Calculating the Number of Moles of Hydrogen Ions

Oct 13, 2025 by magicalceylontours.com 140 views

Introduction

The dissolution of lead(II) carbonate ($$PbCO_3$$) in acidic solutions is an important process in various industrial and environmental applications. In this article, we will explore the quantitative analysis of the dissolution of lead(II) carbonate in 1.00 M H+ solution. We will use the given equation to determine the amount of lead(II) carbonate that will dissolve when 1.00 L of 1.00 M H+ is added to 5.00 g of $$PbCO_3$$.

Theoretical Background

The dissolution of lead(II) carbonate in acidic solutions can be represented by the following equation:

PbCO_3(s) + 2H^+(aq) \rightarrow Pb^{2+}(aq) + H_2O(\ell)

This equation shows that lead(II) carbonate reacts with two moles of hydrogen ions to produce lead(II) ions and water. The reaction is a first-order reaction with respect to lead(II) carbonate and a second-order reaction with respect to hydrogen ions.

Calculating the Number of Moles of Hydrogen Ions

To calculate the number of moles of hydrogen ions, we need to know the concentration of the hydrogen ion solution. The concentration of the hydrogen ion solution is given as 1.00 M, which means that there is 1 mole of hydrogen ions per liter of solution.

Since we are using 1.00 L of 1.00 M H+ solution, the number of moles of hydrogen ions can be calculated as follows:

n_{H^+} = C_{H^+} \times V_{H^+}

n_{H^+} = 1.00 \, M \times 1.00 \, L

n_{H^+} = 1.00 \, mol

Calculating the Number of Moles of Lead(II) Carbonate

To calculate the number of moles of lead(II) carbonate, we need to know the mass of the lead(II) carbonate. The mass of the lead(II) carbonate is given as 5.00 g.

The molar mass of lead(II) carbonate is 267.32 g/mol. We can calculate the number of moles of lead(II) carbonate as follows:

n_{PbCO_3} = \frac{m_{PbCO_3}}{M_{PbCO_3}}

n_{PbCO_3} = \frac{5.00 \, g}{267.32 \, g/mol}

n_{PbCO_3} = 0.0187 \, mol

Calculating the Number of Moles of Lead(II) Ions Produced

According to the equation, 1 mole of lead(II) carbonate produces 1 mole of lead(II) ions. Therefore, the number of moles of lead(II) ions produced is equal to the number of moles of lead(II) carbonate.

n_{Pb^{2+}} = n_{PbCO_3}

n_{Pb^{2+}} = 0.0187 \, mol

Calculating the Number of Moles of Hydrogen Ions Required

According to the equation, 2 moles of hydrogen ions are required to produce 1 mole of lead(II) ions. Therefore, the number of moles of hydrogen ions required is twice the number of moles of lead(II) ions produced.

n_{H^+} = 2 \times n_{Pb^{2+}}

n_{H^+} = 2 \times 0.0187 \, mol

n_{H^+} = 0.0374 \, mol

Calculating the Mass of Lead(II) Carbonate Dissolved

Since the number of moles of hydrogen ions required is 0.0374 mol, and the number of moles of lead(II) carbonate is 0.0187 mol, we can calculate the mass of lead(II) carbonate dissolved as follows:

m_{PbCO_3} = n_{PbCO_3} \times M_{PbCO_3}

m_{PbCO_3} = 0.0187 \, mol \times 267.32 \, g/mol

m_{PbCO_3} = 5.00 \, g

Conclusion

In this article, we have calculated the amount of lead(II) carbonate that will dissolve when 1.00 L of 1.00 M H+ is added to 5.00 g of $$PbCO_3$$. The calculation shows that 5.00 g of lead(II) carbonate will dissolve, which is equal to the initial mass of lead(II) carbonate.

This result is expected, as the reaction is a first-order reaction with respect to lead(II) carbonate and a second-order reaction with respect to hydrogen ions. The number of moles of hydrogen ions required is twice the number of moles of lead(II) ions produced, which is consistent with the stoichiometry of the reaction.

Limitations of the Calculation

The calculation assumes that the reaction is complete and that all of the lead(II) carbonate is dissolved. In reality, the reaction may not be complete, and some of the lead(II) carbonate may remain undissolved. Additionally, the calculation assumes that the concentration of the hydrogen ion solution is constant throughout the reaction. In reality, the concentration of the hydrogen ion solution may decrease as the reaction proceeds.

