Question

If 2.88 g of sodium benzoate is present in 1000 mL of an aqueous solution, then pH of the solution will be (pKa of benzoic acid = 4.2, molar mass of sodium benzoate = 144 g mol$^{-1}$)

• A. 13.1
• B. 6.27
• C. 8.25
• D. 10.2

Answer 1

Answer: (C)

8.25

Answer 2

To find the pH of the solution, we need to consider the hydrolysis of sodium benzoate. Here's a step-by-step breakdown:

1. Calculate moles of sodium benzoate:

   $$\text{Moles} = \frac{2.88\,\text{g}}{144\,\text{g/mol}} = 0.02\,\text{mol}$$

   Since the volume is 1 L, the concentration $[A^-] = 0.02\, \text{M}$.

2. Hydrolysis equilibrium of benzoate ion:

   $$\text{A}^- + \text{H}_2\text{O} \rightleftharpoons \text{HA} + \text{OH}^-$$

3. Determine $K_b$:

   Given $pK_a = 4.2$, so

   $$K_a = 10^{-4.2} \approx 6.31 \times 10^{-5}$$

   and

   $$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{6.31 \times 10^{-5}} \approx 1.58 \times 10^{-10}$$

4. Set up the hydrolysis expression:

   Let the change in concentration due to hydrolysis be $x$. Then, at equilibrium:

   $$K_b = \frac{x^2}{0.02}$$ $$x^2 = (1.58 \times 10^{-10})(0.02) = 3.16 \times 10^{-12}$$ $$x = \sqrt{3.16 \times 10^{-12}} \approx 1.78 \times 10^{-6}\,\text{M}$$

   Here, $x$ is the concentration of $\text{OH}^-$.

5. Calculate $pOH$ and then $pH$:

   $$pOH = -\log (1.78 \times 10^{-6}) \approx 5.75$$ $$pH = 14 - pOH = 14 - 5.75 = 8.25$$

Therefore, the pH of the solution is approximately 8.25.