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

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

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.

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