QUESTIONS
1. A student wishes to determine the enthalpy of neutralisation of hydrochloric acid by sodium hydroxide using a simple calorimetric method. 50cm³ of 1.0 mol/dm³ HCl was measured into an expanded polystyrene cup, and its initial temperature recorded. 50cm³ of 1.0 mol/dm³ NaOH (already at the same initial temperature) was then added quickly, the mixture stirred, and the temperature recorded every 30 seconds for 4 minutes before and after mixing.
The results obtained are shown below:
| Time (s) | −60 | −30 | 0 (mixing) | 30 | 60 | 90 | 120 | 150 | 180 |
|---|---|---|---|---|---|---|---|---|---|
| Temperature (°C) | 25.0 | 25.0 | — | 31.8 | 31.5 | 31.2 | 30.9 | 30.6 | 30.3 |
(a) Explain why temperature readings were taken both before AND after mixing, rather than simply recording the highest temperature reached. (5 marks)
(b) i) Plot a graph of temperature against time, showing readings both before and after mixing. ii) By extrapolating the post-mixing cooling line back to the time of mixing (t=0), determine the theoretical maximum temperature rise that would have occurred if no heat had been lost to the surroundings. (10 marks)
(c) Using your answer from (b), and assuming the specific heat capacity of the solution is 4.2 J g⁻¹K⁻¹ and its density is 1.0 g/cm³, calculate:
i) the heat energy released by the reaction. (5 marks)
ii) the number of moles of water formed in the reaction. (3 marks)
iii) the molar enthalpy of neutralisation, in kJ/mol, stating the correct sign. (7 marks)
(d) State TWO sources of experimental error in this method, and for each, explain how it would affect the calculated enthalpy value (i.e., would it make the result too high or too low in magnitude). (10 marks)
(e) The accepted literature value for the enthalpy of neutralisation of a strong acid by a strong base is approximately −57.3 kJ/mol. Comment on how your calculated value compares to this, and explain why the enthalpy of neutralisation is approximately constant for all strong acid-strong base reactions. (10 marks)
2. A mixture of three amino acids — glycine, alanine, and leucine — was separated using paper chromatography, with a suitable solvent system, and the chromatogram was developed and sprayed with ninhydrin to locate the spots. The results are recorded below:
| Substance | Distance travelled by spot (cm) | Distance travelled by solvent front (cm) |
|---|---|---|
| Unknown mixture, Spot 1 | 3.6 | 12.0 |
| Unknown mixture, Spot 2 | 6.0 | 12.0 |
| Unknown mixture, Spot 3 | 9.6 | 12.0 |
| Reference: Glycine | 3.6 | 12.0 |
| Reference: Alanine | 6.0 | 12.0 |
| Reference: Leucine | 9.6 | 12.0 |
(a) Define the term Rf value, and calculate the Rf value for each of the three spots in the unknown mixture. (9 marks)
(b) Using your calculated Rf values and the reference data provided, identify which amino acid corresponds to each spot in the unknown mixture. (6 marks)
(c) Explain, in terms of intermolecular forces/polarity, why leucine (a non-polar, hydrophobic side chain) travels further up the chromatogram than glycine (minimal side chain) in this particular solvent system. (7 marks)
(d) i) Why is the chromatogram developed in a closed tank rather than in the open air? (4 marks)
ii) Why must the original pencil baseline (where the spots are applied) be drawn above the level of the solvent in the developing tank? (4 marks)
(e) Ninhydrin was used as a locating agent to visualise the spots. Explain why this spraying step is necessary for amino acids, and state what colour change is typically observed. (5 marks)
(f) Suggest ONE other method (other than paper chromatography) that could be used to separate a mixture of amino acids, and briefly state the principle behind it. (5 marks)
ANSWERS
Question 1
(a) Readings are taken both before and after mixing because the reaction is exothermic and heat is continuously lost to the surroundings (via the cup, air, and evaporation) once the maximum temperature is reached; taking readings only at the peak would underestimate the true temperature rise, since some heat is already lost by the time the maximum is recorded. Recording temperatures before mixing establishes a stable baseline, while recording several points after mixing (as the mixture cools) allows extrapolation back to the instant of mixing to estimate the temperature rise that would have occurred without heat loss.
