Unit 9 · Physical Chemistry

pH of Acidic & Alkaline Solutions

Ka, Kb, Kw, degree of ionisation, pH/pOH calculations, salt hydrolysis, buffer solutions, Henderson-Hasselbalch.

9.1

Degree of Ionisation and Acid/Base Strength

Strong vs Weak Acids/BasesStrong: completely ionised in water. HCl, H₂SO₄, HNO₃, NaOH, KOH, Ba(OH)₂.
Weak: partially ionised; equilibrium established. CH₃COOH, HF, NH₃, most organic acids/bases.

Degree of Ionisation α

α = moles ionised / initial moles = fraction ionised For weak acid HA: HA + H₂O ⇌ H₃O⁺ + A⁻ α = [A⁻] / C₀ (where C₀ = initial concentration) Ka = C₀α² / (1−α) ≈ C₀α² (when α << 1) ∴ α = √(Ka/C₀) — dilution increases degree of ionisation
9.2

Acid and Base Dissociation Constants Ka and Kb

Ka (Acid Dissociation Constant)

HA + H₂O ⇌ H₃O⁺ + A⁻ Ka = [H₃O⁺][A⁻] / [HA] (water excluded — pure liquid) pKa = −log Ka Larger Ka (smaller pKa) ↔ stronger acid Examples: HF: Ka = 6.8×10⁻⁴ pKa = 3.17 CH₃COOH: Ka = 1.8×10⁻⁵ pKa = 4.74 H₂CO₃: Ka = 4.3×10⁻⁷ pKa = 6.37 HCN: Ka = 4.9×10⁻¹⁰ pKa = 9.31 H₂O: Ka = 10⁻¹⁶ pKa = 16 (very weak acid)

Kb (Base Dissociation Constant)

B + H₂O ⇌ BH⁺ + OH⁻ Kb = [BH⁺][OH⁻] / [B] pKb = −log Kb Larger Kb (smaller pKb) ↔ stronger base Examples: NH₃: Kb = 1.8×10⁻⁵ pKb = 4.74 CH₃NH₂: Kb = 4.4×10⁻⁴ pKb = 3.36 C₆H⁵NH₂: Kb = 4.3×10⁻¹⁰ pKb = 9.37
9.3

Relationship Between Ka and Kb

Conjugate Acid-Base PairsFor a conjugate acid-base pair (HA and A⁻):
Ka × Kb = Kw = 1.0 × 10⁻¹⁴ (at 25°C)
pKa + pKb = pKw = 14.00 (at 25°C)
Stronger acid ↔ weaker conjugate base, and vice versa.

Example

CH₃COOH has Ka = 1.8×10⁻⁵, pKa = 4.74 Its conjugate base CH₃COO⁻ has: Kb = Kw/Ka = 10⁻¹⁴/1.8×10⁻⁵ = 5.6×10⁻¹° pKb = 14 − 4.74 = 9.26
9.4

Ionic Product of Water Kw

KwH₂O ⇌ H⁺ + OH⁻
Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 25°C
In pure water: [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol/L
Kw increases with temperature (endothermic ionisation); neutral pH changes with T.
9.5

pH and pOH Calculations

Definitions and Formulae

pH = −log[H⁺] pOH = −log[OH⁻] pH + pOH = 14 (at 25°C) Strong acid (complete ionisation): pH = −log Cα (for monoprotic HA; [H⁺] = C) Example: 0.10 M HCl: pH = −log(0.10) = 1.00 Strong base: [OH⁻] = C; pOH = −log C; pH = 14 − pOH Example: 0.050 M NaOH: pOH = 1.30; pH = 12.70 Weak acid (Ka given): HA ⇌ H⁺ + A⁻; [H⁺] = √(Ka × C) (approximation for α << 1) Example: 0.10 M CH₃COOH (Ka = 1.8×10⁻⁵): [H⁺] = √(1.8×10⁻⁵ × 0.10) = √(1.8×10⁻⁶) = 1.34×10⁻³ M pH = 2.87 Weak base: [OH⁻] = √(Kb × C); pOH = −log[OH⁻]; pH = 14 − pOH
9.6

