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In each of the following pairs, indicate which has the higher concentration of \(\mathrm{Cl}^{-}\) ion: (a) \(0.10 \mathrm{M} \mathrm{CaCl}_{2}\) or \(0.15 \mathrm{M} \mathrm{KCl}\) solution, (b) \(100 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{KCl}\) solution or \(400 \mathrm{~mL}\) of \(0.080 \mathrm{M} \mathrm{LiCl}\) solution, (c) \(0.050 \mathrm{M} \mathrm{HCl}\) solution or \(0.020 \mathrm{M} \mathrm{CdCl}_{2}\) solution.

Short Answer

Expert verified
Higher concentration of Cl鈦 ions for each pair: (a) \(0.10 \mathrm{M} \mathrm{CaCl}_{2}\), (b) \(400\,\mathrm{mL}\) of \(0.080\, \mathrm{M} \mathrm{LiCl}\) solution, and (c) \(0.050\, \mathrm{M} \mathrm{HCl}\) solution.

Step by step solution

01

Pair (a) - Molarities

(a) \(0.10 \mathrm{M} \mathrm{CaCl}_{2}\) or \(0.15 \mathrm{M} \mathrm{KCl}\) For \(\mathrm{CaCl}_{2}\), there are 2 Cl鈦 ions in each molecule. For \(\mathrm{KCl}\), there is 1 Cl鈦 ion in each molecule.
02

Pair (a) - Calculating concentration of Cl鈦 ions

For \(\mathrm{CaCl}_{2}\): \(0.10 \mathrm{M}\times2 = 0.20 \mathrm{M}\) Cl鈦 ion. For \(\mathrm{KCl}\): \(0.15 \mathrm{M}\times1= 0.15 \mathrm{M}\) Cl鈦 ion. Higher concentration of Cl鈦 ions: \(0.10 \mathrm{M} \mathrm{CaCl}_{2}\)
03

Pair (b) - Molarities

(b) \(100 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{KCl}\) solution or \(400 \mathrm{~mL}\) of \(0.080 \mathrm{M} \mathrm{LiCl}\) solution. For \(\mathrm{KCl}\), there is 1 Cl鈦 ion in each molecule. For \(\mathrm{LiCl}\), there is 1 Cl鈦 ion in each molecule.
04

Pair (b) - Calculating concentration of Cl鈦 ions in the given volumes

For \(100 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{KCl}\): \(0.10 \mathrm{M}\times1\times100 \mathrm{~mL}= 10\,\mathrm{mmol}\) Cl鈦 ion. For \(400 \mathrm{~mL}\) of \(0.080 \mathrm{M} \mathrm{LiCl}\): \(0.080 \mathrm{M}\times1\times400\mathrm{~mL} = 32\,\mathrm{mmol}\) Cl鈦 ion. Higher concentration of Cl鈦 ions: \(400\,\mathrm{mL}\) of \(0.080\, \mathrm{M} \mathrm{LiCl}\) solution.
05

Pair (c) - Molarities

(c) \(0.050\,\mathrm{M} \mathrm{HCl}\) solution or \(0.020\,\mathrm{M} \mathrm{CdCl}_{2}\) solution. For \(\mathrm{HCl}\), there is 1 Cl鈦 ion in each molecule. For \(\mathrm{CdCl}_{2}\), there are 2 Cl鈦 ions in each molecule.
06

Pair (c) - Calculating concentration of Cl鈦 ions

For \(\mathrm{HCl}\): \(0.050\,\mathrm{M}\times1 = 0.050\,\mathrm{M}\) Cl鈦 ion. For \(\mathrm{CdCl}_{2}\): \(0.020\,\mathrm{M}\times2 = 0.040\,\mathrm{M}\) Cl鈦 ion. Higher concentration of Cl鈦 ions: \(0.050\, \mathrm{M} \mathrm{HCl}\) solution.

