Journey into Solution Chemistry: Unraveling the Mysteries of Polarity, Electrolytes, and Concentrations

Pre-Lab Study Questions

1. Why does an oil and vinegar salad dressing have two separate layers?
   • The separation is due to the difference in polarity between oil (nonpolar) and vinegar (polar).
2. What is meant by mass/mass percent concentration of a solution?
   • Mass/mass percent concentration represents the amount of solute in grams per 100 grams of solution.
3. Why are some electrolytes strong, while others are weak?
   • Strong electrolytes completely dissociate in water, forming ions, while weak electrolytes only partially dissociate.
4. What is molarity?
   • Molarity is a measure of the concentration of a solution, expressed as moles of solute per liter of solution.
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A. Polarity of Solutes and Solvents

Questions and Problems

Q.1 NaCl is soluble in water, but I2 is not. Explain.

• NaCl dissociates into ions in water, interacting with water's polarity. I2, being nonpolar, lacks the interactions needed for solubility.
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Q.2 Write equations for the three solutes tested in A.1. a. HCl + H2O → H3O+ + Cl- b. NH4OH ⇌ NH3 + H2O c. C6H12O6 + O2 → CO2 + H2O

Q.3 Classify the solutes in the following equations. a. XY2(s) → X2+ + 2Y- (aq) - Strong electrolyte b. HX(g) → H+ + X- (aq) - Weak electrolyte c. XYZ(s) ⇌ XYZ(aq) - Nonelectrolyte d. YOH(s) → Y+ + OH- (aq) - Strong electrolyte

C. Electrolytes in Body Fluids

C.1:

• IV Solution: 20 mEq Na, 25 mEq Mg, 17 mEq Cl, 28 mEq S

C.2:

• Cations: Na, Mg
• Anions: Cl, S

C.3:

• Total Charge of Cations: +3
• Total Charge of Anions: -3

C.4:

• Overall Charge: 0

Q.4 What would be the overall charge in any IV solution? Why?

• The overall charge is 0, as the total positive charge from cations equals the total negative charge from anions, maintaining neutrality.

D. Concentrations of a Sodium Chloride Solution

D1 Mass of evaporating dish: 40g D2 Volume of NaCl solution: 10 mL D3 Mass of dish and NaCl solution: 50g

Calculations D.5 Mass of NaCl solution: 50g - 40g = 10g D.6 Mass of dry NaCl salt: 40g - 41g = 9g D.7 Mass/mass percent: (10g / 58.44g) * 100 = 17.11% D.8 Mass/volume percent: (9g / 10g) * 100 = 90% D.9 Moles of NaCl: 9g / 58.44g/mol = 0.154 moles D.10 Volume of sample in liters: 1g = 0.001 L, so 9g = 0.009 L D.11 Molarity of NaCl solution: 0.154 moles / 0.009 L = 17.11 M

Q.5 a. % (m/m): (3.26 / 15.78) * 100 = 20.7% b. % (m/v): (3.26 / 15) * 100 = 21.7% c. Molarity (M): 0.0558 moles / 0.0150 L = 3.72 M

Q.6 How many grams of KI are in 25.0 mL of a 3.0 % (m/v) KI solution?

• 3% of 25 mL = 0.75g

Q.7 How many milliliters of a 2.5 M MgCl2 solution contain 17.5 g MgCl2?

• 17.5g / (237.5g/L * 2.5) = 73.7 mL

The study of solutions, their properties, and their behavior is crucial in chemistry. As we delve deeper into the realm of solution chemistry, we find that the interaction between different substances plays a pivotal role in determining their characteristics. Let's expand on the concepts introduced in the initial questions to provide a more comprehensive understanding.

One fascinating aspect of solution chemistry is the behavior of oil and vinegar in salad dressings. The reason why an oil and vinegar salad dressing tends to form two separate layers lies in the fundamental principles of polarity. Oil, being a nonpolar substance, exhibits weak interactions with the polar molecules present in vinegar. The polar nature of vinegar, primarily composed of water and acetic acid, causes it to form distinct layers when mixed with nonpolar oil. This separation is a manifestation of the intermolecular forces at play between the two substances.

Moving on to concentration terms, mass/mass percent concentration is a crucial parameter in solution chemistry. It is defined as the ratio of the mass of the solute to the mass of the solution, multiplied by 100. This percentage provides a standardized measure of the solute's abundance within the solution, offering a convenient way to express and compare concentrations across different solutions.

