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 concentrations in the final solution of \(\mathrm{NO}_{3}^{-}, \mathrm{Na}^{+}, \mathrm{Sr}^{2+}\), and \(\mathrm{F}^{-}\). \(\left(K_{\mathrm{sp}}\right.\) for \(\left.\mathrm{SrF}_{2}=2.0 \times 10^{-10} .\right)\)

Short answer

NO3- = 0.075 M, Na+ = 0.045 M, Sr2+ ≈ 0.00031 M, F- ≈ 0.00062 M.

Step-by-step solution

Calculate Total Volume

When mixing solutions, we must first determine the final total volume of the mixture. We do this by adding the individual volumes of the solutions being mixed. \[ V_{\text{total}} = V_{\text{NaF}} + V_{\text{Sr(NO}_3\text{)}_2} = 75 \text{ mL} + 25 \text{ mL} = 100 \text{ mL} \]

Determine Initial Moles of Each Ion

Calculate the initial moles of each ion before any reaction takes place. For \(\text{NaF}\):\[ \text{Moles of } \text{Na}^+ = \text{Moles of } \text{F}^- = 75 \times 10^{-3} \text{ L} \times 0.060 \text{ M} = 4.5 \times 10^{-3} \text{ mol} \]For \(\text{Sr(NO}_3\text{)}_2\):\[\text{Moles of } \text{Sr}^{2+} = 25 \times 10^{-3} \text{ L} \times 0.15 \text{ M} = 3.75 \times 10^{-3} \text{ mol} \]\[\text{Moles of } \text{NO}_3^- = 2 \times 3.75 \times 10^{-3} \text{ mol} = 7.5 \times 10^{-3} \text{ mol} \]

Calculate Final Concentrations

Now we will calculate the final concentrations for each ion by dividing the initial moles by the total volume.\(\text{Na}^+\):\[\text{Concentration of } \text{Na}^+ = \frac {4.5 \times 10^{-3} \text{ mol}}{0.100 \text{ L}} = 0.045 \text{ M} \]\(\text{F}^-\):\[\text{Concentration of } \text{F}^- = \frac {4.5 \times 10^{-3} \text{ mol}}{0.100 \text{ L}} = 0.045 \text{ M} \]\(\text{Sr}^{2+}\):\[\text{Concentration of } \text{Sr}^{2+} = \frac {3.75 \times 10^{-3} \text{ mol}}{0.100 \text{ L}} = 0.0375 \text{ M} \]\(\text{NO}_3^-\):\[\text{Concentration of } \text{NO}_3^- = \frac {7.5 \times 10^{-3} \text{ mol}}{0.100 \text{ L}} = 0.075 \text{ M} \]

Check Precipitation Condition with Ksp

We need to check if any reaction occurs, specifically if precipitation of \(\text{SrF}_2\) occurs. This is done by comparing the ionic product to the solubility product \(K_{sp}\).Calculate ionic product for \(\text{SrF}_2\):\[Q = [\text{Sr}^{2+}][\text{F}^-]^2 = (0.0375)(0.045)^2 = 7.59 \times 10^{-5} \]Since \(Q = 7.59 \times 10^{-5} > K_{sp} = 2.0 \times 10^{-10}\), SrF2 will precipitate, and we must reassess concentrations, considering the SrF2 precipitation.

Recalculate Concentrations after Precipitation

Due to the precipitation of \(\text{SrF}_2\), some ions will be removed from solution. The equilibrium involves:\[\text{SrF}_2 (s) \rightleftharpoons \text{Sr}^{2+}(aq) + 2\text{F}^-(aq) \]Setting equilibrium concentrations after precipitation considering Ksp:\[ K_{sp} = [\text{Sr}^{2+}][\text{F}^-]^2 = 2.0 \times 10^{-10} \]If \(x\) is the concentration of \(\text{Sr}^{2+}\) that remains, \(2x\) will be used for \(\text{F}^-\):\[x(2x)^2 = 2.0 \times 10^{-10} \Rightarrow x = [\text{Sr}^{2+}] = 3.1 \times 10^{-4} \text{ M}\]\[[\text{F}^-] = 2 \times 3.1 \times 10^{-4} \text{ M} = 6.2 \times 10^{-4} \text{ M}\]Adjust concentrations of non-precipitating ions, \([\text{Na}^+] = 0.045 \text{ M}\) and \([\text{NO}_3^-] = 0.075 \text{ M}\) remain unchanged.

Key concepts

Solubility Product

In chemical equilibrium, the solubility product (K_{sp}) is a crucial concept that helps us understand whether a precipitate will form when two solutions are mixed. When a salt is dissolved in water, it dissociates into its ions. The solubility product is the equilibrium constant for the dissolution of a sparingly soluble ionic compound. For example, in the case of SrF_2 dissolving in water, the equilibrium can be represented as:\[\text{SrF}_2(s) \rightleftharpoons \text{Sr}^{2+}(aq) + 2\text{F}^-(aq)\]The expression for the solubility product is given by:\[K_{sp} = [\text{Sr}^{2+}][\text{F}^-]^2\]This formula helps us determine at what concentration of ions, a compound will start to precipitate. When the product of the ion concentrations, known as the ionic product, exceeds K_{sp}, the solution is supersaturated, leading to precipitation. In our exercise, the solubility product of SrF_2 is given as 2.0 \times 10^{-10}, which signifies its low solubility in water. Understanding K_{sp} helps predict the formation of solid precipitates under certain conditions.

Ionic Concentration

Ionic concentration refers to the amount of ions in a solution, typically measured in moles per liter (Molarity, M). When two ionic solutions are combined, it's essential to calculate the concentration of each ion in the final mixture to determine reactions such as precipitation.

In our exercise, we combined NaF and Sr(NO_3)_2 solutions. To find the final ionic concentrations, we first calculated the number of moles of each ion present in the respective volumes and initial molarities of the solutions. Calculating the total volume, we then determined the concentrations:

• \([\text{Na}^+] = 0.045 \text{ M}\)
• \([\text{F}^-] = 0.045 \text{ M}\)
• \([\text{Sr}^{2+}] = 0.0375 \text{ M}\)
• \([\text{NO}_3^-] = 0.075 \text{ M}\)

These concentrations tell us how densely packed the ions are within the solution, dictating how likely a solid will form when other solutions are involved in the reaction. Ionic concentration is crucial when assessing chemical reactivity and equilibrium in solution.

Precipitation Reaction

Precipitation reactions occur when ions in a solution combine to form an insoluble product, the precipitate. This process can be predicted by comparing the ionic product (Q) with the solubility product (K_{sp}). If the ionic product Q exceeds K_{sp}, the solution is supersaturated, leading to precipitation.

In the exercise scenario, the reaction was:\[\text{Sr}^{2+}(aq) + \text{2F}^-(aq) \rightarrow \text{SrF}_2(s)\]The ionic product was calculated as:\[Q = [\text{Sr}^{2+}][\text{F}^-]^2 = 7.59 \times 10^{-5}\]Comparing Q to K_{sp} (2.0 \times 10^{-10}), Q was larger, indicating that the conditions were right for SrF_2 to precipitate from the solution.

This principle can be applied to understand various chemical systems, predicting when a solid will form from a mixture of solutions. This process is a key element in fields like chemistry, biology, and environmental science, where precipitations are common.