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)

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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.
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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