The Reaction Between Nitric Acid and Sodium Hydroxide

Oct 13, 2025 by magicalceylontours.com 145 views

Introduction

In this article, we will explore the reaction between nitric acid (HNO3) and sodium hydroxide (NaOH) and calculate the volume of sodium hydroxide required to completely react with a given sample of nitric acid. The reaction between nitric acid and sodium hydroxide is a classic example of a neutralization reaction, where the acid (HNO3) reacts with the base (NaOH) to form a salt (NaNO3) and water (H2O).

The Reaction Between Nitric Acid and Sodium Hydroxide

The reaction between nitric acid and sodium hydroxide can be represented by the following equation:

${ \text{HNO}_3(aq) + \text{NaOH}(aq) \rightarrow \text{NaNO}_3(aq) + \text{H}_2\text{O}(l) }$

In this equation, the nitric acid (HNO3) reacts with the sodium hydroxide (NaOH) to form sodium nitrate (NaNO3) and water (H2O). This reaction is a classic example of a neutralization reaction, where the acid (HNO3) donates a proton (H+), and the base (NaOH) accepts the proton, resulting in the formation of a salt (NaNO3) and water (H2O).

Calculating the Number of Moles of Nitric Acid

To calculate the volume of sodium hydroxide required to completely react with the given sample of nitric acid, we first need to calculate the number of moles of nitric acid present in the sample. The number of moles of nitric acid can be calculated using the formula:

${ \text{Number of moles} = \text{Molarity} \times \text{Volume (in liters)} }$

In this case, the molarity of the nitric acid is 0.100 M, and the volume of the sample is 25 mL, which is equivalent to 0.025 L. Therefore, the number of moles of nitric acid can be calculated as follows:

${ \text{Number of moles} = 0.100 \, \text{M} \times 0.025 \, \text{L} = 0.0025 \, \text{mol} }$

Calculating the Number of Moles of Sodium Hydroxide Required

Since the reaction between nitric acid and sodium hydroxide is a 1:1 reaction, the number of moles of sodium hydroxide required to completely react with the given sample of nitric acid is equal to the number of moles of nitric acid present in the sample. Therefore, the number of moles of sodium hydroxide required is also 0.0025 mol.

Calculating the Volume of Sodium Hydroxide Required

To calculate the volume of sodium hydroxide required to completely react with the given sample of nitric acid, we need to divide the number of moles of sodium hydroxide required by the molarity of the sodium hydroxide. The molarity of the sodium hydroxide is 0.0500 M. Therefore, the volume of sodium hydroxide required can be calculated as follows:

${ \text{Volume} = \frac{\text{Number of moles}}{\text{Molarity}} = \frac{0.0025 \, \text{mol}}{0.0500 \, \text{M}} = 0.05 \, \text{L} }$

Since the volume of the sodium hydroxide required is 0.05 L, which is equivalent to 50 mL, we can conclude that the volume of sodium hydroxide required to completely react with the given sample of nitric acid is 50 mL.

Conclusion

In this article, we explored the reaction between nitric acid (HNO3) and sodium hydroxide (NaOH) and calculated the volume of sodium hydroxide required to completely react with a given sample of nitric acid. The reaction between nitric acid and sodium hydroxide is a classic example of a neutralization reaction, where the acid (HNO3) reacts with the base (NaOH) to form a salt (NaNO3) and water (H2O). We calculated the number of moles of nitric acid present in the sample and used this value to calculate the number of moles of sodium hydroxide required to completely react with the given sample of nitric acid. Finally, we calculated the volume of sodium hydroxide required to completely react with the given sample of nitric acid and found that it is 50 mL.

References

Chemistry: An Atoms First Approach by Steven S. Zumdahl
General Chemistry: Principles and Modern Applications by Linus Pauling

Glossary

Acid: A substance that donates a proton (H+).
Base: A substance that accepts a proton (H+).
Neutralization reaction: A reaction between an acid and a base, resulting in the formation of a salt and water.
Molarity: The number of moles of a substance per liter of solution.
Volume: The amount of space occupied by a substance.

