1. Set up an ICE Table
Acrylic acid (CH₂=CHCOOH) is a weak acid, so it only partially ionizes in water. We can represent its ionization with the following equilibrium:
CH₂=CHCOOH(aq) ⇌ H⁺(aq) + CH₂=CHCOO⁻(aq)
| | CH₂=CHCOOH | H⁺ | CH₂=CHCOO⁻ |
|-------------|--------------|-------|-------------|
| Initial (I) | 0.10 M | 0 | 0 |
| Change (C) | -x | +x | +x |
| Equilibrium (E) | 0.10 - x | x | x |
2. Write the Ka Expression
Ka = [H⁺][CH₂=CHCOO⁻] / [CH₂=CHCOOH]
3. Substitute and Solve for x
0.000056 = (x)(x) / (0.10 - x)
Since Ka is small, we can assume that x is much smaller than 0.10, allowing us to simplify the equation:
0.000056 ≈ (x)(x) / 0.10
x² ≈ 0.0000056
x ≈ √0.0000056 ≈ 0.00237 M
4. Calculate the pH
pH = -log[H⁺] = -log(0.00237) ≈ 2.63
5. Calculate the Percent Dissociation
Percent dissociation = ([H⁺] / [Initial acid]) * 100%
Percent dissociation = (0.00237 / 0.10) * 100% ≈ 2.37%
Therefore:
* The pH of a 0.10 M aqueous solution of acrylic acid is approximately 2.63.
* The percent dissociation of acrylic acid in this solution is approximately 2.37%.
