Question

Using the thermodynamic data provided below, calculate Ksp for Mg(OH)2(5) at 25 AH。f (kJ/mol) So (J/K.mol) Mg(aq) 117.99 -10.5 63.1 2+ -461.96 -229.94 -924.66 OH-(aq) Mg(OH2(5)

Answer 1

Answer

8.90×10^-12

Explanation

i)Solubility equillibrium of Mg(OH)2 is

Mg(OH)2(s) <------> Mg²⁺(aq) + 2OH^-(aq)

ii) ∆H^° for this reaction is calculated by the following formula

∆H° = ∆H°_f(products ) - ∆H°_f(reactants)

∆H° = (1× -461.96kJ/mol)+ (2×-229.94kJ/mol))- ( 1×-924.66kJ/mol)

∆H° = -921.84kJ/mol + 924.66 kJ/mol

∆H° = 2.82kJ/mol

iii) ∆S° for this reaction is calculated as follows

∆S° = S° (products) - S°(reactants)

=((-117.99J/K mol )+ (2×-10.5J/K mol) ) - (63.10J/ K mol))

= -202.09 J/K mol

iv) ∆G° = ∆H° - T∆S°

= 2820J/mol - (298.15K× (-202.09J/ K mol))

= 2820J /mol + 60253J/mol

= 63073J/mol

v) Equillibrium constant K is calculated as follows

∆G° = - RTlnK

lnK = -∆G°/RT

2.303logK = -∆G°/RT

logK = -∆G°/2.303RT

K = 1×10^{-∆G°/2.303RT}

  K = 1× 10- 63073J/mol/(2.303×8.314(J/K mol)×298.15K)

  K = 1× 10^-11.05

  K = 8.91×10^-12

Therefore

The answer is 8.90×10^-12