Salt Hydrolysis

pH of Salt Solutions

Salt TypeExamplepHHydrolysis
Strong acid + strong baseNaCl, KNO₃7No hydrolysis
Strong acid + weak baseNH₄Cl, NH₄NO₃<7 (acidic)NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺
Weak acid + strong baseCH₃COONa, NaHCO₃>7 (basic)CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻
Weak acid + weak baseCH₃COONH₄≈7 (Ka≈Kb)Both ions hydrolyse; pH depends on Ka vs Kb
9.7

Buffer Solutions

BufferA solution that resists significant changes in pH when small amounts of acid or base are added. Contains a weak acid and its conjugate base (or weak base + conjugate acid).

Henderson-Hasselbalch Equation

pH = pKa + log([A⁻]/[HA]) (pOH = pKb + log([BH⁺]/[B]) for basic buffer) Buffer works: Add H⁺: A⁻ + H⁺ → HA (base component absorbs acid — pH barely changes) Add OH⁻: HA + OH⁻ → A⁻ + H₂O (acid component absorbs base) Buffer capacity: maximum when [A⁻]=[HA]; pH = pKa Buffer range: pKa ± 1 (practical range for significant buffering) Example: pH = 4.74 + log([CH₃COO⁻]/[CH₃COOH]) Equal concentrations: pH = 4.74 + log(1) = 4.74 10:1 ratio of A⁻:HA: pH = 4.74 + log(10) = 5.74
9.8

Preparation of Buffer Solutions

Method 1: Mix Weak Acid + Conjugate Base Salt

E.g. Mix CH₃COOH and CH₃COONa. The required ratio is given by Henderson-Hasselbalch.

To prepare pH 5.0 acetate buffer (pKa CH₃COOH = 4.74): 5.0 = 4.74 + log([CH₃COO⁻]/[CH₃COOH]) log ratio = 0.26 → ratio = 10⁰·²⁶ = 1.82 So [CH₃COO⁻]/[CH₃COOH] = 1.82 : 1

Method 2: Weak Acid + Excess Strong Base

Add NaOH to CH₃COOH: NaOH converts part of the acid to its conjugate base. Leave excess acid → buffer system.

Example: 100 mL 0.20 M CH₃COOH + 50 mL 0.20 M NaOH → NaOH neutralises half the acid: Remaining: 0.010 mol CH₃COOH + 0.010 mol CH₃COO⁻ pH = pKa + log(1) = 4.74

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Biological Buffers

Blood pH 7.35–7.45 maintained by: bicarbonate buffer (H₂CO₃/HCO₃⁻, pKa 6.1); haemoglobin buffer; phosphate buffer. Enzymes denature outside pH 7.35–7.45 (acidosis/alkalosis fatal).

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Exercises

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Quiz — Unit 9: pH

Unit 9: pH

25 Qs
Q1

The pH of pure water at 25°C is:

[H⁺]=[OH⁻]=10⁻⁷ M in pure water at 25°C. pH = −log(10⁻⁷) = 7.00. Note: this is only 7 at 25°C. At higher T, Kw increases, neutral pH <7.
Q2

For a weak acid, increasing dilution:

α = √(Ka/C). As C decreases, α increases. Dilute solution → more ionised. This is Ostwald’s dilution law. Ka itself is constant at fixed T.
Q3

Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]). Buffer pH = pKa when:

log([A⁻]/[HA]) = 0 when [A⁻]/[HA] = 1, i.e., [A⁻]=[HA]. This is the maximum buffer capacity point — the buffer can absorb equal amounts of acid and base.
Q4

NH₄Cl solution is acidic because:

NH₄Cl is the salt of a weak base (NH₃) and strong acid (HCl). The NH₄⁺ cation undergoes hydrolysis, releasing H₃O⁺ → acidic solution.
Q5

Ka × Kb for a conjugate acid-base pair equals:

Ka × Kb = Kw for a conjugate pair. This means: stronger the acid (larger Ka), weaker its conjugate base (smaller Kb). Example: HF (Ka=6.8×10⁻⁴) → F⁻ (Kb=1.47×10⁻¹¹).
Q6

