Lesson Notes By Weeks and Term v5 - Grade 12
Chemical Change: acids and bases
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Chemical Change: acids and bases
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Subject: Physical Sciences
Class: Grade 12
Term: 3rd Term
Week: 5
Theme: General lesson support
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Acids and bases are fundamental chemical substances that play crucial roles in our daily lives. From the digestion of food in our stomachs (hydrochloric acid) to the cleaning products we use (often bases), understanding their properties and reactions is essential. In South Africa, the chemical industry relies heavily on the production and use of acids and bases for various purposes, including mining, agriculture (fertilizers), and manufacturing. Improper handling and disposal of these substances can lead to serious environmental problems such as acid mine drainage, which is a significant concern in many parts of the country.
2. 1.
Definitions of Acids and Bases: Arrhenius Definition: Acid:* A substance that produces hydrogen ions (H⁺) in water.
Example: HCl(aq) → H⁺(aq) + Cl⁻(aq)
Base:* A substance that produces hydroxide ions (OH⁻) in water.
Example: NaOH(aq) → Na⁺(aq) + OH⁻(aq)
Limitation:* This definition is limited to aqueous solutions and only considers substances that directly produce H⁺ or OH⁻.
Bronsted-Lowry Definition: Acid:* A proton (H⁺) donor.
Base:* A proton (H⁺) acceptor.
Advantages:* This definition is broader than the Arrhenius definition. It's not limited to aqueous solutions and includes substances that donate or accept protons without necessarily producing H⁺ or OH⁻ directly.
Example:* In the reaction NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq), NH₃ is the base (accepts a proton from H₂O), and H₂O is the acid (donates a proton to NH₃). NH₄⁺ is the conjugate acid of NH₃, and OH⁻ is the conjugate base of H₂
O. Lewis Definition: Acid:* An electron pair acceptor.
Base:* An electron pair donor.
Advantages:* The broadest definition. Includes substances that don't even have hydrogen.
Example:* BF₃ + NH₃ → BF₃NH₃. BF₃ is the Lewis acid (accepts electron pair from N), and NH₃ is the Lewis base (donates electron pair). 2.
2. Strong and Weak Acids and Bases: Strong Acids:* Completely dissociate (ionize) in water.
Examples: HCl, H₂SO₄, HNO₃.
Strong Bases:* Completely dissociate (ionize) in water.
Examples: NaOH, KOH, Ca(OH)₂.
Weak Acids:* Partially dissociate (ionize) in water.
Examples: CH₃COOH (acetic acid), H₂CO₃ (carbonic acid).
Weak Bases:* Partially dissociate (ionize) in water.
Examples: NH₃ (ammonia). 2.3. pH and pOH: pH is a measure of the acidity or alkalinity of a solution. pH = -log₁₀[H⁺], where [H⁺] is the concentration of hydrogen ions in mol/L. pOH = -log₁₀[OH⁻], where [OH⁻] is the concentration of hydroxide ions in mol/
L. In water at 25°C, pH + pOH = 14. pH 7 indicates a basic solution. 2.
4. Acid-Base Reactions: Neutralization and Titration: Neutralization:* The reaction between an acid and a base, producing a salt and water (in most cases).
Example: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
Titration:* A technique used to determine the concentration of an acid or base by reacting it with a solution of known concentration (the standard solution).
Equivalence point:* The point in the titration where the acid and base have completely reacted.
End point:* The point in the titration where the indicator changes color. Ideally, the end point should be as close as possible to the equivalence point.
Indicators:* Substances that change color depending on the pH of the solution.
Examples: phenolphthalein, methyl orange. 2.
5. Titration Calculations: The general formula for titration calculations is: n(acid)/coefficient of acid = n(base)/coefficient of base, where n represents the number of moles. Remember that n = CV, where C is the concentration (mol/L) and V is the volume (L).
Example 1: Calculate the pH of a 0.01 M solution of HCl.
HCl is a strong acid, so it completely dissociates: HCl(aq) → H⁺(aq) + Cl⁻(aq).
Therefore, [H⁺] = 0.01 M.
pH = -log₁₀[0.01] = -log₁₀(10⁻²) =
2. Example 2: Calculate the pH of a 0.005 M solution of NaO
H.
NaOH is a strong base, so it completely dissociates: NaOH(aq) → Na⁺(aq) + OH⁻(aq).
Therefore, [OH⁻] = 0.005 M.
pOH = -log₁₀[0.005] = -log₁₀(5 x 10⁻³) ≈ 2.3
pH = 14 - pOH = 14 - 2.3 = 11.7
Example 3: 25.0 cm³ of a hydrochloric acid solution of unknown concentration is titrated against a 0.10 M sodium hydroxide solution. The endpoint is reached when 20.0 cm³ of the sodium hydroxide solution has been added. Calculate the concentration of the hydrochloric acid solution.
Reaction: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
n(HCl)/1 = n(NaOH)/1
C(HCl) x V(HCl) = C(NaOH) x V(NaOH)
C(HCl) x 0.025 L = 0.10 M x 0.020 L
C(HCl) = (0.10 M x 0.020 L) / 0.025 L = 0.08 M
Guided Practice (With Solutions)
Question 1: Define an acid according to the Bronsted-Lowry definition. Provide an example of a Bronsted-Lowry acid and identify its conjugate base.
Solution: According to the Bronsted-Lowry definition, an acid is a proton (H⁺) donor. An example of a Bronsted-Lowry acid is H₂SO₄ (sulfuric acid). Its conjugate base is HSO₄⁻ (hydrogen sulfate ion), formed after H₂SO₄ donates a proton.
Question 2: Calculate the pH of a 0.001 M solution of nitric acid (HNO₃).