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Chapter 4: Problem 70

In each of the following pairs, indicate which has the higher concentration of \(\mathrm{Cl}^{-}\) ion: \((\mathbf{a}) 0.10 \mathrm{MAlCl}_{3}\) solution or a \(0.25 \mathrm{MLiCl}\) solution, (b) \(150 \mathrm{~mL}\) of a \(0.05 \mathrm{M} \mathrm{MnCl}_{3}\) solution or \(200 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{KCl}\) solution, (c) a \(2.8 M \mathrm{HCl}\) solution or a solution made by dissolving \(23.5 \mathrm{~g}\) of KCl in water to make \(100 \mathrm{~mL}\) of solution.

Short Answer

Expert verified
(a) AlCl3, (b) MnCl3, (c) KCl solution.

Step by step solution

01

Determine Chloride Ions in AlCl3 Solution

For the solution \( ext{0.10 M } ext{AlCl}_3 \), one formula unit of \( ext{AlCl}_3 \) produces three \( ext{Cl}^- \) ions. Thus, the concentration of \( ext{Cl}^- \) ions is \( 0.10 \times 3 = 0.30 ext{ M Cl}^- \).
02

Determine Chloride Ions in LiCl Solution

For the \( 0.25 ext{ M LiCl} \) solution, one formula unit of \( ext{LiCl} \) produces one \( ext{Cl}^- \) ion. Therefore, the concentration of \( ext{Cl}^- \) ions is \( 0.25 ext{ M Cl}^- \).
03

Analyze Part A Solutions

Comparing the two, \( 0.30 ext{ M Cl}^- \) (from \( ext{AlCl}_3 \)) is greater than \( 0.25 ext{ M Cl}^- \) (from \( ext{LiCl} \)). Therefore, \( ext{AlCl}_3 \) has a higher concentration of \( ext{Cl}^- \) ions.
04

Determine Chloride Ions in MnCl3 Solution

For the \( 0.05 ext{ M MnCl}_3 \) solution, each \( ext{MnCl}_3 \) produces three \( ext{Cl}^- \) ions. The effective concentration of \( ext{Cl}^- \) ions is \( 0.05 \times 3 = 0.15 ext{ M Cl}^- \).
05

Determine Total Cl- Moles in MnCl3 Solution

The volume of the \( ext{MnCl}_3 \) solution is \( 150 ext{ mL} = 0.150 ext{ L} \), so the moles of \( ext{Cl}^- \) is \( 0.15 ext{ mol/L} \times 0.150 ext{ L} = 0.0225 ext{ mol Cl}^- \).
06

Determine Chloride Ions in KCl Solution

For the \( 0.10 ext{ M KCl} \) solution, \( ext{KCl} \) yields one \( ext{Cl}^- \) ion per formula unit. Hence, its concentration of \( ext{Cl}^- \) ions is \( 0.10 ext{ M Cl}^- \).
07

Determine Total Cl- Moles in KCl Solution

The volume is \( 200 ext{ mL} = 0.200 ext{ L} \), giving \( 0.10 ext{ mol/L} \times 0.200 ext{ L} = 0.020 ext{ mol Cl}^- \).
08

Analyze Part B Solutions

\( 0.0225 ext{ mol Cl}^- \) from \( ext{MnCl}_3 \) is greater than \( 0.020 ext{ mol Cl}^- \) from \( ext{KCl} \). Therefore, the \( ext{MnCl}_3 \) solution has a higher total amount of \( ext{Cl}^- \) ions.
09

Determine Chloride Ions in HCl Solution

For the \( 2.8 ext{ M HCl} \) solution, \( ext{HCl} \) produces one \( ext{Cl}^- \) ion per formula unit. So, the concentration of \( ext{Cl}^- \) ions is \( 2.8 ext{ M Cl}^- \).
10

Calculate KCl Concentration for Given Mass

First, calculate moles of KCl: Molar mass \( = 39.1 + 35.4 = 74.5 \text{ g/mol} \). Moles \( = \frac{23.5 \text{ g}}{74.5 \text{ g/mol}} = 0.315 \text{ mol} \). Concentration is \( \frac{0.315 \text{ mol}}{0.1 \text{ L}} = 3.15 \text{ M} \).
11

Analyze Part C Solutions

The KCl solution has \( 3.15 ext{ M Cl}^- \) compared to \( 2.8 ext{ M Cl}^- \) from HCl, so it has a higher \( ext{Cl}^- \) concentration.