Electrolytes, substances that conduct electricity when dissolved in water, are classified as strong or weak based on their degree of dissociation. Strong electrolytes, such as salts, acids, and bases, undergo complete dissociation into ions when dissolved in water. This leads to a higher conductivity of the solution. On the other hand, weak electrolytes, like some acids and bases, only partially dissociate, resulting in a lower conductivity. Understanding the distinction between strong and weak electrolytes is crucial for predicting the behavior of solutions in various chemical processes.

Molarity, denoted by the symbol M, represents another vital concept in solution chemistry. It quantifies the concentration of a solute in a solution by expressing the number of moles of solute per liter of solution. Molarity provides a quantitative measure of how much solute is present in a given volume of solution, making it a widely used unit in chemical calculations.

Now, let's explore the interaction between solutes and solvents in more detail. The solubility of a substance depends on its ability to interact with the molecules of the solvent. For instance, potassium permanganate (KMnO4) is soluble in water due to the polar nature of both the solute and the solvent. In contrast, iodine (I2) is insoluble in water because it is a nonpolar substance, and the forces between iodine molecules are stronger than their interaction with water molecules.

Sucrose, a polar molecule, readily dissolves in water, while vegetable oil, a nonpolar substance, does not form a homogeneous solution with water. These examples highlight the significance of intermolecular forces and polarity in determining solubility.

Questions and Problems:

1. NaCl is soluble in water, but I2 is not. This phenomenon can be explained by examining the nature of the solutes. NaCl, an ionic compound, readily dissociates into ions in water, interacting with the polar water molecules through ion-dipole forces. In contrast, iodine (I2) is held together by covalent bonds, and its nonpolar nature prevents effective interactions with water molecules. The dispersion forces between iodine molecules are stronger than their interaction with water, leading to insolubility.
2. Writing chemical equations for the tested solutes: a. HCl + H2O → H3O+ + Cl- b. NH4OH ⇌ NH3 + H2O c. C6H12O6 + O2 → CO2 + H2O
3. Classifying solutes in given equations: a. XY2 (s) → X2+ (aq) + 2Y- (aq) - Strong electrolyte b. HX (g) → H+ (aq) + X- (aq) - Weak electrolyte c. XYZ(s) ⇌ XYZ(aq) - Nonelectrolyte d. YOH(s) → Y+ (aq) + OH- (aq) - Strong electrolyte

C. Electrolytes in Body Fluids:

Understanding the composition of intravenous (IV) solutions is vital in medical contexts. The given IV solution contains 20 mEq Na, 25 mEq Mg, 17 mEq Cl, and 28 mEq S. Categorizing cations (Na, Mg) and anions (Cl, S) helps us determine the overall charge. In this case, the total charge of cations equals the total charge of anions (3 - 3), resulting in an overall charge of 0. This neutrality is essential to maintain physiological balance when administering IV solutions.

D. Concentrations of a Sodium Chloride Solution:

Exploring the concentration of a sodium chloride (NaCl) solution involves various calculations. Starting with the mass of the evaporating dish, volume of the NaCl solution, and the mass of the dish and solution, we can determine key parameters:

D5. Mass of NaCl solution: 50g - 40g = 10g D6. Mass of dry NaCl salt: 40g - 41g = 9g

Calculations: D7. Mass/mass percent: (10g / 58.44g) * 100 = 17.11% D8. Mass/volume percent: (9g / 10g) * 100 = 90% D9. Moles of NaCl: 9g / 58.44g/mol = 0.154 moles D10. Volume of sample in liters: 1g = 0.001 L, so 9g = 0.009 L D11. Molarity of NaCl solution: 0.154 moles / 0.009 L = 17.11 M

Q.5 a. % (m/m): (3.26 / 15.78) * 100 = 20.7% b. % (m/v): (3.26 / 15) * 100 = 21.7% c. Molarity (M): 0.0558 moles / 0.0150 L = 3.72 M

Q.6 How many grams of KI are in 25.0 mL of a 3.0 % (m/v) KI solution?

• A 3.0% (m/v) KI solution implies 3g of KI in 100 mL of solution. Therefore, for 25.0 mL: (3g / 100mL) * 25mL = 0.75g of KI.

Q.7 How many milliliters of a 2.5 M MgCl2 solution contain 17.5 g MgCl2?

• To determine the volume, we can use the formula: Volume (mL) = Mass (g) / Molarity (M). Given that the molar mass of MgCl2 is 95g/mol, the molarity is 2.5 M. Therefore, the volume is (17.5g / 95g/mol * 2.5 M) = 73.7 mL.