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Sulfites (compounds containing the \(\mathrm{SO}_{3}^{2-}\) ions) are used as preservatives in dried fruits and vegetables and in wine making. In an experiment to test for the presence of sulfite in fruit, a student first soaked several dried apricots in water overnight and then filtered the solution to remove all solid particles. She then treated the solution with hydrogen peroxide \(\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)\) to oxidize the sulfite ions to sulfate ions. Finally, the sulfate ions were precipitated by treating the solution with a few drops of a barium chloride \(\left(\mathrm{BaCl}_{2}\right)\) solution. Write a balanced equation for each of the preceding steps.

Short Answer

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Sulfite is oxidized to sulfate and then precipitated as barium sulfate.

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01

Identify the Reaction for Oxidation of Sulfite Ions

In the experiment, sulfite ions (\[\text{SO}_3^{2-}\]) are oxidized to sulfate ions (\[\text{SO}_4^{2-}\]) with hydrogen peroxide (\[\text{H}_2\text{O}_2\]). The balanced chemical equation for this oxidation reaction is:\[\text{SO}_3^{2-} + \text{H}_2\text{O}_2 \rightarrow \text{SO}_4^{2-} + \text{H}_2\text{O}\]
02

Precipitation of Sulfate Ions

After the oxidation, the sulfate ions (\[\text{SO}_4^{2-}\]) are precipitated using a barium chloride (\[\text{BaCl}_2\]) solution. The barium ions (\[\text{Ba}^{2+}\]) react with the sulfate ions to form barium sulfate (\[\text{BaSO}_4\]), which is a solid precipitate. The balanced chemical equation for the precipitation reaction is:\[\text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4(s)\]
03

Combine the Reactions

To summarize the entire process,1. Oxidation of sulfite to sulfate: \[\text{SO}_3^{2-} + \text{H}_2\text{O}_2 \rightarrow \text{SO}_4^{2-} + \text{H}_2\text{O}\]2. Precipitation of sulfate ions with barium: \[\text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4(s)\]These reactions illustrate the sequence that occurs during the test for sulfites in the apricots.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation Reaction
An oxidation reaction is a type of chemical reaction where a substance loses electrons, often associated with gaining oxygen or losing hydrogen. In the context of sulfite testing, the oxidation of sulfite ions (\[\text{SO}_3^{2-}\]) to sulfate ions (\[\text{SO}_4^{2-}\]) is crucial. This transformation is achieved using hydrogen peroxide (\[\text{H}_2\text{O}_2\]), which serves as an oxidizing agent.
  • The sulfite ion undergoes oxidation when it reacts with hydrogen peroxide.
  • During this reaction, an oxygen atom from the hydrogen peroxide is transferred to the sulfite ion, converting it into a sulfate ion.
This process can be represented by the balanced chemical equation:\[\text{SO}_3^{2-} + \text{H}_2\text{O}_2 \rightarrow \text{SO}_4^{2-} + \text{H}_2\text{O}\] This equation shows the systematic process of electron transfer and the transformation involved in going from sulfite to sulfate ions.
Sulfate Precipitation
Precipitation is the process of forming a solid from a solution. After the oxidation step in sulfite testing, the next step involves precipitating the formed sulfate ions (\[\text{SO}_4^{2-}\]). This is achieved by introducing barium chloride (\[\text{BaCl}_2\]) to the solution.
  • When barium ions (\[\text{Ba}^{2+}\]) from barium chloride react with sulfate ions, they form barium sulfate (\[\text{BaSO}_4\]).
  • Barium sulfate is insoluble in water, which means it does not dissolve and instead forms a solid precipitate.
This reaction is visually noticeable as a solid forms in the solution, demonstrating the presence of sulfate ions. The balanced chemical equation for this reaction is:\[\text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4(s)\] The formation of this precipitate is evidence that the oxidation reaction successfully occurred.
Balanced Chemical Equation
A balanced chemical equation accurately reflects the conservation of matter, showing that the number of atoms for each element is the same on both sides of the reaction. Maintaining this balance is important for correctly describing chemical reactions, as it adheres to the law of conservation of mass.
  • In the oxidation of sulfite ions to sulfate, balance is achieved by ensuring all atoms of sulfur, oxygen, and hydrogen are equal on both reactant and product sides.
  • Similarly, in the precipitation of sulfate ions by barium, the equation reflects balance in terms of cations and anions present.
In general, a balanced chemical equation provides a clear representation of the reactants and products, emphasizing the conversion and interactions occurring during the reaction. This not only facilitates understanding of the chemical process but also assists in predicting the outcomes of similar chemical reactions.

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Most popular questions from this chapter

A sample of \(0.6760 \mathrm{~g}\) of an unknown compound containing barium ions \(\left(\mathrm{Ba}^{2+}\right)\) is dissolved in water and treated with an excess of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\). If the mass of the \(\mathrm{BaSO}_{4}\) precipitate formed is \(0.4105 \mathrm{~g}\), what is the percent by mass of Ba in the original unknown compound?

Can the following decomposition reaction be characterized as an acid-base reaction? Explain. $$\mathrm{NH}_{4} \mathrm{Cl}(s) \longrightarrow \mathrm{NH}_{3}(g)+\mathrm{HCl}(g) $$

Calculate the molarity of each of the following solutions: (a) \(29.0 \mathrm{~g}\) of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) in \(545 \mathrm{~mL}\) of solution, (b) \(15.4 \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) in \(74.0 \mathrm{~mL}\) of solution, (c) \(9.00 \mathrm{~g}\) of sodium chloride \((\mathrm{NaCl})\) in \(86.4 \mathrm{~mL}\) of solution.

Calculate the mass of the precipitate formed when \(2.27 \mathrm{~L}\) of \(0.0820 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) is mixed with \(3.06 \mathrm{~L}\) of \(0.0664 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\)

The recommended procedure for preparing a very dilute solution is not to weigh out a very small mass or measure a very small volume of a stock solution. Instead, it is done by a series of dilutions. A sample of \(0.8214 \mathrm{~g}\) of \(\mathrm{KMnO}_{4}\) was dissolved in water and made up to the volume in a \(500-\mathrm{mL}\) volumetric flask. A \(2.000-\mathrm{mL}\) sample of this solution was transferred to a \(1000-\mathrm{mL}\) volumetric flask and diluted to the mark with water. Next, \(10.00 \mathrm{~mL}\) of the diluted solution was transferred to a \(250-\mathrm{mL}\) flask and diluted to the mark with water. (a) Calculate the concentration (in molarity) of the final solution. (b) Calculate the mass of \(\mathrm{KMnO}_{4}\) needed to directly prepare the final solution.

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