3. If 0.0275 L of 3.5 x 102 M Ba(OH)2 solution is needed to neutralize 10.0 mL of nitric acid solution of unknown concentration, what is the molarity of the HNO3? Ba(OH)₂(aq) HNO3(aq) ->> H₂O(l) + + Ba(NO3)₂(aq)
3. If 0.0275 L of 3.5 x 102 M Ba(OH)2 solution is needed to neutralize 10.0 mL of nitric acid solution of unknown concentration, what is the molarity of the HNO3? Ba(OH)₂(aq) HNO3(aq) ->> H₂O(l) + + Ba(NO3)₂(aq)
Chemistry
10th Edition
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
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Problem 3: Acid-Base Neutralization**
**Objective:**
Calculate the molarity of nitric acid (HNO₃) in a solution based on the neutralization reaction with barium hydroxide (Ba(OH)₂).
**Given Data:**
- Volume of Ba(OH)₂ solution: 0.0275 L
- Molarity of Ba(OH)₂ solution: 3.5 x 10⁻² M
- Volume of HNO₃ solution: 10.0 mL
**Chemical Reaction:**
\[ \text{Ba(OH)}_2(\text{aq}) + 2\text{HNO}_3(\text{aq}) \rightarrow 2\text{H}_2\text{O}(\text{l}) + \text{Ba(NO}_3\text{)}_2(\text{aq}) \]
**Steps to Solve:**
1. **Calculate moles of Ba(OH)₂:**
\[ \text{Moles of Ba(OH)}_2 = \text{Molarity} \times \text{Volume} \]
\[ = 3.5 \times 10^{-2} \, \text{M} \times 0.0275 \, \text{L} \]
2. **Use stoichiometry to find moles of HNO₃:**
From the balanced equation, 1 mole of Ba(OH)₂ reacts with 2 moles of HNO₃.
3. **Calculate molarity of HNO₃:**
\[ \text{Molarity} = \frac{\text{Moles of HNO}_3}{\text{Volume of HNO}_3 \, (\text{in liters})} \]
**Note:**
To convert 10.0 mL to liters, use the conversion:
\[ 10.0 \, \text{mL} = 0.010 \, \text{L} \]
**Conclusion:**
This problem demonstrates the use of basic stoichiometry in chemistry to determine the concentration of a solution after a neutralization reaction, highlighting the importance of balanced chemical equations in calculations.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa7d7525f-9c27-4dea-8079-91e82ba48598%2F201f31b2-62a1-4fe5-b796-b0b7b14da134%2Fnoi1l2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 3: Acid-Base Neutralization**
**Objective:**
Calculate the molarity of nitric acid (HNO₃) in a solution based on the neutralization reaction with barium hydroxide (Ba(OH)₂).
**Given Data:**
- Volume of Ba(OH)₂ solution: 0.0275 L
- Molarity of Ba(OH)₂ solution: 3.5 x 10⁻² M
- Volume of HNO₃ solution: 10.0 mL
**Chemical Reaction:**
\[ \text{Ba(OH)}_2(\text{aq}) + 2\text{HNO}_3(\text{aq}) \rightarrow 2\text{H}_2\text{O}(\text{l}) + \text{Ba(NO}_3\text{)}_2(\text{aq}) \]
**Steps to Solve:**
1. **Calculate moles of Ba(OH)₂:**
\[ \text{Moles of Ba(OH)}_2 = \text{Molarity} \times \text{Volume} \]
\[ = 3.5 \times 10^{-2} \, \text{M} \times 0.0275 \, \text{L} \]
2. **Use stoichiometry to find moles of HNO₃:**
From the balanced equation, 1 mole of Ba(OH)₂ reacts with 2 moles of HNO₃.
3. **Calculate molarity of HNO₃:**
\[ \text{Molarity} = \frac{\text{Moles of HNO}_3}{\text{Volume of HNO}_3 \, (\text{in liters})} \]
**Note:**
To convert 10.0 mL to liters, use the conversion:
\[ 10.0 \, \text{mL} = 0.010 \, \text{L} \]
**Conclusion:**
This problem demonstrates the use of basic stoichiometry in chemistry to determine the concentration of a solution after a neutralization reaction, highlighting the importance of balanced chemical equations in calculations.
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