Practice Problem 10

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

Compound              deltaGfo(kJ/mol) 

HCO2(aq)                    -372.3

H+(aq)                          0.00

HCO2-(aq)                   -351.0


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

HCO2H(aqequilibr.gif (63 bytes)  H+(aq) + HCO2-(aq)  

We then calculate the value of  deltaGo for this reaction:

deltaGo = sigmaGfo(products) - sigmaGfo(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  deltaGo and the equilibrium constant for the reaction:

deltaGo = - RT ln K

and solve for the natural log of the equilibrium constant:

Ex21_9a.gif (1097 bytes)

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

Ex21_10.gif (1664 bytes)

We can now calculate the value of the equilibrium constant:

K = e-8.60 = 1.8 x10-4

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

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