The pH of 0.010 M HCl is:

HCl is strong acid, fully ionised: [H⁺] = 0.010 M. pH = −log(0.010) = 2.
Q7

A buffer has pKa = 6.0. Its pH range of effective buffering is:

Buffer range = pKa ± 1 = 5 to 7. Within this range, there is enough of both HA and A⁻ to absorb added acid or base. Outside this range, one component is mostly consumed.
Q8

An acid with pKa = 4 is:

Lower pKa = larger Ka = stronger acid. pKa 4 → Ka = 10⁻⁴; pKa 6 → Ka = 10⁻⁶. Ka(pKa=4) is 100× larger → acid with pKa=4 is stronger.
Q9

Sodium ethanoate (CH₃COONa) solution is basic because:

CH₃COO⁻ is the conjugate base of the weak acid CH₃COOH. In water, it abstracts a proton: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. This produces OH⁻ → basic solution.
Q10

The degree of ionisation of 0.10 M weak acid (Ka = 1.0×10⁻⁴) is:

α = √(Ka/C) = √(10⁻⁴/0.10) = √(10⁻³) = 0.0316 ≈ 3.2%. Only 3.2% ionised at this concentration.
Q11

The pH of blood is:

Normal blood pH = 7.35–7.45. Maintained by bicarbonate buffer. Below 7.35 = acidosis; above 7.45 = alkalosis. Both are medical emergencies if severe.
Q12

The best buffer to use for pH = 5.0 would use an acid with pKa:

Choose buffer with pKa nearest target pH. pKa = 5.0 gives maximum buffer capacity at pH 5.0 (when [A⁻]=[HA]). Range of effectiveness: pKa ± 1 = 4.0–6.0.
Q13

pKa + pKb = 14 is true for:

pKa + pKb = pKw = 14 only for a conjugate pair at 25°C. At other temperatures, pKw changes (pKw = 13.83 at 37°C). The relationship pKa+pKb = pKw always holds for conjugate pairs.
Q14

Kw = [H⁺][OH⁻] = 10⁻¹⁴ at 25°C. This value:

Water ionisation: H₂O ⇌ H⁺ + OH⁻ ΔH > 0 (endothermic). By Le Chatelier, higher T shifts right → more ionisation → larger Kw. At 37°C (body T): Kw = 2.4×10⁻¹⁴, neutral pH = 6.81.
Q15

How should a buffer be prepared at pH 7.4 using H₂PO₄⁻/HPO₄²⁻ (pKa = 7.2)?

pH = 7.2 + log([HPO₄²⁻]/[H₂PO₄⁻]). 7.4−7.2 = 0.2 = log(ratio). ratio = 10²·₀ = 1.58. More conjugate base (HPO₄²⁻) needed.
Q16

Increasing the concentration of a weak acid at fixed T:

As C increases: [H⁺] = √(Ka×C) increases (but less than proportionally). α = √(Ka/C) decreases. So more concentrated acid has lower α but higher [H⁺] and lower pH.
Q17

The pH of 0.10 M NaOH at 25°C is:

[OH⁻] = 0.10 M. pOH = −log(0.10) = 1.0. pH = 14−1 = 13.0.
Q18

A buffer is most effective when:

Maximum buffer capacity when [A⁻]=[HA] and the buffer is concentrated (more moles of buffering species). Concentrated buffer can absorb more acid/base before the ratio changes significantly.
Q19

Hydrolysis of AlCl₃ solution gives a pH < 7 because:

Al³⁺ has high charge density → strongly polarises O–H of coordinated water → [Al(H₂O)₆]³⁺ ⇌ [Al(H₂O)⁵(OH)]²⁺ + H⁺. This hydrolysis produces H⁺ → acidic solution. Similar for Fe³⁺, Cr³⁺.
Q20

The isoelectric point of an amino acid is the pH where:

At the isoelectric point (pI), the amino acid exists predominantly as a zwitterion with equal positive and negative charges → net charge = 0. pI ≈ (pKa1 + pKa2)/2. Amino acids are least soluble and do not migrate in an electric field at pI.
Q21

A 0.10 M solution of Na₂CO₃ is basic because:

