A solution is prepared at 25 °C that is initially 0.11 M in propanoic acid (HC₂H₂CO₂), a weak acid with K=1.3 × 10¯5, and 0.24M in sodium propanoate (NaC₂H-CO₂). Calculate the pH of the solution. Round your answer to 2 decimal places. pH = 0
A solution is prepared at 25 °C that is initially 0.11 M in propanoic acid (HC₂H₂CO₂), a weak acid with K=1.3 × 10¯5, and 0.24M in sodium propanoate (NaC₂H-CO₂). Calculate the pH of the solution. Round your answer to 2 decimal places. pH = 0
Chemistry
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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
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![### pH Calculation of a Buffer Solution
**Problem Statement:**
A solution is prepared at 25 °C that is initially \(0.11 \, M\) in propanoic acid \((HC_3H_5CO_2)\), a weak acid with \(K_a = 1.3 \times 10^{-5}\), and \(0.24 \, M\) in sodium propanoate \((NaC_3H_5CO_2)\). Calculate the pH of the solution. Round your answer to 2 decimal places.
**Solution:**
*Graphic*: There is a textbox for entering the pH value, and two buttons marked "x" (possibly to clear the input) and "↻" (likely to reset the task).
---
**Explanation:**
In this scenario, you have a buffer system consisting of a weak acid (propanoic acid) and its conjugate base (sodium propanoate). To calculate the pH of this buffer solution, you can use the Henderson-Hasselbalch equation, which is given by:
\[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
Where:
- \(\text{p}K_a\) is the negative logarithm of the acid dissociation constant (\(K_a\)): \(\text{p}K_a = -\log(K_a)\)
- \([\text{A}^-]\) is the concentration of the conjugate base (sodium propanoate)
- \([\text{HA}]\) is the concentration of the weak acid (propanoic acid)
1. Calculate \(\text{p}K_a\):
\[ \text{p}K_a = -\log(1.3 \times 10^{-5}) \approx 4.89 \]
2. Using the concentrations given:
- \([\text{A}^-] = 0.24 \, M\)
- \([\text{HA}] = 0.11 \, M\)
3. Substitute these values into the Henderson-Hasselbalch equation:
\[ \text{pH} = 4.89 + \log \left( \frac{0.24}{0.11](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F12c0a0e9-4d98-45bb-8c42-bb6fd08d9097%2F0550f056-538a-4117-bf20-0648cf649c15%2Frs7y988_processed.png&w=3840&q=75)
Transcribed Image Text:### pH Calculation of a Buffer Solution
**Problem Statement:**
A solution is prepared at 25 °C that is initially \(0.11 \, M\) in propanoic acid \((HC_3H_5CO_2)\), a weak acid with \(K_a = 1.3 \times 10^{-5}\), and \(0.24 \, M\) in sodium propanoate \((NaC_3H_5CO_2)\). Calculate the pH of the solution. Round your answer to 2 decimal places.
**Solution:**
*Graphic*: There is a textbox for entering the pH value, and two buttons marked "x" (possibly to clear the input) and "↻" (likely to reset the task).
---
**Explanation:**
In this scenario, you have a buffer system consisting of a weak acid (propanoic acid) and its conjugate base (sodium propanoate). To calculate the pH of this buffer solution, you can use the Henderson-Hasselbalch equation, which is given by:
\[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
Where:
- \(\text{p}K_a\) is the negative logarithm of the acid dissociation constant (\(K_a\)): \(\text{p}K_a = -\log(K_a)\)
- \([\text{A}^-]\) is the concentration of the conjugate base (sodium propanoate)
- \([\text{HA}]\) is the concentration of the weak acid (propanoic acid)
1. Calculate \(\text{p}K_a\):
\[ \text{p}K_a = -\log(1.3 \times 10^{-5}) \approx 4.89 \]
2. Using the concentrations given:
- \([\text{A}^-] = 0.24 \, M\)
- \([\text{HA}] = 0.11 \, M\)
3. Substitute these values into the Henderson-Hasselbalch equation:
\[ \text{pH} = 4.89 + \log \left( \frac{0.24}{0.11
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