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Acetone, \(\mathrm{CH}_{3} \mathrm{COCH}_{3},\) is a nonelectrolyte; hypochlorous acid, \(\mathrm{HClO},\) is a weak electrolyte; and ammonium chloride, \(\mathrm{NH}_{4} \mathrm{Cl},\) is a strong electrolyte. (a) What are the solute particles present in aqueous solutions of each compound? (b) If \(0.1 \mathrm{~mol}\) of each compound is dissolved in solution, which one contains \(0.2 \mathrm{~mol}\) of solute particles, which contains \(0.1 \mathrm{~mol}\) of solute particles, and which contains somewhere between 0.1 and 0.2 mol of solute particles?

Short Answer

Expert verified
In aqueous solutions, acetone (\(CH_3COCH_3\)) has 0.1 mol of solute particles, ammonium chloride (\(NH_4Cl\)) has 0.2 mol of solute particles, and hypochlorous acid (\(HClO\)) has between 0.1 and 0.2 mol of solute particles when 0.1 mol of each compound is dissolved.

Step by step solution

01

Determine solute particles in aqueous solutions

For each compound, we need to decide how it dissociates when it dissolves in water: Acetone (CH3COCH3) is a nonelectrolyte, which means it does not break apart into ions in an aqueous solution. Therefore, the solute particles are simply molecules of CH3COCH3. Hypochlorous acid (HClO) is a weak electrolyte, which means it only partially dissociates into ions. The equilibrium for its ionization is: \[HClO \rightleftharpoons H^+ + ClO^-\] Ammonium chloride (NH4Cl) is a strong electrolyte, which means it completely dissociates into ions in aqueous solution: \[NH_4Cl \rightarrow NH_4^+ + Cl^-\]
02

Determine the number of moles of solute particles for each compound

For each compound, we will calculate the number of moles of solute particles when 0.1 mol of the compound is dissolved. Acetone (CH3COCH3): Since acetone doesn't dissociate in water and remains as intact molecules, there is 0.1 mol of solute particles. Hypochlorous acid (HClO): Since HClO partially dissociates and we don't have any information about its dissociation constant, it's safe to assume that the number of solute particles is between 0.1 and 0.2 mol. Ammonium chloride (NH4Cl): Since NH4Cl completely dissociates into its ions, for each molecule of NH4Cl we get one NH4+ ion and one Cl- ion. So, for 0.1 mol of ammonium chloride, there are 0.2 mol of solute particles.
03

Identify which compound has a certain number of moles of solute particles

Now we can identify the solutions that contain: 0.1 mol of solute particles: Acetone (CH3COCH3). 0.2 mol of solute particles: Ammonium chloride (NH4Cl). Somewhere between 0.1 and 0.2 mol of solute particles: Hypochlorous acid (HClO).

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Key Concepts

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

Nonelectrolytes
Nonelectrolytes like acetone (\(\text{CH}_3\text{COCH}_3\)) do not conduct electricity when dissolved in water. This is because they do not dissociate into ions.
In solution, nonelectrolytes remain as whole molecules.
Therefore, when acetone is dissolved in water, it simply spreads out as acetone molecules without breaking apart.
This characteristic leads to a predictable number of solute particles—equal to the initial quantity of the compound dissolved.
For example, adding 0.1 mol of acetone to water results in 0.1 mol of solute particles.
  • The molecules remain intact.
  • They do not ionize.
  • The solution lacks conductivity because there are no free ions present.
Understanding this behavior helps clarify why nonelectrolytes do not contribute to the ionic balance and conductivity of a solution.
Weak Electrolytes
Weak electrolytes, such as hypochlorous acid (\(\text{HClO}\)), partially dissociate into ions in water.
This partial dissociation leads to a mixed population of both ions and intact molecules in the solution.
The equilibrium process can be represented as:\[\text{HClO} \rightleftharpoons \text{H}^+ + \text{ClO}^-\]Since only some of the molecules dissociate, the number of solute particles is higher than the initial amount, but not as high as it would be if the substance fully dissociated.
  • The equilibrium state depends on the dissociation constant of the substance.
  • This leads to an intermediate level of conductivity.
  • The number of ionized particles is more than 0.1 mol, but less than 0.2 mol for 0.1 mol of \(\text{HClO}\).
This behavior is crucial for understanding the nature of weak acids and bases and their effect on the pH of a solution.
Strong Electrolytes
Strong electrolytes, like ammonium chloride (\(\text{NH}_4\text{Cl}\)), fully dissociate into ions when dissolved in water.
This complete breakdown results in a significantly higher number of solute particles compared to weak electrolytes and nonelectrolytes.
The dissociation process is straightforward:\[\text{NH}_4\text{Cl} \rightarrow \text{NH}_4^+ + \text{Cl}^-\]As every molecule of ammonium chloride separates into two ions, a 0.1 mol solution would give rise to 0.2 mol of total ions.
  • The ions carry electrical charge, enhancing the solution's conductivity.
  • There is no equilibrium between ions and molecules, unlike weak electrolytes.
  • The behavior of strong electrolytes is crucial for predicting ionic interactions and reactions in solution.
This complete dissociation plays a vital role in many chemical reactions, especially in aqueous environments.
Solute Dissociation
Solute dissociation is a crucial concept that determines how substances behave in solution.
It refers to the process where solute particles separate into ions when dissolved.
The degree of dissociation varies across different solutes.
  • Nonelectrolytes remain intact as molecules throughout the solution, resulting in no ion formation.
  • Weak electrolytes partially dissociate, creating an equilibrium with a modest number of ions and remaining molecules.
  • Strong electrolytes dissociate completely, inundating the solution with ions that contribute to conductivity.
Understanding solute dissociation helps in predicting the electrical conductivity and chemical reactivity of solutions
and is essential for a diverse array of applications, from biochemical reactions to industrial processes.

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Most popular questions from this chapter

A \(35.0-\mathrm{mL}\) sample of \(1.00 \mathrm{M} \mathrm{KBr}\) and a \(60.0-\mathrm{mL}\) sample of \(0.600 \mathrm{M} \mathrm{KBr}\) are mixed. The solution is then heated to evaporate water until the total volume is \(50.0 \mathrm{~mL}\). What is the molarity of the KBr in the final solution?

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