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This final lesson ties together everything you have learned about acids, bases, pH, Ka, buffers, titrations, and indicators. The problems here are multi-step, combining concepts from several lessons — exactly like the extended questions you will face in your Edexcel exam.
The key to success is a systematic approach: identify the type of problem, recall the relevant equations, set out your working clearly, and check your answer makes chemical sense.
flowchart TD
A["Read the question: what type of solution?"] --> B{"Strong acid or base?"}
B -->|"Strong acid"| C["[H⁺] = concentration"]
B -->|"Strong base"| D["[OH⁻] = conc, use Kw"]
B -->|"Weak acid"| E["Use [H⁺] = √(Ka x c)"]
B -->|"Buffer"| F["Use Henderson-Hasselbalch"]
B -->|"Mixture"| G["Calculate moles, find excess"]
C --> H["pH = -log[H⁺]"]
D --> H
E --> H
F --> H
G --> I{"What is left after reaction?"}
I -->|"Excess strong acid"| C
I -->|"Excess strong base"| D
I -->|"Weak acid + conjugate base"| F
I -->|"Exact neutralisation"| J["pH = 7 (SA+SB) or calculate from salt hydrolysis"]
Problem 1: 50.0 cm³ of 0.20 mol dm⁻³ ethanoic acid (Ka = 1.74 × 10⁻⁵ mol dm⁻³) is mixed with 25.0 cm³ of 0.20 mol dm⁻³ NaOH. Calculate the pH of the resulting solution.
Solution:
Step 1: Calculate moles.
Step 2: The NaOH reacts with ethanoic acid: CH₃COOH + NaOH → CH₃COONa + H₂O
Step 3: This is a buffer (weak acid + conjugate base). Use Henderson-Hasselbalch:
pH = pKa + log₁₀([A⁻]/[HA])
Since both species are in the same total volume, the ratio of moles equals the ratio of concentrations:
pH = 4.76 + log₁₀(0.005/0.005) pH = 4.76 + log₁₀(1) pH = 4.76
The solution is a buffer at pH 4.76 — the half-equivalence point.
Problem 2: A buffer contains 0.080 mol of CH₃COOH and 0.120 mol of CH₃COO⁻ in 500 cm³ of solution. Ka = 1.74 × 10⁻⁵ mol dm⁻³.
(a) Calculate the initial pH. (b) Calculate the pH after adding 0.010 mol of HCl. (c) Calculate the pH after adding 0.010 mol of NaOH to the original buffer.
Solution (a):
pH = pKa + log₁₀(0.120/0.080) = 4.76 + log₁₀(1.50) = 4.76 + 0.18 = 4.94
Solution (b):
HCl adds 0.010 mol of H⁺, which reacts with CH₃COO⁻:
pH = 4.76 + log₁₀(0.110/0.090) = 4.76 + log₁₀(1.222) = 4.76 + 0.087 = 4.85
pH changed from 4.94 to 4.85 — a decrease of only 0.09 units.
Solution (c):
NaOH adds 0.010 mol of OH⁻, which reacts with CH₃COOH:
pH = 4.76 + log₁₀(0.130/0.070) = 4.76 + log₁₀(1.857) = 4.76 + 0.269 = 5.03
pH changed from 4.94 to 5.03 — an increase of only 0.09 units.
Problem 3: A titration curve is obtained for 25.0 cm³ of a weak acid HA with 0.10 mol dm⁻³ NaOH. The equivalence point occurs at 20.0 cm³ of NaOH. At 10.0 cm³ of NaOH, the pH is 4.20.
(a) Calculate Ka for the acid. (b) Calculate the initial concentration of HA.
Solution (a):
At 10.0 cm³ (half the equivalence volume of 20.0 cm³), this is the half-equivalence point. At the half-equivalence point: pH = pKa = 4.20
Ka = 10⁻⁴·²⁰ = 6.31 × 10⁻⁵ mol dm⁻³
Solution (b):
At the equivalence point, moles of NaOH = moles of HA (1:1 ratio). Moles of NaOH at equivalence = 0.10 × 0.0200 = 2.00 × 10⁻³ mol Moles of HA = 2.00 × 10⁻³ mol [HA] = 2.00 × 10⁻³ / 0.0250 = 0.080 mol dm⁻³
Problem 4: At 37°C (body temperature), Kw = 2.4 × 10⁻¹⁴ mol² dm⁻⁶.
(a) Calculate the pH of pure water at 37°C. (b) Is this water acidic, neutral, or alkaline? (c) Calculate the pH of a 0.010 mol dm⁻³ solution of NaOH at 37°C.
Solution (a):
In pure water, [H⁺] = [OH⁻] = √Kw = √(2.4 × 10⁻¹⁴) = 1.55 × 10⁻⁷ mol dm⁻³ pH = -log₁₀(1.55 × 10⁻⁷) = 6.81
Solution (b):
The water is neutral, even though pH ≠ 7. Neutrality means [H⁺] = [OH⁻], which is satisfied. The pH of neutral water depends on temperature.
