Understanding the Solubility Product Constant

Oct 13, 2025 by recoveryshake.com 34 views

Understanding the Solubility Product Constant

The solubility product constant, denoted as $$K_{sp}$$, is a measure of the equilibrium between a solid ionic compound and its ions in a solution. It is a crucial concept in chemistry, particularly in the study of solubility and chemical equilibrium. In this article, we will explore the solubility product constant of chromium(III) hydroxide, $$Cr(OH)_3$$, and calculate its solubility in $mol \, dm^{-3}$ .

The Solubility Product Constant of Chromium(III) Hydroxide

The solubility product constant of chromium(III) hydroxide, $$K_{sp}$$, is given as $2.7 \times 10^{-31}$ . This value represents the equilibrium constant for the dissolution of chromium(III) hydroxide in water.

Cr(OH)_3 (s) \rightleftharpoons Cr^{3+} (aq) + 3OH^- (aq)

Calculating the Solubility of Chromium(III) Hydroxide

To calculate the solubility of chromium(III) hydroxide, we can use the formula:

K_{sp} = [Cr^{3+}] [OH^-]^3

where $[Cr^{3+}]$ and $[OH^-]$ are the concentrations of chromium(III) ions and hydroxide ions, respectively.

Rearranging the Formula

We can rearrange the formula to solve for $[Cr^{3+}]$ :

[Cr^{3+}] = \frac{K_{sp}}{[OH^-]^3}

Calculating the Concentration of Hydroxide Ions

The concentration of hydroxide ions, $[OH^-]$ , can be calculated using the dissociation constant of water, $K_w$ :

K_w = [H^+] [OH^-] = 10^{-14}

Since the concentration of hydrogen ions, $[H^+]$ , is negligible compared to the concentration of hydroxide ions, we can assume that $[H^+] = 10^{-7}$ .

Calculating the Concentration of Hydroxide Ions

Rearranging the formula for $K_w$ , we get:

[OH^-] = \frac{10^{-14}}{[H^+]} = \frac{10^{-14}}{10^{-7}} = 10^{-7}

Substituting the Values

Substituting the values of $K_{sp}$ and $[OH^-]$ into the formula for $[Cr^{3+}]$ , we get:

[Cr^{3+}] = \frac{2.7 \times 10^{-31}}{(10^{-7})^3} = \frac{2.7 \times 10^{-31}}{10^{-21}} = 2.7 \times 10^{-10}

Calculating the Solubility of Chromium(III) Hydroxide

The solubility of chromium(III) hydroxide is equal to the concentration of chromium(III) ions, $[Cr^{3+}]$ . Therefore, the solubility of chromium(III) hydroxide is:

S = [Cr^{3+}] = 2.7 \times 10^{-10} \, mol \, dm^{-3}

Conclusion

In conclusion, the solubility product constant of chromium(III) hydroxide, $$K_{sp}$$, is $2.7 \times 10^{-31}$ . Using this value, we calculated the solubility of chromium(III) hydroxide in $mol \, dm^{-3}$ , which is $2.7 \times 10^{-10}$ . This result demonstrates the importance of the solubility product constant in understanding the solubility of ionic compounds.

Frequently Asked Questions

Q: What is the solubility product constant?

A: The solubility product constant, denoted as $$K_{sp}$$, is a measure of the equilibrium between a solid ionic compound and its ions in a solution.

Q: How is the solubility product constant calculated?

A: The solubility product constant is calculated using the formula:

K_{sp} = [Cr^{3+}] [OH^-]^3

Q: What is the relationship between the solubility product constant and the solubility of a compound?

A: The solubility product constant is related to the solubility of a compound through the formula:

S = \frac{K_{sp}}{[OH^-]^3}

Q: What is the significance of the solubility product constant in chemistry?

A: The solubility product constant is a crucial concept in chemistry, particularly in the study of solubility and chemical equilibrium.

References

[1] Atkins, P. W., & De Paula, J. (2010). Physical chemistry. Oxford University Press.
[2] Brown, T. E., & LeMay, H. E. (2012). Chemistry: The Central Science. Prentice Hall.
[3] Petrucci, R. H., Harwood, W. S., & Herring, F. G. (2011). General chemistry: Principles and modern applications. Pearson Education.

