Question

A 25.0-mLsample of \(0.050 \mathrm{M}\) barium nitrate solution was mixed with \(25.0 \mathrm{~mL}\) of \(0.050 \mathrm{M}\) 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

In summary, the balanced chemical equation for the reaction between barium nitrate and sodium sulfate is: \( \mathrm{Ba(NO_{3})_{2}(aq) + Na_{2}SO_{4}(aq) \rightarrow BaSO_{4}(s) + 2NaNO_{3}(aq)} \) The solubility product (Ksp) for the precipitate (barium sulfate) under the conditions of the experiment is \( 1.28 \times 10^{-7} \).

Step-by-step solution

a) Write a balanced chemical equation for the reaction that occurred.

The reaction that occurred is a double replacement reaction between barium nitrate and sodium sulfate. The balanced chemical equation for the reaction is: \( \mathrm{Ba(NO_{3})_{2}(aq) + Na_{2}SO_{4}(aq) \rightarrow BaSO_{4}(s) + 2NaNO_{3}(aq)} \)

b1) Find the initial moles of sulfate ions

Since we know the volume and the molarity of the solutions, we can find the initial moles of sulfate ions from sodium sulfate solution. Moles of sulfate ion = Molarity of solution * Volume of solution Moles of sulfate ion = \( 0.050 \mathrm{M} * 0.025 \mathrm{L} = 1.25 \times 10^{-3} \mathrm{mol} \)

b2) Determine the moles of sulfate ions in the filtrate

The radioactive sulfur-35 activity in the filtrate was found to be 250 Bq/mL. The ratio of the initial radioactive sulfur-35 activity drops from \( 1.22\times10^6 \mathrm{Bq/mL}\) to 250 Bq/mL. Therefore, the ratio of the sulfate ions can be determined as: \( \frac{\mathrm{Activity^{before}}}{\mathrm{Activity^{after}}} = \frac{moles \: \mathrm{initial}}{moles \: \mathrm{filtrate}} \) Moles of sulfate ions in the filtrate can be calculated as: Moles of sulfate ions in filtrate \( = \frac{activity_{filtrate}}{activity_{initial}}*moles_{initial} \) Moles of sulfate ions in filtrate \( = \frac{250 \mathrm{Bq/mL}}{1.22 \times 10^{6} \mathrm{Bq/mL}} * 1.25 \times 10^{-3} \mathrm{mol} = 2.56 \times 10^{-7} \mathrm{mol}\)

b3) Calculate the moles of barium ions in the filtrate

The precipitate was formed by capturing all the leftover sulfate ions. So, to calculate the remaining barium ions in filtrate, we can subtract moles of sulfate ions in the filtrate from the initial moles of barium ions. Moles of barium ions in filtrate = Initial moles of barium ions - moles of sulfate ions in filtrate Moles of initial barium ions = Molarity of solution * Volume of solution Moles of initial barium ions = \( 0.050 \mathrm{M} * 0.025 \mathrm{L} = 1.25 \times 10^{-3} \mathrm{mol} \) Moles of barium ions in filtrate = \( 1.25 \times 10^{-3} \mathrm{mol} - 2.56 \times 10^{-7} \mathrm{mol} = 1.25 \times 10^{-3} \mathrm{mol} \)

b4) Calculate the concentrations of the ions in the filtrate

To calculate the concentrations of barium and sulfate ions in the filtrate, divide the moles of ions by the total volume of the solution. The total volume of the solution = Volume of barium nitrate solution + Volume of sodium sulfate solution The total volume of the solution = \( 0.025 \mathrm{L} + 0.025 \mathrm{L} = 0.050 \mathrm{L} \) Concentration of barium ions = Moles of barium ions in filtrate / Total volume of the solution Concentration of barium ions = \( \frac{1.25 \times 10^{-3} \mathrm{mol}}{0.050 \mathrm{L}} = 0.025 \mathrm{M} \) Concentration of sulfate ions = Moles of sulfate ions in filtrate / Total volume of the solution Concentration of sulfate ions = \( \frac{2.56 \times 10^{-7} \mathrm{mol}}{0.050 \mathrm{L}} = 5.12 \times 10^{-6} \mathrm{M} \)

b5) Calculate the Ksp of the precipitate

Now, we have the concentrations of the ions in the filtrate. We can use these concentrations to find the solubility product (Ksp) of the precipitate. The balanced equation for the dissolution of barium sulfate is: \( \mathrm{BaSO_{4}(s) \rightleftharpoons Ba^{2+}(aq) + SO_{4}^{2-}(aq)} \) Ksp = [Ba²⁺] * [SO₄²⁻] Ksp = (0.025 M) * (5.12 * 10⁻⁶ M) = \( 1.28 \times 10^{-7} \)

Key concepts

Solubility Product Constant (Ksp)

The concept of the solubility product constant, or Ksp, is essential in the understanding of solubility equilibria in chemistry. It represents the level at which a compound dissolves in solution. In other words, Ksp is the equilibrium constant for a solid substance dissolving in an aqueous solution. The formula for Ksp is derived from the concentrations of the ions that are in equilibrium with the undissolved solid. For instance, the dissolution of barium sulfate can be represented as:

BaSO4(s) ⇌ Ba²⁺ (aq) + SO₄²⁻ (aq)

A high Ksp value indicates a high solubility of the substance in water, while a low Ksp value denotes poor solubility. To calculate the Ksp, you take the product of the molar concentrations of the dissolved ions, each raised to the power of its coefficient in the balanced equation. For students, calculating Ksp is a matter of first understanding the stoichiometry of the dissolution process, and then measuring or calculating the concentration of ions at equilibrium.

Double Replacement Reaction

A double replacement reaction is a type of chemical reaction where two compounds exchange ions to form two new compounds. This reaction is characterized by the general formula:

AB + CD → AD + CB

In the context of the exercise provided, barium nitrate and sodium sulfate undergo a double replacement reaction. The barium ions (Ba²⁺) replace sodium ions (Na⁺), and vice versa, resulting in the formation of barium sulfate (BaSO4) and sodium nitrate (NaNO3). It's important for students to identify the reactants and predict the products using their knowledge of the reactivity and solubility of the compounds involved. When one of the products is insoluble, as in the case of barium sulfate in this scenario, it precipitates out of the solution, driving the reaction forward.

Radioactive Tracer

Radioactive tracers are substances containing a radioactive atom to allow for the tracing of the substance through a system, such as a chemical reaction or biological pathway. They are widely used in medicine, chemistry, and environmental science. For the given exercise, sulfur-35, a radioactive isotope of sulfur, was used to label sodium sulfate. The radioactive tracer allows the researchers to track and quantify the movement of sulfate ions in the reaction. By measuring the radioactivity before and after the reaction, one can deduce how much of the tracer—and therefore how many sulfate ions—remain in solution. This consequently enables the calculation of the solubility product constant (Ksp) for the resulting precipitate.

Molarity

Molarity, often represented by 'M,' is the measure of concentration of a solution, defined as the number of moles of solute divided by the volume of the solution in liters. It is a very fundamental concept in chemistry, especially when preparing solutions or analyzing reactions that occur in aqueous media. In the exercise, one begins by using the provided molarity of the initial solutions to calculate the amount of each ion before the reaction. Conceptually, understanding molarity helps students to relate moles of substances to the physical volume of a solution, thus bridging the gap between the microscopic scale of atoms and molecules and the macroscopic world we can measure and observe.