Practice Problem 10

Use the following standard-state free energy of formation data to calculate the acid-dissociation equilibrium constant (K_a) at  for formic acid:

Compound              G_f^o(kJ/mol) 

HCO₂(aq)                    -372.3

H⁺(aq)                          0.00

HCO₂^-(aq)                   -351.0

Solution

We can start by writing the equation that corresponds to the acid-dissociation equilibrium for formic acid:

HCO₂H(aq)    H⁺(aq) + HCO₂^-(aq)  

We then calculate the value of  G^o for this reaction:

G^o = G_f^o(products) - G_f^o(reactants)

  = [1 mol H⁺ x 0.00 kJ/mol + 1 mol HCO₂^- x -351.0 kJ/mol] - [1 mol  HCO₂H x -372.3 kJ/mol]

        = 21.3 kJ

We now turn to the relationship between  G^o and the equilibrium constant for the reaction:

G^o = - RT ln K

and solve for the natural log of the equilibrium constant:

Substituting the known value of G^o, R, and T into this equation gives the following result:

We can now calculate the value of the equilibrium constant:

K = e^-8.60 = 1.8 x10^-4

The value of K_a obtained from this calculation agrees with the table value for formic acid, within experimental error.