of 0.559 M sodium phosphate? The Ksp of silver phosphate is 2.8 × 10-18

Answer 1

Solution:

Reaction :

3 AgNO3 (aq) + Na3PO4 (aq) --- > Ag3PO4(s) +3 NaNO3 (aq)

From this reaction mole ratio between reactants = 1 : 1

Calculation of moles of each :

n AgNO3 = volume in L x molarity = 0.337 L x 0.376 M = 0.127 mol AgNO3

n Na3PO4 = 0.475 L x 0.559 M =0.265 mol Na3PO4

Now we calculated limiting reactant.

Moles of AgNO3 required to react with 0.265 mol sodium phosphate

= 0.265 mol sodium phosphate x 1 mol silver nitrate / 1 mol sodium phosphate

= 0.265 mol silver nitrate

But in this reaction we only have 0.127 mol silver nitrate so silver nitrate is limiting reactant.

Limiting reactant consumed completely in the reaction.

Now we have to use ksp value of silver phosphate

Ksp value of silver phosphate = 8.89 E-17

Ksp = [Ag+]³ [PO4^3-]

8.89 E-17 = x³ x

X⁴ = 8.89 E-17

x = 9.71 E-5 mol/L

concentration of Ag+ = 3 x = 3 x 9.71 E-5 M = 2.9 E -4 M

[Ag+] = 2.9 E-4 M