The balanced chemical equation for the dissolution of lead fluoride is:
PbF2(s) ⇌ Pb2+(aq) + 2F-(aq)
The solubility product expression for lead fluoride is:
Ksp = [Pb2+][F-]^2
At equilibrium, the cell potential (Ecell) is related to the Gibbs free energy change (ΔG) by the equation:
ΔG = -nFEcell
where n is the number of moles of electrons transferred in the cell reaction, and F is the Faraday constant (96,485 C/mol).
In this case, the cell reaction is:
Pb(s) + PbF2(s) ⇌ 2Pb2+(aq) + 2F-(aq)
Two moles of electrons are transferred, so n = 2.
The free energy change can be related to the equilibrium constant (K) by the equation:
ΔG = -RT ln(K)
where R is the gas constant (8.314 J/K·mol), and T is the temperature in Kelvin.
We can combine these equations to obtain an expression for Ksp:
Ksp = e^(-ΔG/RT) = e^(nFEcell/RT)
Substituting the given values:
n = 2
F = 96,485 C/mol
Ecell = 7.61×10^-2 V
R = 8.314 J/K·mol
T = 298 K
Ksp = e^(2 * 96,485 C/mol * 7.61×10^-2 V / (8.314 J/K·mol * 298 K))
Ksp = 3.01×10^-8
Therefore, the value of Ksp for lead fluoride at 298 K based on this experiment is 3.01×10^-8.