Question

The exposed electrodes of a light bulb are placed in a solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in an electrical circuit such that the light bulb is glowing. You add a dilute salt solution, and the bulb dims. Which of the following could be the salt in the solution? a. \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) c. \(\mathrm{K}_{2} \mathrm{SO}_{4}\) b. \(\mathrm{NaNO}_{3}\) d. \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) Justify your choices. For those you did not choose, explain why they are incorrect.

Short answer

The correct salt is a. \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\), as adding this salt to the sulfuric acid solution will decrease the conductivity due to the formation of insoluble \(\mathrm{BaSO}_{4}\). The bulb will dim when this salt is added to the solution. The other salts do not have the same effect on the solution, as their ions do not lead to a decrease in conductivity or are not as strong in their reaction.

Step-by-step solution

Identify the ions present in each salt solution

We need to identify the ions present in each salt to determine the effect on the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution. a. \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) - This salt dissociates into \(\mathrm{Ba}^{2+}\) and \(\mathrm{NO}_{3}^{-}\) ions. b. \(\mathrm{NaNO}_{3}\) - This salt dissociates into \(\mathrm{Na}^{+}\) and \(\mathrm{NO}_{3}^{-}\) ions. c. \(\mathrm{K}_{2} \mathrm{SO}_{4}\) - This salt dissociates into \(\mathrm{K}^{+}\) and \(\mathrm{SO}_{4}^{2-}\) ions. d. \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) - This salt dissociates into \(\mathrm{Ca}^{2+}\) and \(\mathrm{NO}_{3}^{-}\) ions.

Analyze the effect of each salt on the sulfuric acid solution

Now, let's analyze the effect of these ions on the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution when added: a. \(\mathrm{Ba}^{2+}\) and \(\mathrm{NO}_{3}^{-}\): In this case, the \(\mathrm{Ba}^{2+}\) ions will react with \(\mathrm{SO}_{4}^{2-}\) ions present in the sulfuric acid solution to form \(\mathrm{BaSO}_{4}\), which is insoluble. This reaction will remove sulfate ions from the solution, reducing its conductivity and dimming the bulb. b. \(\mathrm{Na}^{+}\) and \(\mathrm{NO}_{3}^{-}\): Both of these ions are highly soluble and do not react with the ions in the sulfuric acid solution. Thus, adding this salt will not affect the conductivity of the solution, and the bulb will not dim. c. \(\mathrm{K}^{+}\) and \(\mathrm{SO}_{4}^{2-}\): Both these ions are highly soluble and will not react with the ions present in the sulfuric acid solution. Thus, adding this salt will not affect the conductivity of the solution, and the bulb will not dim. d. \(\mathrm{Ca}^{2+}\) and \(\mathrm{NO}_{3}^{-}\): In this case, the \(\mathrm{Ca}^{2+}\) ions will react with \(\mathrm{SO}_{4}^{2-}\) ions in the sulfuric acid solution to form \(\mathrm{CaSO}_{4}\), which is insoluble. This reaction will remove sulfate ions from the solution, reducing the conductivity and dimming the bulb.

Identify the correct salt and justify the choice

The correct salt is a. \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\), as adding this salt to the sulfuric acid solution will decrease the conductivity due to the formation of insoluble \(\mathrm{BaSO}_{4}\). The bulb will dim when this salt is added to the solution. The other salts (b. \(\mathrm{NaNO}_{3}\), c. \(\mathrm{K}_{2}\mathrm{SO}_{4}\), and d. \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\)) do not have the same effect on the solution. In the case of salts b and c, both of their ions are highly soluble and adding them to the solution won't change the conductivity, so the bulb won't dim. For salt d, even though adding \(\mathrm{Ca}^{2+}\) ions can form insoluble \(\mathrm{CaSO}_{4}\) and reduce the conductivity, the reaction might not be as strong as in the case of salt a. Thus, salt a is the best option to cause the bulb to dim as needed.

Key concepts

Solubility and Precipitation

Solubility and precipitation are key chemical concepts when dealing with solutions. Solubility refers to the ability of a substance to dissolve in a solvent, such as water. It is dependent on various factors, including temperature, pressure, and the chemical nature of the solute and the solvent. Precipitation, on the other hand, occurs when a solute's concentration exceeds its solubility in the solvent, resulting in the formation of a solid, which is called a precipitate. This process is governed by the solubility product constant (Ksp), a unique value for each solute at a given temperature.

In the context of our exercise, when either barium nitrate (Ba(NO3)2) or calcium nitrate (Ca(NO3)2) is added to a sulfuric acid (H2SO4) solution, a reaction takes place that leads to the formation of insoluble barium sulfate (BaSO4) or calcium sulfate (CaSO4), respectively. These compounds precipitate out of the solution, highlighting the concept of solubility and precipitation in action.

Ionic Dissociation

Ionic dissociation is the process where an ionic compound separates into its individual ions when dissolved in a solvent, such as water. This is a crucial process because the free ions in a solution are what make it conduct electricity. The degree of dissociation can vary; strong electrolytes dissociate completely while weak electrolytes only partially dissociate.

In the given exercise, salts such as Ba(NO3)2, NaNO3, K2SO4, and Ca(NO3)2 were considered. These are all examples of strong electrolytes. These salts dissociate entirely into their constituent ions when they dissolve in water. The dissociated ions can then interact with other ions in the solution, sometimes leading to precipitation reactions, as with Ba2+ and SO42- ions forming BaSO4.

Sulfuric Acid

Sulfuric acid (H2SO4) is a strong acid commonly used in various industrial processes. It is also a strong electrolyte, which means it dissociates completely in water to form H+ and SO42- ions. These ions play a significant role in the acid's chemical behavior and its ability to conduct electricity. H2SO4 is also known for its ability to form sulfate salts when reacting with metal cations, which is essential in understanding solubility and precipitation phenomena.

In our exercise, the addition of certain salts to the sulfuric acid solution potentially changes the concentration of the SO42- ions due to the formation of insoluble sulfate salts. The removal of these ions may lead to a measurable change in the solution's conductivity, which is demonstrated by the dimming of the light bulb.

Electrical Conductivity

Electrical conductivity in solutions is an essential concept in chemistry, related to a solution's ability to conduct electric current. This property depends on the presence of ions in the solution. More ions equal more charge carriers, which leads to increased conductivity. When ions are removed from the solution, as with the formation of a precipitate, conductivity decreases.

In our exercise, the presence of two electrolytes – a sulfuric acid solution and a salt – create a conductive environment wherein the light bulb glows due to the free movement of ions. Upon adding a salt that forms a precipitate, some of the free ions are sequestered into the solid state, thus reducing the number of charge carriers and causing the glow of the bulb to dim. This principle is applied to identify the salts that cause a decrease in electrical conductivity through precipitation.