Consider The Titration Of 60.0 ML Of 0.19 \, \text{M} \, \text{H}_3\text{A} \left(K_{a_1}=1.9 \times 10^{-4}, K_{a_2}=1.9 \times 10^{-8}, K_{a_3}=1.6 \times 10^{-11}\right ] Titrated By 0.19 M KOH 0.19 \, \text{M} \, \text{KOH} 0.19 M KOH .a. Calculate

Introduction

In this article, we will delve into the intricacies of titrating a polyprotic acid, specifically $0.19 \, \text{M} \, \text{H}_3\text{A}$, with a strong base, $0.19 \, \text{M} \, \text{KOH}$. The acid in question is a triprotic acid, meaning it can donate three protons (H+ ions) per molecule. Understanding the titration of such an acid is crucial in various fields, including chemistry, biochemistry, and environmental science.

Theoretical Background

Before we proceed with the calculations, let's briefly discuss the theoretical background. A polyprotic acid is an acid that can donate multiple protons. In this case, we have a triprotic acid, $0.19 \, \text{M} \, \text{H}_3\text{A}$, with the following dissociation constants:

• $K_{a_1}=1.9 \times 10^{-4}$
• $K_{a_2}=1.9 \times 10^{-8}$
• $K_{a_3}=1.6 \times 10^{-11}$

These dissociation constants indicate the strength of the acid at each step. The higher the value, the stronger the acid.

Titration Process

The titration process involves adding a strong base, $0.19 \, \text{M} \, \text{KOH}$, to the polyprotic acid, $0.19 \, \text{M} \, \text{H}_3\text{A}$, until the acid is completely neutralized. The reaction can be represented as follows:

$$\text{H}_3\text{A} + 3\text{KOH} \rightarrow \text{K}_3\text{A} + 3\text{H}_2\text{O}$$

Calculations

To calculate the number of moles of acid and base, we can use the following formulas:

• Number of moles of acid: $n_{\text{H}_3\text{A}} = C_{\text{H}_3\text{A}} \times V_{\text{H}_3\text{A}}$
• Number of moles of base: $n_{\text{KOH}} = C_{\text{KOH}} \times V_{\text{KOH}}$

where $C$ is the concentration of the solution and $V$ is the volume of the solution.

Given that the concentration of both the acid and the base is $0.19 \, \text{M}$ and the volume of the acid is $60.0 \, \text{mL}$, we can calculate the number of moles of acid as follows:

$$n_{\text{H}_3\text{A}} = 0.19 \, \text{M} \times 60.0 \, \text{mL} = 11.4 \, \text{mmol}$$

Similarly, we can calculate the number of moles of base as follows:

$$n_{\text{KOH}} = 0.19 \, \text{M} \times 60.0 \, \text{mL} = 11.4 \, \text{mmol}$$

Equivalence Point

The equivalence point is the point at which the acid is completely neutralized by the base. At this point, the number of moles of acid is equal to the number of moles of base.

To calculate the volume of base required to reach the equivalence point, we can use the following formula:

$$V_{\text{KOH}} = \frac{n_{\text{H}_3\text{A}}}{C_{\text{KOH}}}$$

Substituting the values, we get:

$$V_{\text{KOH}} = \frac{11.4 \, \text{mmol}}{0.19 \, \text{M}} = 60.0 \, \text{mL}$$

This means that $60.0 \, \text{mL}$ of $0.19 \, \text{M} \, \text{KOH}$ is required to completely neutralize $60.0 \, \text{mL}$ of $0.19 \, \text{M} \, \text{H}_3\text{A}$.

pH at the Equivalence Point

At the equivalence point, the pH of the solution can be calculated using the following formula:

$$\text{pH} = \frac{1}{3} \log \left( \frac{K_{a_1} K_{a_2} K_{a_3}}{K_{w}^3} \right)$$

where $K_w$ is the dissociation constant of water.

Substituting the values, we get:

$$\text{pH} = \frac{1}{3} \log \left( \frac{(1.9 \times 10^{-4})(1.9 \times 10^{-8})(1.6 \times 10^{-11})}{(1.0 \times 10^{-14})^3} \right)$$

Simplifying, we get:

$$\text{pH} = 7.0$$

This means that the pH of the solution at the equivalence point is 7.0.

