A chemist titrates 220.0 mL of a 0.4703M benzoic acid (HCH₂CO₂) solution with 0.0972M NaOH solution at 25 °C. Calculate the pH at equivalence. The pK of benzoic acid is 4.20. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of NaOH solution added.
A chemist titrates 220.0 mL of a 0.4703M benzoic acid (HCH₂CO₂) solution with 0.0972M NaOH solution at 25 °C. Calculate the pH at equivalence. The pK of benzoic acid is 4.20. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of NaOH solution added.
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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![### Titration of Benzoic Acid
**Problem Statement**
A chemist titrates 220.0 mL of a 0.4703 M benzoic acid (\[HC_6H_5CO_2\]) solution with 0.0972 M NaOH solution at 25 °C. Calculate the pH at equivalence.
The \( pK_a \) of benzoic acid is 4.20.
Round your answer to 2 decimal places.
**Note for Advanced Students:**
You may assume the total volume of the solution equals the initial volume plus the volume of NaOH solution added.
---
This problem requires understanding the concepts of acid-base titration, particularly at the equivalence point where the amount of acid is stoichiometrically equal to the amount of base added. At equivalence, for a weak acid titrated with a strong base, the solution typically contains the conjugate base of the weak acid, which will hydrolyze slightly in water.
To solve, follow these steps:
1. **Determine the moles of benzoic acid and NaOH:**
- Moles of benzoic acid \( (\text{mol} HA) \):
\[ \text{mol HA} = 0.4703 \, M \times 0.2200 \, L = 0.103466 \, \text{mol} \]
- Moles of NaOH needed:
\[ \text{mol NaOH} = \text{mol HA} = 0.103466 \, \text{mol} \]
2. **Volume of NaOH solution needed:**
- Volume \( (V_{\text{NaOH}}) \):
\[ V_{\text{NaOH}} = \frac{\text{mol NaOH}}{0.0972 \, M} = \frac{0.103466 \, \text{mol}}{0.0972 \, M} = 1.0649 \, L \]
Considering significant figures—1.065 L.
3. **Total Volume at Equivalence:**
- Total volume \( V_{\text{total}} \):
\[ V_{\text{total}} = 0.2200 \, L + 1.065 \, L = 1.285 \, L \]
4. **Concentration of the conjugate](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F65f958a7-7c6b-44c4-ab5b-ab922054d7b7%2F206c4780-9f48-40d0-969a-546ad5e6f1a3%2F4xbxtuf_processed.png&w=3840&q=75)
Transcribed Image Text:### Titration of Benzoic Acid
**Problem Statement**
A chemist titrates 220.0 mL of a 0.4703 M benzoic acid (\[HC_6H_5CO_2\]) solution with 0.0972 M NaOH solution at 25 °C. Calculate the pH at equivalence.
The \( pK_a \) of benzoic acid is 4.20.
Round your answer to 2 decimal places.
**Note for Advanced Students:**
You may assume the total volume of the solution equals the initial volume plus the volume of NaOH solution added.
---
This problem requires understanding the concepts of acid-base titration, particularly at the equivalence point where the amount of acid is stoichiometrically equal to the amount of base added. At equivalence, for a weak acid titrated with a strong base, the solution typically contains the conjugate base of the weak acid, which will hydrolyze slightly in water.
To solve, follow these steps:
1. **Determine the moles of benzoic acid and NaOH:**
- Moles of benzoic acid \( (\text{mol} HA) \):
\[ \text{mol HA} = 0.4703 \, M \times 0.2200 \, L = 0.103466 \, \text{mol} \]
- Moles of NaOH needed:
\[ \text{mol NaOH} = \text{mol HA} = 0.103466 \, \text{mol} \]
2. **Volume of NaOH solution needed:**
- Volume \( (V_{\text{NaOH}}) \):
\[ V_{\text{NaOH}} = \frac{\text{mol NaOH}}{0.0972 \, M} = \frac{0.103466 \, \text{mol}}{0.0972 \, M} = 1.0649 \, L \]
Considering significant figures—1.065 L.
3. **Total Volume at Equivalence:**
- Total volume \( V_{\text{total}} \):
\[ V_{\text{total}} = 0.2200 \, L + 1.065 \, L = 1.285 \, L \]
4. **Concentration of the conjugate
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