What is the molar solubility of lead(II) fluoride in 0.10M NaF?

Calculation of Molar Solubility of Lead(II) Fluoride in 0.10 M NaF

Molar Solubility (s) of Lead(II) Fluoride: To calculate the molar solubility of lead(II) fluoride (PbF2) in 0.10 M NaF, we need to use the solubility product constant (Ksp) of PbF2, which is 3.6×10−8.

Solubility Product Constant Expression: The Ksp expression for PbF2 is Ksp = [Pb2+][F-]^2, where [F-] is the concentration of fluoride ions and [Pb2+] is the concentration of lead ions.

Given Concentration of NaF: Since we have 0.10 M NaF, we can assume that all the fluoride ions come from NaF, making [F-] equal to 0.10 M.

Assuming Molar Solubility of PbF2 is s: Let's assume the molar solubility of PbF2 is s, which means the concentration of lead ions [Pb2+] is also s.

Calculation: Substituting the values into the Ksp expression, we get 3.6×10−8 = (s)(0.10)^2. Solving for s gives us s = √(3.6×10−8 / 0.01) ≈ 6×10−5 mol/L.

Therefore, the molar solubility (s) of lead(II) fluoride in 0.10 M NaF is approximately 6×10−5 mol/L.