The Acid Dissociation Reaction

Oct 13, 2025 by ludoplayers7.com 51 views

Introduction

In chemistry, the equilibrium concentration of ions in a solution is a crucial concept that helps us understand the behavior of chemical reactions. In this article, we will determine the equilibrium concentration of $$1 \text{O}_3^{-}$$ ions in a $$0.00450 , \text{M} , \text{H} , \text{Lu}_3$$ solution, given that the acid dissociation constant ($$K_a$$) is $$0.16$$.

Understanding the Problem

The problem involves a solution of $$\text{H} , \text{Lu}_3$$, which is a weak acid. We are given the initial concentration of the acid ($$0.00450 , \text{M}$$) and the acid dissociation constant ($$K_a = 0.16$$). Our goal is to determine the equilibrium concentration of $$1 \text{O}_3^{-}$$ ions in the solution.

The Acid Dissociation Reaction

The acid dissociation reaction for $$\text{H} , \text{Lu}_3$$ can be written as:

\text{H} \, \text{Lu}_3 \rightleftharpoons \text{H}^+ + \text{Lu}_3^-

However, since we are interested in the concentration of $$1 \text{O}_3^{-}$$ ions, we need to consider the reaction:

\text{H} \, \text{Lu}_3 \rightleftharpoons \text{H}^+ + \text{Lu}_3^{3-}

The Acid Dissociation Constant ($$K_a$$)

The acid dissociation constant ($$K_a$$) is a measure of the strength of an acid. It is defined as the ratio of the concentrations of the products to the concentration of the reactant:

K_a = \frac{[\text{H}^+][\text{Lu}_3^{3-}]}{[\text{H} \, \text{Lu}_3]}

Determining the Equilibrium Concentration

To determine the equilibrium concentration of $$1 \text{O}_3^{-}$$ ions, we can use the acid dissociation constant ($$K_a$$) and the initial concentration of the acid ($$0.00450 , \text{M}$$). We can start by assuming that the initial concentration of the acid is equal to the equilibrium concentration of the acid ($$[\text{H} , \text{Lu}_3]_0 = [\text{H} , \text{Lu}_3]_e$$).

We can then use the acid dissociation constant ($$K_a$$) to determine the equilibrium concentrations of the products ($$[\text{H}^+]$$ and $$[\text{Lu}_3^{3-}]$$):

K_a = \frac{[\text{H}^+][\text{Lu}_3^{3-}]}{[\text{H} \, \text{Lu}_3]}

0.16 = \frac{[\text{H}^+][\text{Lu}_3^{3-}]}{0.00450}

Solving for the Equilibrium Concentrations

To solve for the equilibrium concentrations, we can start by assuming that the equilibrium concentration of the acid is equal to the initial concentration ($$[\text{H} , \text{Lu}_3]_e = [\text{H} , \text{Lu}_3]_0 = 0.00450 , \text{M}$$).

We can then use the acid dissociation constant ($$K_a$$) to determine the equilibrium concentrations of the products ($$[\text{H}^+]$$ and $$[\text{Lu}_3^{3-}]$$):

0.16 = \frac{[\text{H}^+][\text{Lu}_3^{3-}]}{0.00450}

[\text{H}^+][\text{Lu}_3^{3-}] = 0.16 \times 0.00450

[\text{H}^+][\text{Lu}_3^{3-}] = 0.00072

Using the Quotient Rule

We can use the quotient rule to determine the equilibrium concentrations of the products ($$[\text{H}^+]$$ and $$[\text{Lu}_3^{3-}]$$):

\frac{[\text{H}^+]}{[\text{H} \, \text{Lu}_3]} = \sqrt{K_a}

\frac{[\text{H}^+]}{0.00450} = \sqrt{0.16}

[\text{H}^+] = 0.00450 \times \sqrt{0.16}

[\text{H}^+] = 0.00072

Determining the Equilibrium Concentration of $$1 \text{O}_3^{-}$$ Ions

Now that we have determined the equilibrium concentrations of the products ($$[\text{H}^+]$$ and $$[\text{Lu}_3^{3-}]$$), we can use the acid dissociation constant ($$K_a$$) to determine the equilibrium concentration of $$1 \text{O}_3^{-}$$ ions:

K_a = \frac{[\text{H}^+][\text{Lu}_3^{3-}]}{[\text{H} \, \text{Lu}_3]}

0.16 = \frac{0.00072 \times [\text{Lu}_3^{3-}]}{0.00450}

[\text{Lu}_3^{3-}] = \frac{0.16 \times 0.00450}{0.00072}

[\text{Lu}_3^{3-}] = 0.00133

Conclusion

In this article, we have determined the equilibrium concentration of $$1 \text{O}_3^{-}$$ ions in a $$0.00450 , \text{M} , \text{H} , \text{Lu}_3$$ solution, given that the acid dissociation constant ($$K_a$$) is $$0.16$$. We have used the acid dissociation constant ($$K_a$$) and the initial concentration of the acid ($$0.00450 , \text{M}$$) to determine the equilibrium concentrations of the products ($$[\text{H}^+]$$ and $$[\text{Lu}_3^{3-}]$$). We have then used the acid dissociation constant ($$K_a$$) to determine the equilibrium concentration of $$1 \text{O}_3^{-}$$ ions.

