Problem 70 - Chemistry: The Central Science in SI Units, Global Edition

Question

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

Step-by-Step Solution

Answer

(a) The $0.10\,\mathrm{M}\,\mathrm{AlCl}_{3}$ solution has a higher concentration of $\mathrm{Cl}^-$ ions. (b) The $150\,\text{mL}$ of the $0.05\,\mathrm{M}\,\mathrm{MnCl}_{3}$ solution has a higher amount of $\mathrm{Cl}^-$ ions. (c) The solution made by dissolving $23.5\,\mathrm{g}$ of $\mathrm{KCl}$ in water to make $100\,\mathrm{mL}$ of solution has a higher concentration of $\mathrm{Cl}^-$ ions.

Step 1: Part (a): Comparing 0.10 M AlCl3 and 0.25 M LiCl

We need to find the concentration of \(\mathrm{Cl}^-\) ions in each solution. For \(0.10\,\mathrm{M}\,\mathrm{AlCl}_{3}\): A mole of \(\text{AlCl}_3\) dissociates into one \(\text{Al}^{3+}\) ion and three \(\text{Cl}^-\) ions. $$\text{AlCl}_3 \rightarrow \text{Al}^{3+} + 3\,\text{Cl}^-$$ So the concentration of \(\text{Cl}^-\) ions will be \(0.10\,\mathrm{M}\times 3=0.30\,\mathrm{M}\). For \(0.25\,\mathrm{M}\,\mathrm{LiCl}\): A mole of \(\text{LiCl}\) dissociates into one \(\text{Li}^+\) ion and one \(\text{Cl}^-\) ion. $$\text{LiCl} \rightarrow \text{Li}^+ + \text{Cl}^-$$ So the concentration of \(\text{Cl}^-\) ions will be \(0.25\,\mathrm{M}\). Comparing both concentrations, the \(0.10\,\mathrm{M}\,\mathrm{AlCl}_{3}\) solution has a higher concentration of \(\mathrm{Cl}^-\) ions.

Step 2: Part (b): Comparing 150 mL of 0.05 M MnCl3 and 200 mL of 0.10 M KCl

First, find the moles of \(\mathrm{Cl}^-\) ions in each solution. For \(150\,\text{mL}\,\text{ of }\,0.05\,\text{M}\,\text{MnCl}_3\): A mole of \(\text{MnCl}_3\) dissociates into one \(\text{Mn}^{2+}\) ion and three \(\text{Cl}^-\) ions. $$\text{MnCl}_3 \rightarrow \text{Mn}^{2+} + 3\,\text{Cl}^-$$ Moles of \(\mathrm{Cl}^-\) ions = (volume in L) \(\times\) (concentration of MnCl3) \(\times\) (number of \(\mathrm{Cl}^-\) ions): $$\frac{150}{1000}\,\text{L} \times 0.05\,\text{M} \times 3 = 0.0225\,\text{ moles of }\text{Cl}^-$$ For \(200\,\text{mL}\,\text{ of }\,0.10\,\text{M}\,\text{KCl}\): A mole of \(\text{KCl}\) dissociates into one \(\text{K}^+\) ion and one \(\text{Cl}^-\) ion. $$\text{KCl} \rightarrow \text{K}^+ + \text{Cl}^-$$ Moles of \(\mathrm{Cl}^-\) ions = (volume in L) \(\times\) (concentration of KCl): $$\frac{200}{1000}\,\text{L} \times 0.10\,\text{M} = 0.020\,\text{ moles of }\text{Cl}^-$$ Comparing moles of \(\mathrm{Cl}^-\) ions, the 150 mL of the \(0.05\,\mathrm{M}\,\mathrm{MnCl}_{3}\) solution has a higher amount of \(\mathrm{Cl}^-\) ions.