Solution (c):
[OH⁻] = 0.010 mol dm⁻³ [H⁺] = Kw / [OH⁻] = (2.4 × 10⁻¹⁴) / 0.010 = 2.4 × 10⁻¹² mol dm⁻³ pH = -log₁₀(2.4 × 10⁻¹²) = 11.62
Note: at 37°C, this is lower than the 12.00 you would calculate at 25°C for the same NaOH concentration because Kw is larger.
Problem 5: A student prepares a buffer by mixing 100 cm³ of 0.50 mol dm⁻³ propanoic acid (Ka = 1.35 × 10⁻⁵) with 25.0 cm³ of 1.00 mol dm⁻³ NaOH.
(a) Show that this produces a buffer solution and calculate its pH. (b) The student adds 5.00 × 10⁻³ mol of HCl to the buffer. Calculate the new pH. (c) What indicator should be used if this buffer were to be titrated to the equivalence point with NaOH?
Solution (a):
Moles of propanoic acid = 0.50 × 0.100 = 0.050 mol Moles of NaOH = 1.00 × 0.0250 = 0.025 mol
The NaOH reacts: CH₃CH₂COOH + NaOH → CH₃CH₂COONa + H₂O
Remaining propanoic acid = 0.050 - 0.025 = 0.025 mol Produced propanoate = 0.025 mol
Both a weak acid (0.025 mol CH₃CH₂COOH) and its conjugate base (0.025 mol CH₃CH₂COO⁻) are present. This is a buffer.
pH = pKa + log₁₀(0.025/0.025) = -log₁₀(1.35 × 10⁻⁵) + 0 = 4.87
Solution (b):
After adding 5.00 × 10⁻³ mol HCl:
pH = 4.87 + log₁₀(0.020/0.030) = 4.87 + log₁₀(0.667) = 4.87 - 0.18 = 4.69
Solution (c):
This would be a WA/SB titration. The steep section is pH 7–11. Phenolphthalein (pH 8.3–10.0) is suitable.
Problem 6: A student has 0.20 mol dm⁻³ ethanoic acid (Ka = 1.74 × 10⁻⁵). They dilute it tenfold. Calculate the pH before and after dilution, and explain why the pH does not change by exactly 1 unit.
Before: [H⁺] = √(1.74 × 10⁻⁵ × 0.20) = √(3.48 × 10⁻⁶) = 1.87 × 10⁻³, pH = 2.73
After (0.020 mol dm⁻³): [H⁺] = √(1.74 × 10⁻⁵ × 0.020) = √(3.48 × 10⁻⁷) = 5.90 × 10⁻⁴, pH = 3.23
The pH changed by 0.50, not 1.00. For weak acids, the [H⁺] depends on √c (from the formula [H⁺] = √(Ka·c)). A tenfold decrease in c gives a √10 = 3.16-fold decrease in [H⁺], which corresponds to a pH increase of log₁₀(√10) = 0.50 units.
| Pitfall | How to avoid it |
|---|---|
| Not identifying the solution type | Ask: strong acid? Weak acid? Buffer? Mixture? |
| Using concentrations when mixing | Always calculate moles first |
| Forgetting Kw for bases | [H⁺] = Kw/[OH⁻], then pH = -log[H⁺] |
| Ignoring H₂SO₄ is diprotic | [H⁺] = 2 × [H₂SO₄] |
| Not recognising buffer formation | Weak acid + less-than-stoichiometric strong base = buffer |
| Assuming pH 7 is neutral at all temperatures | Neutral = [H⁺] = [OH⁻], depends on Kw(T) |
| Rounding too early | Keep 4+ s.f. in intermediate steps |
| Forgetting to state the weak acid assumption | Write: "Assuming dissociation is negligible, [HA] ≈ c" |
| Equation | Application |
|---|---|
| pH = -log₁₀[H⁺] | All pH calculations |
| [H⁺] = 10⁻ᵖᴴ | Finding [H⁺] from pH |
| Kw = [H⁺][OH⁻] | Linking acid and base (= 1.00 × 10⁻¹⁴ at 25°C) |
| Ka = [H⁺][A⁻]/[HA] | Weak acid equilibrium |
| [H⁺] = √(Ka × c) | pH of weak acid |
| Ka = [H⁺]²/c | Ka from pH |
| pH = pKa + log([A⁻]/[HA]) | Buffer pH (Henderson-Hasselbalch) |
| pKa = -log₁₀Ka | Converting between Ka and pKa |
| n = c × V | Moles from concentration and volume |
Edexcel 9CH0 specification, Topic 12: Acid-base equilibria, all sub-strands 12.1–12.5 are tested synoptically in this lesson. You should be able to integrate Bronsted-Lowry classification (12.1), pH and Kw (12.2), Ka and weak-acid pH (12.3), buffer composition and Henderson-Hasselbalch (12.4), and titration / indicator selection (12.5) into multi-step problems linked also to organic acids (Topic 17), thermodynamics (Topic 16), and equilibrium (Topic 11) (refer to the official specification document for exact wording). Examined heavily on Paper 3 as 12-mark synoptic stems combining identification, calculation, curve interpretation, and indicator choice.