Future Work

Future work could involve experimental verification of the calculation. This could involve measuring the amount of lead(II) carbonate dissolved in a 1.00 L of 1.00 M H+ solution and comparing the result to the calculated value.

References

[1] "Lead(II) Carbonate" in CRC Handbook of Chemistry and Physics, 97th ed., CRC Press, 2016.
[2] "Acid-Base Chemistry" in Chemistry: An Atoms First Approach, 2nd ed., Cengage Learning, 2013.

Glossary

Lead(II) Carbonate: A white solid compound with the formula PbCO3.
Hydrogen Ion: A positively charged ion with the formula H+.
Lead(II) Ion: A positively charged ion with the formula Pb2+.
Molar Mass: The mass of one mole of a substance.
Molarity: The concentration of a solution in moles per liter.

Q: What is the dissolution of lead(II) carbonate in acidic solution?

A: The dissolution of lead(II) carbonate in acidic solution is a chemical reaction in which lead(II) carbonate reacts with hydrogen ions to produce lead(II) ions and water.

Q: What is the equation for the dissolution of lead(II) carbonate in acidic solution?

A: The equation for the dissolution of lead(II) carbonate in acidic solution is:

PbCO_3(s) + 2H^+(aq) \rightarrow Pb^{2+}(aq) + H_2O(\ell)

Q: What is the role of hydrogen ions in the dissolution of lead(II) carbonate?

A: Hydrogen ions play a crucial role in the dissolution of lead(II) carbonate. They react with lead(II) carbonate to produce lead(II) ions and water.

Q: How many moles of hydrogen ions are required to dissolve 1 mole of lead(II) carbonate?

A: According to the equation, 2 moles of hydrogen ions are required to dissolve 1 mole of lead(II) carbonate.

Q: What is the molar mass of lead(II) carbonate?

A: The molar mass of lead(II) carbonate is 267.32 g/mol.

Q: How many grams of lead(II) carbonate will dissolve when 1.00 L of 1.00 M H+ is added to 5.00 g of PbCO3?

A: According to the calculation, 5.00 g of lead(II) carbonate will dissolve when 1.00 L of 1.00 M H+ is added to 5.00 g of PbCO3.

Q: What are the limitations of the calculation?

A: The calculation assumes that the reaction is complete and that all of the lead(II) carbonate is dissolved. In reality, the reaction may not be complete, and some of the lead(II) carbonate may remain undissolved. Additionally, the calculation assumes that the concentration of the hydrogen ion solution is constant throughout the reaction. In reality, the concentration of the hydrogen ion solution may decrease as the reaction proceeds.

Q: What is the future work for this topic?

A: Future work could involve experimental verification of the calculation. This could involve measuring the amount of lead(II) carbonate dissolved in a 1.00 L of 1.00 M H+ solution and comparing the result to the calculated value.

Q: What are the references for this topic?

A: The references for this topic are:

[1] "Lead(II) Carbonate" in CRC Handbook of Chemistry and Physics, 97th ed., CRC Press, 2016.
[2] "Acid-Base Chemistry" in Chemistry: An Atoms First Approach, 2nd ed., Cengage Learning, 2013.

Q: What is the glossary for this topic?

A: The glossary for this topic includes:

Lead(II) Carbonate: A white solid compound with the formula PbCO3.
Hydrogen Ion: A positively charged ion with the formula H+.
Lead(II) Ion: A positively charged ion with the formula Pb2+.
Molar Mass: The mass of one mole of a substance.
Molarity: The concentration of a solution in moles per liter.

Q: What are the applications of this topic?

A: The applications of this topic include:

Industrial Processes: The dissolution of lead(II) carbonate in acidic solution is an important process in various industrial applications, such as the production of lead-based batteries and the treatment of wastewater.
Environmental Science: The dissolution of lead(II) carbonate in acidic solution is also an important process in environmental science, as it can affect the pH of natural waters and the availability of lead ions to aquatic organisms.

Q: What are the safety precautions for this topic?

A: The safety precautions for this topic include:

Handling Chemicals: When handling chemicals, it is essential to wear protective clothing, including gloves, goggles, and a lab coat.
Working with Acids: When working with acids, it is essential to handle them with care, as they can cause burns and other injuries.
Proper Disposal: When disposing of chemicals, it is essential to follow proper procedures to prevent environmental contamination.