Step 3: Part (c): Comparing 2.8 M HCl and 23.5 g KCl in 100 mL solution

First, find the concentration of \(\mathrm{Cl}^-\) ions in each solution. For \(2.8\,\mathrm{M}\,\mathrm{HCl}\): A mole of \(\mathrm{HCl}\) dissociates into one \(\mathrm{H}^+\) ion and one \(\mathrm{Cl}^-\) ion. $$\mathrm{HCl} \rightarrow \mathrm{H}^+ + \mathrm{Cl}^-$$ So the concentration of \(\mathrm{Cl}^-\) ions will be \(2.8\,\mathrm{M}\). For the solution made by dissolving \(23.5\,\mathrm{g}\) of \(\mathrm{KCl}\) in water to make \(100\,\mathrm{mL}\) of solution: Calculate the concentration of \(\mathrm{KCl}\): Molar mass of \(\mathrm{KCl} = 39.1 (\mathrm{K}) + 35.5 (\mathrm{Cl}) = 74.6\,\text{g/mol}\). Moles of \(\mathrm{KCl} = \frac{23.5\,\mathrm{g}}{74.6\,\text{g/mol}} = 0.315\,\text{moles}\). Concentration of KCl = \(\frac{0.315\,\text{moles}}{0.1\,\text{L}} = 3.15\,\mathrm{M}\). A mole of \(\mathrm{KCl}\) dissociates into one \(\text{K}^+\) ion and one \(\text{Cl}^-\) ion. $$\mathrm{KCl} \rightarrow \mathrm{K}^+ + \mathrm{Cl}^-$$ So the concentration of \(\mathrm{Cl}^-\) ions will also be \(3.15\,\text{M}\). Comparing both concentrations, the solution made by dissolving \(23.5\,\mathrm{g}\) of \(\mathrm{KCl}\) in water to make \(100\,\mathrm{mL}\) of solution has a higher concentration of \(\mathrm{Cl}^-\) ions.

Key Concepts

Molarity

Molarity is a way to express the concentration of a solution, specifically how many moles of solute are present in one liter of solution. It is denoted by the letter "M" and is calculated in moles per liter (mol/L). Understanding molarity is essential because it helps determine the proportion of a solute in the solution, which is necessary for predicting and controlling chemical reactions.

For example, when you see "0.10 M AlCl₃," it means there are 0.10 moles of aluminum chloride dissolved in one liter of the solution. In multiple solutions, comparing molarity gives us a direct understanding of which solution is more or less concentrated. To calculate, divide the number of moles of solute by the volume of the solution in liters. Remember, a higher molarity indicates a higher presence of the solute per unit volume.

Dissociation of Ionic Compounds

When ionic compounds dissolve in water, they split into their individual ions in a process called dissociation. This is important in understanding how molecules interact in a solution. Ionic compounds such as salts separate into cations (positive ions) and anions (negative ions), both of which can participate in various chemical processes depending on their concentration.

Take for instance the ionic compound AlCl₃ used in the solution analysis. Upon dissolving, each AlCl₃ unit dissociates into one Al³⁺ cation and three Cl⁻ anions. Therefore, the concentration of chloride ions, Cl⁻, in the solution will be three times the molarity of the initial AlCl₃ solution. Such relationships allow chemists to predict the availability of ions and adjust chemical solution formulations accordingly. Knowing how many ions form from dissociation is crucial because reactions often depend on these charges.

Chemical Solutions Analysis

Analyzing chemical solutions involves interpreting the concentration and interaction of components within a solution. This often requires calculating the concentration of ions and understanding the dissociation of compounds in the solution to make comparisons, as exemplified in the different chemical solution scenarios given.

Methods to analyze involve steps like calculating molarity, evaluating the dissociation equation, and comparing mole contributions. For instance, if you want to find the amount of Cl⁻ ions from KCl in a given solution, you first determine the number of moles of KCl being dissolved, then use the dissociation information that each KCl gives one Cl⁻ to equate it to Cl⁻ moles in the solution.

In homework exercises like these, key data like the volume and weight of solutes, alongside their molecular weights, are essential for accurate solution analysis. Whether comparing solutions of different substances or figuring out which has a higher concentration of a specific ion, chemical solution analysis is the process of breaking down these complex mixtures into understandable and manageable chemical relationships.