Question

an acid-base titration was preformed to determine the Ksp of a slightly soluble salt. the Ksp was experimentally determined by 80°C, and the detaG° was determined to be 25.28 KJ/mol. if deltaS° is -115 J/mol•k, what is delta H in KJ/mol for this slightly soluble salt?

Answer 1

∆G° = ∆H° - T∆S°

Given that,

∆G° = 25.28 kJ/mol

∆S° = -115 J / mol.K × ( 1 kJ/1000 J)

∆S° = -0.115 kJ/mol.K

T = 80+273 = 353 K

∆H° = ∆G° + T∆S°

∆H° = 25.28 kJ/mol + ( 353 K × (-0.115 kJ/mol.K)

∆H° 25.28 kJ/mol - 40.595 kJ/mol

∆H° = -15.315 kJ/mol

Hence ∆H° of the salt is -15.32 kJ/mol