Problem 45 - Chemistry and Chemical Reactivity

Question

When \(250 \mathrm{mg}\) of \(\mathrm{SrF}_{2},\) strontium fluoride, is added to 1.00 L of water, the salt dissolves to a very small extent. $$\operatorname{SrF}_{2}(s) \rightleftarrows \operatorname{Sr}^{2+}(a q)+2 F^{-}(a q)$$ At equilibrium, the concentration of \(\mathrm{Sr}^{2+}\) is found to be \(1.03 \times 10^{-3} \mathrm{M}\). What is the value of \(K_{\mathrm{sp}}\) for \(\mathrm{SrF}_{2} ?\)

Step-by-Step Solution

Answer

The value of \(K_{sp}\) for \(\mathrm{SrF}_{2}\) is \(4.37 \times 10^{-9}\).

Step 1: Understanding the Reaction

Strontium fluoride, \(\text{SrF}_2\), dissolves in water to form one \(\text{Sr}^{2+}\) ion and two \(\text{F}^{-}\) ions. The equilibrium expression for the reaction is: \[ \text{SrF}_2 (s) \rightleftarrows \text{Sr}^{2+} (aq) + 2F^{-} (aq) \] The solubility product \(K_{sp}\) is based on the equilibrium concentrations of the ions formed.

Step 2: Writing the Expression for Ksp

The expression for the solubility product \(K_{sp}\) is: \[ K_{sp} = [\text{Sr}^{2+}][F^{-}]^2 \] We express \(K_{sp}\) in terms of concentrations of the ions at equilibrium.

Step 3: Identify Known Concentrations

We are given that the concentration of \(\text{Sr}^{2+}\) at equilibrium is \(1.03 \times 10^{-3} \text{ M}\). Since two \(F^{-}\) ions are produced for every \(\text{Sr}^{2+}\), the concentration of \(F^{-}\) is \(2 \times 1.03 \times 10^{-3} \text{ M}\).

Step 4: Calculate the Ksp Value

Substitute the equilibrium concentrations into the \(K_{sp}\) expression: \[ K_{sp} = (1.03 \times 10^{-3}) \times (2 \times 1.03 \times 10^{-3})^2 \]. Calculate each term: \( (1.03 \times 10^{-3}) = 1.03 \times 10^{-3} \) and \( (2 \times 1.03 \times 10^{-3})^2 = 4.24 \times 10^{-6} \). Hence, \[ K_{sp} = (1.03 \times 10^{-3}) \times (4.24 \times 10^{-6}) = 4.37 \times 10^{-9} \]

Key Concepts

Solubility Product Constant

The solubility product constant, abbreviated as \( K_{sp} \), is a crucial concept in chemistry, particularly in understanding precipitation reactions and solubility. It represents the equilibrium constant for the dissolution of a sparingly soluble ionic compound. When a substance such as strontium fluoride (\( \text{SrF}_2 \)) dissolves in water, it separates into its constituent ions.At equilibrium, the rate at which the solid dissolves equals the rate at which the ions in solution combine to form the solid again. The solubility product expression for a compound like \( \text{SrF}_2 \), which dissociates into \( \text{Sr}^{2+} \) and \( 2 \text{F}^{-} \), is given by the formula:\[ K_{sp} = [\text{Sr}^{2+}][\text{F}^{-}]^2 \]In this equation:

• \([\text{Sr}^{2+}]\) represents the equilibrium concentration of strontium ions.
• \([\text{F}^{-}]\) is the equilibrium concentration of fluoride ions, squared because two fluoride ions are produced per formula unit.

The value of \( K_{sp} \) provides insight into the solubility of the compound. A smaller \( K_{sp} \) indicates lower solubility, while a larger value suggests higher solubility.

Equilibrium Concentration

Equilibrium concentration refers to the concentration of each ion in a saturated solution at equilibrium, where no further dissolution or precipitation occurs. During the dissolution of \( \text{SrF}_2 \) in water, the compound dissociates as:\[ \text{SrF}_2 (s) \rightleftarrows \text{Sr}^{2+} (aq) + 2\text{F}^{-} (aq) \]The equilibrium concentration helps us calculate \( K_{sp} \) by determining how much of each ion is present in the solution at equilibrium. For \( \text{SrF}_2 \):- The concentration of \( \text{Sr}^{2+} \) is given as \( 1.03 \times 10^{-3} \) M.- Since two \( \text{F}^{-} \) ions are produced per \( \text{Sr}^{2+} \), their concentration is \( 2 \times 1.03 \times 10^{-3} \) M.These concentrations are used in the \( K_{sp} \) equation to determine the solubility product, showing the relationship between the concentration of dissolved ions and the solid in equilibrium.

Dissolution Reaction

A dissolution reaction describes the process through which an ionic compound dissolves in a solvent, typically water, breaking into its individual ions. For \( \text{SrF}_2 \), the dissolution reaction is:\[ \text{SrF}_2 (s) \rightleftarrows \text{Sr}^{2+} (aq) + 2\text{F}^{-} (aq) \]In simpler terms, this reaction illustrates how a solid ionic compound transitions into dissolved ions:

• Strontium ions \((\text{Sr}^{2+})\)
• Fluoride ions \((\text{F}^{-})\), with two fluoride ions generated per formula unit of \( \text{SrF}_2 \)

The dissolution process can reach a state of dynamic equilibrium, where the rate of dissolution equals the rate of precipitation. In practice, only a small quantity of \( \text{SrF}_2 \) dissolves due to its low solubility, dictated by its \( K_{sp} \). Understanding the dissolution reaction is essential for predicting how substances will behave when mixed in solutions. This knowledge is not only foundational in academic settings but also has practical implications in fields like pharmaceuticals and environmental science.

Chemical Equilibrium

Chemical equilibrium is a state where the rate of the forward reaction matches the rate of the reverse reaction, resulting in no net change in the concentration of reactants and products. In the context of \( \text{SrF}_2 \) dissolving in water:\[ \text{SrF}_2 (s) \rightleftarrows \text{Sr}^{2+} (aq) + 2\text{F}^{-} (aq) \]At equilibrium, the amounts of dissolved \( \text{Sr}^{2+} \) and \( \text{F}^{-} \) remain constant, even though individual molecules continue to dissolve and precipitate. This establishes a dynamic balance.**Key Points:**

• The concentrations of \( \text{Sr}^{2+} \) and \( \text{F}^{-} \) ions at equilibrium are determined by the system's \( K_{sp} \).
• Equilibrium doesn’t mean equal concentrations, but rather that the processes occurring in the solution and solid are balanced.

Understanding chemical equilibrium helps in predicting the solubility of compounds and in performing calculations related to concentration changes and precipitation.