What is the pH of a 0.24 M solution of sodium propionate,NaC₃H₅O₂,at 25°C? (For propionic acid,HC₃H₅O₂,K_a = 1.3 × 10^-5 at 25°C.)

Question

Quizplus · Accepted Answer

Sodium propionate (NaC₃H₅O₂) is the salt of a weak acid (propionic acid, HC₃H₅O₂) and dissociates completely in water to give C₃H₅O₂⁻ ions and Na⁺ ions. The C₃H₅O₂⁻ ions will partially react with water to produce HC₃H₅O₂ and OH⁻ ions, making the solution basic. To find the pH, we first calculate the pOH using the formula for the hydrolysis of the anion, then convert pOH to pH.Given:- $K_a$ for HC₃H₅O₂ = $1.3 \times 10^{-5}$- $K_w$ (water ionization constant) = $1.0 \times 10^{-14}$ at 25°C- Concentration of sodium propionate = 0.24 MFirst, calculate the $K_b$ for C₃H₅O₂⁻ using the relationship $K_w = K_a \times K_b$:$$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.3 \times 10^{-5}} = 7.69 \times 10^{-10}$$Then, use the $K_b$ to find the concentration of OH⁻ produced by the hydrolysis of C₃H₅O₂⁻. For the hydrolysis reaction:$$C_3H_5O_2^- + H_2O \rightleftharpoons HC_3H_5O_2 + OH^-$$Assuming $x$ is the concentration of OH⁻ produced, the equilibrium expression is:$$K_b = \frac{x^2}{0.24 - x} \approx \frac{x^2}{0.24}$$Solving for $x$ gives:$$x^2 = K_b \times 0.24 = 7.69 \times 10^{-10} \times 0.24$$$$x^2 = 1.8456 \times 10^{-10}$$$$x = \sqrt{1.8456 \times 10^{-10}} \approx 1.3585 \times 10^{-5}$$The pOH is calculated from the concentration of OH⁻:$$pOH = -\log(1.3585 \times 10^{-5}) \approx 4.87$$Finally, convert pOH to pH:$$pH = 14 - pOH = 14 - 4.87 = 9.13$$Therefore, the pH of the solution is approximately 9.13.

What Is the PH of a 0