References

[1] Atkins, P. W., & De Paula, J. (2010). Physical chemistry. Oxford University Press.
[2] Brown, T. E., LeMay, H. E., Bursten, B. E., & Murphy, C. (2012). Chemistry: The Central Science. Pearson Education.
[3] Chang, R. (2010). Chemistry. McGraw-Hill.

Appendix

The following is a list of the equations used in this article:

$K_a = \frac{[\text{H}^+][\text{Lu}_3^{3-}]}{[\text{H} \, \text{Lu}_3]}$
$0.16 = \frac{[\text{H}^+][\text{Lu}_3^{3-}]}{0.00450}$
$[\text{H}^+][\text{Lu}_3^{3-}] = 0.00072$
$\frac{[\text{H}^+]}{[\text{H} \, \text{Lu}_3]} = \sqrt{K_a}$
$[\text{H}^+] = 0.00450 \times \sqrt{0.16}$
$[\text{Lu}_3^{3-}] = \frac{0.16 \times 0.00450}{0.00072}$$<br/>$

Q: What is the acid dissociation constant ($$K_a$$) and how is it used to determine the equilibrium concentration of ions in a solution?

A: The acid dissociation constant ($$K_a$$) is a measure of the strength of an acid. It is defined as the ratio of the concentrations of the products to the concentration of the reactant. In the case of the $$\text{H} , \text{Lu}_3$$ solution, the acid dissociation constant ($$K_a$$) is $$0.16$$. We can use the acid dissociation constant ($$K_a$$) to determine the equilibrium concentrations of the products ($$[\text{H}^+]$$ and $$[\text{Lu}_3^{3-}]$$) and the equilibrium concentration of $$1 \text{O}_3^{-}$$ ions.

Q: How do we determine the equilibrium concentrations of the products ($$[\text{H}^+]$$ and $$[\text{Lu}_3^{3-}]$$) in the $$\text{H} , \text{Lu}_3$$ solution?

A: We can use the acid dissociation constant ($$K_a$$) and the initial concentration of the acid ($$0.00450 , \text{M}$$) to determine the equilibrium concentrations of the products ($$[\text{H}^+]$$ and $$[\text{Lu}_3^{3-}]$$). We can start by assuming that the equilibrium concentration of the acid is equal to the initial concentration ($$[\text{H} , \text{Lu}_3]_e = [\text{H} , \text{Lu}_3]_0 = 0.00450 , \text{M}$$). We can then use the acid dissociation constant ($$K_a$$) to determine the equilibrium concentrations of the products ($$[\text{H}^+]$$ and $$[\text{Lu}_3^{3-}]$$).

Q: How do we determine the equilibrium concentration of $$1 \text{O}_3^{-}$$ ions in the $$\text{H} , \text{Lu}_3$$ solution?

A: We can use the acid dissociation constant ($$K_a$$) and the equilibrium concentrations of the products ($$[\text{H}^+]$$ and $$[\text{Lu}_3^{3-}]$$) to determine the equilibrium concentration of $$1 \text{O}_3^{-}$$ ions. We can use the equation:

K_a = \frac{[\text{H}^+][\text{Lu}_3^{3-}]}{[\text{H} \, \text{Lu}_3]}

to determine the equilibrium concentration of $$1 \text{O}_3^{-}$$ ions.

Q: What is the significance of the acid dissociation constant ($$K_a$$) in determining the equilibrium concentration of ions in a solution?

A: The acid dissociation constant ($$K_a$$) is a measure of the strength of an acid. It is used to determine the equilibrium concentrations of the products ($$[\text{H}^+]$$ and $$[\text{Lu}_3^{3-}]$$) and the equilibrium concentration of $$1 \text{O}_3^{-}$$ ions. The acid dissociation constant ($$K_a$$) is a crucial concept in chemistry and is used to understand the behavior of chemical reactions.

Q: How do we apply the acid dissociation constant ($$K_a$$) to real-world problems?

A: The acid dissociation constant ($$K_a$$) is used in a variety of real-world applications, including:

Determining the pH of a solution
Understanding the behavior of chemical reactions
Designing chemical processes and equipment
Developing new materials and products

Q: What are some common mistakes to avoid when determining the equilibrium concentration of ions in a solution?

A: Some common mistakes to avoid when determining the equilibrium concentration of ions in a solution include:

Failing to consider the acid dissociation constant ($$K_a$$)
Not using the correct initial concentration of the acid
Not using the correct equilibrium concentrations of the products
Not considering the effects of temperature and pressure on the equilibrium concentration of ions

Q: How do we ensure the accuracy and reliability of our calculations when determining the equilibrium concentration of ions in a solution?

A: To ensure the accuracy and reliability of our calculations, we should:

Use the correct values for the acid dissociation constant ($$K_a$$) and the initial concentration of the acid
Use the correct equilibrium concentrations of the products
Consider the effects of temperature and pressure on the equilibrium concentration of ions
Use a consistent and systematic approach to our calculations

Conclusion

In this Q&A article, we have discussed the acid dissociation constant ($$K_a$$) and its role in determining the equilibrium concentration of ions in a solution. We have also discussed how to determine the equilibrium concentrations of the products ($$[\text{H}^+]$$ and $$[\text{Lu}_3^{3-}]$$) and the equilibrium concentration of $$1 \text{O}_3^{-}$$ ions. We have also discussed some common mistakes to avoid and how to ensure the accuracy and reliability of our calculations.