(b) i) (Graph: temperature (y-axis, °C) vs time (x-axis, s); pre-mixing line is flat at 25.0°C; a near-vertical rise occurs at t=0 (mixing); post-mixing points show a gently declining line from 31.8°C at t=30s down to 30.3°C at t=180s.)
ii) Extrapolating the post-mixing cooling line (using points 30–180s) back to t=0:
Rate of cooling ≈ (31.8−30.3)/(180−30) = 1.5/150 = 0.01°C/s
Extrapolated temperature at t=0: 31.8 + (0.01×30) = 31.8+0.3 = ≈32.1°C
Theoretical maximum temperature rise = 32.1 − 25.0 = 7.1°C
(c) i) Total volume = 50+50 = 100cm³; mass = 100×1.0 = 100g
Heat released, Q = mcΔT = 100 × 4.2 × 7.1 = 2982 J ≈ 2.98 kJ
ii) moles HCl = 1.0 × (50/1000) = 0.05 mol; moles NaOH = 1.0 × (50/1000) = 0.05 mol (equal, exactly reacting)
Since HCl + NaOH → NaCl + H₂O (1:1 ratio), moles of water formed = 0.05 mol
iii) Molar enthalpy of neutralisation = Q/moles = 2.98/0.05 = 59.6 kJ/mol
Since the reaction is exothermic (heat released, temperature rose), the enthalpy change is given a negative sign:
ΔH = −59.6 kJ/mol
(d) Two sources of error:
- Heat loss to the surroundings/cup during mixing (before the first post-mixing reading is taken) — this is only partially corrected by extrapolation, so some heat loss is still unaccounted for, making the calculated ΔH too low in magnitude (less negative than the true value).
- Heat capacity of the polystyrene cup itself is ignored in the calculation (only the solution's heat capacity is used) — since the cup also absorbs some heat, ignoring this makes the calculated heat released, and hence ΔH, too low in magnitude (an underestimate of the true enthalpy change).
(e) The calculated value (−59.6 kJ/mol) is reasonably close to, and slightly greater in magnitude than, the accepted literature value (−57.3 kJ/mol) — this is somewhat unusual (experimental values are typically lower in magnitude due to heat loss), and any excess may be attributed to extrapolation/rounding effects in this particular data set, but the values are of the same order of magnitude, confirming the method is broadly valid within experimental error.
The enthalpy of neutralisation is approximately constant for all strong acid-strong base reactions because, in dilute aqueous solution, strong acids and strong bases are essentially fully ionised (dissociated), so the overall net ionic reaction is always the same regardless of which specific strong acid/base is used:
H⁺(aq) + OH⁻(aq) → H₂O(l)
Since this is the only reaction actually occurring (the spectator ions, e.g. Na⁺ and Cl⁻, do not participate), the enthalpy change per mole of water formed is essentially identical for any strong acid-strong base combination.
Question 2
(a) Rf value: The retardation factor (Rf) is defined as the ratio of the distance travelled by a solute (spot) from the baseline to the distance travelled by the solvent front, measured from the same baseline, under identical experimental conditions.
Rf = distance travelled by spot / distance travelled by solvent front
Spot 1: Rf = 3.6/12.0 = 0.30
Spot 2: Rf = 6.0/12.0 = 0.50
Spot 3: Rf = 9.6/12.0 = 0.80
(b) Comparing Rf values to the reference amino acids:
Glycine: Rf = 3.6/12.0 = 0.30 → matches Spot 1 = Glycine
Alanine: Rf = 6.0/12.0 = 0.50 → matches Spot 2 = Alanine
Leucine: Rf = 9.6/12.0 = 0.80 → matches Spot 3 = Leucine
(c) Leucine has a bulky, non-polar (hydrophobic) hydrocarbon side chain (−CH₂CH(CH₃)₂), making the molecule overall less polar and less soluble in a polar stationary phase (water held on the paper); it therefore has a greater affinity for the (typically less polar/more organic) mobile solvent phase, causing it to travel further up the paper. Glycine, having only a hydrogen atom as its side chain, is the most polar of the three amino acids, giving it a stronger affinity for the polar stationary phase (water bound to the cellulose paper) and a weaker affinity for the mobile phase, causing it to travel the shortest distance.
(d) i) The chromatogram is developed in a closed tank to maintain a solvent-vapour-saturated atmosphere, preventing evaporation of the solvent from the paper during the run; without this, uneven/inconsistent solvent flow and distorted, non-reproducible Rf values would result.
ii) The baseline must be drawn above the level of the solvent in the tank so that the applied spots (amino acids) do not directly dissolve into and diffuse away in the bulk solvent reservoir; if submerged, the samples would simply wash off into the solvent rather than being carried up the paper by capillary action, and no separation would occur.
(e) Amino acids are generally colourless in visible light, so a locating agent is necessary to make the separated spots visible for identification and Rf calculation. Ninhydrin reacts specifically with the free amino (−NH₂) groups of amino acids to form a coloured complex, typically producing a purple/blue colouration (Ruhemann's purple) at each spot's location, allowing the spots to be marked, circled, and measured.
(f) Another method: Electrophoresis. This technique separates amino acids based on differences in their net electric charge (and hence their migration rate) when placed in an electric field at a given pH; amino acids with different isoelectric points will carry different net charges at a fixed pH and migrate at different rates/directions toward the anode or cathode, allowing them to be separated and identified.