Question

What is the lead (II) ion concentration in a solution prepared by mixing 353 mL of 0.448 M lead (II) nitrate with 461 mL of 0.316 M sodium fluoride? The Ksp of lead (II) fluoride is 3.6 × 10-8

Answer 1

Ksp = [Pb2+] [F-]²

After mixing,

total volume = 461+353 = 814 mL

[F-] = (0.316/461)814 = 0.5579 M

Ksp = [Pb2+] [F-]²

3.6 × 10^-8 = [Pb2+] [0.5579]²

[Pb2+]=11.56 × 10^-8 M

Pb²⁺ + 2F- give PbF2 (s)

Concentration of Pb2+ = (0.448/353) 814 = 1.033 M

As per the balnced equation one Pb2+, reacts with 2 F-. Thus Limiting reagent is F-

maximum concentration of Pb2+ unreacted = 0.754 M

Solubility of PbF2 is 11.56 × 10^-8 M

total amount of lead ion = 0.754 (solubility can be negleted)