Solutions and Solution Chemistry (Chapter 13) – Study Notes

Solutions and Solution Chemistry

Introduction to Solutions

Solutions are homogeneous mixtures composed of two or more substances. The substance present in the greatest amount is called the solvent, while the other substances are solutes. Understanding the properties and calculations related to solutions is essential in chemistry.

• Solvent: The component of a solution present in the largest amount; typically a liquid.

• Solute: The component(s) present in lesser amounts; dissolved in the solvent.

• Homogeneous mixture: A mixture with uniform composition throughout.

• Electrolyte: A substance that produces ions when dissolved in water, allowing the solution to conduct electricity.

• Nonelectrolyte: A substance that does not produce ions in solution.

Types of Solutions and Solubility

The ability of a solute to dissolve in a solvent depends on the nature of both substances. Solubility is affected by factors such as temperature, pressure, and the chemical properties of the solute and solvent.

• Ionic solutes typically dissolve in polar solvents (e.g., water).

• Saturated solution: Contains the maximum amount of solute that can dissolve under given conditions.

• Unsaturated solution: Contains less solute than the maximum amount that can dissolve.

• Supersaturated solution: Contains more solute than is normally possible at a given temperature.

Example: A sugar solution with 13.75 g of sugar dissolved in 87.5 g of water is an example of a weak electrolyte solution.

Electrolytes: Strong vs. Weak

Electrolytes are classified based on their ability to dissociate into ions in solution.

• Strong electrolytes: Completely dissociate into ions (e.g., NaCl, KBr, HCl).

• Weak electrolytes: Partially dissociate into ions (e.g., acetic acid, sugar solutions).

• Nonelectrolytes: Do not dissociate into ions (e.g., glucose, ethanol).

Example: NaCl is a strong electrolyte because it fully dissociates in water.

Concentration Units

Concentration expresses the amount of solute in a given amount of solvent or solution. Common units include percent by mass and molarity.

• Percent by mass (%): $\text{Percent by mass} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\%$

• Molarity (M): $M = \frac{\text{moles of solute}}{\text{liters of solution}}$

Example Calculation: If 33.2 g of ammonium carbonate is dissolved in 395 g of water, the percent by mass is:

• Total mass of solution = 33.2 g + 395 g = 428.2 g

• Percent by mass = $(\frac{33.2}{428.2}) \times 100\% = 7.75\%$

Calculating Mass of Solute in Solution

To find the mass of solute in a solution with a known percent by mass:

• Mass of solute = $(\text{percent by mass} \times \text{mass of solution}) / 100$

Example: A solution contains 2.06% NaCl by mass. In 6.57 g of solution, mass of NaCl = $(2.06 \times 6.57) / 100 = 0.135$ g.

Preparation of Solutions

To prepare a solution of a specific concentration, calculate the required mass of solute and volume of solvent.

• For percent by mass: $\text{mass of solute} = \text{percent by mass} \times \text{total mass of solution} / 100$

• For molarity: $\text{moles of solute} = M \times \text{volume of solution (L)}$

Example: To prepare 8.0 g of 8.15% NaCl solution, calculate the mass of water needed:

• Let $x$ = mass of water

• Total mass = $8.0 + x$

• Set up equation: $\frac{8.0}{8.0 + x} \times 100 = 8.15$

• Solve for $x$

Dilution of Solutions

When a solution is diluted, the amount of solute remains constant, but the volume increases, decreasing the concentration.

• Use the dilution equation: $M_1V_1 = M_2V_2$

• $M_1$ = initial molarity, $V_1$ = initial volume

• $M_2$ = final molarity, $V_2$ = final volume

Example: If 1.5 L of 2.08 M KNO3 is diluted by adding 1.5 L of water, the new molarity is:

• Total volume = 1.5 L + 1.5 L = 3.0 L

• $M_2 = \frac{2.08 \times 1.5}{3.0} = 1.04$ M

Neutralization Reactions

Neutralization occurs when an acid reacts with a base to produce water and a salt. The stoichiometry of the reaction is used to calculate the volumes and concentrations required.

• General reaction: $\text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water}$

• Example: $\text{NaOH(aq)} + \text{HCl(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}$

• To find the volume of base needed: $M_1V_1 = M_2V_2$ (where $M_1$ and $V_1$ are for acid, $M_2$ and $V_2$ for base)

Example Calculation: To neutralize 55 mL of 0.0855 M HCl with 0.150 M NaOH:

• Moles of HCl = $0.0855 \times 0.055 = 0.00470$ mol

• Volume of NaOH needed = $\frac{0.00470}{0.150} = 0.0313$ L = 31.3 mL

Periodic Table Reference

The periodic table is a useful tool for identifying elements, their symbols, and atomic masses, which are necessary for solution calculations.

• Atomic mass is used to convert between grams and moles.

• Element symbols are used in chemical equations and formulas.

Summary Table: Solution Concentration Calculations

Type of Calculation	Formula	Example
Percent by mass	$\frac{\text{mass of solute}}{\text{mass of solution}} \times 100\%$	7.75% for 33.2 g solute in 428.2 g solution
Molarity	$\frac{\text{moles of solute}}{\text{liters of solution}}$	2.58 M for 0.129 mol solute in 0.05 L solution
Dilution	$M_1V_1 = M_2V_2$	1.04 M after diluting 1.5 L of 2.08 M solution to 3.0 L
Neutralization	$M_1V_1 = M_2V_2$	31.3 mL of 0.150 M NaOH to neutralize 55 mL of 0.0855 M HCl

Additional info: These notes expand upon quiz questions and provide academic context for solution chemistry, including definitions, formulas, and example calculations relevant to Chapter 13 of an introductory chemistry course.