Lesson Notes By Weeks and Term v5 - Grade 10
Reactions in aqueous solution and stoichiometry – Week 6 focus
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Reactions in aqueous solution and stoichiometry – Week 6 focus
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Subject: Physical Sciences
Class: Grade 10
Term: 2nd Term
Week: 6
Theme: General lesson support
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Reactions in aqueous solutions are fundamental to many natural processes and industrial applications. Understanding how substances react when dissolved in water (aqueous solutions) and the quantitative relationships (stoichiometry) governing these reactions is crucial. In South Africa, this knowledge is vital for understanding water purification processes, managing industrial waste, analyzing soil composition for agriculture, and addressing issues related to acid mine drainage. Stoichiometry allows us to predict the amounts of reactants and products involved in chemical reactions, enabling us to optimize resource use and minimize waste.
Aqueous Solutions: An aqueous solution is a solution in which the solvent is water. Water is an excellent solvent for many ionic and polar covalent compounds due to its polarity.
Electrolytes and Non-Electrolytes: Electrolyte: A substance that dissolves in water to produce ions and conduct electricity.
Non-Electrolyte: A substance that dissolves in water but does not produce ions and therefore does not conduct electricity.
Strong Electrolyte: An electrolyte that completely dissociates (ionizes) into ions in water. Examples include strong acids (e.g., HCl, H 2 SO 4 , HNO 3 ), strong bases (e.g., NaOH, KOH), and soluble ionic compounds (e.g., NaCl, KNO 3 ).
Weak Electrolyte: An electrolyte that only partially dissociates into ions in water. Examples include weak acids (e.g., CH 3 COOH (acetic acid), H 2 CO 3 (carbonic acid)) and weak bases (e.g., NH 3 (ammonia)).
Types of Reactions in Aqueous Solutions: Precipitation Reactions: Reactions in which two soluble ionic compounds react to form an insoluble ionic compound (a precipitate). To predict whether a precipitate will form, we use solubility rules (these will be given in tests/exams).
Acid-Base Neutralization Reactions: Reactions between an acid and a base. In aqueous solutions, these reactions typically produce a salt and water.
Redox Reactions: Reactions involving the transfer of electrons. These can be identified by changes in oxidation states of the reacting species. (Will be covered in more detail later but introduce the concept).
Writing Balanced Chemical Equations: Molecular Equation: Shows the complete chemical formulas of all reactants and products.
Ionic Equation: Shows all soluble ionic compounds as separate ions in the solution.
Net Ionic Equation: Shows only the ions that are directly involved in the reaction (i.e., the ions that form a precipitate, water, or a gas). Spectator ions (ions that do not participate in the reaction) are omitted.
Molarity (Concentration): Molarity (M) is defined as the number of moles of solute per liter (or cubic decimeter) of solution. M = moles of solute / volume of solution (in dm 3 ) n = MV where n is moles, M is molarity, and V is volume in dm 3 .
Stoichiometry of Reactions in Solution: Stoichiometry is the study of the quantitative relationships between reactants and products in chemical reactions. Use the balanced chemical equation to determine the mole ratio between reactants and products. Use molarity and volume to calculate the moles of reactants in solution. Use the mole ratio to calculate the moles of product formed or reactant required. Convert moles of product to mass (using molar mass) or volume (using molarity).
Example 1: Precipitation Reaction
Write the balanced molecular, ionic, and net ionic equations for the reaction between aqueous solutions of lead(II) nitrate, Pb(NO 3 ) 2 (aq), and potassium iodide, KI(aq).
Solution:
Molecular Equation: Pb(NO 3 ) 2 (aq) + 2KI(aq) → PbI 2 (s) + 2KNO 3 (aq)
Ionic Equation: Pb 2+ (aq) + 2NO 3 - (aq) + 2K + (aq) + 2I - (aq) → PbI 2 (s) + 2K + (aq) + 2NO 3 - (aq)
Net Ionic Equation: Pb 2+ (aq) + 2I - (aq) → PbI 2 (s)
Explanation: Lead(II) iodide (PbI 2 ) is insoluble, forming a precipitate. Potassium ions (K + ) and nitrate ions (NO 3 - ) are spectator ions.
