Lesson Notes By Weeks and Term v5 - Grade 10

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Performance objectives

• Define aqueous solution, solute, solvent, and electrolyte.
• Write balanced chemical equations for reactions in aqueous solutions.
• Perform stoichiometric calculations involving concentration and volume.
• Predict the formation of a precipitate and identify spectator ions.
• Apply stoichiometry to problems in a South African context.

Lesson summary

Reactions in aqueous solutions are fundamental to understanding chemistry and how it impacts our daily lives. An aqueous solution is simply a solution where water is the solvent. Many important chemical reactions, both natural and industrial, occur in water. From the chemical reactions that allow plants to grow in our gardens to the treatment of polluted water sources, understanding reactions in aqueous solutions and stoichiometry is crucial. Stoichiometry, which deals with the quantitative relationships between reactants and products in a chemical reaction, allows us to predict how much of a substance we need to react with another substance and how much product we will obtain.

Lesson notes

2.1 Aqueous Solutions: The Watery World of Chemistry An aqueous solution is a solution in which water is the solvent. The solvent is the substance that dissolves the solute, which is the substance being dissolved.

Solvent: Water (H₂O)

Solute: Can be ionic compounds (like NaCl), covalent compounds (like sugar), or other substances. Electrolytes are substances that, when dissolved in water, produce ions and can conduct electricity. Strong electrolytes dissociate completely into ions, while weak electrolytes only partially dissociate. Non-electrolytes do not form ions and do not conduct electricity.

Strong Electrolytes: NaCl (table salt), HCl (hydrochloric acid), NaOH (sodium hydroxide)

Weak Electrolytes: CH₃COOH (acetic acid – found in vinegar)

Non-Electrolytes: C₁₂H₂₂O₁₁ (sucrose – table sugar), ethanol (alcohol) 2.2 Types of Reactions in Aqueous Solutions Precipitation Reactions: These reactions involve the formation of an insoluble solid (precipitate) when two aqueous solutions are mixed. The ability to predict precipitation relies on understanding solubility rules.

Solubility Rules (Simplified): Most nitrate (NO₃⁻) salts are soluble. Most sodium (Na⁺), potassium (K⁺), and ammonium (NH₄⁺) salts are soluble. Most chloride (Cl⁻), bromide (Br⁻), and iodide (I⁻) salts are soluble, except those of silver (Ag⁺), lead (Pb²⁺), and mercury (Hg₂²⁺). Most sulfate (SO₄²⁻) salts are soluble, except those of barium (Ba²⁺), strontium (Sr²⁺), lead (Pb²⁺), calcium (Ca²⁺), and silver (Ag⁺). Most hydroxide (OH⁻) and sulfide (S²⁻) salts are insoluble, except those of sodium (Na⁺), potassium (K⁺), and ammonium (NH₄⁺). Calcium, strontium, and barium hydroxides are moderately soluble.

Acid-Base Neutralization Reactions: These reactions involve the reaction of an acid and a base to form a salt and water.

Acid: A substance that donates a proton (H⁺).

Base: A substance that accepts a proton (H⁺).

Example: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) Redox Reactions (Oxidation-Reduction Reactions): These reactions involve the transfer of electrons between reactants.

Oxidation: Loss of electrons.

Reduction: Gain of electrons.

Example: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) (Zinc is oxidized, Copper is reduced) 2.3 Writing Balanced Chemical Equations Balancing chemical equations is crucial for stoichiometry. The number of atoms of each element must be the same on both sides of the equation. State symbols indicate the physical state of each substance: (s) – solid (l) – liquid (g) – gas (aq) – aqueous solution

Example: Unbalanced: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

Balanced: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq) (This equation is already balanced!) 2.4 Stoichiometry: The Quantitative Study of Chemical Reactions Stoichiometry involves using mole ratios from balanced chemical equations to calculate the amounts of reactants and products.

Key Concepts: Mole (mol): The SI unit for the amount of substance. 1 mole = 6.022 x 10²³ particles (Avogadro's number).

Molar Mass (M): The mass of one mole of a substance (g/mol). Calculated by adding the atomic masses of all the atoms in the chemical formula.

Concentration (c): The amount of solute dissolved in a given volume of solution (mol/dm³ or M).

Volume (V): Usually measured in dm³ (liters). n = m/M: Number of moles (n) = mass (m) / molar mass (M) c = n/V: Concentration (c) = number of moles (n) / volume (V)

Worked example

Example 1: Precipitation Reaction and Stoichiometry

Problem: When 20.0 cm³ of a 0.100 mol/dm³ solution of lead(II) nitrate (Pb(NO₃)₂) is mixed with 30.0 cm³ of a 0.200 mol/dm³ solution of potassium iodide (KI), a yellow precipitate of lead(II) iodide (PbI₂) forms. Calculate the mass of PbI₂ formed.

Solution:

Write the balanced chemical equation:

Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq)

Calculate the number of moles of each reactant:

n(Pb(NO₃)₂) = cV = (0.100 mol/dm³)(20.0 cm³)(1 dm³/1000 cm³) = 0.00200 mol

n(KI) = cV = (0.200 mol/dm³)(30.0 cm³)(1 dm³/1000 cm³) = 0.00600 mol

Determine the limiting reactant: