An analytical chemist is titrating 73.6 mL of a 0.8200M solution of benzoic acid (HC H, CO₂) with a 0.3800M solution of KOH. The p K of benzoic acid is 4.20. Calculate the pH of the acid solution after the chemist has added 177. mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. D Round your answer to 2 decimal places. olo Ar pH = 0 X B
An analytical chemist is titrating 73.6 mL of a 0.8200M solution of benzoic acid (HC H, CO₂) with a 0.3800M solution of KOH. The p K of benzoic acid is 4.20. Calculate the pH of the acid solution after the chemist has added 177. mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. D Round your answer to 2 decimal places. olo Ar pH = 0 X B
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Titration Challenge: Determining pH after Addition of KOH**
An analytical chemist is titrating 73.6 mL of a 0.8200 M solution of benzoic acid (HC6H5CO2) with a 0.3800 M solution of KOH. The \( pK_a \) of benzoic acid is 4.20. Determine the pH of the acid solution after the chemist has added 177. mL of the KOH solution to it.
**Note for advanced students:** You may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added.
**Instructions**: Round your answer to 2 decimal places.
In the input field provided, enter the calculated pH value.
### Example Calculation:
1. **Determine initial moles of benzoic acid:**
\[ \text{Volume of acid} \times \text{Concentration of acid} = 0.0736 \, \text{L} \times 0.8200 \, \text{mol/L} = 0.060352 \, \text{mol} \]
2. **Determine moles of KOH added:**
\[ \text{Volume of KOH} \times \text{Concentration of KOH} = 0.177 \, \text{L} \times 0.3800 \, \text{mol/L} = 0.06726 \, \text{mol} \]
3. **Perform stoichiometry to find moles of acid remaining and moles of salt formed:**
\[ 0.060352 \, \text{mol acid} - 0.06726 \, \text{mol KOH} = -0.006908 \, \text{mol acid remaining (indicating excess base)} \]
4. **Calculate the concentration of excess OH\(^-\) ions in the final solution:**
\[ 0.006908 \, \text{mol OH}^-/ (0.0736 \, \text{L} + 0.177 \, \text{L}) = 0.0290088 \, \text{mol/L} \]
5. **Determine the pOH of the solution:**
\[ \text{pOH} = -\log(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6081883-2ab6-472f-be82-1598250102e1%2F15f5b85f-7455-4f3b-9570-8e42ee0b9bd8%2Fbmrjw8b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Titration Challenge: Determining pH after Addition of KOH**
An analytical chemist is titrating 73.6 mL of a 0.8200 M solution of benzoic acid (HC6H5CO2) with a 0.3800 M solution of KOH. The \( pK_a \) of benzoic acid is 4.20. Determine the pH of the acid solution after the chemist has added 177. mL of the KOH solution to it.
**Note for advanced students:** You may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added.
**Instructions**: Round your answer to 2 decimal places.
In the input field provided, enter the calculated pH value.
### Example Calculation:
1. **Determine initial moles of benzoic acid:**
\[ \text{Volume of acid} \times \text{Concentration of acid} = 0.0736 \, \text{L} \times 0.8200 \, \text{mol/L} = 0.060352 \, \text{mol} \]
2. **Determine moles of KOH added:**
\[ \text{Volume of KOH} \times \text{Concentration of KOH} = 0.177 \, \text{L} \times 0.3800 \, \text{mol/L} = 0.06726 \, \text{mol} \]
3. **Perform stoichiometry to find moles of acid remaining and moles of salt formed:**
\[ 0.060352 \, \text{mol acid} - 0.06726 \, \text{mol KOH} = -0.006908 \, \text{mol acid remaining (indicating excess base)} \]
4. **Calculate the concentration of excess OH\(^-\) ions in the final solution:**
\[ 0.006908 \, \text{mol OH}^-/ (0.0736 \, \text{L} + 0.177 \, \text{L}) = 0.0290088 \, \text{mol/L} \]
5. **Determine the pOH of the solution:**
\[ \text{pOH} = -\log(
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