The molarity of an aqueous solution of sodium hydroxide, NaOH, is determined by titration against a 0.173 M nitric acid, HN03, solution. HNO3 + NaOH → NaNO3 + H2O If 34.0 mL of the base are required to neutralize 25.6 mL of nitric acid, what is the molarity of the sodium hydroxide solution? Molarity M
The molarity of an aqueous solution of sodium hydroxide, NaOH, is determined by titration against a 0.173 M nitric acid, HN03, solution. HNO3 + NaOH → NaNO3 + H2O If 34.0 mL of the base are required to neutralize 25.6 mL of nitric acid, what is the molarity of the sodium hydroxide solution? Molarity M
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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 \( \text{NaOH} \):
\[ \text{Moles of } \text{NaOH} = \text{Moles of } \text{HNO}_3 \]
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa527ffdc-16df-4a1f-9953-96956a75d05e%2F55614bd2-aa4a-419c-aeb0-ba48f89e7b80%2F1kbvot.png&w=3840&q=75)
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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