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Chapter 4: Problem 31

Which of the following solutions of strong electrolytes contains the largest number of moles of chloride ions: \(100.0 \mathrm{~mL}\) of \(0.30 \mathrm{M} \mathrm{AlCl}_{3}, 50.0 \mathrm{~mL}\) of \(0.60 \mathrm{M} \mathrm{MgCl}_{2}\), or \(200.0 \mathrm{~mL}\) of \(0.40 \mathrm{M} \mathrm{NaCl} ?\)

Short Answer

Expert verified
The solution with the largest number of moles of chloride ions is \(100.0 \mathrm{~mL}\) of $0.30 \mathrm{M} \mathrm{AlCl}_{3}$, which contains 0.09 moles of Cl鈦�.

Step by step solution

01

Identify the dissociation of the strong electrolytes

Each of the strong electrolytes given will dissociate completely in the solution, producing respective ions. We can write the dissociation reaction as: AlCl鈧� 鈫� Al鲁鈦� + 3Cl鈦� MgCl鈧� 鈫� Mg虏鈦� + 2Cl鈦� NaCl 鈫� Na鈦� + Cl鈦�
02

Calculate the moles of chloride ions produced by each electrolyte

To find the moles of chloride ions produced by each electrolyte, we'll follow this formula: moles of Cl鈦� = [concentration (M) * volume (L)] * (moles of Cl鈦� / moles of electrolyte) Make sure to convert the given volume from mL to L by dividing the volume by 1000.
03

Calculate the moles of chloride ions in the AlCl鈧� solution

Using the given concentration and volume of AlCl鈧� solution: moles of Cl鈦� = [0.30 M * (100.0 mL / 1000)] * 3 moles of Cl鈦� = 0.09 moles
04

Calculate the moles of chloride ions in the MgCl鈧� solution

Using the given concentration and volume of MgCl鈧� solution: moles of Cl鈦� = [0.60 M * (50.0 mL / 1000)] * 2 moles of Cl鈦� = 0.06 moles
05

Calculate the moles of chloride ions in the NaCl solution

Using the given concentration and volume of NaCl solution: moles of Cl鈦� = [0.40 M * (200.0 mL / 1000)] * 1 moles of Cl鈦� = 0.08 moles
06

Compare the moles of chloride ions for each solution

Now that we've calculated the moles of chloride ions for each electrolyte solution, we can compare them to see which one has the highest value. AlCl鈧�: 0.09 moles Cl鈦� MgCl鈧�: 0.06 moles Cl鈦� NaCl: 0.08 moles Cl鈦� Out of these values, the AlCl鈧� solution contains the largest number of moles of chloride ions (0.09 moles). Therefore, \(100.0 \mathrm{~mL}\) of $0.30 \mathrm{M} \mathrm{AlCl}_{3}$ contains the largest number of moles of chloride ions.

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Key Concepts

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

Moles Calculation
When dealing with chemical solutions, calculating the number of moles is fundamental. It involves understanding concentrations and volumes of solutes to derive the quantity of ions or molecules present. Here鈥檚 how you can simplify this calculation:
  • First, convert the volume of the solution from milliliters (mL) to liters (L) by dividing by 1000. This is crucial as molarity is defined in terms of liters.
  • Next, use the molarity (M), which indicates how many moles of solute are present in one liter of solution.
  • The formula for number of moles is: \[ \text{Moles} = \text{Molarity (M)} \times \text{Volume (L)} \]
For instance, to find the moles of chloride ions in a solution: \[ 0.30 \text{ M} \times \left( \frac{100.0 \text{ mL}}{1000} \right) = 0.03 \text{ moles of } \text{AlCl}_3 \] Applying stoichiometry further helps in calculating the moles for the ions produced after dissociation.
Dissociation Reaction
Strong electrolytes like sodium chloride (NaCl), aluminum chloride (AlCl鈧�), and magnesium chloride (MgCl鈧�) fully dissociate in water. This means they split into their constituent ions completely. Understanding this reaction is key to analyzing solutions:
  • For NaCl, it dissociates into \[ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \]
  • In AlCl鈧�, each molecule produces three chloride ions:\[ \text{AlCl}_3 \rightarrow \text{Al}^{3+} + 3\text{Cl}^- \]
  • MgCl鈧� dissociates yielding:\[ \text{MgCl}_2 \rightarrow \text{Mg}^{2+} + 2\text{Cl}^- \]
Each initial electrolyte molecule results in multiple ions forming in the solution, contributing to the total chloride ion count.
Chloride Ions
Chloride ions (\[ \text{Cl}^- \]) are commonly found in solutions of strong electrolytes. They play a significant role in determining properties like conductivity. When it comes to calculating these from strong electrolytes, always consider the dissociation pattern:
  • Identify how many chloride ions are generated per mole of the compound. For example, AlCl鈧� produces three times as many, whereas NaCl provides only one per formula unit.
  • Use the dissociation formula to understand the stoichiometry; multiply the moles of compound by the number of Cl鈦� ions per mole of compound.
From the given examples:- AlCl鈧� solution in the exercise results in the highest production of chloride ions.- The calculation shows 0.09 moles of Cl鈦� ions from AlCl鈧�, 0.06 from MgCl鈧�, and 0.08 from NaCl.Thus, understanding chloride ion production helps compare electrolytic solutions efficiently.

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Most popular questions from this chapter

You are given a solid that is a mixture of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\). A \(0.205-g\) sample of the mixture is dissolved in water. An excess of an aqueous solution of \(\mathrm{BaCl}_{2}\) is added. The \(\mathrm{BaSO}_{4}\) that is formed is filtered, dried, and weighed. Its mass is \(0.298 \mathrm{~g}\). What mass of \(\mathrm{SO}_{4}^{2-}\) ion is in the sample? What is the mass percent of \(\mathrm{SO}_{4}{ }^{2-}\) ion in the sample? What are the percent compositions by mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) in the sample?

A mixture contains only sodium chloride and potassium chloride. A \(0.1586-\mathrm{g}\) sample of the mixture was dissolved in water. It took \(22.90 \mathrm{~mL}\) of \(0.1000 \mathrm{M} \mathrm{AgNO}_{3}\) to completely precipitate all the chloride present. What is the composition (by mass percent) of the mixture?

A \(6.50-\mathrm{g}\) sample of a diprotic acid requires \(137.5 \mathrm{~mL}\) of a \(0.750 \mathrm{M} \mathrm{NaOH}\) solution for complete neutralization. Determine the molar mass of the acid.

Many oxidation-reduction reactions can be balanced by inspection. Try to balance the following reactions by inspection. In each reaction, identify the substance reduced and the substance oxidized. a. \(\mathrm{Al}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{CH}_{4}(g)+\mathrm{S}(s) \rightarrow \mathrm{CS}_{2}(l)+\mathrm{H}_{2} \mathrm{~S}(g)\) c. \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) d. \(\mathrm{Cu}(s)+\mathrm{Ag}^{+}(a q) \rightarrow \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)\)

Write the balanced formula, complete ionic, and net ionic equations for each of the following acid-base reactions. a. \(\mathrm{HNO}_{3}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \rightarrow\) b. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{KOH}(a q) \rightarrow\) c. \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{HCl}(a q) \rightarrow\)
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