Example 2: Acid-Base Neutralization Reaction
Write the balanced molecular and net ionic equations for the reaction between hydrochloric acid, HCl(aq), and sodium hydroxide, NaOH(aq).
Solution:
Molecular Equation: HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l)
Ionic Equation: H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq) → Na + (aq) + Cl - (aq) + H 2 O(l)
Net Ionic Equation: H + (aq) + OH - (aq) → H 2 O(l)
Explanation: Sodium ions (Na + ) and chloride ions (Cl - ) are spectator ions. The net ionic equation represents the fundamental neutralization reaction.
Example 3: Stoichiometry Calculation
What mass of silver chloride (AgCl) will precipitate when 100.0 cm 3 of 0.200 mol/dm 3 silver nitrate (AgNO 3 ) solution is mixed with excess sodium chloride (NaCl) solution?
Solution:
Balanced Equation: AgNO 3 (aq) + NaCl(aq) → AgCl(s) + NaNO 3 (aq)
Step 1: Calculate moles of AgNO 3 :
Volume of AgNO 3 = 100.0 cm 3 = 0.100 dm 3
Molarity of AgNO 3 = 0.200 mol/dm 3
Moles of AgNO 3 = M × V = 0.200 mol/dm 3 × 0.100 dm 3 = 0.0200 mol
Step 2: Determine moles of AgCl formed:
From the balanced equation, 1 mole of AgNO 3 produces 1 mole of AgCl.
Therefore, moles of AgCl = 0.0200 mol
Step 3: Calculate mass of AgCl:
Molar mass of AgCl = 107.87 g/mol (Ag) + 35.45 g/mol (Cl) = 143.32 g/mol
Mass of AgCl = moles × molar mass = 0.0200 mol × 143.32 g/mol = 2.87 g
Answer: 2.87 g of silver chloride will precipitate.
Example 4: Calculating concentration after a reaction
25 cm 3 of 0.1 mol/dm 3 H 2 SO 4 is reacted with excess Zinc. What will the concentration of the remaining solution be if the final volume is 50 cm 3 ?
Solution:
Balanced Equation: H 2 SO 4 (aq) + Zn(s) -> ZnSO 4 (aq) + H 2 (g)
Step 1: Calculate moles of H 2 SO 4 that reacted:
Volume of H 2 SO 4 = 25 cm 3 = 0.025 dm 3
Molarity of H 2 SO 4 = 0.1 mol/dm 3
Moles of H 2 SO 4 = M × V = 0.1 mol/dm 3 × 0.025 dm 3 = 0.0025 mol
Step 2: Determine moles of H + that reacted:
H 2 SO 4 is a strong acid and completely dissociates in solution producing 2H + ions for every 1 H 2 SO 4 molecule.
Therefore, moles of H + reacted = 2 x 0.0025 mol = 0.005 mol
Step 3: Since we start off with a strong acid, nearly all the sulfate ions will exist in solution.
Therefore, the only way we can calculate the concentration of the remaining solution is by determining the concentration of the sulfate ions. This is assuming that the excess zinc doesn't react any further or introduce any other ions into the solution.
Step 4: The concentration of the sulfate ions will be equal to the initial concentration of the acid solution and the moles of sulfate ions present will be equal to the moles of the acid.
Therefore we can determine the new concentration by using the total volume of the final solution (50 cm 3 ).
Step 5: Calculate the concentration of the resulting solution.
Concentration = moles / Volume = 0.0025 mol / 0.050 dm 3 = 0.05 mol/dm 3
Answer: 0.05 mol/dm 3 .
Guided Practice (With Solutions)
Question 1:
Identify the spectator ions in the reaction between aqueous solutions of sodium chloride (NaCl) and silver nitrate (AgNO 3 ).
