What concentration of the lead ion, pb2+, must be exceeded to precipitate pbf2 from a solution that is 1.00×10−2 m in the fluori
de ion, f−? ksp for lead(ii) fluoride is 3.3×10−8 ?
1 answer:
Answer is: concentration
of Pb²⁺ must be exceeded is 3.3·10⁻⁴ M.
<span>
Chemical reaction : </span>Pb²⁺(aq) + 2F⁻(aq) → PbF₂(s).
<span>
Ksp(PbF</span>₂) = 3.3·10⁻⁸.<span>
c(F</span>⁻) = 0.01 M.<span>
Ksp(PbF</span>₂) = c(Pb²⁺) ·
c(F⁻)².<span>
c(Pb²</span>⁺) = Ksp(PbF₂) ÷ c(Cl⁻)².<span>
c(Pb²</span>⁺) = 3.3·10⁻⁸ ÷ (0.01 M)².<span>
c(Pb²</span>⁺) = 0.000000033 M³ ÷ 0.0001 M².<span>
c(Pb²</span>⁺) = 0.00033 M = 3.3·10⁻⁴ M.
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