Question (12 marks, synoptic):
A student is given a 25.0 cm³ aliquot of an unknown weak monoprotic acid HA at unknown concentration. She titrates it with 0.100 mol dm⁻³ NaOH(aq), monitoring pH continuously. The resulting curve has the following key features:
(a) Calculate the concentration of HA. (1)
(b) Calculate Ka of HA. Identify HA from a list of common acids: methanoic (pKa = 3.75), ethanoic (4.76), propanoic (4.87), benzoic (4.20). (2)
(c) Verify the initial pH by calculation from Ka and the concentration of HA. (3)
(d) Calculate the [A⁻]/[HA] ratio in a buffer that contains 50.0 cm³ of HA at the original concentration mixed with 25.0 cm³ of 0.100 mol dm⁻³ NaOH. State the resulting pH. (3)
(e) Suggest a suitable indicator for the original titration with reasoning. (2)
(f) The student wants to prepare a buffer of pH 5.20 starting from the same HA. Calculate the [A⁻]/[HA] ratio required. (1)
Solution with mark scheme:
(a) V_eq = 25.0 cm³ at 0.100 mol dm⁻³ NaOH; moles NaOH = 2.50 × 10⁻³ mol; moles HA = 2.50 × 10⁻³ mol (1:1); [HA] = 2.50 × 10⁻³ / 0.0250 = 0.100 mol dm⁻³.
B1 — concentration = 0.100 mol dm⁻³.
(b) pH = pKa at half-equivalence, so pKa = 4.76 → Ka = 10⁻⁴·⁷⁶ = 1.74 × 10⁻⁵ mol dm⁻³. Identifies as ethanoic acid.
M1 — Ka calculation; A1 — identification.
(c) For a weak acid: [H⁺]² = Ka × c = 1.74 × 10⁻⁵ × 0.100 = 1.74 × 10⁻⁶ → [H⁺] = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³ → pH = −log₁₀(1.32 × 10⁻³) = 2.88.
The given initial pH is 3.05, slightly higher than calculated. Possible reasons: (i) the concentration of HA is slightly less than 0.100 mol dm⁻³ in practice (e.g. 0.075 mol dm⁻³ would give pH 3.05); (ii) the % dissociation assumption may need correction; (iii) experimental pH meter reading uncertainty (±0.05).
M1 — calculation; M1 — comparison; A1 — at least one plausible source of discrepancy.
(d) Moles HA initially = 0.0500 dm³ × 0.100 mol dm⁻³ = 5.00 × 10⁻³ mol. Moles NaOH = 0.0250 × 0.100 = 2.50 × 10⁻³ mol.
NaOH neutralises half: HA → A⁻, so moles A⁻ formed = 2.50 × 10⁻³ mol; moles HA remaining = 5.00 × 10⁻³ − 2.50 × 10⁻³ = 2.50 × 10⁻³ mol. Ratio [A⁻]/[HA] = 1.
pH = pKa + log₁₀(1) = pKa = 4.76.
M1 — moles tracking; M1 — ratio = 1; A1 — pH = 4.76.
(e) Equivalence pH 8.72; phenolphthalein (pKa 9.3, range 8.2–10.0) is the suitable indicator because its transition range encompasses the equivalence pH and falls within the steep portion of the curve. Methyl orange (pKa 3.7) and bromothymol blue (pKa 7.1) would change in the buffering region or before the steep jump.
M1 — phenolphthalein; A1 — justification including pKa match and steep-portion argument.
(f) log₁₀([A⁻]/[HA]) = pH − pKa = 5.20 − 4.76 = 0.44 → [A⁻]/[HA] = 10⁰·⁴⁴ = 2.75.
B1 — ratio = 2.75 (or 2.8 to 2 s.f.).
Total: 12 marks (M9 A3, split as shown).
Question (8 marks, synoptic): Lactic acid (CH₃CH(OH)COOH, pKa = 3.86) accumulates in muscle tissue during anaerobic respiration. Resting muscle pH is 7.0; vigorous exercise can lower muscle pH to 6.4 before fatigue.
(a) Calculate the [lactate⁻]/[lactic acid] ratio at resting pH 7.0 and at fatigued pH 6.4. (3)
(b) Use these ratios to comment on the effectiveness of the lactic-acid / lactate buffer system as a regulator of muscle pH during exercise. (2)
(c) The bicarbonate buffer (HCO₃⁻ / H₂CO₃, pKa ≈ 6.35) also operates in muscle. Compare its [HCO₃⁻]/[H₂CO₃] ratio at pH 7.0 and 6.4. Which buffer system is more effective at the relevant pH? (3)
Mark scheme decomposition by AO:
(a)
(b)
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