Problem 10 - Chemistry An Atoms First Approach

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. \(\operatorname{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) Justify your choices. For those you did not choose, explain why they are incorrect.

Step-by-Step Solution

Answer

The correct salt that could cause the light bulb to dim is Ba(NO3)2. The Ba²⁺ ions can form a precipitate when reacting with the sulfate ions in the H₂SO₄ solution, reducing the overall number of free ions and consequently reducing the conductivity of the solution.

Step 1: List the dissociated ions of each salt

First, we need to list the ions that each given salt dissociates into when added to the solution: a. Ba(NO3)2 -> Ba²⁺ + 2NO₃⁻ b. NaNO3 -> Na⁺ + NO₃⁻ c. K2SO4 -> 2K⁺ + SO₄²⁻ d. Ca(NO3)2 -> Ca²⁺ + 2NO₃⁻

Step 2: Analyze the interactions between ions and H2SO4 solution

Now, let's analyze how the ions from each salt will interact with the H2SO4 solution: a. The Ba²⁺ ion could form a precipitate with the sulfate ions already present in the H2SO4 solution: Ba²⁺ + SO₄²⁻ -> BaSO4(s). The formation of a precipitate would decrease the overall number of free ions in the solution, reducing its conductivity. b. The addition of Na⁺ and NO₃⁻ would simply increase the number of free ions in the solution, potentially increasing its conductivity. Therefore, NaNO3 would not cause the light bulb to dim. c. Similar to NaNO3, K2SO4 would add more K⁺ and SO₄²⁻ ions, possibly increasing the conductivity of the solution. It would not cause the light bulb to dim. d. The addition of Ca²⁺ and NO₃⁻ from Ca(NO3)2 would increase the number of free ions in the solution, potentially increasing its conductivity. It would not cause the light bulb to dim.

Step 3: Determine the correct salt

Based on the analysis, the correct salt that could cause the light bulb to dim is: a. Ba(NO3)2 The Ba²⁺ ions can form a precipitate when reacting with the sulfate ions in the H₂SO₄ solution, reducing the overall number of free ions and consequently reducing the conductivity of the solution.

Key Concepts

Ionic Interactions

Ionic interactions are the forces that hold ions together in a compound. When ionic compounds dissolve in water, they split into their component ions, which can interact in the solution. This is particularly handy in chemistry to predict the behavior of solutions.
In the original exercise, we are dealing with compounds like \(\text{Ba(NO}_{3}\text{)}_{2}\), which dissociates into Ba²⁺ and NO₃⁻ ions.
These ions, when present in a solution, can interact with other ions.

• Oppositely charged ions attract each other, potentially forming a precipitate.
• Like-charged ions repel each other, helping maintain the solution balance.

In the exercise, the interaction between Ba²⁺ and SO₄²⁻ ions results in the formation of a precipitate (BaSO₄). These ionic interactions help us understand why the number of free ions decreases, reducing the solution's overall conductivity.

Electrolyte Conductivity

Electrolyte conductivity refers to a solution's ability to conduct electricity. This is mostly dependent on the concentration of ions present in the solution. The more free moving ions there are, the better the solution can carry an electrical current.
In the context of our exercise, we are working with H₂SO₄, a solution with high ionic content because of the dissociation of H⁺ and SO₄²⁻ ions. The presence or absence of specific ions can dramatically alter the solution's conductivity:

• Adding Ba(NO₃)₂ reduces conductivity by forming a precipitate, decreasing free ions.
• On the contrary, adding compounds like NaNO₃ or K₂SO₄ enhances conductivity as they increase the ion concentration.

This is why the light bulb dims upon adding Ba(NO₃)₂, as there are fewer ions available to conduct electricity.

Precipitate Formation

Precipitate formation occurs when two ions in solution form an insoluble compound, which then settles out of the solution. This process effectively removes ions from the solution, altering its properties, such as conductivity.
In our original problem, Ba²⁺ ions from Ba(NO₃)₂ react with SO₄²⁻ ions from H₂SO₄ to form BaSO₄, a solid precipitate: \(\text{Ba}^{2+} + \text{SO}_{4}^{2-} \rightarrow \text{BaSO}_{4(s)}\).
This reaction causes the overall ionic concentration to drop, as ions become part of the solid precipitate and can no longer move freely in the solution. This makes the solution less conductive, resulting in the dimming of the light bulb. Precipitation is a common method in chemistry for driving reactions that selectively remove specific ions from a solution.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They show the reactants, products, and their respective quantities needed to uphold the law of conservation of mass.
In our exercise's context, chemical equations illustrate ionic dissociations and interactions that lead to precipitate formation. Using Ba(NO₃)₂ as an example: \[ \text{Ba(NO}_{3}\text{)}_{2(s)} \rightarrow \text{Ba}^{2+} (aq) + 2\text{NO}_{3}^{-} (aq) \].
This equation shows how Ba(NO₃)₂ dissociates in water. Additionally, when writing chemical equations for precipitation reactions, it's crucial to denote the state of matter (e.g., (s) for solids and (aq) for aqueous) to understand solution dynamics and solubility rules better. Such notations help interpret the outcome in an experiment or exercise, like the dimming effect in the given light bulb scenario.