Question

A 25.0 -mL sample of 0.050\(M\) barium nitrate solution was mixed with 25.0 mL of 0.050\(M\) sodium sulfate solution labeled with radioactive sulfur-35. The activity of the initial sodium sulfate solution was \(1.22 \times 10^{6} \mathrm{Bq} / \mathrm{mL}\) . After the resultant precipitate was removed by filtration, the remaining filtrate was found to have an activity of 250 \(\mathrm{Bq} / \mathrm{mL}\) . (a) Write a balanced chemical equation for the reaction that occurred. (b) Calculate the \(K_{s p}\) for the precipitate under the conditions of the experiment.

Short answer

The balanced chemical equation for the reaction between barium nitrate and sodium sulfate is: \[Ba^{2+} (aq) + SO_4^{2-} (aq) \rightarrow BaSO_4 (s)\]. The solubility product constant, \(K_{sp}\), for the BaSO\(_4\) precipitate under the conditions of the experiment is \(9.83 \times 10^{-6}\).

Step-by-step solution

Write the balanced chemical equation

The reaction between barium nitrate and sodium sulfate can be represented as follows: \[Ba(NO_3)_2 (aq) + Na_2SO_4 (aq) \rightarrow 2NaNO_3 (aq) + BaSO_4 (s)\] The balanced chemical equation for this reaction is: \[Ba^{2+} (aq) + SO_4^{2-} (aq) \rightarrow BaSO_4 (s)\]

Calculate moles of solute

Based on the initial volume and concentration of solutions, we can calculate the moles of barium nitrate and sodium sulfate. Moles of Ba(NO\(_3\))\(_2\) = volume × concentration Moles of Ba(NO\(_3\))\(_2\) = 25.0 mL × 0.050 M Moles of Ba(NO\(_3\))\(_2\) = 0.00125 mol Moles of Na\(_2\)SO\(_4\) = volume × concentration Moles of Na\(_2\)SO\(_4\) = 25.0 mL × 0.050 M Moles of Na\(_2\)SO\(_4\) = 0.00125 mol

Determine concentration of ions in the filtrate

After the reaction, the remaining filtrate only contains radioactive sulfate ions (labeled with sulfur-35). Activity in the filtrate = 250 Bq/mL Initial activity of Na\(_2\)SO\(_4\) = \(1.22 \times 10^{6}\) Bq/mL Sulfate ions in the filtrate = \(\frac{250 \text{Bq/mL}}{1.22 \times 10^{6} \text{Bq/mL}} \times 0.50 \,\mathrm{M}\) = \(2.05 \times 10^{-4}\) M.

Calculate concentration of barium ions in the precipitate

Since the balanced chemical equation has a 1:1 ratio of barium and sulfate ions, the concentration of barium ions in the precipitate can be determined by subtracting the concentration of sulfate ions in the filtrate from the initial barium nitrate solution. Concentration of Ba\(^{2+}\) in the precipitate = Initial Concentration - Concentration in the filtrate Concentration of Ba\(^{2+}\) in the precipitate = 0.050 M - \(2.05\times 10^{-4}\) M Concentration of Ba\(^{2+}\) in the precipitate = \(4.795\times 10^{-2}\) M

Calculate K\(_{sp}\) of the precipitate

Now that we have the concentration of both ions in the precipitate, we can calculate the solubility product constant K\(_{sp}\) for BaSO\(_4\) by using the expression K\(_{sp} = [Ba^{2+}] \times [SO_4^{2-}]\). K\(_{sp} = (4.795\times 10^{-2}\,\mathrm{M}) \times (2.05\times 10^{-4}\,\mathrm{M})\) K\(_{sp} = 9.83 \times 10^{-6}\) So, the solubility product constant K\(_{sp}\) for the BaSO\(_4\) precipitate under the conditions of the experiment is \(9.83 \times 10^{-6}\).

Key concepts

Chemical Reactions

In chemical reactions, substances undergo transformation resulting in the formation of new products. In the exercise above, we observe a classic example of a precipitation reaction where two soluble salts interact to form an insoluble product, a precipitate. Specifically, barium nitrate and sodium sulfate are mixed. This leads to the formation of barium sulfate ( BaSO_4 ), which precipitates out of the solution.

Understanding chemical reactions requires writing balanced chemical equations. These equations use chemical symbols to represent what happens during the reaction, including the quantities of relevant compounds. They help in maintaining the law of conservation of mass, which states that matter cannot be created or destroyed during a reaction. Thus, knowing how to write and balance these equations is crucial in understanding the nature of chemical interactions and precipitate formation.

Radioactive Isotopes

Radioactive isotopes are variants of elements that possess unstable nuclei, leading to radiation emission as they decay to stabilize. In the exercise, sulfur-35, a radioactive isotope, is used as a label for the sulfate ions. This labeling helps track the sulfate ions throughout the reaction process because these isotopes can be detected due to their radioactive nature.

The decay can be measured using a unit called Becquerel (Bq), where one Bq equals one decay per second. Radioactive isotopes are integral in experiments like this as they provide a unique method to monitor and measure reaction components at specific stages. By observing the change in active levels before and after precipitation, we gain insights into its concentration and behavior in chemical processes.

Concentration Calculations

Concentration calculations are fundamental in chemistry as they provide information on the amount of solute present in a given volume of solvent. In the context of the exercise, concentration calculations are employed to determine the moles of reactants and the concentration of ions in the filtrate.

Initially, the concentrations of barium nitrate and sodium sulfate were both 0.050 M, and their interaction allowed us to compute the moles of each substance. The shift in activity levels indicated by the radioactive sulfur hinted at the final concentrations. Concentration calculations further involve changes in the resultant mixture, leading to the computation of the precipitate's solubility product constant ( K_{sp} ). Accurately measuring these concentrations is essential for predicting the extent of reactions and the quantities of products formed.

Precipitation Reactions

Precipitation reactions occur when two solutions containing soluble salts are combined, resulting in an insoluble product precipitating from the solution. These reactions are crucial in various scientific applications, including separating and identifying substances.

In the exercise, the reaction between barium nitrate and sodium sulfate leads to the formation of barium sulfate, a classic precipitate. This solid can be removed from the solution via filtration, allowing for further analysis of the remaining substances in the solution.

The formation of a precipitate is particularly useful for determining reaction components' solubility and concentration, as demonstrated in the calculation of the solubility product constant ( K_{sp} ). The K_{sp} value helps predict the extent to which a compound will dissolve in solution, providing valuable insights into the chemical behavior of the components involved.