Consider the titration of 15.75 mL of 0.549 M potassium propanoate, CH3CH;CO,K, with 0.207 M HNO3. What is the pH of the solution after 39.83 mL of the HNO3 has been added? (pK, of CH3CH;COH = 4.691)
Consider the titration of 15.75 mL of 0.549 M potassium propanoate, CH3CH;CO,K, with 0.207 M HNO3. What is the pH of the solution after 39.83 mL of the HNO3 has been added? (pK, of CH3CH;COH = 4.691)
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Chapter16: Acid-base Equilibria
Section: Chapter Questions
Problem 16.149QP: A solution of weak base is titrated to the equivalence point with a strong acid. Which one of the...
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![**Titration Calculation Example**
Consider the titration of 15.75 mL of 0.549 M potassium propanoate, CH₃CH₂CO₂K, with 0.207 M HNO₃. What is the pH of the solution after 39.83 mL of the HNO₃ has been added? (pKa of CH₃CH₂CO₂H = 4.691)
---
*This problem explores the titration of a weak base, potassium propanoate, with a strong acid, nitric acid (HNO₃). The objective is to determine the pH of the solution after a specific volume of HNO₃ has been added.*
1. **Understand the Reactants and Products**:
- Potassium propanoate (CH₃CH₂CO₂K) dissociates in water into CH₃CH₂CO₂⁻ and K⁺.
- HNO₃ is a strong acid that dissociates completely in water into H⁺ and NO₃⁻.
2. **Initial Moles Calculation**:
- Calculate the initial moles of CH₃CH₂CO₂K:
\[
\text{Moles of CH₃CH₂CO₂K} = 15.75 \text{ mL} \times 0.549 \text{ M} = 8.65 \text{ mmol}
\]
- Calculate the moles of HNO₃ added:
\[
\text{Moles of HNO₃} = 39.83 \text{ mL} \times 0.207 \text{ M} = 8.25 \text{ mmol}
\]
3. **Reaction Completion**:
- This is a stoichiometric reaction between CH₃CH₂CO₂⁻ and H⁺:
\[
\text{CH₃CH₂CO₂⁻} + \text{H⁺} \rightarrow \text{CH₃CH₂CO₂H}
\]
- After reaction, the remaining moles of CH₃CH₂CO₂⁻ = 8.65 mmol - 8.25 mmol = 0.40 mmol.
- Moles of CH₃CH₂CO₂H formed = 8.25 mmol.
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Transcribed Image Text:**Titration Calculation Example**
Consider the titration of 15.75 mL of 0.549 M potassium propanoate, CH₃CH₂CO₂K, with 0.207 M HNO₃. What is the pH of the solution after 39.83 mL of the HNO₃ has been added? (pKa of CH₃CH₂CO₂H = 4.691)
---
*This problem explores the titration of a weak base, potassium propanoate, with a strong acid, nitric acid (HNO₃). The objective is to determine the pH of the solution after a specific volume of HNO₃ has been added.*
1. **Understand the Reactants and Products**:
- Potassium propanoate (CH₃CH₂CO₂K) dissociates in water into CH₃CH₂CO₂⁻ and K⁺.
- HNO₃ is a strong acid that dissociates completely in water into H⁺ and NO₃⁻.
2. **Initial Moles Calculation**:
- Calculate the initial moles of CH₃CH₂CO₂K:
\[
\text{Moles of CH₃CH₂CO₂K} = 15.75 \text{ mL} \times 0.549 \text{ M} = 8.65 \text{ mmol}
\]
- Calculate the moles of HNO₃ added:
\[
\text{Moles of HNO₃} = 39.83 \text{ mL} \times 0.207 \text{ M} = 8.25 \text{ mmol}
\]
3. **Reaction Completion**:
- This is a stoichiometric reaction between CH₃CH₂CO₂⁻ and H⁺:
\[
\text{CH₃CH₂CO₂⁻} + \text{H⁺} \rightarrow \text{CH₃CH₂CO₂H}
\]
- After reaction, the remaining moles of CH₃CH₂CO₂⁻ = 8.65 mmol - 8.25 mmol = 0.40 mmol.
- Moles of CH₃CH₂CO₂H formed = 8.25 mmol.
4.
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