### Titration of Potassium Propanoate with Nitric Acid**Problem Statement:**Consider the titration of 17.50 mL of 0.459 M potassium propanoate, \( \text{CH}_3\text{CH}_2\text{CO}_2\text{K} \), with 0.107 M \( \text{HNO}_3 \). What is the pH of the solution after 39.38 mL of the \( \text{HNO}_3 \) has been added?Given: \( \text{pK}_a \) of \( \text{CH}_3\text{CH}_2\text{CO}_2\text{H} \) = 4.691### Explanation and Calculation:To determine the pH of the solution after adding 39.38 mL of 0.107 M \( \text{HNO}_3 \) to 17.50 mL of 0.459 M potassium propanoate, follow these steps:**1. Moles of potassium propanoate (\( \text{CH}_3\text{CH}_2\text{CO}_2\text{K} \)) initially:**\[ \text{Moles} = \text{Concentration} \times \text{Volume} \]\[ \text{Initial moles of } \text{CH}_3\text{CH}_2\text{CO}_2\text{K} = 0.459 \, M \times 0.0175 \, L = 0.0080325 \, \text{moles} \]**2. Moles of \( \text{HNO}_3 \) added:**\[ \text{Moles} = \text{Concentration} \times \text{Volume} \]\[ \text{Moles of } \text{HNO}_3 = 0.107 \, M \times 0.03938 \, L = 0.00421566 \, \text{moles} \]**3. Reaction between \( \text{HNO}_3 \) and \( \text{CH}_3\text{CH}_2\text{CO}_2\text{K} \):**\[ \text{CH}_3\text{CH}_2\text{

Millimoles of propanoic acid form =0.107*39.38=4 .21 Millimoles of salt remain=17.50*0.459 -4.21…

Answered: Consider the titration of 17.50 mL of…

Consider the titration of 17.50 mL of 0.459 M potassium propanoate, CH3CH2CO2K, with 0.107 M HNO3. What is the p of the solution after 39.38 mL of the HNO3 has been added? (pK, of CH3CH2CO,H = 4.691)

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Question

### Titration of Potassium Propanoate with Nitric Acid

**Problem Statement:**

Consider the titration of 17.50 mL of 0.459 M potassium propanoate, \( \text{CH}_3\text{CH}_2\text{CO}_2\text{K} \), with 0.107 M \( \text{HNO}_3 \). What is the pH of the solution after 39.38 mL of the \( \text{HNO}_3 \) has been added?

Given: \( \text{pK}_a \) of \( \text{CH}_3\text{CH}_2\text{CO}_2\text{H} \) = 4.691

### Explanation and Calculation:

To determine the pH of the solution after adding 39.38 mL of 0.107 M \( \text{HNO}_3 \) to 17.50 mL of 0.459 M potassium propanoate, follow these steps:

**1. Moles of potassium propanoate (\( \text{CH}_3\text{CH}_2\text{CO}_2\text{K} \)) initially:**
\[ \text{Moles} = \text{Concentration} \times \text{Volume} \]
\[ \text{Initial moles of } \text{CH}_3\text{CH}_2\text{CO}_2\text{K} = 0.459 \, M \times 0.0175 \, L = 0.0080325 \, \text{moles} \]

**2. Moles of \( \text{HNO}_3 \) added:**
\[ \text{Moles} = \text{Concentration} \times \text{Volume} \]
\[ \text{Moles of } \text{HNO}_3 = 0.107 \, M \times 0.03938 \, L = 0.00421566 \, \text{moles} \]

**3. Reaction between \( \text{HNO}_3 \) and \( \text{CH}_3\text{CH}_2\text{CO}_2\text{K} \):**
\[ \text{CH}_3\text{CH}_2\text{

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