Understanding the Equilibrium
* HF is a weak acid: It only partially ionizes in water.
* Equilibrium: The reaction reaches a point where the rates of the forward and reverse reactions are equal.
* Ka: The acid dissociation constant, which represents the extent to which an acid ionizes in solution.
Setting Up the ICE Table
We'll use an ICE (Initial, Change, Equilibrium) table to track the concentrations of the species involved:
| Species | Initial (I) | Change (C) | Equilibrium (E) |
|---|---|---|---|
| HF | | -x | |
| H+ | 0 | +x | |
| F- | 0 | +x | |
Calculations
1. Find [H+] from pH:
* pH = -log[H+]
* 5.4 = -log[H+]
* [H+] = 10^-5.4 M
2. Determine the change (x):
* Since [H+] = x at equilibrium, x = 10^-5.4 M
3. Fill in the ICE table:
* We know [F-] at equilibrium is 3.5 M, and this is equal to the change (x) plus the initial concentration of F-:
* 3.5 = x + [F-]initial
* [F-]initial = 3.5 - x = 3.5 - 10^-5.4 ≈ 3.5 M
* Now we can fill in the rest of the table:
| Species | Initial (I) | Change (C) | Equilibrium (E) |
|---|---|---|---|
| HF | 1.2 x 10^-6 | -10^-5.4 | 1.2 x 10^-6 - 10^-5.4 |
| H+ | 0 | +10^-5.4 | 10^-5.4 |
| F- | 3.5 | +10^-5.4 | 3.5 |
4. Calculate Ka:
* Ka = ([H+][F-])/[HF]
* Ka = (10^-5.4 * 3.5) / (1.2 x 10^-6 - 10^-5.4)
* Ka ≈ 7.2 x 10^-4
Therefore, the Ka value for HF at this temperature is approximately 7.2 x 10^-4.
