1. Write the balanced chemical equation for the dissociation of sodium propionate in water:
NaC3H5O2(aq) → Na+(aq) + C3H5O2-(aq)
2. The propionate ion (C3H5O2-) is a weak base and will undergo hydrolysis in water, producing hydroxide ions (OH-) and increasing the pH of the solution. The reaction for the hydrolysis of propionate ion is:
C3H5O2-(aq) + H2O(l) ⇌ HC3H5O2(aq) + OH-(aq)
3. The equilibrium constant for the hydrolysis of propionate ion is given by the base dissociation constant (Kb) of propionic acid (HC3H5O2):
Kb = [HC3H5O2][OH-]/[C3H5O2-]
4. The value of Kb for propionic acid is 1.34 × 10^-5 at 25°C.
5. Assuming that the hydrolysis of propionate ion is negligible (which is a reasonable assumption for a 0.067 M solution), the concentration of hydroxide ions produced by hydrolysis can be calculated using the Kb expression:
[OH-] = Kb[C3H5O2-]/[HC3H5O2] ≈ Kb[C3H5O2-]
6. Substituting the value of Kb and the concentration of propionate ion into the equation above, we get:
[OH-] ≈ (1.34 × 10^-5)(0.067) ≈ 9.0 × 10^-7 M
7. Finally, we can calculate the pH of the solution using the following equation:
pH = -log[H3O+] = -log(Kw/[OH-])
where Kw is the ion product constant for water (Kw = 1.0 × 10^-14 at 25°C).
8. Substituting the value of [OH-] into the equation above, we get:
pH = -log(1.0 × 10^-14/9.0 × 10^-7) ≈ 8.75
Therefore, the pH of a 0.067 M aqueous solution of sodium propionate is approximately 8.75.
