Problem 134 - General Chemistry

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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. \( \mathrm{CaCl}_{2}: \frac{1.0}{110.98} \approx 0.0090 \text{ moles} \)

Step 2: 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 \( \mathrm{Cl}^- \).

Step 3: 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 \( \mathrm{CaCl}_{2} \), \[ \frac{0.0180}{0.250} = 0.0720 \text{ mol/L} \].

Step 4: 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.

Key Concepts

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.