CO₃²⁻ is the conjugate base of the weak acid HCO₃⁻ (pKa = 10.3). Strong hydrolysis: CO₃²⁻+H₂O ⇌ HCO₃⁻+OH⁻ → basic. Na₂CO₃ solutions are significantly basic (pH ~11–12 at 0.1 M).
Q22

The pH of an ammonia buffer (pKb(NH₃)=4.74) at [NH₃]=[NH₄⁺] is:

For basic buffer: pOH = pKb + log([BH⁺]/[B]) = 4.74 + log(1) = 4.74. pH = 14−4.74 = 9.26. Also: pKa(NH₄⁺) = 14−4.74 = 9.26. pH = pKa when [NH₃]=[NH₄⁺].
Q23

For a strong acid-strong base neutralisation, the salt formed:

Strong acid + strong base → salt with no tendency to hydrolyse (cation from strong base, anion from strong acid: neither accepts/donates proton to water appreciably). Example: NaCl, KNO₃ → pH = 7 at 25°C.
Q24

At the half-equivalence point of a weak acid titration with strong base:

At half-equivalence: half the acid converted to conjugate base → [HA]=[A⁻] → Henderson-Hasselbalch: pH = pKa + log(1) = pKa. This is how pKa is determined experimentally from a titration curve.
Q25

A weak acid HA (0.20 M, Ka = 5.0×10⁻⁶). [H⁺] = ?

[H⁺] = √(Ka×C) = √(5.0×10⁻⁶×0.20) = √(1.0×10⁻⁶) = 1.0×10⁻³ M. pH = 3.00.

Unit 9 Quiz — pH & Buffers (25 Questions)

Select one answer each
Q1

The pH of a solution is defined as:

pH = –log₁₀[H⁺] = –log₁₀[H₃O⁺]. A tenfold increase in [H⁺] decreases pH by 1 unit.
Q2

The ionic product of water (Kw) at 25°C is:

Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 25°C. For pure water, [H⁺] = [OH⁻] = 1×10⁻⁷, giving pH = 7.
Q3

A weak acid HA with Ka = 1×10⁻⁵ and concentration 0.1 mol/L has [H⁺] approximately:

[H⁺] = √(Ka × C) = √(1×10⁻⁵ × 0.1) = √(1×10⁻⁶) = 1×10⁻³ mol/L (weak acid approximation).
Q4

A buffer solution resists changes in pH when:

Buffers contain a weak acid and its conjugate base. Added H⁺ reacts with A⁻; added OH⁻ reacts with HA — pH changes minimally.
Q5

The Henderson-Hasselbalch equation is:

pH = pKa + log([A⁻]/[HA]). When [A⁻] = [HA], log(1) = 0, so pH = pKa — the buffer's optimal pH.
Q6

The pKa of ethanoic acid is 4.75. A buffer made with equal concentrations of CH₃COOH and CH₃COO⁻ has pH:

pH = pKa + log(1/1) = pKa + 0 = 4.75. Maximum buffer capacity occurs when [A⁻] = [HA].
Q7

pKw = pKa + pKb for a conjugate acid-base pair. If Ka(CH₃COOH) = 1.8×10⁻⁵, then Kb(CH₃COO⁻) is:

Kw = Ka × Kb → Kb = Kw/Ka = 1×10⁻¹⁴ / 1.8×10⁻⁵ = 5.6×10⁻¹⁰.
Q8

At the half-equivalence point in an acid-base titration:

At half-equivalence, half the acid is neutralised: [HA] = [A⁻]. By Henderson-Hasselbalch: pH = pKa + log(1) = pKa.
Q9

The pH of a 0.01 mol/L HCl solution is:

HCl is a strong acid: fully dissociates. [H⁺] = 0.01 = 10⁻² mol/L. pH = –log(10⁻²) = 2.
Q10

The pH of a 0.001 mol/L NaOH solution at 25°C is:

[OH⁻] = 10⁻³. pOH = 3. pH = 14 – 3 = 11. (At 25°C, pH + pOH = 14.)
Q11

A polyprotic acid releases:

H₂SO₄, H₃PO₄, H₂CO₃ are polyprotic. Each ionisation step has its own Ka, with Ka₁ > Ka₂ > Ka₃.
Q12

The pKa of a strong acid like HCl is:

Strong acids fully dissociate (Ka very large → pKa very negative, e.g. HCl pKa ≈ –7). Not meaningful to quote pKa.
Q13

An amino acid at its isoelectric point (pI) has:

At pI, the molecule has equal + and – charges — net zero. For glycine, pI = (pKa1 + pKa2)/2 = (2.34 + 9.60)/2 ≈ 6.0.
Q14

Increasing temperature increases Kw, which means:

Higher T: Kw increases → [H⁺] = [OH⁻] > 10⁻⁷ → pH of pure water < 7, but solution is still neutral (equal concentrations).
Q15

The degree of ionisation (α) of a weak acid increases when:

Ka = cα²/(1–α). For dilute solutions: α ≈ √(Ka/c). Lower c → higher α. Dilution increases % ionisation.
Q16

A 0.1 mol/L solution of weak acid has pH 3. The degree of ionisation is:

pH=3 → [H⁺]=10⁻³. α = [H⁺]/C = 10⁻³/0.1 = 10⁻² = 1%.
Q17

Which of the following is amphiprotic?

HCO₃⁻ can donate H⁺ (acting as acid → CO₃²⁻) or accept H⁺ (acting as base → H₂CO₃). Amphiprotic/amphoteric.
Q18

The salt of a weak acid and strong base (e.g. CH₃COONa) forms an alkaline solution because:

The conjugate base A⁻ of a weak acid accepts H⁺ from water, generating OH⁻ → alkaline solution.
Q19

A blood buffer system uses:

Blood pH (~7.4) maintained by H₂CO₃ ⇌ HCO₃⁻ + H⁺. CO₂ from respiration links this to the lungs.
Q20

pOH + pH = 14 at 25°C because:

Kw = 10⁻¹⁴ → logKw = –14 → –logKw = 14. Since pH = –log[H⁺] and pOH = –log[OH⁻]: pH + pOH = pKw = 14.
Q21

Salt hydrolysis occurs when:

Weak acid/strong base salts: A⁻ + H₂O → HA + OH⁻ (alkaline). Strong acid/weak base salts: BH⁺ → B + H⁺ (acidic).
Q22

The common ion effect states that adding a common ion to a weak acid equilibrium:

Adding CH₃COO⁻ (from CH₃COONa) shifts CH₃COOH ⇌ CH₃COO⁻ + H⁺ to the left → less H⁺ → higher pH.
Q23

Zwitterion formation in amino acids involves:

At physiological pH, the amino group gains H⁺ (→ NH₃⁺) and carboxyl group loses H⁺ (→ COO⁻) — net charge zero.
Q24

A solution with pH 5 is how many times more acidic than one with pH 7?

Each pH unit = 10× change in [H⁺]. pH 5 → [H⁺] = 10⁻⁵; pH 7 → [H⁺] = 10⁻⁷. Ratio = 10⁻⁵/10⁻⁷ = 100.
Q25

Acid rain has pH below 5.6 due to dissolved:

Industrial combustion emits SO₂ and NOₓ → H₂SO₄, HNO₃ in atmosphere. CO₂ gives weak carbonic acid (normal pH 5.6).
📝 Go to Unit Test →
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Unit Test — 50 marks

Section A — Short Answer

30 marks
Q1 [5 marks]

(a) Define Ka, pKa, and explain what a larger Ka value means. [2] (b) Calculate the pH of 0.20 M propanoic acid (Ka = 1.35×10⁻⁵). [3]

(a) Ka = [H⁺][A⁻]/[HA]: equilibrium constant for acid ionisation in water. pKa = −log(Ka). Larger Ka → more ionised → stronger acid; smaller pKa. (b) [H⁺] = √(1.35×10⁻⁵×0.20) = √(2.70×10⁻¹°) = 1.64×10⁻⁵ M. pH = −log(1.64×10⁻⁵) = 4.78.
Q2 [5 marks]

