QuestionJune 20, 2025

Calculate the pH at 25^circ C of a 0.68 M solution of sodium benzoate (NaC_(6)H_(5)CO_(2)) Note that benzoic acid (HC_(6)H_(5)CO_(2)) is a weak acid with a pK_(a) of 4.20 . Round your answer to 1 decimal place.

Calculate the pH at 25^circ C of a 0.68 M solution of sodium benzoate (NaC_(6)H_(5)CO_(2)) Note that benzoic acid (HC_(6)H_(5)CO_(2)) is a weak acid with a pK_(a) of 4.20 . Round your answer to 1 decimal place.
Calculate the pH at 25^circ C of a 0.68 M solution of sodium benzoate (NaC_(6)H_(5)CO_(2)) Note that benzoic acid (HC_(6)H_(5)CO_(2)) is a weak acid with a pK_(a) of 4.20 . Round your answer to 1 decimal place.

Solution
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Answer

9.4 Explanation 1. Identify the reaction Sodium benzoate (NaC_{6}H_{5}CO_{2}) dissociates in water to form C_{6}H_{5}CO_{2}^-, which acts as a base. 2. Use the hydrolysis equation The hydrolysis of C_{6}H_{5}CO_{2}^- is: C_{6}H_{5}CO_{2}^- + H_2O \rightleftharpoons HC_{6}H_{5}CO_{2} + OH^-. 3. Calculate K_b Use **K_w = K_a \cdot K_b**. Given pK_a = 4.20, K_a = 10^{-4.20}, and K_w = 1.0 \times 10^{-14}. K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{10^{-4.20}} = 10^{-9.80}. 4. Set up the equilibrium expression For C_{6}H_{5}CO_{2}^-: K_b = \frac OH^-]^2} C_{6}H_{5}CO_{2}^-] - [OH^- . Assume [OH^-] \ll 0.68 M. 5. Solve for [OH^-] [OH^-] = \sqrt{K_b \cdot [C_{6}H_{5}CO_{2}^- = \sqrt{10^{-9.80} \times 0.68}. [OH^-] = \sqrt{6.8 \times 10^{-10}} \approx 2.61 \times 10^{-5} M. 6. Calculate pOH pOH = -\log[OH^-] = -\log(2.61 \times 10^{-5}) \approx 4.58. 7. Calculate pH **pH + pOH = 14**. Thus, pH = 14 - 4.58 = 9.42.

Explanation

1. Identify the reaction<br /> Sodium benzoate $(NaC_{6}H_{5}CO_{2})$ dissociates in water to form $C_{6}H_{5}CO_{2}^-$, which acts as a base.<br />2. Use the hydrolysis equation<br /> The hydrolysis of $C_{6}H_{5}CO_{2}^-$ is: $C_{6}H_{5}CO_{2}^- + H_2O \rightleftharpoons HC_{6}H_{5}CO_{2} + OH^-$.<br />3. Calculate $K_b$<br /> Use **$K_w = K_a \cdot K_b$**. Given $pK_a = 4.20$, $K_a = 10^{-4.20}$, and $K_w = 1.0 \times 10^{-14}$.<br /> $K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{10^{-4.20}} = 10^{-9.80}$.<br />4. Set up the equilibrium expression<br /> For $C_{6}H_{5}CO_{2}^-$: $K_b = \frac{[OH^-]^2}{[C_{6}H_{5}CO_{2}^-] - [OH^-]}$. Assume $[OH^-] \ll 0.68$ M.<br />5. Solve for $[OH^-]$<br /> $[OH^-] = \sqrt{K_b \cdot [C_{6}H_{5}CO_{2}^-]} = \sqrt{10^{-9.80} \times 0.68}$.<br /> $[OH^-] = \sqrt{6.8 \times 10^{-10}} \approx 2.61 \times 10^{-5}$ M.<br />6. Calculate $pOH$<br /> $pOH = -\log[OH^-] = -\log(2.61 \times 10^{-5}) \approx 4.58$.<br />7. Calculate pH<br /> **$pH + pOH = 14$**. Thus, $pH = 14 - 4.58 = 9.42$.
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