What Is The Concentration Of An $H_2SO_4$ Solution If 10.00 ML Is Required To Neutralize 40.00 ML Of 0.500 M $NaOH$?

Introduction

In chemistry, the concentration of a solution is a measure of the amount of solute present in a given volume of the solution. It is an essential concept in various fields, including chemistry, biology, and environmental science. In this article, we will discuss how to calculate the concentration of an $H_2SO_4$ solution using the information provided by a neutralization reaction between $H_2SO_4$ and $NaOH$.

Neutralization Reaction

A neutralization reaction is a chemical reaction between an acid and a base that results in the formation of a salt and water. In this reaction, $H_2SO_4$ (sulfuric acid) reacts with $NaOH$ (sodium hydroxide) to form $Na_2SO_4$ (sodium sulfate) and water.

$$H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$$

Stoichiometry of the Reaction

To calculate the concentration of the $H_2SO_4$ solution, we need to use the stoichiometry of the reaction. The balanced equation shows that 1 mole of $H_2SO_4$ reacts with 2 moles of $NaOH$. This means that the mole ratio of $H_2SO_4$ to $NaOH$ is 1:2.

Calculating the Number of Moles of $NaOH$

We are given that 40.00 mL of 0.500 M $NaOH$ is required to neutralize the $H_2SO_4$ solution. To calculate the number of moles of $NaOH$, we can use the formula:

$$n = C \times V$$

where $n$ is the number of moles, $C$ is the concentration in moles per liter (M), and $V$ is the volume in liters (L).

$$n_{NaOH} = 0.500 \, M \times 0.04000 \, L = 0.02000 \, mol$$

Calculating the Number of Moles of $H_2SO_4$

Since the mole ratio of $H_2SO_4$ to $NaOH$ is 1:2, the number of moles of $H_2SO_4$ is half the number of moles of $NaOH$.

$$n_{H_2SO_4} = \frac{1}{2} \times n_{NaOH} = \frac{1}{2} \times 0.02000 \, mol = 0.01000 \, mol$$

Calculating the Concentration of the $H_2SO_4$ Solution

We are given that 10.00 mL of the $H_2SO_4$ solution is required to neutralize the $NaOH$ solution. To calculate the concentration of the $H_2SO_4$ solution, we can use the formula:

$$C = \frac{n}{V}$$

where $C$ is the concentration in moles per liter (M), $n$ is the number of moles, and $V$ is the volume in liters (L).

$$C_{H_2SO_4} = \frac{0.01000 \, mol}{0.01000 \, L} = 1.000 \, M$$

Conclusion

In this article, we discussed how to calculate the concentration of an $H_2SO_4$ solution using the information provided by a neutralization reaction between $H_2SO_4$ and $NaOH$. We used the stoichiometry of the reaction to calculate the number of moles of $H_2SO_4$ and then used the formula to calculate the concentration of the $H_2SO_4$ solution. The concentration of the $H_2SO_4$ solution was found to be 1.000 M.

Frequently Asked Questions

• What is the concentration of an $H_2SO_4$ solution if 10.00 mL is required to neutralize 40.00 mL of 0.500 M $NaOH$?
• How do you calculate the concentration of a solution using a neutralization reaction?
• What is the mole ratio of $H_2SO_4$ to $NaOH$ in the neutralization reaction?

References

• Chemistry: An Atoms First Approach, by Steven S. Zumdahl
• General Chemistry: Principles and Modern Applications, by Linus Pauling
• Chemistry: The Central Science, by Theodore L. Brown

Further Reading

• Neutralization Reactions: A Comprehensive Review
• Stoichiometry of Chemical Reactions: A Guide to Calculations
• Concentration of Solutions: A Tutorial
  

Introduction

In our previous article, we discussed how to calculate the concentration of an $H_2SO_4$ solution using the information provided by a neutralization reaction between $H_2SO_4$ and $NaOH$. In this article, we will answer some frequently asked questions related to the concentration of an $H_2SO_4$ solution.

Q: What is the concentration of an $H_2SO_4$ solution if 10.00 mL is required to neutralize 40.00 mL of 0.500 M $NaOH$?

A: To calculate the concentration of the $H_2SO_4$ solution, we need to use the stoichiometry of the reaction. The balanced equation shows that 1 mole of $H_2SO_4$ reacts with 2 moles of $NaOH$. This means that the mole ratio of $H_2SO_4$ to $NaOH$ is 1:2. We can calculate the number of moles of $NaOH$ using the formula:

$$n = C \times V$$

where $n$ is the number of moles, $C$ is the concentration in moles per liter (M), and $V$ is the volume in liters (L).

$$n_{NaOH} = 0.500 \, M \times 0.04000 \, L = 0.02000 \, mol$$

Since the mole ratio of $H_2SO_4$ to $NaOH$ is 1:2, the number of moles of $H_2SO_4$ is half the number of moles of $NaOH$.

$$n_{H_2SO_4} = \frac{1}{2} \times n_{NaOH} = \frac{1}{2} \times 0.02000 \, mol = 0.01000 \, mol$$

We can then calculate the concentration of the $H_2SO_4$ solution using the formula:

$$C = \frac{n}{V}$$

where $C$ is the concentration in moles per liter (M), $n$ is the number of moles, and $V$ is the volume in liters (L).

$$C_{H_2SO_4} = \frac{0.01000 \, mol}{0.01000 \, L} = 1.000 \, M$$

Q: How do you calculate the concentration of a solution using a neutralization reaction?

