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Chapter 4: Problem 134

Three \(1.0-\mathrm{g}\) samples of \(\mathrm{PbCl}_{2}, \mathrm{KCl}\), and \(\mathrm{CaCl}_{2}\) are placed in separate \(500-\mathrm{mL}\) beakers. In each case, enough \(25^{\circ} \mathrm{C}\) water is added to bring the total volume of the mixture to \(250 \mathrm{~mL}\). Each of the mixtures is then stirred for five minutes. Which of the mixtures will have the highest concentration of chloride (Cl \(^{-}\) ) ion?

Short Answer

Expert verified
The \( \mathrm{CaCl}_{2} \) solution will have the highest chloride concentration.

Step by step solution

01

Calculate Moles of Each Compound

To find the moles, we need the molar mass of each compound. For \( \mathrm{PbCl}_{2} \), the molar mass is approximately 278.1 g/mol. For \( \mathrm{KCl} \), it's about 74.55 g/mol, and for \( \mathrm{CaCl}_{2} \), it's about 110.98 g/mol. Calculate moles for each: 1. \( \mathrm{PbCl}_{2}: \frac{1.0}{278.1} \approx 0.0036 \text{ moles} \)2. \( \mathrm{KCl}: \frac{1.0}{74.55} \approx 0.0134 \text{ moles} \)3. \( \mathrm{CaCl}_{2}: \frac{1.0}{110.98} \approx 0.0090 \text{ moles} \)
02

Determine Chloride Ion Moles Produced

Each compound dissociates differently in water, producing chloride ions:1. \( \mathrm{PbCl}_{2} \to \mathrm{Pb}^{2+} + 2\mathrm{Cl}^- \) thereby producing \(2 \times 0.0036 = 0.0072\) moles of \( \mathrm{Cl}^- \).2. \( \mathrm{KCl} \to \mathrm{K}^+ + \mathrm{Cl}^- \) thereby producing \(0.0134\) moles of \( \mathrm{Cl}^- \).3. \( \mathrm{CaCl}_{2} \to \mathrm{Ca}^{2+} + 2\mathrm{Cl}^- \) thereby producing \(2 \times 0.0090 = 0.0180\) moles of \( \mathrm{Cl}^- \).
03

Calculate Chloride Ion Concentration

Use the volume of the solution to find the concentration:1. For \( \mathrm{PbCl}_{2} \), \[ \frac{0.0072}{0.250} = 0.0288 \text{ mol/L} \]2. For \( \mathrm{KCl} \), \[ \frac{0.0134}{0.250} = 0.0536 \text{ mol/L} \]3. For \( \mathrm{CaCl}_{2} \), \[ \frac{0.0180}{0.250} = 0.0720 \text{ mol/L} \].
04

Compare the Concentrations

Compare the concentrations of chloride ions:1. \( \mathrm{PbCl}_{2}: 0.0288 \text{ mol/L} \)2. \( \mathrm{KCl}: 0.0536 \text{ mol/L} \)3. \( \mathrm{CaCl}_{2}: 0.0720 \text{ mol/L} \)The \( \mathrm{CaCl}_{2} \) solution has the highest chloride ion concentration.