Prepare a phosphate buffer at pH 7.0 using H₂PO₄⁻ (pKa = 7.2) and HPO₄²⁻. (a) Calculate the required ratio [HPO₄²⁻]/[H₂PO₄⁻]. [2] (b) Describe how to make 500 mL of this buffer with total phosphate concentration 0.10 M. [3]

(a) pH = 7.2 + log([HPO₄²⁻]/[H₂PO₄⁻]); 7.0−7.2 = −0.2 = log(ratio); ratio = 10⁻²·₀ = 0.631. [H₂PO₄⁻]/[HPO₄²⁻] = 1.58:1. (b) Total phosphate = 0.10 M in 500 mL = 0.050 mol. Let x = mol HPO₄²⁻; 0.050−x = mol H₂PO₄⁻. Ratio x/(0.050−x) = 0.631 → x = 0.631(0.050−x) → x(1+0.631) = 0.0316 → x = 0.0193 mol HPO₄²⁻; 0.0307 mol H₂PO₄⁻. Dissolve in water and make up to 500 mL.
Q3 [5 marks]

Explain salt hydrolysis. Give examples and equations for: (a) an acidic salt [2]; (b) a basic salt [2]; (c) a neutral salt [1].

(a) Acidic salt: NH₄Cl (strong acid HCl + weak base NH₃). NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺. NH₄⁺ donates H⁺ → pH < 7. (b) Basic salt: CH₃COONa (weak acid CH₃COOH + strong base NaOH). CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. CH₃COO⁻ accepts H⁺ from water → OH⁻ produced → pH > 7. (c) Neutral salt: NaCl (strong acid HCl + strong base NaOH). Na⁺ and Cl⁻ neither donate nor accept protons to/from water → pH = 7.
Q4 [5 marks]

(a) Describe how a blood buffer maintains pH 7.4. [3] (b) Explain acidosis and alkalosis: causes and physiological consequences. [2]

(a) Blood buffer: CO₂(aq)+H₂O ⇌ H₂CO₃ ⇌ HCO₃⁻+H⁺ ([H₂CO₃]/[HCO₃⁻] ≈ 1:20). pH = 6.1+log(20) = 7.4. If H⁺ added (e.g. exercise, lactic acid): HCO₃⁻+H⁺→H₂CO₃→CO₂+H₂O (exhaled by lungs). If OH⁻ added: H₂CO₃+OH⁻→HCO₃⁻+H₂O. Kidneys regulate [HCO₃⁻] (slower, hours). (b) Acidosis (pH <7.35): caused by excess CO₂ (respiratory acidosis: hypoventilation, COPD) or excess organic acids (metabolic acidosis: DKA, renal failure). Consequences: enzyme dysfunction, cardiac arrhythmia, coma. Alkalosis (pH >7.45): hyperventilation, excess antacids. Consequences: muscle spasms, tetany, seizures.
Q5 [5 marks]

Determine the pH of: (a) 0.050 M HNO₃ [1]; (b) 0.050 M KOH [1]; (c) 0.050 M NH₃ (Kb=1.8×10⁻⁵) [2]; (d) equal volumes of 0.10 M HCl and 0.10 M NaOH mixed [1].

(a) HNO₃ strong: [H⁺]=0.050; pH = −log(0.050) = 1.30. (b) KOH strong: [OH⁻]=0.050; pOH=1.30; pH = 14−1.30 = 12.70. (c) [OH⁻] = √(1.8×10⁻⁵×0.050) = √(9×10⁻¹¹) = 9.49×10⁻⁶; pOH=5.02; pH = 14−5.02 = 8.98. (d) Equal volumes 0.10 M HCl + 0.10 M NaOH: neutralise exactly → NaCl solution → pH = 7.00.

