Question 14.9: Calculating Ka and pKa for a Weak Acid from the pH of the So......
Calculating K_a and pK_a 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 K_a and pK_a 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 H_3O^+ concentration, its value can be calculated from the pH. Finally, substitute the equilibrium concentrations into the equilibrium equation to obtain the value of K_a and take the negative log of K_a to obtain the pK_a.
*A very small concentration of H_3O^+ is present initially because of the dissociation of water.
We can calculate the value of x from the pH:
x=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\operatorname{antilog}(-\mathrm{pH})=10^{-\mathrm{pH}}=10^{-2.036}=9.20 \times 10^{-3} \mathrm{M}The other equilibrium concentrations are
\begin{aligned}& {\left[\mathrm{F}^{-}\right]=x=9.20 \times 10^{-3} \mathrm{M}} \\& {[\mathrm{HF}]=0.250-x=0.250-0.00920=0.241 \mathrm{M}}\end{aligned}Substituting these concentrations into the equilibrium equation gives the value of K_a:
K_{\mathrm{a}}=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{F}^{-}\right]}{[\mathrm{HF}]}=\frac{(x)(x)}{(0.250-x)}=\frac{\left(9.20 \times 10^{-3}\right)\left(9.20 \times 10^{-3}\right)}{0.241}=3.51 \times 10^{-4} \\ \mathrm{p} K_{\mathrm{a}}=-\log K_{\mathrm{a}}=-\log \left(3.51 \times 10^{-4}\right)=3.455BALLPARK CHECK
Because the pH is about 2, [ H_3O^+ ] 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 pK_a is between 3 and 4. The ballpark check and the solution agree.