Not accounting for the stoichiometry of the reaction

Oct 13, 2025 by sorusgentrification.com 11 views

What is the Concentration of Fluoride Ions when Dissolved in a Solution Containing 0.0500 M of $Sr^{2+}$ Ions?

Understanding the Problem

The problem requires us to find the concentration of fluoride ions ( $F^{-}$ ) when $SrF_{2(s)}$ is dissolved in a solution containing 0.0500 M of $Sr^{2+}$ ions. To solve this problem, we need to use the concept of solubility product constant ( $K_{sp}$ ) and the equilibrium expression for the dissolution of $SrF_{2(s)}$ .

The Equilibrium Expression

The dissolution of $SrF_{2(s)}$ can be represented by the following equilibrium expression:

SrF_{2(s)} \rightleftharpoons Sr^{2+}(aq) + 2 F^{-}_{(aq)} \quad K_{sp} = 4.33 \times 10^{-9}

In this expression, $K_{sp}$ is the solubility product constant, which is a measure of the equilibrium between the dissolved ions and the solid compound.

Calculating the Concentration of Fluoride Ions

To calculate the concentration of fluoride ions, we can use the equilibrium expression and the given concentration of $Sr^{2+}$ ions. Let's assume that the concentration of $F^{-}$ ions is $x$ M. Since the dissolution of $SrF_{2(s)}$ produces 2 moles of $F^{-}$ ions for every 1 mole of $Sr^{2+}$ ions, the concentration of $F^{-}$ ions will be twice the concentration of $Sr^{2+}$ ions.

Setting Up the Equation

We can set up an equation using the equilibrium expression and the given concentration of $Sr^{2+}$ ions:

K_{sp} = [Sr^{2+}][F^{-}]^2

Substituting the given values, we get:

4.33 \times 10^{-9} = (0.0500)(x)^2

Solving for x

To solve for $x$ , we can rearrange the equation and take the square root of both sides:

x = \sqrt{\frac{4.33 \times 10^{-9}}{0.0500}}

Simplifying the expression, we get:

x = \sqrt{8.66 \times 10^{-8}}

Taking the square root, we get:

x = 9.30 \times 10^{-4}

Conclusion

Therefore, the concentration of fluoride ions when dissolved in a solution containing 0.0500 M of $Sr^{2+}$ ions is $9.30 \times 10^{-4}$ M.

Understanding the Significance of the Result

The result indicates that the concentration of fluoride ions is very low compared to the concentration of $Sr^{2+}$ ions. This is because the solubility product constant ( $K_{sp}$ ) is very small, indicating that the dissolution of $SrF_{2(s)}$ is limited.

Real-World Applications

The concept of solubility product constant ( $K_{sp}$ ) and the equilibrium expression for the dissolution of $SrF_{2(s)}$ has important real-world applications in various fields, including:

Water Treatment: The solubility product constant ( $K_{sp}$ ) is used to predict the solubility of minerals in water, which is important for water treatment processes.
Environmental Science: The equilibrium expression for the dissolution of $SrF_{2(s)}$ is used to predict the fate and transport of strontium and fluoride ions in the environment.
Materials Science: The solubility product constant ( $K_{sp}$ ) is used to predict the solubility of materials in various solvents, which is important for materials science applications.

Conclusion

In conclusion, the concentration of fluoride ions when dissolved in a solution containing 0.0500 M of $Sr^{2+}$ ions is $9.30 \times 10^{-4}$ M. The result indicates that the concentration of fluoride ions is very low compared to the concentration of $Sr^{2+}$ ions, which is due to the small solubility product constant ( $K_{sp}$ ). The concept of solubility product constant ( $K_{sp}$ ) and the equilibrium expression for the dissolution of $SrF_{2(s)}$ has important real-world applications in various fields.
Q&A: Solubility Product Constant ( $K_{sp}$ ) and the Dissolution of $SrF_{2(s)}$

Q: What is the solubility product constant ( $K_{sp}$ )?

A: The solubility product constant ( $K_{sp}$ ) is a measure of the equilibrium between the dissolved ions and the solid compound. It is a numerical value that represents the concentration of ions in a solution at equilibrium.

