Final answer:
To find the pH of a 0.250 M NaC₂H₃O₂ solution, we need to consider the dissociation of NaC₂H₃O₂ in water. The concentration of OH⁻ ions can be found by considering the dissociation of water and the equilibrium expression. Plugging the concentration of OH⁻ ions into the pH equation gives the pH value. The pH is found to be 13.4.
Step-by-step explanation:
To find the pH of a 0.250 M NaC₂H₃O₂ solution, we need to consider the dissociation of NaC₂H₃O₂ in water. NaC₂H₃O₂ is the sodium salt of acetic acid (CH₃COOH). It dissociates in water to form Na⁺ and the conjugate base of acetic acid, C₂H₃O₂⁻. Since C₂H₃O₂⁻ is the conjugate base of a weak acid, it will react with water to produce hydroxide ions (OH⁻). The reaction can be represented as:
C₂H₃O₂⁻ + H₂O ⇌ CH₃COOH + OH⁻
To calculate the pH, we need to find the concentration of OH⁻ ions and then use the equation pH = -log[H⁺]. The concentration of OH⁻ ions can be found by considering the dissociation of water and the equilibrium expression for the reaction:
Kw = [H⁺][OH⁻] = 1.0 x 10⁻¹⁴
Since NaC₂H₃O₂ is a strong electrolyte, we can assume that its concentration remains unchanged after dissociation. Thus, the concentration of C₂H₃O₂⁻ in the solution is 0.250 M. Since the reaction is in equilibrium, the concentration of OH⁻ is the same as the concentration of acetic acid, which is 0.250 M. Therefore, we can plug this value into the equilibrium expression and solve for [H⁺]:
1.0 x 10⁻¹⁴ = [H⁺] x 0.250
[H⁺] = 4.0 x 10⁻¹⁴ M
Plugging this value into the pH equation gives:
pH = -log(4.0 x 10⁻¹⁴) = 13.4