Question

When \(250 \mathrm{mg}\) of \(\mathrm{SrF}_{2},\) strontium fluoride, is added to \(1.00 \mathrm{L}\) of water, the salt dissolves to a very small extent. $$\mathrm{SrF}_{2}(\mathrm{s}) \rightleftarrows \mathrm{Sr}^{2+}(\mathrm{aq})+2 \mathrm{F}^{-}(\mathrm{aq})$$ At equilibrium, the concentration of \(\mathrm{Sr}^{2+}\) is found to be \(1.0 \times 10^{-3} \mathrm{M} .\) What is the value of \(K_{\mathrm{sp}}\) for \(\mathrm{SrF}_{2} ?\)

Short answer

The value of \(K_{\mathrm{sp}}\) for \(\mathrm{SrF}_2\) is \(4.0 \times 10^{-9}\).

Step-by-step solution

Understanding the Dissolution Equation

The dissolution of strontium fluoride (\(\mathrm{SrF}_2\)) in water can be represented by the equation: \[\mathrm{SrF}_{2}(\mathrm{s}) \rightleftarrows \mathrm{Sr}^{2+}(\mathrm{aq}) + 2 \mathrm{F}^{-}(\mathrm{aq})\]This represents the dissociation of solid \(\mathrm{SrF}_2\) into one \(\mathrm{Sr}^{2+}\) ion and two \(\mathrm{F}^{-}\) ions in solution. The equilibrium constant for this process is known as the solubility product, \(K_{\mathrm{sp}}\).

Writing the Expression for Ksp

The solubility product, \(K_{\mathrm{sp}}\), is the product of the concentrations of the ions at equilibrium, each raised to the power of their stoichiometric coefficients. For the dissolution of \(\mathrm{SrF}_2\), the \(K_{\mathrm{sp}}\) expression is given by:\[K_{\mathrm{sp}} = [\mathrm{Sr}^{2+}][\mathrm{F}^-]^2\]

Finding the Equilibrium Concentrations

We are given that the equilibrium concentration of \(\mathrm{Sr}^{2+}\) is \(1.0 \times 10^{-3}\,\mathrm{M}\). Since the stoichiometry of the reaction shows that for each \(1\,\mathrm{mol}\) of \(\mathrm{SrF}_2\) that dissolves, \(2\,\mathrm{mol}\) of \(\mathrm{F}^-\) ions are formed, the equilibrium concentration of \(\mathrm{F}^-\) is:\[[\mathrm{F}^-] = 2 \times [\mathrm{Sr}^{2+}] = 2 \times 1.0 \times 10^{-3}\,\mathrm{M} = 2.0 \times 10^{-3}\,\mathrm{M}\]

Calculating Ksp

Substitute the equilibrium concentrations into the \(K_{\mathrm{sp}}\) expression:\[K_{\mathrm{sp}} = [\mathrm{Sr}^{2+}][\mathrm{F}^-]^2 = (1.0 \times 10^{-3})(2.0 \times 10^{-3})^2\]Calculate the value:\[K_{\mathrm{sp}} = (1.0 \times 10^{-3}) \times (4.0 \times 10^{-6}) = 4.0 \times 10^{-9}\]

Key concepts

Chemical Equilibrium

In chemical equilibrium, a system reaches a state where the concentrations of reactants and products remain constant over time. This doesn't necessarily mean that the concentrations are equal, but that their rates of formation and decomposition are balanced.

• In the case of strontium fluoride (\(\mathrm{SrF}_2\)) dissolving in water, it forms a dynamic equilibrium, denoted by the double-headed arrow \(\rightleftarrows\).
• Even as \(\mathrm{SrF}_2\) continues to dissolve, the \(\mathrm{Sr}^{2+}\) and \(\mathrm{F}^-\) ions can also recombine to form the solid, so the processes are ongoing and balanced.

The state of equilibrium for a dissolving salt like \(\mathrm{SrF}_2\) is typically described using its solubility product, \(K_{sp}\), which provides a quantitative measure of the extent to which the salt can dissolve.

Ionic Dissolution

Ionic dissolution is the process where an ionic solid, like strontium fluoride \(\mathrm{SrF}_2\), separates into its constituent ions when dissolved in water.

• This process involves breaking the ionic bonds in the solid, allowing the strontium \(\mathrm{Sr}^{2+}\) and fluoride \(\mathrm{F}^-\) ions to be released into the aqueous solution.
• Once dissolved, these ions become surrounded by water molecules, making them solvated, which helps maintain their presence in the solution.

The dissolution process reaches equilibrium when the rate of the ions entering the solution equals the rate of ions recombining to form the solid, and it is this equilibrium that is quantified by the solubility product \(K_{sp}\).

Stoichiometry

Stoichiometry is the area of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction.

• For the dissolution of \(\mathrm{SrF}_2\), the stoichiometric relationship is crucial to determine the concentrations of ions at equilibrium.
• As per the dissolution reaction: \(\mathrm{SrF}_{2} \,() \rightleftarrows \, \mathrm{Sr}^{2+} \,() + 2 \mathrm{F}^{-} \, ()\), we see that for every mole of \(\mathrm{SrF}_2\) that dissolves, one mole of \(\mathrm{Sr}^{2+}\) and two moles of \(\mathrm{F}^-\) are produced.

This relationship allows us to calculate that if the concentration of \(\mathrm{Sr}^{2+}\) is \(1.0 \times 10^{-3}\,\mathrm{M}\), the concentration of \(\mathrm{F}^-\) would be twice that, \(2.0 \times 10^{-3}\,\mathrm{M}\).

Chemical Reactions

Chemical reactions, like the dissolution of \(\mathrm{SrF}_2\), involve the rearrangement of atoms to form new substances. In simple terms, a chemical reaction requires breaking some bonds and forming new ones, leading to changes in the identity of the substances involved.

• The chemical equation \(\mathrm{SrF}_{2}(\mathrm{s}) \rightleftarrows \mathrm{Sr}^{2+}(\mathrm{aq}) + 2 \mathrm{F}^{-}(\mathrm{aq})\) shows the transformation of solid strontium fluoride into its ionic components in a water medium.
• During this dissolution, the solid does not react chemically but undergoes a phase change and forms ions.

Understanding chemical reactions like this focus on not just the changes in composition, but also examining how energy and entropy influence the reaction and the equilibrium balance it achieves.