Question

1. 200 mL of an aqueous solution contains 0.030 M concentrations of both Pb2+ and Ag+. If 100 mL of 6.0 x 10-2 M NaCl is added to this solution will a precipitate form? If so, what will the precipitate be? The Ksp values for PbCl2 and AgCl are 1.7 x 10-5 and 1.8 x 10-

Answer 1

No. of moles of Pb²⁺ = Molarity x Volume

= 0.03 M x 2 L = 0.06 mol

No. of moles of Ag²⁺ = Molarity x Volume

= 0.03 M x 2 L = 0.06 mol

No. of moles of Cl^- = Molarity x Volume

= 0.06 M x 1 L = 0.06 mol

Total volume of solution = 200 mL + 100 mL = 300 mL = 3 L

[Pb²⁺] = No. of moles / Volume (L) = 0.06 mol / 3 L = 0.02 M

[Ag²⁺] = No. of moles / Volume (L) = 0.06 mol / 3 L = 0.02 M

[Cl^-] = No. of moles / Volume (L) = 0.06 mol / 3 L = 0.02 M

PbCl₂ (aq) $\rightleftharpoons$ Pb²⁺ (aq) + 2 Cl^- (aq)

AgCl (aq) $\rightleftharpoons$ Ag⁺ (aq) + Cl^- (aq)

Reaction quotient = Q_PbCl2 = [Pb²⁺][Cl^-]²

Putting the values we get, Q_PbCl2 = (0.02)(0.02)² = 8 x 10^-6

Ksp of PbCl₂ = 1.7 x 10^-5 = 17 x 10^-6

Since Ksp>Q, hence the precipitate PbCl₂ will not form.

Reaction quotient = Q_AgCl = [Ag⁺][Cl^-]

Putting the values we get, Q_AgCl = (0.02)(0.02)= 4 x 10^-4

Ksp of AgCl= 1.8 x 10^-10

Since Q>Ksp, hence the precipitate AgClwill form.