Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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![**Problem Statement:**
A 25.0 mL solution of HNO₃ is neutralized with 25.4 mL of 0.250 M Ba(OH)₂. What is the concentration of the original HNO₃ solution?
**Explanation:**
To solve this problem, we need to use the concept of neutralization and the balanced chemical equation between HNO₃ and Ba(OH)₂:
\[ 2 \, \text{HNO}_3 + \text{Ba(OH)}_2 \rightarrow \text{Ba(NO}_3)_2 + 2 \, \text{H}_2\text{O} \]
### Steps to Solve:
1. **Calculate Moles of Ba(OH)₂:**
- Volume of Ba(OH)₂ = 25.4 mL = 0.0254 L
- Molarity of Ba(OH)₂ = 0.250 M
\[
\text{Moles of Ba(OH)₂} = \text{Molarity} \times \text{Volume} = 0.250 \, \text{mol/L} \times 0.0254 \, \text{L} = 0.00635 \, \text{mol}
\]
2. **Use Stoichiometry to Determine Moles of HNO₃:**
- According to the balanced equation, 1 mole of Ba(OH)₂ reacts with 2 moles of HNO₃.
\[
\text{Moles of HNO₃} = 0.00635 \, \text{mol Ba(OH)₂} \times 2 = 0.0127 \, \text{mol HNO₃}
\]
3. **Calculate the Concentration of HNO₃:**
- Volume of HNO₃ = 25.0 mL = 0.025 L
\[
\text{Concentration of HNO₃} = \frac{\text{Moles of HNO₃}}{\text{Volume of HNO₃}} = \frac{0.0127 \, \text{mol}}{0.025 \, \text{L}} = 0.508 \, \text{M}
\]
**Summary:**
The](https://content.bartleby.com/qna-images/question/6895f9cd-8521-4c9a-bc62-df11c4ac3cae/a17b22fc-abf2-4546-8935-d2dd2768c221/8pz5zfc_processed.jpeg)
Transcribed Image Text:**Problem Statement:**
A 25.0 mL solution of HNO₃ is neutralized with 25.4 mL of 0.250 M Ba(OH)₂. What is the concentration of the original HNO₃ solution?
**Explanation:**
To solve this problem, we need to use the concept of neutralization and the balanced chemical equation between HNO₃ and Ba(OH)₂:
\[ 2 \, \text{HNO}_3 + \text{Ba(OH)}_2 \rightarrow \text{Ba(NO}_3)_2 + 2 \, \text{H}_2\text{O} \]
### Steps to Solve:
1. **Calculate Moles of Ba(OH)₂:**
- Volume of Ba(OH)₂ = 25.4 mL = 0.0254 L
- Molarity of Ba(OH)₂ = 0.250 M
\[
\text{Moles of Ba(OH)₂} = \text{Molarity} \times \text{Volume} = 0.250 \, \text{mol/L} \times 0.0254 \, \text{L} = 0.00635 \, \text{mol}
\]
2. **Use Stoichiometry to Determine Moles of HNO₃:**
- According to the balanced equation, 1 mole of Ba(OH)₂ reacts with 2 moles of HNO₃.
\[
\text{Moles of HNO₃} = 0.00635 \, \text{mol Ba(OH)₂} \times 2 = 0.0127 \, \text{mol HNO₃}
\]
3. **Calculate the Concentration of HNO₃:**
- Volume of HNO₃ = 25.0 mL = 0.025 L
\[
\text{Concentration of HNO₃} = \frac{\text{Moles of HNO₃}}{\text{Volume of HNO₃}} = \frac{0.0127 \, \text{mol}}{0.025 \, \text{L}} = 0.508 \, \text{M}
\]
**Summary:**
The
Expert Solution
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Step 1: Given :
Volume of HNO3 = 25.0 ml
Volume of Ba(OH)2 = 25.4 ml
molarity of Ba(OH)2 = 0.250 M
We need to calculate the concentration of the original HNO3 solution.
Step by stepSolved in 3 steps with 1 images
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