Q: How is the solubility product constant ( $K_{sp}$ ) related to the dissolution of $SrF_{2(s)}$ ?

A: The solubility product constant ( $K_{sp}$ ) is related to the dissolution of $SrF_{2(s)}$ through the equilibrium expression:

SrF_{2(s)} \rightleftharpoons Sr^{2+}(aq) + 2 F^{-}_{(aq)} \quad K_{sp} = 4.33 \times 10^{-9}

Q: What is the significance of the solubility product constant ( $K_{sp}$ ) in the dissolution of $SrF_{2(s)}$ ?

A: The solubility product constant ( $K_{sp}$ ) is significant in the dissolution of $SrF_{2(s)}$ because it determines the concentration of ions in the solution at equilibrium. A small $K_{sp}$ value indicates that the dissolution of $SrF_{2(s)}$ is limited, resulting in a low concentration of ions in the solution.

Q: How is the concentration of fluoride ions ( $F^{-}$ ) related to the concentration of $Sr^{2+}$ ions?

A: The concentration of fluoride ions ( $F^{-}$ ) is related to the concentration of $Sr^{2+}$ ions through the equilibrium expression:

K_{sp} = [Sr^{2+}][F^{-}]^2

Q: What is the concentration of fluoride ions ( $F^{-}$ ) when dissolved in a solution containing 0.0500 M of $Sr^{2+}$ ions?

A: The concentration of fluoride ions ( $F^{-}$ ) when dissolved in a solution containing 0.0500 M of $Sr^{2+}$ ions is $9.30 \times 10^{-4}$ M.

Q: What are some real-world applications of the solubility product constant ( $K_{sp}$ ) and the equilibrium expression for the dissolution of $SrF_{2(s)}$ ?

A: Some real-world applications of the solubility product constant ( $K_{sp}$ ) and the equilibrium expression for the dissolution of $SrF_{2(s)}$ include:

Water Treatment: The solubility product constant ( $K_{sp}$ ) is used to predict the solubility of minerals in water, which is important for water treatment processes.
Environmental Science: The equilibrium expression for the dissolution of $SrF_{2(s)}$ is used to predict the fate and transport of strontium and fluoride ions in the environment.
Materials Science: The solubility product constant ( $K_{sp}$ ) is used to predict the solubility of materials in various solvents, which is important for materials science applications.

Q: What are some common mistakes to avoid when working with the solubility product constant ( $K_{sp}$ ) and the equilibrium expression for the dissolution of $SrF_{2(s)}$ ?

A: Some common mistakes to avoid when working with the solubility product constant ( $K_{sp}$ ) and the equilibrium expression for the dissolution of $SrF_{2(s)}$ include:

Not considering the units of the solubility product constant ( $K_{sp}$ ): The solubility product constant ( $K_{sp}$ ) has units of concentration, which must be considered when using it in calculations.
Not accounting for the stoichiometry of the reaction: The equilibrium expression for the dissolution of $SrF_{2(s)}$ involves a 2:1 ratio of $F^{-}$ ions to $Sr^{2+}$ ions, which must be accounted for when using the solubility product constant ( $K_{sp}$ ) in calculations.

Q: What are some tips for working with the solubility product constant ( $K_{sp}$ ) and the equilibrium expression for the dissolution of $SrF_{2(s)}$ ?

A: Some tips for working with the solubility product constant ( $K_{sp}$ ) and the equilibrium expression for the dissolution of $SrF_{2(s)}$ include:

Use a calculator to perform calculations: The solubility product constant ( $K_{sp}$ ) and the equilibrium expression for the dissolution of $SrF_{2(s)}$ involve complex calculations, which can be performed using a calculator.
Check the units of the solubility product constant ( $K_{sp}$ ): The solubility product constant ( $K_{sp}$ ) has units of concentration, which must be checked when using it in calculations.
Account for the stoichiometry of the reaction: The equilibrium expression for the dissolution of $SrF_{2(s)}$ involves a 2:1 ratio of $F^{-}$ ions to $Sr^{2+}$ ions, which must be accounted for when using the solubility product constant ( $K_{sp}$ ) in calculations.