Acids, Bases and Buffers

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your WAEC exam.

Arrhenius Theory:

• Acid: Substance that produces H⁺ ions in aqueous solution
• Base: Substance that produces OH⁻ ions in aqueous solution

Example: $\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-$; $\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-$

Bronsted-Lowry Theory (WAEC Focus):

• Acid: Proton (H⁺) donor
• Base: Proton (H⁺) acceptor
• Conjugate acid-base pair: differ by one proton

Examples:

• $\text{HCl}$ (acid) donates H⁺ → $\text{Cl}^-$ (conjugate base)
• $\text{NH}_3$ (base) accepts H⁺ → $\text{NH}_4^+$ (conjugate acid)

Strong vs Weak Acids/Bases:

Type	Degree of Ionisation	pH Range (0.1M)	Examples
Strong acid	~100%	1-2	HCl, H₂SO₄, HNO₃
Weak acid	Partial	3-5	CH₃COOH, H₂CO₃
Strong base	~100%	12-13	NaOH, KOH
Weak base	Partial	10-11	NH₃, Na₂CO₃

⚡ WAEC Tip: Strong acids fully ionise; weak acids partially ionise. For a 0.1M strong acid like HCl, [H⁺] = 0.1M, so pH = 1. For a 0.1M weak acid like CH₃COOH with Ka = 1.8 × 10⁻⁵, pH is higher (around 3).

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

pH and pOH Calculations:

$$pH = -\log[H^+]$$

$$pOH = -\log[OH^-]$$

$$pH + pOH = 14 \text{ (at 25°C)}$$

pH of Strong Acid/Base: For 0.01M HCl: $[H^+] = 0.01$, so $pH = -\log(10^{-2}) = 2$

For 0.1M NaOH: $[OH^-] = 0.1$, so $pOH = 1$, therefore $pH = 14 - 1 = 13$

pH of Weak Acid: $$[H^+] = \sqrt{K_a \times C}$$

Where $K_a$ = acid dissociation constant, $C$ = concentration

Example: Find pH of 0.1M CH₃COOH (Ka = 1.8 × 10⁻⁵) $$[H^+] = \sqrt{1.8 \times 10^{-5} \times 0.1} = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3}$$ $$pH = -\log(1.34 \times 10^{-3}) = 2.87$$

Buffer Solutions:

A buffer resists pH change on addition of small amounts of acid or base.

Types:

1. Weak acid + its salt (e.g., CH₃COOH/CH₃COONa)
2. Weak base + its salt (e.g., NH₃/NH₄Cl)

Henderson-Hasselbalch Equation: $$pH = pK_a + \log\frac{[\text{salt}]}{[\text{acid}]}$$

Worked Example: A buffer solution contains 0.1M CH₃COOH and 0.1M CH₃COONa. Calculate pH. (Ka = 1.8 × 10⁻⁵)

Solution: $pK_a = -\log(1.8 \times 10^{-5}) = 4.74$ $pH = 4.74 + \log\frac{0.1}{0.1} = 4.74 + 0 = 4.74$

Neutralisation Reactions:

$$\text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water}$$

Examples:

• $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$
• $\text{H}_2\text{SO}_4 + 2\text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O}$

Salt Hydrolysis:

• Salt of strong acid + weak base → acidic solution (e.g., NH₄Cl)
• Salt of weak acid + strong base → basic solution (e.g., Na₂CO₃)
• Salt of strong acid + strong base → neutral solution (e.g., NaCl)

⚡ Common Mistake: Students confuse pH with [H⁺]. Remember: lower pH = higher [H⁺]. A pH of 1 has [H⁺] = 10⁻¹ = 0.1M, while pH of 4 has [H⁺] = 10⁻⁴ = 0.0001M.

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🔴 Extended — Deep Study (3mo+)

Comprehensive theory for serious exam preparation.

Strength of Acids - Ka Values:

For polyprotic acids (acids with multiple ionisable protons):

• H₂SO₄: First proton strong, second weak (Ka₂ = 1.2 × 10⁻²)
• H₂CO₃: Ka₁ = 4.3 × 10⁻⁷, Ka₂ = 5.6 × 10⁻¹¹
• H₃PO₄: Ka₁ = 7.5 × 10⁻³, Ka₂ = 6.2 × 10⁻⁸, Ka₃ = 4.8 × 10⁻¹³

Weak Base Dissociation Constant (Kb): $$K_b = \frac{[OH^-]^2}{[\text{base}]}$$

For conjugate acid-base pair: $$K_a \times K_b = K_w = 10^{-14} \text{ (at 25°C)}$$

Buffer Capacity: Maximum amount of acid/base that can be added before pH changes significantly.

• Higher concentration → higher buffer capacity
• Maximum buffer capacity at $pH = pK_a$ (when $[\text{salt}] = [\text{acid}]$)

Indicators and Titration:

Indicator	pH Range	Colour Change
Methyl orange	3.1-4.4	Red → Yellow
Methyl red	4.4-6.2	Red → Yellow
Bromothymol blue	6.0-7.6	Yellow → Blue
Phenolphthalein	8.3-10.0	Colourless → Pink

Choosing the Right Indicator:

• Strong acid vs Strong base: Any indicator works (equivalence at pH = 7)
• Strong acid vs Weak base: Methyl orange (equivalence at pH < 7)
• Weak acid vs Strong base: Phenolphthalein (equivalence at pH > 7)

Acid-Base Titration Curves:

For weak acid + strong base:

• Initial pH ~3 (weak acid alone)
• Rises gradually as NaOH added
• Sharp increase near equivalence point
• Final pH ~8-9 (weak conjugate base)

⚡ WAEC Previous Year Pattern:

Year	Question	Concept
2023	Calculate pH of buffer	Henderson-Hasselbalch
2022	Identify conjugate pairs	Bronsted-Lowry
2021	Titration curve	Strong vs weak acid

Volumetric Analysis (WAEC Focus):

$$C_1V_1 = C_2V_2 \text{ (for neutralisation)}$$

Where $C$ = concentration, $V$ = volume

For acid-base titration: $$\frac{M_A \times V_A}{M_B \times V_B} = \frac{n_A}{n_B}$$

Where $n$ = number of moles of H⁺ or OH⁻ per formula unit

Example: 25.0 cm³ of 0.1M HCl is titrated with NaOH. What volume of 0.1M NaOH is required?

Solution: $n_{HCl} = 1$ (monoprotic), $n_{NaOH} = 1$ $C_1V_1 = C_2V_2$ $0.1 \times 25 = 0.1 \times V_2$ $V_2 = 25,\text{cm}^3$

Rainwater and pH:

• Normal rainwater pH ~5.6 (slightly acidic due to dissolved CO₂)
• Acid rain pH < 5.6 (caused by SO₂ and NOₓ from fossil fuel combustion)
• SO₂ + H₂O → H₂SO₄ (sulfuric acid)
• NO₂ + H₂O → HNO₃ (nitric acid)

⚡ Exam Strategy: For buffer calculations, remember that the salt concentration equals the concentration of the conjugate base (for weak acid buffers). The weak acid concentration is the initial concentration minus the small amount that dissociates (which is often negligible).

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