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 fluoride ion, f−? ksp for lead(ii) fluoride is 3.3×10−8 ?

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.

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

lol for the pointsss