QuestionJune 20, 2025

Calculate the pH at 25^circ C of a 0.68M 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. pH=square

Calculate the pH at 25^circ C of a 0.68M 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. pH=square
Calculate the pH at 25^circ C of a 0.68M 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. pH=square

Solution
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Answer

9.8 Explanation 1. Determine the K_b of Sodium Benzoate Use the relation K_w = K_a \cdot K_b. Given K_w = 1.0 \times 10^{-14} and pK_a = 4.20, calculate K_a = 10^{-4.20}. Then, K_b = \frac{K_w}{K_a}. 2. Calculate Hydroxide Ion Concentration [OH^-] Use the formula for weak base dissociation: **K_b = \frac OH^-]^2} Base **. Solve for [OH^-] using K_b from Step 1 and the concentration of sodium benzoate (0.68 M). 3. Calculate pOH Use **pOH = -\log[OH^-]** to find the pOH from the hydroxide ion concentration. 4. Calculate pH Use **pH + pOH = 14** to find the pH.

Explanation

1. Determine the $K_b$ of Sodium Benzoate<br /> Use the relation $K_w = K_a \cdot K_b$. Given $K_w = 1.0 \times 10^{-14}$ and $pK_a = 4.20$, calculate $K_a = 10^{-4.20}$. Then, $K_b = \frac{K_w}{K_a}$.<br />2. Calculate Hydroxide Ion Concentration $[OH^-]$<br /> Use the formula for weak base dissociation: **$K_b = \frac{[OH^-]^2}{[Base]}$**. Solve for $[OH^-]$ using $K_b$ from Step 1 and the concentration of sodium benzoate (0.68 M).<br />3. Calculate pOH<br /> Use **$pOH = -\log[OH^-]$** to find the pOH from the hydroxide ion concentration.<br />4. Calculate pH<br /> Use **$pH + pOH = 14$** to find the pH.
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