Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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**Determining the Molarity of a Sodium Hydroxide Solution by Titration**

The molarity of an aqueous solution of **sodium hydroxide (NaOH)** is determined by titration against a 0.173 M **nitric acid (HNO₃)** solution.

**Balanced Chemical Equation:**

\[
\text{HNO}_3 + \text{NaOH} \rightarrow \text{NaNO}_3 + \text{H}_2\text{O}
\]

**Problem Statement:**

If **34.0 mL** of the base (NaOH) are required to neutralize **25.6 mL** of the nitric acid solution, what is the molarity of the sodium hydroxide solution?

**Calculation:**

To find the molarity of the sodium hydroxide solution, use the formula for molarity and the stoichiometry of the balanced chemical equation.

Given:
- Volume of \( \text{HNO}_3 \) = 25.6 mL
- Molarity of \( \text{HNO}_3 \) = 0.173 M
- Volume of \( \text{NaOH} \) = 34.0 mL

First, calculate the moles of \( \text{HNO}_3 \):

\[ \text{Moles of } \text{HNO}_3 = \text{Molarity of } \text{HNO}_3 \times \frac{\text{Volume of } \text{HNO}_3 \text{ (in L)}}{1000} \]

\[ \text{Moles of } \text{HNO}_3 = 0.173 \text{ M} \times \frac{25.6 \text{ mL}}{1000} \]

\[ \text{Moles of } \text{HNO}_3 = 0.173 \times 0.0256 \]

\[ \text{Moles of } \text{HNO}_3 = 0.0044288 \text{ moles} \]

Since the reaction is a 1:1 molar ratio between \( \text{HNO}_3 \) and \( \text{NaOH} \):

\[ \text{Moles of } \text{NaOH} = \text{Moles of } \text{HNO}_3 \]

\[
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Transcribed Image Text:**Determining the Molarity of a Sodium Hydroxide Solution by Titration** The molarity of an aqueous solution of **sodium hydroxide (NaOH)** is determined by titration against a 0.173 M **nitric acid (HNO₃)** solution. **Balanced Chemical Equation:** \[ \text{HNO}_3 + \text{NaOH} \rightarrow \text{NaNO}_3 + \text{H}_2\text{O} \] **Problem Statement:** If **34.0 mL** of the base (NaOH) are required to neutralize **25.6 mL** of the nitric acid solution, what is the molarity of the sodium hydroxide solution? **Calculation:** To find the molarity of the sodium hydroxide solution, use the formula for molarity and the stoichiometry of the balanced chemical equation. Given: - Volume of \( \text{HNO}_3 \) = 25.6 mL - Molarity of \( \text{HNO}_3 \) = 0.173 M - Volume of \( \text{NaOH} \) = 34.0 mL First, calculate the moles of \( \text{HNO}_3 \): \[ \text{Moles of } \text{HNO}_3 = \text{Molarity of } \text{HNO}_3 \times \frac{\text{Volume of } \text{HNO}_3 \text{ (in L)}}{1000} \] \[ \text{Moles of } \text{HNO}_3 = 0.173 \text{ M} \times \frac{25.6 \text{ mL}}{1000} \] \[ \text{Moles of } \text{HNO}_3 = 0.173 \times 0.0256 \] \[ \text{Moles of } \text{HNO}_3 = 0.0044288 \text{ moles} \] Since the reaction is a 1:1 molar ratio between \( \text{HNO}_3 \) and \( \text{NaOH} \): \[ \text{Moles of } \text{NaOH} = \text{Moles of } \text{HNO}_3 \] \[
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