Answer:
K s p = 3.37 × 10 ⁻⁸
Step-by-step explanation:
Anode (oxidation) half reaction:
P b (s) → Pb²⁺(aq) + 2e⁻ E° = 0.130 V
Cathode (reduction) half reaction:
PbSO₄(s) + 2e- → Pb(s) + SO₄²⁻(aq) Eº = -0.351 V
Adding the two half reactions together gives a net reaction equivalent to the solubility equilibrium of lead (II) sulfate salt:
PbSO₄(s) → Pb²⁺ + SO₄²⁻(aq)
Ksp = [Pb²⁺] [SO₄²⁻]
E ° c e l l = 0.130 - 0.351 = − 0.221 V
using the Nernst equation; -RTlnK = − n F E ° c el l
where R is molar gas constant, T is temperature, n is number of moles of electron, F is Faraday's constant,
-RTlnK = − ( 2 m o l ) ( 96485 C / m o l ) ( − 0.221 V )
-RTlnK = -42646 J / m o l
ln K = lnK s p = -( 42646 / R T)
lnK s p = -( 426466/ 8.314 × 298.15 )
Ksp = e∧(-17.204)
K s p = 3.37 × 10 ⁻⁸