What is the [H3O+] and the pH of a benzoic acid-benzoate buffer that consists of 0.12 M C6H₂COOH and 0.43 M C6H5COONa? (K₂ of benzoic acid = 6.3 × 10-5) Be sure to report your answer to the correct number of significant figures. [H₂O+]= pH x 10 M
What is the [H3O+] and the pH of a benzoic acid-benzoate buffer that consists of 0.12 M C6H₂COOH and 0.43 M C6H5COONa? (K₂ of benzoic acid = 6.3 × 10-5) Be sure to report your answer to the correct number of significant figures. [H₂O+]= pH x 10 M
Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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![**Buffer pH Calculation**
**Problem Statement:**
Determine the concentration of hydronium ions \([\text{H}_3\text{O}^+]\) and the pH of a benzoic acid-benzoate buffer with the following concentrations:
- 0.12 M \( \text{C}_6\text{H}_5\text{COOH} \) (benzoic acid)
- 0.43 M \( \text{C}_6\text{H}_5\text{COONa} \) (sodium benzoate)
**Given:**
- \( K_a \) of benzoic acid = \( 6.3 \times 10^{-5} \)
**Instructions:**
Calculate \([\text{H}_3\text{O}^+]\) and pH, ensuring that your answers reflect the correct number of significant figures.
**Calculations:**
\[ [\text{H}_3\text{O}^+] = \, \boxed{\phantom{00}} \times 10^{\boxed{\phantom{00}}} \, M \]
\[ \text{pH} = \, \boxed{\phantom{00}} \]
This exercise will involve using the Henderson-Hasselbalch equation for buffer solutions:
\[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
Where:
- \( \text{p}K_a = -\log(K_a) \)
- \([\text{A}^-]\) is the concentration of the benzoate ion
- \([\text{HA}]\) is the concentration of benzoic acid
**Note:** The calculations here must take into account the significant figures based on the given concentrations and \( K_a \) value.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F52d8280a-7ac0-487b-875a-b87cea38d519%2F1706eed6-0d5d-4551-ae4b-f408da11bcc5%2Fhlc00jd_processed.png&w=3840&q=75)
Transcribed Image Text:**Buffer pH Calculation**
**Problem Statement:**
Determine the concentration of hydronium ions \([\text{H}_3\text{O}^+]\) and the pH of a benzoic acid-benzoate buffer with the following concentrations:
- 0.12 M \( \text{C}_6\text{H}_5\text{COOH} \) (benzoic acid)
- 0.43 M \( \text{C}_6\text{H}_5\text{COONa} \) (sodium benzoate)
**Given:**
- \( K_a \) of benzoic acid = \( 6.3 \times 10^{-5} \)
**Instructions:**
Calculate \([\text{H}_3\text{O}^+]\) and pH, ensuring that your answers reflect the correct number of significant figures.
**Calculations:**
\[ [\text{H}_3\text{O}^+] = \, \boxed{\phantom{00}} \times 10^{\boxed{\phantom{00}}} \, M \]
\[ \text{pH} = \, \boxed{\phantom{00}} \]
This exercise will involve using the Henderson-Hasselbalch equation for buffer solutions:
\[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
Where:
- \( \text{p}K_a = -\log(K_a) \)
- \([\text{A}^-]\) is the concentration of the benzoate ion
- \([\text{HA}]\) is the concentration of benzoic acid
**Note:** The calculations here must take into account the significant figures based on the given concentrations and \( K_a \) value.
Expert Solution
Step 1
Recall the data from the question,
Molarity of C6H5COOH=0.12 M
Molarity of C6H5COONa=0.43 M
Ka of benzoic acid=6.3*10-5
Molarity of hydronium ion = ?
pH=?
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