Question

What is the minimum mass of Na₂CO₃ that must be added to 39.7 mL of a 5.3 × 10^-4M AgNO₃ solution in order for precipitation to occur?

For Ag₂CO_3. K_sp = 8.6 × 10^–12

A.1.1 × 10^-3 g

B.2.2 × 10^-3 g

C.3.1 × 10^-4 g

D.6.8 × 10^-8 g

E.1.3 × 10^-4 g

Answer 1

K_sp for Ag₂CO₃ = 8.6 10^-12

8.6 10^-12 = [Ag⁺]²[CO₃^2-]

8.6 10^-12 = [5.3 10^-4]²[CO₃^2-]

[CO₃^2-] = 3.06 10^-5 M

No. of moles of Na₂CO₃ = M V = 3.06 10^-5 0.0397 = 1.22 10^-6 mol

Molecular Weight of Na₂CO₃ = 106 g/mol

Mass of Na₂CO₃ = 1.22 10^-6 106 = 1.3 10^-4 g

So the answer is E.