Constant-boiling HCl can be used as a primary standard for acid-base titrations. A 50.00 mL sample of constant-boiling HCl with a concentration of 0.1296 M was collected and titrated to an endpoint with 38.47 mL of a Ba(OH), solution. What is the molarity of the Ba(OH), solution? concentration: M

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**Acid-Base Titration Standardization Exercise**

Constant-boiling HCl can be used as a primary standard for acid–base titrations. A 50.00 mL sample of constant-boiling HCl with a concentration of 0.1296 M was collected and titrated to an endpoint with 38.47 mL of a Ba(OH)₂ solution. What is the molarity of the Ba(OH)₂ solution?

**Concentration:**  
[Text Box] M

---

For calculations, remember that in a titration, the moles of HCl will equal the moles of Ba(OH)₂ reacting according to their balanced equation. Use the formula:

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

Where:
- \(\text{M}_1\) and \(\text{V}_1\) are the molarity and volume of HCl.
- \(\text{M}_2\) and \(\text{V}_2\) are the molarity and volume of Ba(OH)₂.

Please perform the necessary calculations to find the molarity of the Ba(OH)₂ solution.
Transcribed Image Text:**Acid-Base Titration Standardization Exercise** Constant-boiling HCl can be used as a primary standard for acid–base titrations. A 50.00 mL sample of constant-boiling HCl with a concentration of 0.1296 M was collected and titrated to an endpoint with 38.47 mL of a Ba(OH)₂ solution. What is the molarity of the Ba(OH)₂ solution? **Concentration:** [Text Box] M --- For calculations, remember that in a titration, the moles of HCl will equal the moles of Ba(OH)₂ reacting according to their balanced equation. Use the formula: \[ \text{M}_1 \times \text{V}_1 = \text{M}_2 \times \text{V}_2 \] Where: - \(\text{M}_1\) and \(\text{V}_1\) are the molarity and volume of HCl. - \(\text{M}_2\) and \(\text{V}_2\) are the molarity and volume of Ba(OH)₂. Please perform the necessary calculations to find the molarity of the Ba(OH)₂ solution.
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