Lesson Notes By Weeks and Term v5 - Grade 10

Lesson Video

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Performance objectives

• Define aqueous solutions, electrolytes, and concentration.
• Write balanced chemical equations with state symbols.
• Perform stoichiometric calculations using molarity.
• Identify precipitation and acid-base reactions.
• Apply the concept of limiting reactants in calculations.

Lesson summary

Reactions in aqueous solutions are fundamental to many chemical processes occurring all around us, from the digestion of food in our bodies to the purification of water we drink. In South Africa, understanding these reactions is especially important for addressing challenges related to water quality, mining effluent treatment, and agricultural practices. Many industrial processes, vital to the South African economy, rely on aqueous solutions and the precise control of chemical reactions within them. Stoichiometry, the quantitative relationship between reactants and products in a chemical reaction, provides the tools to predict and control these processes efficiently.

Lesson notes

2.1 Aqueous Solutions An aqueous solution is a solution in which the solvent is water. Water is an excellent solvent because it is a polar molecule. Its polarity allows it to interact with and dissolve many ionic and polar compounds.

Solute: The substance being dissolved in the solvent.

Examples: Salt (NaCl), Sugar (C 12 H 22 O 11 )

Solvent: The substance doing the dissolving. In an aqueous solution, the solvent is always water (H 2 O).

Electrolyte: A substance that, when dissolved in water, conducts electricity. Electrolytes contain ions that are free to move and carry an electrical charge.

Strong Electrolytes: These substances completely dissociate into ions when dissolved in water.

Examples: Strong acids (HCl, H 2 SO 4 ), strong bases (NaOH, KOH), and soluble ionic compounds (NaCl, KNO 3 ).

Weak Electrolytes: These substances only partially dissociate into ions when dissolved in water.

Examples: Weak acids (CH 3 COOH – acetic acid), weak bases (NH 3 – ammonia).

Non-electrolytes: Substances that do not form ions when dissolved in water and therefore do not conduct electricity.

Examples: Sugar (C 12 H 22 O 11 ), ethanol (C 2 H 5 OH). 2.2 Chemical Equations with State Symbols Chemical equations represent chemical reactions. It's crucial to include state symbols to indicate the physical state of each substance involved. (s): Solid (l): Liquid (g): Gas (aq): Aqueous (dissolved in water)

Example: ``` AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq) ``` This equation shows that silver nitrate (AgNO 3 ) in aqueous solution reacts with sodium chloride (NaCl) in aqueous solution to form solid silver chloride (AgCl) and sodium nitrate (NaNO 3 ) in aqueous solution. Note the equation is balanced, meaning there are the same number of each type of atom on both sides of the arrow. 2.3 Molarity (Concentration) Molarity (M) is a measure of the concentration of a solution. It is defined as the number of moles of solute per liter of solution. ``` Molarity (M) = moles of solute / liters of solution M = n / V ``` Where: M = Molarity (mol/L or M) n = number of moles (mol) V = Volume of the solution (L)

Example 1: What is the molarity of a solution prepared by dissolving 10.0 g of NaCl in enough water to make 500.0 mL of solution?

Calculate the number of moles of NaCl: Molar mass of NaCl = 22.99 g/mol (Na) + 35.45 g/mol (Cl) = 58.44 g/mol moles of NaCl (n) = mass / molar mass = 10.0 g / 58.44 g/mol = 0.171 mol Convert volume to liters: Volume (V) = 500.0 mL = 0.5000 L Calculate Molarity: Molarity (M) = n / V = 0.171 mol / 0.5000 L = 0.342 M Therefore, the molarity of the solution is 0.342 M. 2.4 Stoichiometry in Aqueous Solutions Stoichiometry allows us to determine the amount of reactants and products involved in a chemical reaction. When dealing with aqueous solutions, we often use molarity to relate volume and number of moles.

Example 2: What volume of 0.200 M NaOH solution is needed to completely react with 25.0 mL of 0.150 M H 2 SO 4 solution?

Write the balanced chemical equation: ``` H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l) ``` Determine the number of moles of H 2 SO 4 : moles of H 2 SO 4 (n) = Molarity x Volume = 0.150 mol/L x 0.0250 L = 0.00375 mol Use the stoichiometry of the balanced equation to find the moles of NaOH required: From the balanced equation, 1 mole of H 2 SO 4 reacts with 2 moles of NaOH. moles of NaOH = 2 x moles of H 2 SO 4 = 2 x 0.00375 mol = 0.00750 mol Calculate the volume of 0.200 M NaOH solution required: Volume of NaOH (V) = moles / Molarity = 0.00750 mol / 0.200 mol/L = 0.0375 L Convert volume to milliliters (if required): Volume of NaOH (V) = 0.0375 L x 1000 mL/L = 37.5 mL Therefore, 37.5 mL of 0.200 M NaOH solution is needed. 2.5 Precipitation Reactions A precipitation reaction is a reaction in which two aqueous solutions are mixed, and an insoluble solid (the precipitate) forms.

Example 3: Predict what will happen when aqueous solutions of lead(II) nitrate, Pb(NO₃)₂, and potassium iodide, KI, are mixed.

Write the possible products: PbI₂ and KNO₃ Determine the solubility of the products using solubility rules: PbI₂ is insoluble (most iodide compounds are soluble, except those of Pb 2+ , Ag + , and Hg 2 2+ ) KNO₃ is soluble (all nitrate compounds are soluble). Write the balanced chemical equation with state symbols: ``` Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq) ``` A yellow precipitate of lead(II) iodide (PbI₂) will form. 2.6 Acid-Base Neutralization Reactions An acid-base neutralization reaction is a reaction between an acid and a base. In aqueous solutions, these reactions typically produce a salt and water.

Example 4: Hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH). Write the balanced chemical equation for this reaction. ``` HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) ``` 2.7 Limiting Reactants The limiting reactant is the reactant that is completely consumed in a chemical reaction. It determines the amount of product that can be formed.