1. Understand the Chemistry
* NaF is the salt of a weak acid (HF) and a strong base (NaOH).
* When NaF dissolves in water, it dissociates completely into Na+ and F- ions.
* The F- ion is the conjugate base of the weak acid HF, and it will undergo hydrolysis (reaction with water) to produce OH- ions, making the solution basic.
2. Set up the Hydrolysis Equilibrium
The hydrolysis reaction for F- is:
F- (aq) + H2O (l) ⇌ HF (aq) + OH- (aq)
3. Use the Kb Expression
We need to find the Kb (base dissociation constant) for the fluoride ion. We can calculate it using the Kw (ion product constant of water) and Ka (acid dissociation constant) for HF:
* Kw = Ka * Kb
* Kb = Kw / Ka
The Ka for HF is 6.8 x 10^-4. Kw is 1.0 x 10^-14.
* Kb = (1.0 x 10^-14) / (6.8 x 10^-4) = 1.47 x 10^-11
4. Set Up an ICE Table
| | F- | HF | OH- |
|-----|---------|----------|-----------|
| I | 0.89 M | 0 | 0 |
| C | -x | +x | +x |
| E | 0.89-x | x | x |
5. Solve for x (the Hydroxide Concentration)
Kb = [HF][OH-] / [F-]
1.47 x 10^-11 = x * x / (0.89 - x)
Since Kb is very small, we can assume x is negligible compared to 0.89:
1.47 x 10^-11 ≈ x^2 / 0.89
x^2 ≈ 1.31 x 10^-11
x ≈ 3.62 x 10^-6 M (This is the concentration of OH-)
6. Calculate pOH and pH
* pOH = -log[OH-] = -log(3.62 x 10^-6) ≈ 5.44
* pH + pOH = 14
* pH = 14 - pOH ≈ 8.56
Therefore, the pH of a 0.89 M NaF solution is approximately 8.56.
