1. Understand the Chemistry
* Sodium acetate (NaCH₃COO) is the salt of a weak acid (acetic acid, CH₃COOH) and a strong base (sodium hydroxide, NaOH).
* When sodium acetate dissolves in water, it dissociates completely into sodium ions (Na⁺) and acetate ions (CH₃COO⁻).
* The acetate ions will react with water in an equilibrium reaction to produce hydroxide ions (OH⁻) and acetic acid (CH₃COOH), making the solution slightly basic.
2. Set up the Equilibrium
The relevant equilibrium reaction is:
CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq)
3. Use the Kb Expression
We need to use the base dissociation constant (Kb) for the acetate ion to calculate the hydroxide ion concentration:
Kb = [CH₃COOH][OH⁻] / [CH₃COO⁻]
* We can find the Kb value for acetate in a table (look up the Ka for acetic acid and use the relationship Kb * Ka = Kw, where Kw = 1.0 x 10⁻¹⁴). The Kb for acetate is approximately 5.6 x 10⁻¹⁰.
4. Set up an ICE Table
We'll use an ICE (Initial, Change, Equilibrium) table to determine the equilibrium concentrations:
| | CH₃COO⁻ | CH₃COOH | OH⁻ |
|-----------|----------|----------|---------|
| Initial | 0.01 M | 0 | 0 |
| Change | -x | +x | +x |
| Equilibrium| 0.01 - x | x | x |
5. Solve for x (the [OH⁻])
Substitute the equilibrium concentrations into the Kb expression:
5.6 x 10⁻¹⁰ = (x)(x) / (0.01 - x)
Since Kb is very small, we can assume that x is much smaller than 0.01, so we can simplify the equation:
5.6 x 10⁻¹⁰ ≈ x² / 0.01
Solve for x:
x² ≈ 5.6 x 10⁻¹²
x ≈ 2.37 x 10⁻⁶ M (this is the [OH⁻])
6. Calculate pOH
pOH = -log[OH⁻] = -log(2.37 x 10⁻⁶) ≈ 5.63
7. Calculate pH
pH + pOH = 14
pH = 14 - pOH = 14 - 5.63 ≈ 8.37
Therefore, the pH of a 0.01 M solution of sodium acetate is approximately 8.37.
