Solution Properties and Calculations

Overview

This lecture covers the fundamental properties of solutions, focusing on calculating solution concentrations using molarity and performing dilution calculations with the dilution equation.

Properties of Solutions

Solutions are homogeneous mixtures with uniform composition and properties throughout.
The component present in the largest amount is the solvent; all other components are solutes.
Aqueous solutions use water as the solvent.
Concentration refers to the relative amount of solute in a solution, described as dilute (low concentration) or concentrated (high concentration).

Molarity and Calculations

Molarity (M) is the amount of solute (in moles) per liter of solution: M = mol solute / L solution.
To find molarity, convert solution volume to liters as needed.
Moles of solute can be calculated using molarity: mol = M × L.
Given mass of solute, convert mass to moles using molar mass before finding molarity.
To find the mass of solute in a solution, use: mass = (M × L) × molar mass.
For greater accuracy, retain guard digits in intermediate calculations or perform all steps at once to avoid rounding errors.

Example Problems & Solutions

Example 1: Calculated molarity of soft drink by dividing moles of sucrose by solution volume in liters (0.375 M).
Example 2: Found moles of sugar in sip by multiplying molarity by volume in liters (0.004 mol).
Example 3: Calculated molarity of vinegar (acetic acid) by converting 25.2 g to moles before dividing by solution volume (0.839 M).
Example 4: Found grams of NaCl in solution by first calculating moles, then multiplying by NaCl’s molar mass (77.4 g).
Example 5: Determined required solution volume for given solute mass by converting grams to moles, then dividing by molarity (1.50 L).

Dilution and the Dilution Equation

Dilution reduces solution concentration by adding solvent but leaves moles of solute unchanged.
The dilution equation: C₁V₁ = C₂V₂, where C is concentration and V is volume before (1) and after (2) dilution.
Units of concentration and volume must be consistent for the equation to work.
Example applications include preparing laboratory reagents or diluting stock solutions for use.

Example Dilution Problems

Example: Diluting 0.850 L of 5.00-M Cu(NO₃)₂ to 1.80 L yields a 2.36 M solution.
Example: 11 mL of 0.45 M HBr diluted to 0.041 L (41 mL) makes a 0.12 M solution.
Example: To prepare 5.00 L of 0.100 M KOH from 1.59 M KOH stock, use 0.314 L of stock solution.

Key Terms & Definitions

Solution — Homogeneous mixture with uniform composition throughout.
Solvent — Component in the greatest amount; dissolves the solute.
Solute — Component dissolved in solvent, present in lesser amounts.
Aqueous Solution — Solution where water is the solvent.
Molarity (M) — Moles of solute per liter of solution (mol/L).
Dilution — Process of reducing solution concentration by adding solvent.
Stock Solution — A concentrated solution that can be diluted for use.

Action Items / Next Steps

Practice molarity and dilution problems for various solutes and solvents.
Review definitions and calculation steps for molarity and dilution.
Complete homework problems as assigned in the textbook or by the instructor.