Section B — Extended Response

20 marks
Q6 [10 marks]

(a) Explain the concept of buffer capacity. Discuss how the concentration of the buffer components and the ratio [A⁻]/[HA] affect buffer capacity and buffering range. [4] (b) A biochemist needs to prepare 1.00 L of an acetate buffer (pKa 4.74) at pH 5.00 with an ionic strength of 0.10 M total acetate. Describe the preparation in detail. [3] (c) Compare the behaviour of a buffer and a neutral salt solution (NaCl) when 0.010 mol HCl is added to 1.00 L of each. [3]

(a) Buffer capacity: maximum amount of acid/base buffer can absorb before significant pH change. Depends on: (1) Concentration: more concentrated buffer has more moles of HA and A⁻ → can absorb more acid/base before ratio changes significantly. Doubling concentration doubles capacity. (2) Ratio [A⁻]/[HA]: buffer capacity is highest when [A⁻]=[HA] (pH=pKa); equal amounts available to absorb either acid or base. At extreme ratios (10:1 or 1:10), buffer can only absorb one direction effectively. Buffering range is pKa ± 1 (ratio 1:10 to 10:1). (b) Preparation: pH = 4.74 + log([CH₃COO⁻]/[CH₃COOH]); 5.00−4.74 = 0.26 = log(ratio); ratio = 10²·²⁶ = 1.82. Total acetate = 0.10 mol. Let x = mol CH₃COO⁻; 0.10−x = mol CH₃COOH. x/(0.10−x) = 1.82 → x = 0.1820−1.82x → 2.82x = 0.1820 → x = 0.0645 mol CH₃COO⁻ (as NaCH₃COO, M=82: 5.29 g). 0.0355 mol CH₃COOH (M=60: 2.13 g). Dissolve both in ~900 mL water, adjust with drops of HCl/NaOH if needed, make up to 1.00 L with volumetric flask. Check pH. (c) Buffer (0.10 M acetate, pH 5.00): 0.010 mol HCl added. CH₃COO⁻ reacts: 0.010 mol CH₃COO⁻ consumed, 0.010 mol CH₃COOH formed. New ratio = (0.0645−0.010)/(0.0355+0.010) = 0.0545/0.0455 = 1.20. New pH = 4.74+log(1.20) = 4.74+0.079 = 4.82. Change = only 0.18 pH units. NaCl solution (pH=7.00): 0.010 mol HCl added to 1.00 L. [H⁺] = 0.010 M. pH = 2.00. Change = 5 pH units. Conclusion: buffer dramatically stabilises pH; NaCl has zero buffering capacity.
Q7 [10 marks]

(a) Distinguish between strong and weak acids with reference to Ka, degree of ionisation, and conductivity. Give specific examples and data. [4] (b) A student measures the conductivity of 0.10 M solutions of HCl, CH₃COOH, and NaCl. Predict the order of conductivity and explain. [3] (c) Describe a simple experiment to determine the Ka of a weak acid, including calculations. [3]

(a) Strong acid (HCl): Ka = very large (≈10⁶+); degree of ionisation = 1.00 (100%); all HA ionises to H⁺ + A⁻; [H⁺] = C(acid). Weak acid (CH₃COOH): Ka = 1.8×10⁻⁵ (small); α ≈ 1.3% at 0.10 M; equilibrium heavily to left; [H⁺] << C(acid). Data: 0.10 M HCl: [H⁺]=0.10 M, pH=1.00. 0.10 M CH₃COOH: [H⁺]=1.34×10⁻³ M, pH=2.87. Conductivity reflects total ion concentration. (b) Order: HCl > NaCl > CH₃COOH. HCl: fully ionised, has both H⁺ (very mobile, proton hopping) and Cl⁻; H⁺ has highest molar conductivity. NaCl: fully ionised to Na⁺ + Cl⁻; same concentration of ions as HCl but Na⁺ less mobile than H⁺. CH₃COOH: only ~1.3% ionised → very few ions → very low conductivity. (c) Experiment: measure pH of 0.10 M CH₃COOH with pH meter → e.g. pH=2.87. [H⁺] = 10⁻²·⁸⁷ = 1.35×10⁻³ M. ICE table: [CH₃COOH] at eq = 0.10−1.35×10⁻³ ≈ 0.0987 M. Ka = [H⁺][CH₃COO⁻]/[CH₃COOH] = (1.35×10⁻³)²/0.0987 = 1.85×10⁻⁶/0.0987 = 1.85×10⁻⁵. Alternatively, titration curve half-equivalence point method: half-way to equivalence, pH = pKa directly.
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