Solutions: Properties, Concentrations, and Calculations (Chapter 13 Study Notes)

Solutions and Their Properties

Definition of Solutions

A solution is a homogeneous mixture composed of two or more substances. In a solution, the solute is the substance that is dissolved, and the solvent is the substance that does the dissolving (usually present in greater amount).

• Solvent: The major component of a solution (e.g., water in saltwater).

• Solute: The minor component(s) dissolved in the solvent (e.g., NaCl in saltwater).

• Aqueous solution: A solution in which water is the solvent.

Example: Saltwater is a solution where salt (NaCl) is the solute and water is the solvent.

Types of Solutions

• Strong Electrolyte Solution: Contains solutes that dissociate completely into ions in water (e.g., NaCl(aq)).

• Weak Electrolyte Solution: Contains solutes that partially dissociate into ions (e.g., acetic acid in water).

• Nonelectrolyte Solution: Contains solutes that do not form ions in solution (e.g., sugar in water).

Example: Saltwater is a strong electrolyte solution; sugar water is a nonelectrolyte solution.

Solubility and Solution Formation

Solubility Principles

Solubility is the maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature.

• "Like dissolves like": Polar solutes dissolve in polar solvents; nonpolar solutes dissolve in nonpolar solvents.

• Ionic compounds typically dissolve in polar solvents like water.

Example: NaCl dissolves in water, but not in oil.

Saturated, Unsaturated, and Supersaturated Solutions

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

• Saturated Solution: Contains the maximum amount of dissolved solute at equilibrium.

• Supersaturated Solution: Contains more solute than can theoretically dissolve at a given temperature; unstable and can precipitate excess solute.

Example: If 76 g of sodium acetate (molar mass = 82 g/mol) can dissolve in 100 g water, a solution with 80 g in 100 g water is supersaturated.

Concentration Units

Mass Percent

Mass percent expresses the mass of solute in 100 g of solution.

• Formula: $\text{Mass percent} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\%$

Example: Dissolving 2.45 g of sugar in 200.0 g water: $\text{Mass percent} = \frac{2.45}{2.45 + 200.0} \times 100\% = 1.21\%$

Molarity (M)

Molarity is the number of moles of solute per liter of solution.

• Formula: $M = \frac{\text{moles of solute}}{\text{liters of solution}}$

Example: Dissolving 10.7 g NaCl (molar mass = 58.44 g/mol) in 0.250 L water: $\text{moles NaCl} = \frac{10.7}{58.44} = 0.183$ mol $M = \frac{0.183}{0.250} = 0.732$ M

Calculating Moles from Molarity

• Formula: $\text{moles} = M \times \text{volume (L)}$

Example: How many moles of KOH are in 750. mL of 5.00 M KOH? $\text{moles} = 5.00 \times 0.750 = 3.75$ mol

Solution Stoichiometry and Dilution

Dilution Calculations

When diluting a solution, the amount of solute remains constant before and after dilution.

• Formula: $M_1V_1 = M_2V_2$

Example: What is the final concentration if 25.0 mL of 2.00 M solution is diluted to 0.100 M? $2.00 \times 0.0250 = 0.100 \times V_2$ $V_2 = 0.500$ L

Preparing Solutions

• To prepare a solution of a given molarity, dissolve the calculated mass of solute in enough solvent to reach the desired volume.

Example: To make 50.0 mL of 2.45 M KCl, calculate grams needed: $\text{moles} = 2.45 \times 0.0500 = 0.123$ mol $\text{mass} = 0.123 \times 74.55 = 9.16$ g

Electrolytes and Precipitation Reactions

Strong, Weak, and Nonelectrolytes

• Strong electrolytes: Dissociate completely in water (e.g., NaCl, KBr).

• Weak electrolytes: Partially dissociate (e.g., acetic acid).

• Nonelectrolytes: Do not dissociate (e.g., sugar).

Example: NaCl(aq) is a strong electrolyte; C6H12O6 (glucose) is a nonelectrolyte.

Precipitation Reactions

When two solutions are mixed, an insoluble product (precipitate) may form if the ions combine to form an insoluble compound.

• General reaction: $\text{AB}(aq) + \text{CD}(aq) \rightarrow \text{AD}(s) + \text{CB}(aq)$

Example: $\text{Na}_2\text{SO}_4(aq) + \text{BaCl}_2(aq) \rightarrow \text{BaSO}_4(s) + 2\text{NaCl}(aq)$

Solution Concentration and Ion Calculations

Ion Concentration in Solution

For ionic compounds, the concentration of ions depends on the formula and the molarity of the solution.

• Example: 0.15 M BaCl2 yields 0.15 M Ba2+ and 0.30 M Cl- ions.

Solution	K+ Ion Concentration
1.0 M K2SO4	2.0 M
1.0 M KMnO4	1.0 M
1.0 M K3PO4	3.0 M

Example: 1.0 M K3PO4 yields the highest K+ concentration (3.0 M).

Acid-Base and Neutralization Calculations

Neutralization Reactions

Acids and bases react to form water and a salt. The stoichiometry of the reaction is important for calculations.

• Example reaction: $\text{H}_2\text{SO}_4(aq) + 2\text{NaOH}(aq) \rightarrow \text{Na}_2\text{SO}_4(aq) + 2\text{H}_2\text{O}(l)$

To determine the volume of base needed to neutralize an acid, use the balanced equation and molarity values.

Summary Table: Key Solution Concepts

Concept	Definition/Formula	Example
Mass Percent	$\frac{\text{mass solute}}{\text{mass solution}} \times 100\%$	2.45 g sugar in 202.45 g solution = 1.21%
Molarity (M)	$\frac{\text{moles solute}}{\text{L solution}}$	0.183 mol NaCl in 0.250 L = 0.732 M
Dilution	$M_1V_1 = M_2V_2$	25.0 mL of 2.00 M to 0.100 M: $V_2 = 0.500$ L
Ion Concentration	Multiply molarity by number of ions	0.15 M BaCl2 = 0.30 M Cl-

Additional info: These notes are based on a pre-test covering Chapter 13 (Solutions) in an introductory chemistry course. All calculations and concepts are foundational for understanding solution chemistry, including concentration units, solution preparation, and stoichiometry.