**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)

HCO_{2}(*aq*)
-372.3

H^{+}(*aq*)
0.00

HCO_{2}^{-}(*aq*)
-351.0

**Solution**

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

HCO_{2}H(*aq*) H^{+}(*aq*) +
HCO_{2}^{-}(*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 HCO_{2}^{-} x
-351.0 kJ/mol] - [1 mol HCO_{2}H 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.