Question 15.9: The pH of 0.250 M HF is 2.036. What are the values of Ka and......
The pH of 0.250 M HF is 2.036. What are the values of K_{a} and pK_{a} for hydrofluoric acid?
STRATEGY
Use the general procedure for solving equilibrium problems described in Section 14.5. First, write the balanced equation for the dissociation equilibrium. Then, add known quantities to the “initial,” “change,” and “equilibrium” rows in the equilibrium table. Define x as the concentration of HF that dissociates. Because x equals the H_{3}O^{+} 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}.
| IDENTIFY | |
| Known | Unknown |
| Concentration of HF (0.250 M) | K_{a}, pK_{a} |
| pH = 2.036 | |
We can calculate the value of x from the pH:
[H_{3}O^{+}]=antilog (-pH)=10^{-pH}=10^{-2.036}=9.20\times 10^{-3} MThe other equilibrium concentrations are
[F^{-}]=x=9.20\times 10^{-3} M[HF] = 0.250 – x = 0.250 – 0.009 20 = 0.241 M
Substituting these concentrations into the equilibrium equation gives the value of K_{a}:
K_{a}=\frac{[H_{3}O^{+}][F^{-}]}{[HF]}=\frac{(x)(x)}{(0.250 – x)}=\frac{(9.20\times 10^{-3})(9.20\times 10^{-3})}{0.241}=3.51\times 10^{-4}pK_{a}=- log K_{a}=-log (3.51\times 10^{-4})=3.455
| HF(aq) | + | H_{2}O(l) | \xrightleftharpoons{} | H_{3}O^{+}(aq) | + | F^{-}(aq) | |
| Initial concentration (M) | 0.250 | ∼0* | 0 | ||||
| Change (M) | -x | +x | +x | ||||
| Equilibrium concentration (M) | (0.250 – x) | x | x | ||||
| *A very small concentration of H_{3}O^{+} is present initially because of the dissociation of water. | |||||||