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Key Concepts

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

Molar Mass Calculation
Understanding how to calculate molar mass is a fundamental chemistry skill. Molar mass is the mass of one mole of a substance and is generally expressed in grams per mole (g/mol).
To calculate the molar mass of a compound, you sum the atomic masses of all the atoms present in the molecular formula.
  • For instance, in \( \mathrm{PbCl}_{2} \), the molar mass consists of the atomic mass of one lead (Pb) atom and two chloride (Cl) atoms.
    If the atomic masses are approximately 207.2 for Pb and 35.45 for each Cl, then the calculation is: 207.2 + (2 \times 35.45) = 278.1 \text{ g/mol}.
  • Similarly, the molar mass of \( \mathrm{KCl} \) is calculated by adding the atomic masses of potassium (K) and chlorine (Cl): 39.10 + 35.45 = 74.55 \text{ g/mol}.
  • For \( \mathrm{CaCl}_{2} \), calcium (Ca) has an atomic mass of about 40.08 g, and the formula calculates to: 40.08 + (2 \times 35.45) = 110.98 \text{ g/mol}.
Once you know the molar mass, you can find the number of moles in a given sample by dividing its mass by its molar mass, as shown in the exercise solution.
Dissociation of Ionic Compounds
When ionic compounds are dissolved in water, they dissociate into their constituent ions.
This dissociation is crucial because it determines the amount of each ion produced, further affecting the solution's concentration.
  • \( \mathrm{PbCl}_{2} \) dissociates into one \( \mathrm{Pb}^{2+} \) ion and two \( \mathrm{Cl}^{-} \) ions. Hence, if 0.0036 moles of \( \mathrm{PbCl}_{2} \) dissociate, it will produce \( 2 \times 0.0036 = 0.0072 \) moles of \( \mathrm{Cl}^{-} \) ions.
  • \( \mathrm{KCl} \) dissociates simpler into one \( \mathrm{K}^{+} \) ion and one \( \mathrm{Cl}^{-} \) ion, yielding \( 0.0134 \) moles of \( \mathrm{Cl}^{-} \) ions from an equivalent amount of compound.
  • \( \mathrm{CaCl}_{2} \) dissociates to form one \( \mathrm{Ca}^{2+} \) ion and two \( \mathrm{Cl}^{-} \) ions, leading to \( 2 \times 0.0090 = 0.0180 \) moles of \( \mathrm{Cl}^{-} \) ions.
The dissociation process directly influences the chloride ion concentration in the solution.
Mole Concept
The mole concept is a central idea in chemistry that helps relate the mass of a substance to the number of its particles.
A mole is defined as \( 6.022 \times 10^{23} \) entities (atoms, molecules, etc.) of a substance.
  • Moles allow chemists to count particles by weighing them. It simplifies using large numbers for chemical reactions.
  • Here, you calculate moles from mass using the formula: \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}.\
  • In the exercise, for each compound, you start with a 1.0 g sample, and by using the molar mass, you found the number of moles. These moles tell how many formula units are present, which is crucial for determining the resulting ion concentrations after dissociation.
Chemical Concentration Comparison
Concentration tells you how much of a substance is present in a specific volume of solution and is typically expressed as molarity (mol/L).
To find this, you use the formula: \text{Concentration (mol/L)} = \frac{\text{moles of solute}}{\text{volume of solution (L)}}.\
  • In this exercise, each solution is diluted to a total volume of 0.250 liters.
  • For \( \mathrm{PbCl}_{2} \), \( \mathrm{KCl} \), and \( \mathrm{CaCl}_{2} \), you've already calculated the moles of \( \mathrm{Cl}^{-} \) ions produced.
    Their chloride ion concentrations are calculated as:\[ \mathrm{PbCl}_{2}: \, 0.0288 \text{ mol/L} \,; \mathrm{KCl}: \, 0.0536 \text{ mol/L} \,; \mathrm{CaCl}_{2}: \, 0.0720 \text{ mol/L} \]
  • Comparing these values, \( \mathrm{CaCl}_{2} \) has the highest chloride ion concentration.
Thus, concentration comparisons help determine which solution has more ions present, aiding in understanding solution behaviors.

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Most popular questions from this chapter

Obtain the oxidation number for the element noted in each of the following. a. Cr in \(\mathrm{CrO}_{3}\) b. \(\mathrm{Hg}\) in \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) c. Ga in \(\mathrm{Ga}(\mathrm{OH})_{3}\) d. \(\mathrm{P}\) in \(\mathrm{Na}_{3} \mathrm{PO}_{4}\)

A metal, \(\mathrm{M}\), was converted to the chloride, \(\mathrm{MCl}_{2} .\) Then a solution of the chloride was treated with silver nitrate to give silver chloride crystals, which were filtered from the solution. \(\mathrm{MCl}_{2}(a q)+2 \mathrm{AgNO}_{3}(a q) \longrightarrow \mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{AgCl}(s)\) If \(2.434 \mathrm{~g}\) of the metal gave \(7.964 \mathrm{~g}\) of silver chloride, what is the atomic weight of the metal? What is the metal?

A \(1.345-\mathrm{g}\) sample of a compound of barium and oxygen was dissolved in hydrochloric acid to give a solution of barium ion, which was then precipitated with an excess of potassium chromate to give \(2.012 \mathrm{~g}\) of barium chromate, \(\mathrm{BaCrO}_{4}\). What is the formula of the compound?

Elemental bromine is the source of bromine compounds. The element is produced from certain brine solutions that occur naturally. These brines are essentially solutions of calcium bromide that, when treated with chlorine gas, yield bromine in a displacement reaction. What are the molecular equation and net ionic equation for the reaction? A solution containing \(40.0 \mathrm{~g}\) of calcium bromide requires \(14.2 \mathrm{~g}\) of chlorine to react completely with it, and \(22.2 \mathrm{~g}\) of calcium chloride is produced in addition to whatever bromine is obtained. How many grams of calcium bromide are required to produce \(10.0\) pounds of bromine?

A \(10.0-\mathrm{mL}\) sample of potassium iodide solution was analyzed by adding an excess of silver nitrate solution to produce silver iodide crystals, which were filtered from the solution. \(\mathrm{KI}(a q)+\mathrm{AgNO}_{3}(a q) \longrightarrow \mathrm{KNO}_{3}(a q)+\operatorname{AgI}(s)\) If \(2.183 \mathrm{~g}\) of silver iodide was obtained, what was the molarity of the original KI solution?
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