Introduction

In our previous article, we explored the reaction between nitric acid (HNO3) and sodium hydroxide (NaOH) and calculated the volume of sodium hydroxide required to completely react with a given sample of nitric acid. In this article, we will answer some frequently asked questions related to the reaction between nitric acid and sodium hydroxide.

Q: What is the reaction between nitric acid and sodium hydroxide?

A: The reaction between nitric acid and sodium hydroxide is a neutralization reaction, where the acid (HNO3) donates a proton (H+), and the base (NaOH) accepts the proton, resulting in the formation of a salt (NaNO3) and water (H2O).

Q: What is the equation for the reaction between nitric acid and sodium hydroxide?

A: The equation for the reaction between nitric acid and sodium hydroxide is:

${ \text{HNO}_3(aq) + \text{NaOH}(aq) \rightarrow \text{NaNO}_3(aq) + \text{H}_2\text{O}(l) }$

Q: How do you calculate the number of moles of nitric acid present in a sample?

A: To calculate the number of moles of nitric acid present in a sample, you need to multiply the molarity of the nitric acid by the volume of the sample in liters. The formula for this calculation is:

${ \text{Number of moles} = \text{Molarity} \times \text{Volume (in liters)} }$

Q: How do you calculate the number of moles of sodium hydroxide required to completely react with a given sample of nitric acid?

A: Since the reaction between nitric acid and sodium hydroxide is a 1:1 reaction, the number of moles of sodium hydroxide required to completely react with a given sample of nitric acid is equal to the number of moles of nitric acid present in the sample.

Q: How do you calculate the volume of sodium hydroxide required to completely react with a given sample of nitric acid?

A: To calculate the volume of sodium hydroxide required to completely react with a given sample of nitric acid, you need to divide the number of moles of sodium hydroxide required by the molarity of the sodium hydroxide. The formula for this calculation is:

${ \text{Volume} = \frac{\text{Number of moles}}{\text{Molarity}} }$

Q: What is the volume of sodium hydroxide required to completely react with a 25 mL sample of 0.100 M nitric acid?

A: To calculate the volume of sodium hydroxide required to completely react with a 25 mL sample of 0.100 M nitric acid, you need to follow the steps outlined above. The number of moles of nitric acid present in the sample is 0.0025 mol, and the molarity of the sodium hydroxide is 0.0500 M. Therefore, the volume of sodium hydroxide required is 0.05 L, which is equivalent to 50 mL.

Q: What are some common mistakes to avoid when calculating the volume of sodium hydroxide required to completely react with a given sample of nitric acid?

A: Some common mistakes to avoid when calculating the volume of sodium hydroxide required to completely react with a given sample of nitric acid include:

Failing to calculate the number of moles of nitric acid present in the sample
Failing to calculate the number of moles of sodium hydroxide required to completely react with the given sample of nitric acid
Failing to calculate the volume of sodium hydroxide required to completely react with the given sample of nitric acid
Using the wrong molarity of the sodium hydroxide
Using the wrong volume of the sample

Conclusion

In this article, we answered some frequently asked questions related to the reaction between nitric acid and sodium hydroxide. We also outlined the steps to calculate the volume of sodium hydroxide required to completely react with a given sample of nitric acid. By following these steps and avoiding common mistakes, you can accurately calculate the volume of sodium hydroxide required to completely react with a given sample of nitric acid.

References

Chemistry: An Atoms First Approach by Steven S. Zumdahl
General Chemistry: Principles and Modern Applications by Linus Pauling

Glossary

Acid: A substance that donates a proton (H+).
Base: A substance that accepts a proton (H+).
Neutralization reaction: A reaction between an acid and a base, resulting in the formation of a salt and water.
Molarity: The number of moles of a substance per liter of solution.
Volume: The amount of space occupied by a substance.