Final answer:
The pH of a 0.250 M NaC₂H₃O₂ solution is approximately 9.12, indicating a basic solution due to the presence of the acetate ion, which is the conjugate base of a weak acid.
Step-by-step explanation:
To calculate the pH of a 0.250 M NaC₂H₃O₂ solution, we must recognize that sodium acetate (NaC₂H₃O₂) dissociates completely in water to give Na⁺ and C₂H₃O₂⁻. The acetate ion (C₂H₃O₂⁻) is the conjugate base of acetic acid (CH₃COOH), and it will react with water to form CH₃COOH and hydroxide ions (OH⁻), thus increasing the pH of the solution. The reaction is as follows: C₂H₃O₂⁻ + H₂O → CH₃COOH + OH⁻
Because sodium acetate is a salt of a weak acid (acetic acid) and a strong base (NaOH), the solution will be basic. We can use the given Ka value for acetic acid to find the Kb of the acetate ion:
Kb = Kw / Ka
Where Kw is the ion-product constant for water (1.0 x 10⁺¹⁴ at 25°C).
Then, we can calculate the concentration of OH⁻ formed and use this to find the pOH of the solution, and from that, the pH.
The calculation shows that the pH of the 0.250 M NaC₂H₃O₂ solution is approximately 9.12, which corresponds to option (c).