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Key Concepts

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

Molarity
Molarity is a key concept in chemistry that refers to the concentration of a specific component in a solution. It is defined as the number of moles of solute present per liter of solution. Molarity, denoted by the symbol "M," is a convenient way to express concentration because it directly correlates to the amount of substance that reacts or behaves in solution.
  • For example, a 1 M solution of sodium chloride (NaCl) means there is 1 mole of NaCl dissolved in 1 liter of solution.
  • To calculate molarity, use the formula: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]
Understanding molarity allows students to make quantitative predictions about chemical reactions and to compare the concentrations of various solutions easily.
It is a crucial skill for solving problems, such as comparing concentrations of chloride ions in different chemical solutions.
Chemical Solutions
Chemical solutions are homogeneous mixtures composed of two or more substances. Typically, a solution consists of a solute, which is the substance being dissolved, and a solvent, the substance doing the dissolving. Water is the most common solvent.
  • Solutions are characterized by their uniform composition and properties throughout.
  • The concentration of a solution can be measured using different units, including molarity, which indicates the strength of the solution.
Understanding the nature of chemical solutions is vital when discussing concentrations because it implies how the solute particles are distributed in the solvent.
In chemical solutions, determining the solution's molarity assists researchers and students in knowing how a solute behaves under various forces like temperature and pressure.
Ionic Compounds
Ionic compounds are chemical compounds composed of ions held together by electrostatic forces termed ionic bonding. These compounds usually form when metals react with non-metals. Upon dissolving in water, ionic compounds such as salt (sodium chloride) dissociate into their respective ions.
  • An ionic compound will form a solution containing free-moving ions, making it an electrolyte and highly conductive of electricity.
  • The characteristic of dissociation is critical for understanding solutions with ionic compounds, as it determines the concentration of ions like \( \text{Cl}^- \).
  • For instance, when aluminum chloride (AlCl3) dissolves in water, it produces three chloride ions per formula unit. Consequently, even a low molarity of such a compound can result in a relatively high concentration of chloride ions, compared to a simpler ionic compound like LiCl, which releases only one chloride ion per molecule.
Grasping how ionic compounds dissociate in solution aids in understanding differences in ion concentrations across various chemical solutions.
This knowledge is particularly useful when comparing solutions based on their chemical formulas and intended applications.

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Most popular questions from this chapter

Citric acid, \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7},\) is a triprotic acid. It occurs naturally in citrus fruits like lemons and has applications in food flavouring and preservatives. A solution containing an unknown concentration of the acid is titrated with KOH. It requires \(23.20 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{KOH}\) solution to titrate all three acidic protons in \(100.00 \mathrm{~mL}\) of the citric acid solution. Write a balanced net ionic equation for the neutralization reaction, and calculate the molarity of the citric acid solution.

The labels have fallen off three bottles containing powdered samples of metals; one contains zinc, one lead, and the other platinum. You have three solutions at your disposal: \(1 \mathrm{M}\) sodium nitrate, \(1 M\) nitric acid, and \(1 M\) nickel nitrate. How could you use these solutions to determine the identities of each metal powder?

Write balanced net ionic equations for the reactions that occur in each of the following cases. Identify the spectator ion or ions in each reaction. (a) \(\mathrm{Ba}(\mathrm{OH})_{2}(a q)+\mathrm{FeCl}_{3}(a q) \longrightarrow\) (b) \(\mathrm{ZnCl}_{2}(a q)+\mathrm{Cs}_{2} \mathrm{CO}_{3}(a q) \longrightarrow\) (c) \(\mathrm{Na}_{2} \mathrm{~S}(a q)+\operatorname{CoSO}_{4}(a q) \longrightarrow\)

The metal cadmium tends to form Cd \(^{2+}\) ions. The following observations are made: (i) When a strip of zinc metal is placed in \(\mathrm{CdCl}_{2}(a q),\) cadmium metal is deposited on the strip. (ii) When a strip of cadmium metal is placed in \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q),\) nickel metal is deposited on the strip. (a) Write net ionic equations to explain each of the preceding observations. (b) Which elements more closely define the position of cadmium in the activity series? (c) What experiments would you need to perform to locate more precisely the position of cadmium in the activity series?

Calculate the concentration of each ion in the following solutions obtained by mixing: (a) \(32.0 \mathrm{~mL}\) of \(0.30 \mathrm{M} \mathrm{KMnO}_{4}\) with \(15.0 \mathrm{~mL}\) of \(0.60 \mathrm{M} \mathrm{KMnO}_{4}\) (b) \(60.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{ZnCl}_{2}\) with \(5.0 \mathrm{~mL}\) of \(0.200 \mathrm{MZn}\left(\mathrm{NO}_{3}\right)_{2},(\mathbf{c}) 4.2 \mathrm{~g}\) of \(\mathrm{CaCl}_{2}\) in \(150.0 \mathrm{~mL}\) of \(0.02 M \mathrm{KCl}\) solution. Assume that the volumes are additive.
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