Conclusion

In conclusion, the titration of $60.0 \, \text{mL}$ of $0.19 \, \text{M} \, \text{H}_3\text{A}$ with $0.19 \, \text{M} \, \text{KOH}$ requires $60.0 \, \text{mL}$ of base to completely neutralize the acid. The pH of the solution at the equivalence point is 7.0.

References

• Atkins, P. W., & De Paula, J. (2010). Physical chemistry. Oxford University Press.
• Chang, R. (2010). Physical chemistry for the life sciences. Cambridge University Press.
• Levine, I. N. (2014). Physical chemistry. McGraw-Hill Education.

Appendix

The following table summarizes the calculations:

Number of moles of acid	11.4 mmol
Number of moles of base	11.4 mmol
Volume of base required	60.0 mL
pH at the equivalence point	7.0

Introduction

In our previous article, we discussed the titration of a polyprotic acid, specifically $0.19 \, \text{M} \, \text{H}_3\text{A}$, with a strong base, $0.19 \, \text{M} \, \text{KOH}$. In this article, we will address some of the frequently asked questions related to this topic.

Q: What is a polyprotic acid?

A: A polyprotic acid is an acid that can donate multiple protons (H+ ions) per molecule. In this case, we have a triprotic acid, $0.19 \, \text{M} \, \text{H}_3\text{A}$, which can donate three protons.

Q: What is the significance of the dissociation constants (Ka) in a polyprotic acid?

A: The dissociation constants (Ka) indicate the strength of the acid at each step. The higher the value, the stronger the acid. In this case, we have the following dissociation constants:

• $K_{a_1}=1.9 \times 10^{-4}$
• $K_{a_2}=1.9 \times 10^{-8}$
• $K_{a_3}=1.6 \times 10^{-11}$

Q: How do you calculate the number of moles of acid and base in a titration reaction?

A: To calculate the number of moles of acid and base, we can use the following formulas:

• Number of moles of acid: $n_{\text{H}_3\text{A}} = C_{\text{H}_3\text{A}} \times V_{\text{H}_3\text{A}}$
• Number of moles of base: $n_{\text{KOH}} = C_{\text{KOH}} \times V_{\text{KOH}}$

where $C$ is the concentration of the solution and $V$ is the volume of the solution.

Q: What is the equivalence point in a titration reaction?

A: The equivalence point is the point at which the acid is completely neutralized by the base. At this point, the number of moles of acid is equal to the number of moles of base.

Q: How do you calculate the pH at the equivalence point in a titration reaction?

A: To calculate the pH at the equivalence point, we can use the following formula:

$$\text{pH} = \frac{1}{3} \log \left( \frac{K_{a_1} K_{a_2} K_{a_3}}{K_{w}^3} \right)$$

where $K_w$ is the dissociation constant of water.

Q: What is the significance of the pH at the equivalence point in a titration reaction?

A: The pH at the equivalence point indicates the pH of the solution when the acid is completely neutralized by the base. In this case, the pH at the equivalence point is 7.0.

Q: What are some common applications of titration reactions?

A: Titration reactions have various applications in chemistry, biochemistry, and environmental science. Some common applications include:

• Determining the concentration of a substance
• Identifying the presence of a substance
• Measuring the pH of a solution
• Monitoring the progress of a chemical reaction

Conclusion

In conclusion, the titration of a polyprotic acid with a strong base is a complex process that requires careful consideration of the dissociation constants, the number of moles of acid and base, and the pH at the equivalence point. By understanding these concepts, we can better appreciate the significance of titration reactions in various fields.

References

• Atkins, P. W., & De Paula, J. (2010). Physical chemistry. Oxford University Press.
• Chang, R. (2010). Physical chemistry for the life sciences. Cambridge University Press.
• Levine, I. N. (2014). Physical chemistry. McGraw-Hill Education.

Appendix

The following table summarizes the calculations:

Number of moles of acid	11.4 mmol
Number of moles of base	11.4 mmol
Volume of base required	60.0 mL
pH at the equivalence point	7.0

Note: The values in the table are calculated using the formulas and values provided in the article.