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What are the equilibrium concentrations of [tex]\text{Pb}^{2+}[/tex] and [tex]\text{F}^-[/tex] in a saturated solution of lead(II) fluoride [tex]\text{PbF}_2[/tex] if the [tex]K_{sp}[/tex] of [tex]\text{PbF}_2[/tex] is [tex]3.20 \times 10^{-8}[/tex]?

A) [tex]\text{Pb}^{2+} = 2.00 \times 10^{-3} \text{ M}[/tex], [tex]\text{F}^- = 4.00 \times 10^{-3} \text{ M}[/tex]

B) [tex]\text{Pb}^{2+} = 4.00 \times 10^{-3} \text{ M}[/tex], [tex]\text{F}^- = 8.00 \times 10^{-3} \text{ M}[/tex]

C) [tex]\text{Pb}^{2+} = 8.00 \times 10^{-3} \text{ M}[/tex], [tex]\text{F}^- = 4.00 \times 10^{-3} \text{ M}[/tex]

D) [tex]\text{Pb}^{2+} = 1.00 \times 10^{-3} \text{ M}[/tex], [tex]\text{F}^- = 2.00 \times 10^{-3} \text{ M}[/tex]

Answer :

Final answer:

The equilibrium concentrations of Pb²⁺ and F⁻ in a saturated solution of lead(II) fluoride PbF₂ are found to be A) Pb²⁺ = 2.00✖ 10⁻ ³M and F⁻ = 4.00 ✖10⁻ ³M.

Explanation:

The equilibrium concentrations of Pb²⁺ and F⁻ in a saturated solution of lead(II) fluoride PbF₂ can be determined using the given Kₛₚ value. According to the dissociation equation, for every mole of PbF₂ that dissociates, 1 mol of Pb²+ and 2 mol of F⁻ are produced.

Applying the equilibrium calculations, the correct equilibrium concentrations are Pb²⁺ = 2.00✖ 10⁻ ³M and F⁻ = 4.00 ✖10⁻ ³M, which matches option A.

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