Answer is: concentration of Pb²⁺ must be exceeded is 3.3·10⁻⁴ M.
Chemical reaction : Pb²⁺(aq) + 2F⁻(aq) → PbF₂(s).
Ksp(PbF₂) = 3.3·10⁻⁸.
c(F⁻) = 0.01 M.
Ksp(PbF₂) = c(Pb²⁺) ·
c(F⁻)².
c(Pb²⁺) = Ksp(PbF₂) ÷ c(Cl⁻)².
c(Pb²⁺) = 3.3·10⁻⁸ ÷ (0.01 M)².
c(Pb²⁺) = 0.000000033 M³ ÷ 0.0001 M².
c(Pb²⁺) = 0.00033 M = 3.3·10⁻⁴ M.
The concentration of Pb²⁺ must be exceeded is 3.3 × 10⁻⁴ m.
Ksp value of PbF₂ = 3.3 × 10⁻⁸
Concentration of fluoride ion = 1.00 × 10⁻²
It is required to calculate the concentration of lead ion.
The solubility product constant, Ksp, is the equilibrium constant for a solid substance dissolving in an aqueous solution.
A chemical reaction between lead ion and fluoride ion occurs as
Chemical reaction :
Pb²⁺(aq) + 2F⁻(aq) → PbF₂(s).
Ksp(PbF₂) = 3.3 × 10⁻⁸
Ksp(PbF₂) = c(Pb²⁺) · c(F⁻)²
c(Pb²⁺) = Ksp(PbF₂) ÷ c(Cl⁻)²
c(Pb²⁺) = 3.3×10⁻⁸ ÷ (0.01 m)²
c(Pb²⁺) = 0.000000033 m³ ÷ 0.0001 m²
c(Pb²⁺) = 0.00033 m = 3.3·10⁻⁴ m
Hence, the concentration of Pb²⁺ must be exceeded is 3.3 × 10⁻⁴ m.
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