Calculating Ka and pKa for a Weak Acid from the pH of the Solution
The pH of 0.250 M HF is 2.036. What are the values of Ka and pKa for hydrofluoric acid?
STRATEGY
First, write the balanced equation for the dissociation equilibrium. Then, define x as the concentration of HF that dissociates and make the usual table under the balanced equation (Steps 1 and 2 of Figure 13.6 on page 522). Because x equals the H3O+ concentration, its value can be calculated from the pH. Finally, substitute the equilibrium concentrations into the equilibrium equation to obtain the value of Ka and take the negative log of Ka to obtain the pKa.
*A very small concentration of H3O+ is present initially because of the dissociation of water.
We can calculate the value of x from the pH:
x=[H3O+]=antilog(−pH)=10−pH=10−2.036=9.20×10−3MThe other equilibrium concentrations are
[F−]=x=9.20×10−3M[HF]=0.250−x=0.250−0.00920=0.241MSubstituting these concentrations into the equilibrium equation gives the value of Ka:
Ka=[HF][H3O+][F−]=(0.250−x)(x)(x)=0.241(9.20×10−3)(9.20×10−3)=3.51×10−4pKa=−logKa=−log(3.51×10−4)=3.455BALLPARK CHECK
Because the pH is about 2, [ H3O+ ] and [ F− ] are about 10−2 M and [HF] is about 0.25 M (0.250 M – 10−2 M). The value of Ka is therefore about (10−2 )(10−2 )/0.25, or 4 × 10−4 , and the pKa is between 3 and 4. The ballpark check and the solution agree.