The Reaction Between HNO3 and NaOH

Oct 13, 2025 by magicalceylontours.com 4 views

Introduction

In this article, we will explore the reaction between nitric acid (HNO3) and sodium hydroxide (NaOH) and calculate the volume of NaOH required to completely react with a 25 mL sample of 0.100 M HNO3. This calculation is essential in understanding the stoichiometry of the reaction and the amount of reagents required for a complete reaction.

The Reaction Between HNO3 and NaOH

The reaction between HNO3 and NaOH 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 balanced chemical equation for this reaction is:

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

Calculating the Number of Moles of HNO3

To calculate the number of moles of HNO3, we can use the formula:

${ \text{moles} = \text{molarity} \times \text{volume (in liters)} }$

Given that the molarity of HNO3 is 0.100 M and the volume is 25 mL (or 0.025 L), we can calculate the number of moles as follows:

${ \text{moles of HNO}_3 = 0.100 \, \text{M} \times 0.025 \, \text{L} = 0.0025 \, \text{mol} }$

Calculating the Number of Moles of NaOH Required

Since the reaction is a 1:1 ratio, the number of moles of NaOH required is equal to the number of moles of HNO3. Therefore, the number of moles of NaOH required is also 0.0025 mol.

Calculating the Volume of NaOH Required

To calculate the volume of NaOH required, we can use the formula:

${ \text{volume} = \frac{\text{moles}}{\text{molarity}} }$

Given that the number of moles of NaOH required is 0.0025 mol and the molarity of NaOH is also 0.100 M, we can calculate the volume as follows:

${ \text{volume of NaOH} = \frac{0.0025 \, \text{mol}}{0.100 \, \text{M}} = 0.025 \, \text{L} }$

Converting the Volume from Liters to Milliliters

To convert the volume from liters to milliliters, we can multiply by 1000 (since there are 1000 mL in 1 L):

${ 0.025 \, \text{L} \times 1000 \, \frac{\text{mL}}{\text{L}} = 25 \, \text{mL} }$

Conclusion

In this article, we calculated the volume of NaOH required to completely react with a 25 mL sample of 0.100 M HNO3. The calculation involved determining the number of moles of HNO3, calculating the number of moles of NaOH required, and then calculating the volume of NaOH required. The final answer is 25 mL.

References

Chemical Equations and Stoichiometry. (n.d.). Retrieved from https://www.chemguide.co.uk/physical/acidbase/stoichiometry.html
Molarity and Concentration. (n.d.). Retrieved from https://www.chemguide.co.uk/physical/concentration/molarity.html

Introduction

In our previous article, we explored the reaction between nitric acid (HNO3) and sodium hydroxide (NaOH) and calculated the volume of NaOH required to completely react with a 25 mL sample of 0.100 M HNO3. In this article, we will answer some frequently asked questions related to this topic.

Q: What is the balanced chemical equation for the reaction between HNO3 and NaOH?

A: The balanced chemical equation for the reaction between HNO3 and NaOH is:

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

Q: How do I calculate the number of moles of HNO3?

A: To calculate the number of moles of HNO3, you can use the formula:

${ \text{moles} = \text{molarity} \times \text{volume (in liters)} }$

Given that the molarity of HNO3 is 0.100 M and the volume is 25 mL (or 0.025 L), you can calculate the number of moles as follows:

${ \text{moles of HNO}_3 = 0.100 \, \text{M} \times 0.025 \, \text{L} = 0.0025 \, \text{mol} }$

Q: How do I calculate the number of moles of NaOH required?

A: Since the reaction is a 1:1 ratio, the number of moles of NaOH required is equal to the number of moles of HNO3. Therefore, the number of moles of NaOH required is also 0.0025 mol.

Q: How do I calculate the volume of NaOH required?

A: To calculate the volume of NaOH required, you can use the formula:

${ \text{volume} = \frac{\text{moles}}{\text{molarity}} }$

Given that the number of moles of NaOH required is 0.0025 mol and the molarity of NaOH is also 0.100 M, you can calculate the volume as follows:

${ \text{volume of NaOH} = \frac{0.0025 \, \text{mol}}{0.100 \, \text{M}} = 0.025 \, \text{L} }$

Q: How do I convert the volume from liters to milliliters?

A: To convert the volume from liters to milliliters, you can multiply by 1000 (since there are 1000 mL in 1 L):

${ 0.025 \, \text{L} \times 1000 \, \frac{\text{mL}}{\text{L}} = 25 \, \text{mL} }$

Q: What is the significance of the balanced chemical equation in this reaction?

A: The balanced chemical equation is essential in understanding the stoichiometry of the reaction and the amount of reagents required for a complete reaction.

Q: What are some common mistakes to avoid when calculating the volume of NaOH required?

A: Some common mistakes to avoid when calculating the volume of NaOH required include:

Not converting the volume from liters to milliliters
Not using the correct molarity of NaOH
Not calculating the number of moles of NaOH required correctly

Conclusion

In this article, we answered some frequently asked questions related to the reaction between HNO3 and NaOH and calculated the volume of NaOH required to completely react with a 25 mL sample of 0.100 M HNO3. We hope this article has been helpful in clarifying any doubts you may have had.

References

Chemical Equations and Stoichiometry. (n.d.). Retrieved from https://www.chemguide.co.uk/physical/acidbase/stoichiometry.html
Molarity and Concentration. (n.d.). Retrieved from https://www.chemguide.co.uk/physical/concentration/molarity.html