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The problem involves a neutralization reaction between nitric acid (HNO₃) and barium hydroxide (Ba(OH)₂):

\[ 2 \text{HNO}_3 + \text{Ba(OH)}_2 \rightarrow 2 \text{H}_2\text{O} + \text{Ba(NO}_3)_2 \]

The balanced equation shows that 2 moles of HNO₃ react with 1 mole of Ba(OH)₂. To find the concentration of the HNO₃ solution, we'll use the titration formula based on molarity and volume:

\[
\text{M}_1 \times \text{V}_1 = \text{M}_2 \times \text{V}_2
\]

Where:
- \( M_1 \) is the molarity of HNO₃ (what we need to find),
- \( V_1 \) is the volume of HNO₃ solution (0.105 L),
- \( M_2 \) is the molarity of Ba(OH)₂ (0.200 M),
- \( V_2 \) is the volume of Ba(OH)₂ used (32.1 mL = 0.0321 L).

Since the reaction ratio is 2:1 (2 moles of HNO₃ per 1 mole of Ba(OH)₂), we modify the formula:

\[
M_1 \times V_1 = 2 \times M_2 \times V_2
\]

Now we can plug in the values:

\[
M_1 \times 0.105 = 2 \times 0.200 \times 0.0321
\]

\[
M_1 \times 0.105 = 0.01284
\]

\[
M_1 = \frac{0.01284}{0.105} = 0.122 M
\]

Thus, the concentration of the HNO₃ solution is **0.122 M**.

Would you like more details or have any questions? Here are some related problems:

1. What would the concentration be if the volume of Ba(OH)₂ was halved?
2. How does the stoichiometry affect titration calculations?
3. What is the significance of a 1:2 reaction ratio?
4. How would the calculation change if a different base was used?
5. What is the molarity of a solution if the amount of Ba(OH)₂ was different?

**Tip:** Always ensure volumes are converted to liters when using molarity equations in titration problems.

A 0.105 L sample of an unknown HNO₃ solution required 32.1 mL of 0.200 M Ba(OH)₂ for complete neutralization. What is the concentration of the HNO₃ solution?

Calculate the concentration of an unknown HNO₃ solution using a neutralization reaction with 32.1 mL of 0.200 M Ba(OH)₂. The problem involves stoichiometry, molarity calculations, and understanding the reaction between an acid and a base.

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