Understanding the Solubility Product Constant and Solubility of Chromium(III) Hydroxide

In our previous article, we explored the solubility product constant of chromium(III) hydroxide, $$K_{sp}$$, and calculated its solubility in $mol \, dm^{-3}$ . In this article, we will answer some frequently asked questions related to the solubility product constant and the solubility of chromium(III) hydroxide.

Q: What is the solubility product constant?

A: The solubility product constant, denoted as $$K_{sp}$$, is a measure of the equilibrium between a solid ionic compound and its ions in a solution.

Q: How is the solubility product constant calculated?

A: The solubility product constant is calculated using the formula:

K_{sp} = [Cr^{3+}] [OH^-]^3

where $[Cr^{3+}]$ and $[OH^-]$ are the concentrations of chromium(III) ions and hydroxide ions, respectively.

Q: What is the relationship between the solubility product constant and the solubility of a compound?

A: The solubility product constant is related to the solubility of a compound through the formula:

S = \frac{K_{sp}}{[OH^-]^3}

Q: What is the significance of the solubility product constant in chemistry?

A: The solubility product constant is a crucial concept in chemistry, particularly in the study of solubility and chemical equilibrium.

Q: How is the solubility of a compound affected by the solubility product constant?

A: The solubility of a compound is affected by the solubility product constant in the following way:

If the solubility product constant is high, the compound is more soluble.
If the solubility product constant is low, the compound is less soluble.

Q: What is the relationship between the solubility product constant and the pH of a solution?

A: The solubility product constant is related to the pH of a solution through the formula:

K_{sp} = [Cr^{3+}] [OH^-]^3

Since the concentration of hydroxide ions, $[OH^-]$ , is related to the pH of a solution, the solubility product constant is also related to the pH of a solution.

Q: How can the solubility product constant be used to predict the solubility of a compound?

A: The solubility product constant can be used to predict the solubility of a compound by using the formula:

S = \frac{K_{sp}}{[OH^-]^3}

This formula allows us to calculate the solubility of a compound based on its solubility product constant and the concentration of hydroxide ions in the solution.

Q: What are some common applications of the solubility product constant?

A: The solubility product constant has many common applications in chemistry, including:

Predicting the solubility of ionic compounds
Understanding the behavior of chemical systems
Designing chemical reactions and processes

Q: What are some common mistakes to avoid when working with the solubility product constant?

A: Some common mistakes to avoid when working with the solubility product constant include:

Failing to account for the concentration of hydroxide ions
Failing to use the correct formula for the solubility product constant
Failing to consider the pH of the solution

Conclusion

In conclusion, the solubility product constant is a crucial concept in chemistry, particularly in the study of solubility and chemical equilibrium. By understanding the solubility product constant and its relationship to the solubility of a compound, we can better predict the behavior of chemical systems and design more effective chemical reactions and processes.

Frequently Asked Questions

Q: What is the solubility product constant?

A: The solubility product constant, denoted as $$K_{sp}$$, is a measure of the equilibrium between a solid ionic compound and its ions in a solution.

Q: How is the solubility product constant calculated?

A: The solubility product constant is calculated using the formula:

K_{sp} = [Cr^{3+}] [OH^-]^3

Q: What is the relationship between the solubility product constant and the solubility of a compound?

A: The solubility product constant is related to the solubility of a compound through the formula:

S = \frac{K_{sp}}{[OH^-]^3}

Q: What is the significance of the solubility product constant in chemistry?

A: The solubility product constant is a crucial concept in chemistry, particularly in the study of solubility and chemical equilibrium.

References

[1] Atkins, P. W., & De Paula, J. (2010). Physical chemistry. Oxford University Press.
[2] Brown, T. E., & LeMay, H. E. (2012). Chemistry: The Central Science. Prentice Hall.
[3] Petrucci, R. H., Harwood, W. S., & Herring, F. G. (2011). General chemistry: Principles and modern applications. Pearson Education.