Solution:
Molecular Equation: NaCl(aq) + AgNO 3 (aq) → AgCl(s) + NaNO 3 (aq)
Ionic Equation: Na + (aq) + Cl - (aq) + Ag + (aq) + NO 3 - (aq) → AgCl(s) + Na + (aq) + NO 3 - (aq)
Net Ionic Equation: Ag + (aq) + Cl - (aq) → AgCl(s)
Spectator Ions: Na + (aq) and NO 3 - (aq)
Explanation: Sodium and nitrate ions remain in solution unchanged throughout the reaction.
Question 2:
What volume of 0.150 mol/dm 3 hydrochloric acid (HCl) is required to neutralize 25.0 cm 3 of 0.200 mol/dm 3 sodium hydroxide (NaOH)?
Solution:
Balanced Equation: HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l)
Step 1: Calculate moles of NaOH:
Volume of NaOH = 25.0 cm 3 = 0.0250 dm 3
Molarity of NaOH = 0.200 mol/dm 3
Moles of NaOH = M × V = 0.200 mol/dm 3 × 0.0250 dm 3 = 0.00500 mol
Step 2: Determine moles of HCl required:
From the balanced equation, 1 mole of HCl reacts with 1 mole of NaOH.
Therefore, moles of HCl = 0.00500 mol
Step 3: Calculate volume of HCl:
Molarity of HCl = 0.150 mol/dm 3
Volume of HCl = moles / M = 0.00500 mol / 0.150 mol/dm 3 = 0.0333 dm 3
Volume of HCl = 0.0333 dm 3 × 1000 cm 3 /dm 3 = 33.3 cm 3
Answer: 33.3 cm 3 of HCl is required.
Question 3:
A solution is prepared by dissolving 5.85 g of sodium chloride (NaCl) in enough water to make 500.0 cm 3 of solution. Calculate the molarity of the solution.
Solution:
Step 1: Calculate moles of NaCl:
Molar mass of NaCl = 22.99 g/mol (Na) + 35.45 g/mol (Cl) = 58.44 g/mol
Moles of NaCl = mass / molar mass = 5.85 g / 58.44 g/mol = 0.100 mol
Step 2: Calculate molarity of the solution:
Volume of solution = 500.0 cm 3 = 0.500 dm 3
Molarity of NaCl = moles / volume = 0.100 mol / 0.500 dm 3 = 0.200 mol/dm 3
Answer: The molarity of the solution is 0.200 mol/dm 3 .
Independent Practice (Questions Only)
Question 1: Classify the following substances as strong electrolytes, weak electrolytes, or non-electrolytes: sugar (C 12 H 22 O 11 ), potassium chloride (KCl), acetic acid (CH 3 COOH), and ethanol (C 2 H 5 OH).
Question 2: Write the balanced net ionic equation for the reaction between aqueous solutions of barium chloride (BaCl 2 ) and sodium sulfate (Na 2 SO 4 ).
Question 3: What mass of lead(II) chromate (PbCrO 4 ) will precipitate when 50.0 cm 3 of 0.100 mol/dm 3 lead(II) nitrate (Pb(NO 3 ) 2 ) solution is mixed with 50.0 cm 3 of 0.100 mol/dm 3 potassium chromate (K 2 CrO 4 ) solution?
Question 4: Calculate the concentration of chloride ions (Cl - ) in a solution prepared by dissolving 11.1 g of calcium chloride (CaCl 2 ) in enough water to make 250.0 cm 3 of solution.
Question 5: What volume of 0.500 mol/dm 3 sulfuric acid (H 2 SO 4 ) is required to completely neutralize 40.0 cm 3 of 0.250 mol/dm 3 potassium hydroxide (KOH) solution?
Question 6: 30 cm 3 of 0.2 mol/dm 3 BaCl 2 is mixed with 50 cm 3 of 0.1 mol/dm 3 Na 2 SO 4 . What mass of precipitate will form?
Question 7: A 50 cm 3 sample of hydrochloric acid is titrated with a 0.1 mol/dm 3 solution of NaOH. 45 cm 3 of NaOH is required to neutralize the solution. Calculate the concentration of the hydrochloric acid.
Question 8: If 200 cm 3 of water is added to 50 cm 3 of 1.0 mol/dm 3 NaCl solution, what is the concentration of the new solution?