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Key Concepts

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

Molarity
Molarity is an essential concept in chemistry, particularly when dealing with solutions. It is a way to express the concentration of a solution in terms of the amount of substance per volume of solution. Specifically, molarity (symbolized as M) is defined as the number of moles of solute present in one liter of solution. This can be written mathematically as:\[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}\]Understanding molarity is crucial for accurately preparing solutions and performing concentration calculations. It allows chemists to predict how solutions will react in various conditions. Moreover, molarity directly influences the reaction rate and equilibrium in chemical processes. For students, grasping this concept means being able to communicate about substances in a language that is standardized across the scientific community.
For example, in the exercise above, when comparing solutions like \(0.10\ \mathrm{M}\ \mathrm{CaCl}_{2}\) and \(0.15\ \mathrm{M}\ \mathrm{KCl}\), knowing the molarity helps determine the amount of chloride ions contributed by each compound, as each formula unit provides different ions.
Chemical solutions
Chemical solutions consist of solutes dissolved in solvents, creating homogeneous mixtures at the molecular level. In these solutions, the solute is the substance that gets dissolved, while the solvent is the material that dissolves the solute.
Chemical solutions are foundational to various processes in chemistry, ranging from simple dissolution to complex reaction mixtures. The composition of a solution, notably including its molarity, directly affects its properties, such as boiling and freezing points, vapor pressure, and osmotic pressure.
  • Solute: The substance dissolved, like \(\mathrm{CaCl}_{2}\), \(\mathrm{KCl}\), or \(\mathrm{HCl}\) in the examples.
  • Solvent: Typically, the substance in greater amount, often water in laboratory solutions.
    • In a laboratory setting, chemical solutions can be manipulated by adjusting concentrations to study effects on chemical reactions or to prepare reagents precisely for experimental needs. For instance, by understanding the role of each component, we can predict the behavior of solutions like those in the exercise, determining which has higher solute concentrations and thus more intense chemical activities.
Concentration calculations
Concentration calculations are a vital aspect of chemistry that involve determining how much solute is present in a given volume of solution. These calculations allow chemists to modify and understand the properties of solutions, making them integral to experimental designs and industrial applications.
The formulas used in concentration calculations depend on the type of concentration measurement needed. For molarity, as highlighted earlier, we calculate based on moles per liter. But in particular scenarios, such as the exercise, we also consider the stoichiometry of the dissolved solutes:
  • For ionic compounds like \(\mathrm{CaCl}_{2}\), the formula unit's stoichiometric coefficient tells us that it releases two \(\mathrm{Cl}^{-}\) ions per formula unit.
  • Conversely, \(\mathrm{KCl}\) releases only one \(\mathrm{Cl}^{-}\) ion per formula unit.
By multiplying the molarity by the number of ions produced in solution (as with \(\mathrm{CdCl}_{2}\) releasing two \(\mathrm{Cl}^{-}\) ions), we accurately determine the concentration of specific ions. Practicing such calculations enhances one's ability to makes predictions and adjustments in both experimental and theoretical contexts, making it a cornerstone of chemical analysis and synthesis.

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Most popular questions from this chapter

Determine the oxidation number for the indicated element in each of the following compounds: (a) Ti in \(\mathrm{TiO}_{2}\) (b) Sn in \(\mathrm{SnCl}_{3}^{-}\), (c) \(\mathrm{C}\) in \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\), (d) \(\mathrm{N}\) in \(\mathrm{N}_{2} \mathrm{H}_{4}\), (e) \(\mathrm{N}\) in \(\mathrm{HNO}_{2}\), (f) \(\mathrm{Cr}\) in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\)

The arsenic in a 1.22-g sample of a pesticide was converted to \(\mathrm{AsO}_{4}{ }^{3-}\) by suitable chemical treatment. It was then titrated using \(\mathrm{Ag}^{+}\) to form \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) as a precipitate. (a) What is the oxidation state of \(\mathrm{As}\) in \(\mathrm{AsO}_{4}{ }^{3-}\) ? (b) Name \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) by analogy to the corresponding compound containing phosphorus in place of arsenic. (c) If it took \(25.0 \mathrm{~mL}\) of \(0.102 \mathrm{M} \mathrm{Ag}^{+}\) to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

\(\mathrm{HCl}, \mathrm{HBr}\), and \(\mathrm{HI}\) are strong acids, yet \(\mathrm{HF}\) is a weak acid. What does this mean in terms of the extent to which these substances are ionized in solution?

Separate samples of a solution of an unknown salt are treated with dilute solutions of \(\mathrm{HBr}, \mathrm{H}_{2} \mathrm{SO}_{4}\), and \(\mathrm{NaOH}\). A precipitate forms in all three cases. Which of the following cations could the solution contain: \(\mathrm{K}^{+} ; \mathrm{Pb}^{2+} ; \mathrm{Ba}^{2+}\) ?

(a) Suppose you prepare \(500 \mathrm{~mL}\) of a \(0.10 \mathrm{M}\) solution of some salt and then spill some of it. What happens to the concentration of the solution left in the container? (b) Suppose you prepare \(500 \mathrm{~mL}\) of a \(0.10 \mathrm{M}\) aqueous solution of some salt and let it sit out, uncovered, for a long time, and some water evaporates. What happens to the concentration of the solution left in the container? (c) \(\mathrm{A}\) certain volume of a \(0.50 \mathrm{M}\) solution contains \(4.5 \mathrm{~g}\) of a salt. What mass of the salt is present in the same volume of a \(2.50 \mathrm{M}\) solution?

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