Answer to Question #237099 in General Chemistry for Nana

Question

What mass of NaF must be added to 1.0 L of 0.0025 M Pb²⁺ to initiate precipitation of PbF₂(s)? (K_sp of PbF₂ is 3.3 × 10^-8; assume no volume change on addition of NaF.) (Please give your answer with 2 significant figures.)

Answer 1

Equilibrium: PbF2(s) ←→ Pb2+(aq) + 2 F-(aq)

Ksp = 3.3X10^-8 = [Pb2+][F-]^2

Since [Pb2+] = 3.4X10^-3,

Ksp = 3.3X10^-8 = (3.4X10^-3)[F-]^2

[F-] = 3.1X10^-3 M

The molar mass of NaF is 42.0 g/mol, and each mole NaF provides 1 mol F-. So,

3.1X10^-3 mol/L F- X 1 L X 42.0 g/mol = 0.13 g NaF