Problem 39 - Chemistry: An Atoms-Focused Approach

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

Question: Arrange the given solutions in order of their ability to conduct electricity: (a) 1.0 M NaCl, (b) 1.2 M KCl, (c) 1.0 M Na2SO4, and (d) 0.75 M LiCl. Answer: The order of conductivity, starting with the most conductive, is (c) 1.0 M Na2SO4, (b) 1.2 M KCl, (a) 1.0 M NaCl, and (d) 0.75 M LiCl.

Step 1: Analyze the ions in each solution

We will first look at the ions present in each solution. For this, we need to consider how many ions are present in each substance. (a) \(1.0 M \ \mathrm{NaCl}\): Sodium chloride dissociates into one sodium ion and one chloride ion in water. (b) \(1.2 M \ \mathrm{KCl}\): Potassium chloride dissociates into one potassium ion and one chloride ion in water. (c) \(1.0 M \ \mathrm{Na}_{2} \mathrm{SO}_{4}\): Sodium sulfate dissociates into two sodium ions and one sulfate ion in water. (d) \(0.75 M\) LiCl: Lithium chloride dissociates into one lithium ion and one chloride ion in water.

Step 2: Calculate the concentration of ions in each solution

To compare their conductivity, we need to calculate the concentration of ions in each solution. (a) \(1.0 \mathrm{M} \mathrm{NaCl}\): \(1.0 + 1.0 = 2.0 \mathrm{M}\) total ions (b) \(1.2 \mathrm{M} \mathrm{KCl}\): \(1.2 + 1.2 = 2.4 \mathrm{M}\) total ions (c) \(1.0 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\): \((1.0 \times 2) + 1.0 = 3.0 \mathrm{M}\) total ions (d) \(0.75 M\) LiCl: \(0.75 + 0.75 = 1.5 \mathrm{M}\) total ions

Step 3: Rank the solutions based on their ion concentrations

Now we can rank the solutions based on their ion concentrations: 1. \(1.0 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) (3.0 M total ions) 2. \(1.2 \mathrm{M} \mathrm{KCl}\) (2.4 M total ions) 3. \(1.0 \mathrm{M} \mathrm{NaCl}\) (2.0 M total ions) 4. \(0.75 M\) LiCl (1.5 M total ions) So the order of conductivity, starting with the most conductive, is (c) \(1.0 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\), (b) \(1.2 \mathrm{M} \mathrm{KCl}\), (a) \(1.0 \mathrm{M} \mathrm{NaCl}\), and (d) \(0.75 M\) LiCl.