Question

14. What is the pH of a 0.24 M solution of sodium propionate, NaC3H502, at 25°C? (For

propionic acid, HC3H502, Ka = 1.3 × 10 - at 25°C.)

Answer 1

9.1

Step-by-step explanation:

Step 1: Calculate the basic dissociation constant of propionate ion (Kb)

Sodium propionate is a strong electrolyte that dissociates according to the following equation.

NaC₃H₅O₂ ⇒ Na⁺ + C₃H₅O₂⁻

Propionate is the conjugate base of propionic acid according to the following equation.

C₃H₅O₂⁻ + H₂O ⇄ HC₃H₅O₂ + OH⁻

We can calculate Kb for propionate using the following expression.

Ka × Kb = Kw

Kb = Kw/Ka = 1.0 × 10⁻¹⁴/1.3 × 10⁻⁵ = 7.7 × 10⁻¹⁰

Step 2: Calculate the concentration of OH⁻

The concentration of the base (Cb) is 0.24 M. We can calculate [OH⁻] using the following expression.

[OH⁻] = √(Kb × Cb) = √(7.7 × 10⁻¹⁰ × 0.24) = 1.4 × 10⁻⁵ M

Step 3: Calculate the concentration of H⁺

We will use the following expression.

Kw = [H⁺] × [OH⁻]

[H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴/1.4 × 10⁻⁵ = 7.1 × 10⁻¹⁰ M

Step 4: Calculate the pH of the solution

We will use the definition of pH.

pH = -log [H⁺] = -log 7.1 × 10⁻¹⁰ = 9.1