What volume of 0.0432 M calcium hydroxide is required to neutralize 34.20 mL of 0.0355 M nitric acid? Volume mL %3D

Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter4: Chemical Reactions In Solution
Section: Chapter Questions
Problem 4.91QE
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**Question:**

What volume of 0.0432 M calcium hydroxide is required to neutralize 34.20 mL of 0.0355 M nitric acid?

**Answer:**

Volume = ____ mL

---

**Explanation:**

This problem involves a stoichiometric calculation for a neutralization reaction between calcium hydroxide (\( \text{Ca(OH)}_2 \)), a base, and nitric acid (\( \text{HNO}_3 \)), an acid. The reaction can be represented as:

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

The goal is to find the volume of the calcium hydroxide solution needed to completely neutralize the given volume and concentration of nitric acid. Use the formula:

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

Where:
- \( \text{M}_1 \) and \( \text{V}_1 \) are the molarity and volume of \(\text{HNO}_3\), respectively.
- \( \text{M}_2 \) and \( \text{V}_2 \) are the molarity and volume of \(\text{Ca(OH)}_2\), respectively.
- \( \text{n}_1 \) and \( \text{n}_2 \) are the stoichiometric coefficients (from the balanced equation).

Here’s how you would solve it:

1. Calculate the moles of nitric acid.
2. Use the stoichiometric coefficients to find the moles of calcium hydroxide needed.
3. Calculate the volume of the calcium hydroxide solution required using its molarity.
Transcribed Image Text:**Question:** What volume of 0.0432 M calcium hydroxide is required to neutralize 34.20 mL of 0.0355 M nitric acid? **Answer:** Volume = ____ mL --- **Explanation:** This problem involves a stoichiometric calculation for a neutralization reaction between calcium hydroxide (\( \text{Ca(OH)}_2 \)), a base, and nitric acid (\( \text{HNO}_3 \)), an acid. The reaction can be represented as: \[ \text{Ca(OH)}_2 + 2 \text{HNO}_3 \rightarrow \text{Ca(NO}_3\text{)}_2 + 2 \text{H}_2\text{O} \] The goal is to find the volume of the calcium hydroxide solution needed to completely neutralize the given volume and concentration of nitric acid. Use the formula: \[ \text{M}_1 \times \text{V}_1 \times \text{n}_1 = \text{M}_2 \times \text{V}_2 \times \text{n}_2 \] Where: - \( \text{M}_1 \) and \( \text{V}_1 \) are the molarity and volume of \(\text{HNO}_3\), respectively. - \( \text{M}_2 \) and \( \text{V}_2 \) are the molarity and volume of \(\text{Ca(OH)}_2\), respectively. - \( \text{n}_1 \) and \( \text{n}_2 \) are the stoichiometric coefficients (from the balanced equation). Here’s how you would solve it: 1. Calculate the moles of nitric acid. 2. Use the stoichiometric coefficients to find the moles of calcium hydroxide needed. 3. Calculate the volume of the calcium hydroxide solution required using its molarity.
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