Problem 43 - Chemistry The Science in Context

Question

Rank the following solutions on the basis of their ability to conduct electricity, starting with the most conductive: (a) \(1.0 \mathrm{M} \mathrm{NaCl} ;\) (b) \(1.2 \mathrm{M} \mathrm{KCl} ;\) (c) \(1.0 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) (d) \(0.75 M\) LiCl.

Step-by-Step Solution

Answer

Based on the total concentration of ions calculated for each solution, rank them in terms of their ability to conduct electricity.

Step 1: Calculate ions produced by each solution

For each solution, we need to find the number of ions produced by each solute when dissolved in water. (a) \(1.0 \mathrm{M}\) NaCl: Sodium chloride (\(\mathrm{NaCl}\)) dissociates into one sodium ion (\(\mathrm{Na}^{+}\)) and one chloride ion (\(\mathrm{Cl}^{-}\)) per molecule. $$\mathrm{NaCl} \rightarrow \mathrm{Na}^{+} + \mathrm{Cl}^{-}$$ (b) \(1.2 \mathrm{M}\) KCl: Potassium chloride (\(\mathrm{KCl}\)) dissociates into one potassium ion (\(\mathrm{K}^{+}\)) and one chloride ion (\(\mathrm{Cl}^{-}\)) per molecule. $$\mathrm{KCl} \rightarrow \mathrm{K}^{+} + \mathrm{Cl}^{-}$$ (c) \(1.0 \mathrm{M}\) Na\(_2\)SO\(_4\): Sodium sulfate (\(\mathrm{Na}_{2}\mathrm{SO}_{4}\)) dissociates into two sodium ions (\(2\mathrm{Na}^{+}\)) and one sulfate ion (\(\mathrm{SO}_{4}^{2-}\)) per molecule. $$\mathrm{Na}_{2}\mathrm{SO}_{4} \rightarrow 2\mathrm{Na}^{+} + \mathrm{SO}_{4}^{2-}$$ (d) \(0.75 \mathrm{M}\) LiCl: Lithium chloride (\(\mathrm{LiCl}\)) dissociates into one lithium ion (\(\mathrm{Li}^{+}\)) and one chloride ion (\(\mathrm{Cl}^{-}\)) per molecule. $$\mathrm{LiCl} \rightarrow \mathrm{Li}^{+} + \mathrm{Cl}^{-}$$

Step 2: Calculate the total concentration of ions in each solution

To find the total concentration of ions in each solution, we need to sum the concentrations of all ions produced by the solute. (a) \(1.0 \mathrm{M}\) NaCl: Total concentration of ions \(= 1 + 1 = 2\) (b) \(1.2 \mathrm{M}\) KCl: Total concentration of ions \(= 1.2 + 1.2 = 2.4\) (c) \(1.0 \mathrm{M}\) Na\(_2\)SO\(_4\): Total concentration of ions \(= 2 + 1 = 3.0\) (d) \(0.75 \mathrm{M}\) LiCl: Total concentration of ions \(= 0.75 + 0.75 = 1.5\)

Step 3: Rank the solutions based on their conductivity

The higher the total concentration of ions in a solution, the more conductive it will be. So, the ranking is as follows: 1. \(1.0 \mathrm{M}\) Na\(_2\)SO\(_4\) (3.0 total concentration of ions) 2. \(1.2 \mathrm{M}\) KCl (2.4 total concentration of ions) 3. \(0.75 \mathrm{M}\) LiCl (1.5 total concentration of ions) 4. \(1.0 \mathrm{M}\) NaCl (2.0 total concentration of ions)

Key Concepts

Ion Dissociation

Ion dissociation is a crucial process involved when compounds dissolve in water. It is the process through which an ionic compound separates into its individual ions. This happens because the water molecules surround and interact with the ions, effectively breaking apart the ionic bonds holding them together. For example:

• Sodium chloride (NaCl) dissociates into sodium ions (Na⁺) and chloride ions (Cl^-).
• Potassium chloride (KCl) splits into potassium ions (K⁺) and chloride ions (Cl^-).
• Sodium sulfate (Na₂SO₄) dissociates into two sodium ions (2Na⁺) and one sulfate ion (SO₄^2-).
• Lithium chloride (LiCl) separates into lithium ions (Li⁺) and chloride ions (Cl^-).

This dissociation process is essential because it produces free ions in the solution, which are necessary for conducting electricity. The more ions that are produced, the higher the potential electrical conductivity of the solution.

Ionic Concentration

Ionic concentration refers to the total concentration of ions that are present in a solution after ion dissociation has occurred. This is very important when considering the conductivity of a solution because conductivity depends on the availability of ions to carry an electric charge. To calculate ionic concentration, you must consider both the concentration of the solution and the number of ions produced per formula unit that dissolves. For instance:

• For a 1.0 M NaCl solution, it dissociates into 1 Na⁺ and 1 Cl^-, resulting in a total ionic concentration of 2 M.
• For a 1.2 M KCl solution, KCl dissociates into 1 K⁺ and 1 Cl^-, yielding a total ionic concentration of 2.4 M.
• For a 1.0 M Na₂SO₄ solution, it dissociates into 2 Na⁺ and 1 SO₄^2-, giving a total ionic concentration of 3.0 M.
• For a 0.75 M LiCl solution, it dissociates into 1 Li⁺ and 1 Cl^-, providing a total ionic concentration of 1.5 M.

The sum of all these ions indicates how conductive the solution might be. Higher ionic concentration usually means better conductivity.

Solution Ranking

When ranking solutions based on their ability to conduct electricity, it's directly correlated with their total ionic concentration. Conductivity increases with more ions available to move and carry an electrical charge. Here's the ranking based on the total concentration of ions:

• 1.0 M Na₂SO₄: With a total ion concentration of 3.0 M, it's the most conductive of the four solutions.
• 1.2 M KCl: Its total ion concentration is 2.4 M, placing it second in conductivity.
• 1.0 M NaCl: The total ion concentration is 2.0 M, ranking third.
• 0.75 M LiCl: Its total ion concentration is 1.5 M, making it the least conductive of the solutions listed.

Understanding these rankings aids in predicting how well a solution will conduct electricity, crucial for many applications in chemistry and industry. The greater the ion concentration, the higher the solution's conductivity.