25.00 ml of an unknown molarity Acid HA are neutralized by 24.78 ml of 0.105 M Base BOH according to the reaction HA (aq) + BOH (aq) → BA (aq) + H20 (1) Explain how to determine the number of moles of acid that reacted.
25.00 ml of an unknown molarity Acid HA are neutralized by 24.78 ml of 0.105 M Base BOH according to the reaction HA (aq) + BOH (aq) → BA (aq) + H20 (1) Explain how to determine the number of moles of acid that reacted.
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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![### Neutralization of Acid by Base
**Problem Statement:**
25.00 ml of an unknown molarity Acid HA are neutralized by 24.78 ml of 0.105 M Base BOH according to the reaction:
\[ \text{HA (aq) + BOH (aq) → BA (aq) + H}_2\text{O (l)} \]
Explain how to determine the number of moles of acid that reacted.
**Solution:**
To determine the number of moles of acid (HA) that reacted, follow these steps:
1. **Determine the moles of BOH used:**
- Use the volume and molarity of the BOH solution.
- Volume of BOH = 24.78 ml = 0.02478 L
- Molarity of BOH = 0.105 M (moles per liter)
\[
\text{Moles of BOH} = \text{Molarity} \times \text{Volume (in liters)}
\]
\[
\text{Moles of BOH} = 0.105 \, \text{M} \times 0.02478 \, \text{L}
\]
\[
\text{Moles of BOH} = 0.002602
\]
2. **Use the stoichiometry of the reaction:**
- The balanced chemical equation is:
\[ \text{HA (aq) + BOH (aq) → BA (aq) + H}_2\text{O (l)} \]
- The stoichiometric ratio is 1:1 for HA and BOH.
3. **Determine the moles of HA:**
- Since the ratio is 1:1, the moles of HA will be equal to the moles of BOH.
\[
\text{Moles of HA} = \text{Moles of BOH} = 0.002602
\]
**Conclusion:**
The number of moles of the acid HA that reacted is 0.002602 moles.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F19b0aed3-745e-4879-8577-ea35399669d0%2Fa4e8c933-30d2-4f4b-a44d-5f524cb3c094%2Fclryy32_processed.png&w=3840&q=75)
Transcribed Image Text:### Neutralization of Acid by Base
**Problem Statement:**
25.00 ml of an unknown molarity Acid HA are neutralized by 24.78 ml of 0.105 M Base BOH according to the reaction:
\[ \text{HA (aq) + BOH (aq) → BA (aq) + H}_2\text{O (l)} \]
Explain how to determine the number of moles of acid that reacted.
**Solution:**
To determine the number of moles of acid (HA) that reacted, follow these steps:
1. **Determine the moles of BOH used:**
- Use the volume and molarity of the BOH solution.
- Volume of BOH = 24.78 ml = 0.02478 L
- Molarity of BOH = 0.105 M (moles per liter)
\[
\text{Moles of BOH} = \text{Molarity} \times \text{Volume (in liters)}
\]
\[
\text{Moles of BOH} = 0.105 \, \text{M} \times 0.02478 \, \text{L}
\]
\[
\text{Moles of BOH} = 0.002602
\]
2. **Use the stoichiometry of the reaction:**
- The balanced chemical equation is:
\[ \text{HA (aq) + BOH (aq) → BA (aq) + H}_2\text{O (l)} \]
- The stoichiometric ratio is 1:1 for HA and BOH.
3. **Determine the moles of HA:**
- Since the ratio is 1:1, the moles of HA will be equal to the moles of BOH.
\[
\text{Moles of HA} = \text{Moles of BOH} = 0.002602
\]
**Conclusion:**
The number of moles of the acid HA that reacted is 0.002602 moles.
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