Chapter 5: Introduction to Solutions and Aqueous Reactions – Study Notes

Chapter 5: Introduction to Solutions and Aqueous Reactions

Overview

This chapter introduces the fundamental concepts of solutions, their properties, and the calculations involved in preparing and analyzing solutions in the context of aqueous reactions. It covers terminology, molarity, dilution, stoichiometry of solution reactions, and the behavior of ions in solution, including electrolytes, acids, and bases.

Calculations Involving Solutions

Terminology

• Solution: A homogeneous mixture of two or more substances.

• Solvent: The major component of the mixture; the substance in which the solute is dissolved.

• Solute: The minor component of the mixture; the substance that is dissolved.

• Dilute solution: A solution with a low concentration of solute.

• Concentrated solution: A solution with a high concentration of solute.

A. Molar Concentration

Molarity (M) is a common way to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution.

• Formula:

$\text{Molarity (M)} = \frac{\text{amount of solute (mol)}}{\text{volume of solution (L)}}$

• Preparation of a Solution: To prepare a solution of specified molarity, dissolve the calculated amount of solute in a volumetric flask and add solvent up to the desired volume.

• Example: To prepare 1.00 L of 1.00 M NaCl, dissolve 58.44 g NaCl in water and dilute to 1.00 L.

B. Using Molarity as a Conversion Factor

Molarity can be used to convert between moles of solute and volume of solution.

• Conversion Factor: Use the molarity to relate moles and liters.

• Example: How many liters of 0.125 M NaOH solution contain 0.255 mol NaOH?

$\text{Volume} = \frac{\text{moles of solute}}{\text{molarity}} = \frac{0.255\ \text{mol}}{0.125\ \text{mol/L}} = 2.04\ \text{L}$

C. Diluting Solutions

Stock solutions are often prepared as concentrated solutions and then diluted to lower concentrations as needed. When a solution is diluted, the number of moles of solute remains constant; only the volume changes.

• Formula for Dilution:

$M_1V_1 = M_2V_2$

• Example: To prepare 500 mL of 0.100 M HNO3 from a 15.9 M stock, use the formula above to calculate the required volume of stock solution.

Stoichiometry of Solution Reactions

Conceptual Plan

Stoichiometry in solution reactions involves using molarity and volume to determine the amounts of reactants and products. The general approach is:

• Convert volume of solution to moles using molarity.

• Use stoichiometric coefficients from the balanced chemical equation to relate moles of reactants and products.

• Convert moles back to volume or mass as needed.

Example 1

Reaction: $2\text{KCl}(aq) + \text{Pb(NO}_3)_2(aq) \rightarrow \text{PbCl}_2(s) + 2\text{KNO}_3(aq)$

• Given: 0.150 L of 0.175 M Pb(NO3)2

• Find: Volume of 0.150 M KCl needed and mass of PbCl2 produced.

• Steps:

  1. Calculate moles of Pb(NO3)2: $0.150\ \text{L} \times 0.175\ \text{mol/L} = 0.02625\ \text{mol}$

  2. Use stoichiometry: $2\ \text{mol KCl} : 1\ \text{mol Pb(NO}_3)_2$

  3. Calculate moles of KCl needed: $0.02625\ \text{mol} \times 2 = 0.0525\ \text{mol}$

  4. Calculate volume of KCl solution: $0.0525\ \text{mol} / 0.150\ \text{mol/L} = 0.350\ \text{L}$

  5. Calculate mass of PbCl2 produced: $0.02625\ \text{mol} \times 278.1\ \text{g/mol} = 7.30\ \text{g}$

Example 2

Reaction: $2\text{KOH}(aq) + \text{NiSO}_4(aq) \rightarrow \text{Ni(OH)}_2(s) + \text{K}_2\text{SO}_4(aq)$

• Given: 100.0 mL of 0.200 M KOH and 200.0 mL of 0.175 M NiSO4

• Find: Mass of Ni(OH)2 produced.

• Steps:

  1. Calculate moles of KOH: $0.100\ \text{L} \times 0.200\ \text{mol/L} = 0.0200\ \text{mol}$

  2. Calculate moles of NiSO4: $0.200\ \text{L} \times 0.175\ \text{mol/L} = 0.0350\ \text{mol}$

  3. Determine limiting reagent using stoichiometry.

  4. Calculate theoretical yield of Ni(OH)2 based on limiting reagent.

Ions in Solution and Ionic Equations

Electrolytes

Electrolytes are substances that dissolve in water to form ions, allowing the solution to conduct electricity.

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

• Weak electrolytes: Dissociate partially into ions (e.g., weak acids and bases).

• Nonelectrolytes: Dissolve in water but do not form ions (e.g., sugar).

Ionization and Dissociation

• Dissociation: When ionic compounds dissolve, their ions separate and become surrounded by water molecules.

• Example: $\text{Na}_2\text{SO}_4(aq) \rightarrow 2\text{Na}^+(aq) + \text{SO}_4^{2-}(aq)$

• Polyatomic ions remain intact during dissociation.

Acids and Bases

• Acids: Taste sour, turn blue litmus red, and produce H+ ions in aqueous solution.

• Bases: Taste bitter, feel slippery, turn red litmus blue, and produce OH- ions in aqueous solution.

• Arrhenius definition: Acids produce H+ ions; bases produce OH- ions in water.

• Polyprotic acids: Acids that have more than one ionizable proton.

• Strong acids: Ionize completely in water (e.g., HCl, HBr, HI, HNO3, HClO3).

• Weak acids: Ionize partially in water (e.g., acetic acid).

• Strong bases: Ionic compounds that dissociate completely to give OH- ions (e.g., NaOH, KOH).

• Weak bases: Molecular compounds that react with water to produce OH- ions (e.g., NH3).

Autoionization of Water:

$\text{H}_2\text{O}(l) \rightleftharpoons \text{H}^+(aq) + \text{OH}^-(aq)$

Water is amphoteric, meaning it can act as both an acid and a base.

Table: Strong Acids and Bases

Strong Acids	Strong Bases
Hydrochloric acid, HCl	Lithium hydroxide, LiOH
Hydrobromic acid, HBr	Sodium hydroxide, NaOH
Hydroiodic acid, HI	Potassium hydroxide, KOH
Nitric acid, HNO3	Barium hydroxide, Ba(OH)2
Chloric acid, HClO3	Calcium hydroxide, Ca(OH)2
Additional info: All strong acids and bases dissociate completely in water.	Strontium hydroxide, Sr(OH)2