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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.

Short Answer

Expert verified
The correct salt that could cause the light bulb to dim is Ba(NO3)2. The Ba虏鈦 ions can form a precipitate when reacting with the sulfate ions in the H鈧係O鈧 solution, reducing the overall number of free ions and consequently reducing the conductivity of the solution.

Step by step solution

01

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 -> Ba虏鈦 + 2NO鈧冣伝 b. NaNO3 -> Na鈦 + NO鈧冣伝 c. K2SO4 -> 2K鈦 + SO鈧劼测伝 d. Ca(NO3)2 -> Ca虏鈦 + 2NO鈧冣伝
02

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 Ba虏鈦 ion could form a precipitate with the sulfate ions already present in the H2SO4 solution: Ba虏鈦 + SO鈧劼测伝 -> 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鈦 and NO鈧冣伝 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鈦 and SO鈧劼测伝 ions, possibly increasing the conductivity of the solution. It would not cause the light bulb to dim. d. The addition of Ca虏鈦 and NO鈧冣伝 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.
03

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 Ba虏鈦 ions can form a precipitate when reacting with the sulfate ions in the H鈧係O鈧 solution, reducing the overall number of free ions and consequently reducing the conductivity of the solution.

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Key Concepts

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

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 Ba虏鈦 and NO鈧冣伝 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 Ba虏鈦 and SO鈧劼测伝 ions results in the formation of a precipitate (BaSO鈧). 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 H鈧係O鈧, a solution with high ionic content because of the dissociation of H鈦 and SO鈧劼测伝 ions. The presence or absence of specific ions can dramatically alter the solution's conductivity:
  • Adding Ba(NO鈧)鈧 reduces conductivity by forming a precipitate, decreasing free ions.
  • On the contrary, adding compounds like NaNO鈧 or K鈧係O鈧 enhances conductivity as they increase the ion concentration.
This is why the light bulb dims upon adding Ba(NO鈧)鈧, 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, Ba虏鈦 ions from Ba(NO鈧)鈧 react with SO鈧劼测伝 ions from H鈧係O鈧 to form BaSO鈧, 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(NO鈧)鈧 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(NO鈧)鈧 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.

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Most popular questions from this chapter

Chlorisondamine chloride \(\left(\mathrm{C}_{14} \mathrm{H}_{20} \mathrm{Cl}_{6} \mathrm{N}_{2}\right)\) is a drug used in the treatment of hypertension. A \(1.28-\mathrm{g}\) sample of a medication containing the drug was treated to destroy the organic material and to release all the chlorine as chloride ion. When the filtered solution containing chloride ion was treated with an excess of silver nitrate, 0.104 g silver chloride was recovered. Calculate the mass percent of chlorisondamine chloride in the medication, assuming the drug is the only source of chloride.

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