A: To calculate the concentration of a solution using a neutralization reaction, you need to follow these steps:

1. Write the balanced equation for the neutralization reaction.
2. Determine the mole ratio of the acid to the base.
3. Calculate the number of moles of the acid using the formula:

$$n = C \times V$$

where $n$ is the number of moles, $C$ is the concentration in moles per liter (M), and $V$ is the volume in liters (L). 4. Calculate the number of moles of the base using the mole ratio. 5. Calculate the concentration of the solution using the formula:

$$C = \frac{n}{V}$$

where $C$ is the concentration in moles per liter (M), $n$ is the number of moles, and $V$ is the volume in liters (L).

Q: What is the mole ratio of $H_2SO_4$ to $NaOH$ in the neutralization reaction?

A: The mole ratio of $H_2SO_4$ to $NaOH$ in the neutralization reaction is 1:2. This means that 1 mole of $H_2SO_4$ reacts with 2 moles of $NaOH$.

Q: How do you calculate the number of moles of $H_2SO_4$ in a neutralization reaction?

A: To calculate the number of moles of $H_2SO_4$ in a neutralization reaction, you need to follow these steps:

1. Write the balanced equation for the neutralization reaction.
2. Determine the mole ratio of the acid to the base.
3. Calculate the number of moles of the base using the formula:

$$n = C \times V$$

where $n$ is the number of moles, $C$ is the concentration in moles per liter (M), and $V$ is the volume in liters (L). 4. Calculate the number of moles of the acid using the mole ratio.

Q: How do you calculate the concentration of a solution using the number of moles and volume?

A: To calculate the concentration of a solution using the number of moles and volume, you can use the formula:

$$C = \frac{n}{V}$$

where $C$ is the concentration in moles per liter (M), $n$ is the number of moles, and $V$ is the volume in liters (L).

Q: What is the concentration of a solution if 20.00 mL of 0.500 M $HCl$ is required to neutralize 40.00 mL of 0.500 M $NaOH$?

A: To calculate the concentration of the solution, we need to use the stoichiometry of the reaction. The balanced equation shows that 1 mole of $HCl$ reacts with 1 mole of $NaOH$. This means that the mole ratio of $HCl$ to $NaOH$ is 1:1. We can calculate the number of moles of $NaOH$ using the formula:

$$n = C \times V$$

where $n$ is the number of moles, $C$ is the concentration in moles per liter (M), and $V$ is the volume in liters (L).

$$n_{NaOH} = 0.500 \, M \times 0.04000 \, L = 0.02000 \, mol$$

Since the mole ratio of $HCl$ to $NaOH$ is 1:1, the number of moles of $HCl$ is equal to the number of moles of $NaOH$.

$$n_{HCl} = n_{NaOH} = 0.02000 \, mol$$

We can then calculate the concentration of the solution using the formula:

$$C = \frac{n}{V}$$

where $C$ is the concentration in moles per liter (M), $n$ is the number of moles, and $V$ is the volume in liters (L).

$$C_{HCl} = \frac{0.02000 \, mol}{0.02000 \, L} = 1.000 \, M$$

Conclusion

In this article, we answered some frequently asked questions related to the concentration of an $H_2SO_4$ solution. We discussed how to calculate the concentration of a solution using a neutralization reaction and provided examples to illustrate the concept. We also answered questions related to the mole ratio of $H_2SO_4$ to $NaOH$ and how to calculate the number of moles of $H_2SO_4$ in a neutralization reaction.

Frequently Asked Questions

• What is the concentration of an $H_2SO_4$ solution if 10.00 mL is required to neutralize 40.00 mL of 0.500 M $NaOH$?
• How do you calculate the concentration of a solution using a neutralization reaction?
• What is the mole ratio of $H_2SO_4$ to $NaOH$ in the neutralization reaction?
• How do you calculate the number of moles of $H_2SO_4$ in a neutralization reaction?
• How do you calculate the concentration of a solution using the number of moles and volume?
• What is the concentration of a solution if 20.00 mL of 0.500 M $HCl$ is required to neutralize 40.00 mL of 0.500 M $NaOH$?

References

• Chemistry: An Atoms First Approach, by Steven S. Zumdahl
• General Chemistry: Principles and Modern Applications, by Linus Pauling
• Chemistry: The Central Science, by Theodore L. Brown

Further Reading

• Neutralization Reactions: A Comprehensive Review
• Stoichiometry of Chemical Reactions: A Guide to Calculations
• Concentration of Solutions: A Tutorial