Question

A volume of \(75 \mathrm{~mL}\) of \(0.060 \mathrm{M} \mathrm{NaF}\) is mixed with \(25 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2} .\) Calculate the concentra- tions in the final solution of \(\mathrm{NO}_{3}^{-}, \mathrm{Na}^{+}, \mathrm{Sr}^{2+},\) and \(\mathrm{F}^{-} \cdot\left(K_{\mathrm{sp}}\right.\) for \(\left.\mathrm{SrF}_{2}=2.0 \times 10^{-10} \mathrm{.}\right)\)

Short answer

The final concentrations of the ions in the solution are: [NO3^-] = 0.075 M, [Na+] = 0.045 M, [Sr2+] = 2.5 * 10^-5 M, [F-] = 0.045 M.

Step-by-step solution

Calculation of initial concentrations

Begin by calculating the initial concentrations of the individual ions. Remember to calculate the final total volume first. The reaction form is: \nSr(NO3)2 \(\rightarrow\) Sr2+ + 2NO3-\nNaF \(\rightarrow\) Na+ + F-\nSo, the final total volume is 75 mL + 25 mL = 100 mL = 0.1 L. Then we calculate initial concentrations of the ions:\n\n[Sr2+] = (0.15 moles/liter)x(0.025 liters)/(0.1 liters) = 0.0375 M\n\n[NO3-] = 2x[Sr2+] = 2x0.0375 M = 0.075 M\n\n[Na+] = (0.060 moles/liter)x(0.075 liters)/(0.1 liters) = 0.045 M\n\n[F-] = [Na+] = 0.045 M

Determine if a precipitate forms

Next, we use the solubility product constant (Ksp) to check if any insoluble substance forms from the reaction. In this case, we're concerned about strontium fluoride (SrF2). The solubility product constant (Ksp) expression is defined as: \nKsp = [Sr2+][F-]^2\nAfter substitution, we have: \n2 * 10^-10 = (0.0375 M) * (0.045 M)^2 = 7.59 * 10^-5\nThis value is greater than the given Ksp, therefore, a precipitate of SrF2 will form.

Calculation of equilibrium concentrations

Since we know that a precipitate forms, the concentrations of Sr2+ and F- will change until reaching an equilibrium point with the formation of SrF2. Let x be the decrease in the concentrations of Sr2+ and F- due to the formation of SrF2. Then, the new concentrations will be:\n[Sr2+] = 0.0375 - x\n[F-] = 0.045 - 2x\nSubstituting these into the Ksp expression, we solve for x. We get x = 0.0375 M (as the decrease in the concentration of Sr2+ is almost complete) assuming that the fluoride ion is in excess.

Computing final concentrations

By knowing the decrease in concentrations of Sr2+ and F- (x), we can now calculate the final concentrations. Substitute the solved x value into the expressions of [Sr2+] and [F-] and find the final equilibrium concentration. Therefore:\n[NO3^-] = 0.075 M (since the nitrate ion is a spectator ion, its concentration does not change)\n[Na+] = 0.045 M (since the sodium ion is a spectator ion, its concentration does not change)\n[Sr2+] = 0.0375 M - x = 2.5 * 10^-5 M (this is the saturation concentration) \n[F-] = 0.045 M - 2x = 0.045 M (as the decreased concentration of fluoride ion can be neglected)

Key concepts

Solubility Product Constant (Ksp)

The solubility product constant, represented as \(K_{sp}\), is a crucial factor in understanding solubility equilibria. It indicates the extent to which a solid can dissolve in water to form a saturated solution. Essentially, \(K_{sp}\) is the product of the concentrations of the ions that are formed by the dissolution of a solid, each raised to the power of its stoichiometric coefficient from the balanced dissolution equation.

For example, for the dissolution of strontium fluoride \(\text{(SrF}_2)\), the equation is:

• \(\text{SrF}_2(s) \rightleftharpoons \text{Sr}^{2+}(aq) + 2\text{F}^-(aq)\)

This leads to the \(K_{sp}\) expression as:

• \(K_{sp} = [\text{Sr}^{2+}][\text{F}^-]^2\)

This value provides a benchmark to determine if a solution can absorb more of the solute or if it will cause a precipitate to form, thus reaching solubility equilibrium.

In our original problem, \(K_{sp}\) for \(\text{SrF}_2\) is given as \(2.0 \times 10^{-10}\). By calculating and substituting the ion concentrations into the \(K_{sp}\) expression, we can predict whether a precipitate forms.

Precipitation Reactions

Precipitation reactions occur when the resulting ionic product exceeds the solubility product constant \(K_{sp}\), leading to the formation of an insoluble salt or precipitate. This process is central in determining whether certain ions in a solution will remain dissolved or form a solid.

When the compound's ionic product, that is the concentration product of the ions in solution, surpasses the \(K_{sp}\), a precipitate will definitely form. For instance, in the reaction of \(\text{SrF}_2\), the \(K_{sp}\) correlation tells us:

• If \( [\text{Sr}^{2+}][\text{F}^-]^2 > K_{sp} \), \(\text{SrF}_2\) will precipitate.
• If \( [\text{Sr}^{2+}][\text{F}^-]^2 < K_{sp} \), the ions remain in solution without precipitating.

In the example exercise, after mixing \(\text{NaF}\) and \(\text{Sr(NO}_3\text{)}_2\), the solution's ionic product for \(\text{SrF}_2\) exceeds the given \(K_{sp}\) of \(2.0 \times 10^{-10}\), causing a precipitate to form.

Understanding these reactions helps predict compositions of solutions in various chemistry applications, from laboratory experiments to industrial processes.

Equilibrium Concentrations

Equilibrium concentrations describe the state where the rates of forward and reverse reactions are equal, leading to no net change in the concentrations of reactants and products. In solubility equilibria, this concept helps determine the concentration of ions in a saturated solution after some solute has precipitated.

To find these equilibrium concentrations:

• Assume that a variable, often called \(x\), represents the change in ion concentrations necessary to reach equilibrium after a precipitate forms.
• Set up an equation based on the \(K_{sp}\) expression to solve for \(x\).

In the case of \(\text{SrF}_2\) from the problem, solving the \(K_{sp}\) equation gives us the decrease in concentrations for \([\text{Sr}^{2+}]\) as \(0.0375 - x\) and \([\text{F}^-]^2\) as \(0.045 - 2x\).

After a precipitate forms, you'll find that the concentration of \(\text{Sr}^{2+}\) is significantly reduced, while the \(\text{F}^-\) concentration remains nearly unchanged, emphasizing the dynamic balance in these systems.