Four 0.100 mol kg⁻¹ Solutions and a Benzoic Acid Anomaly
An examiner gives a class four aqueous solutions, each prepared to be 0.100 mol kg⁻¹ in solute: (A) glucose, (B) NaCl, (C) CaCl2 and (D) ethanoic acid (Ka = 1.8 × 10⁻⁵). Students must arrange them in order of boiling point without measuring. One group writes A < B < C < D; another writes D < A < B < C. The teacher says neither list is correct and that one of the four cannot be placed confidently without a measured value. In a second exercise the class dissolves 1.20 g of benzoic acid (formula mass 122 g mol⁻¹) in 25.0 g of benzene; the freezing point depression recorded is 1.62 K, whereas the value calculated from the formula mass with Kf = 5.12 K kg mol⁻¹ is about 2.01 K. A student calls the gap 'experimental error' and is told that the discrepancy is real and structurally informative.
1. Give the correct boiling-point order, using van't Hoff factors, and identify the solution that cannot be placed confidently.
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Use ΔTb = Kb·m·i with m the same (0.100 mol kg⁻¹) for all four, so the order follows i.
(A) Glucose: non-electrolyte, i = 1, ΔTb = 0.512 × 0.100 × 1 = 0.0512 K. (B) NaCl: strong electrolyte, i ≈ 2, ΔTb ≈ 0.102 K. (C) CaCl2: strong electrolyte, ideal i = 3, ΔTb ≈ 0.154 K (slightly less in practice). (D) Ethanoic acid: weak acid, i = 1 + α, with α ≈ √(Ka/c) = √(1.8 × 10⁻⁵/0.100) = √(1.8 × 10⁻⁴) ≈ 0.013. So i ≈ 1.013 and ΔTb ≈ 0.0519 K.
Order: A < D < B < C. Solution D is the one that cannot be placed confidently: its predicted elevation exceeds glucose's by only about 0.0007 K, far below the precision of a school ebulliometer, and i depends on Ka and on the exact concentration. Both student lists fail: the first puts the weak acid above CaCl2, the second puts it below glucose.
2. Interpret the benzoic acid result quantitatively. What is the van't Hoff factor, and what does it tell you about the species present?
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Expected molality: n = 1.20/122 = 9.84 × 10⁻³ mol in 0.0250 kg of benzene, so m = 0.3934 mol kg⁻¹. Expected ΔTf = 5.12 × 0.3934 = 2.01 K. Observed ΔTf = 1.62 K, so i = observed/expected = 1.62/2.01 ≈ 0.806.
A value below 1 means fewer particles than formula units, i.e. association, not experimental error. Carboxylic acids dimerise in non-polar solvents through two hydrogen bonds: 2 C6H5COOH ⇌ (C6H5COOH)2. If α is the fraction dimerised, i = 1 − α + α/2 = 1 − α/2, giving α = 2(1 − 0.806) ≈ 0.39. About 39% of the benzoic acid is present as dimer under these conditions.
This also explains the apparent molar mass: moles calculated from ΔTf would be 2.01/1.62 ≈ 1.24 times too small, giving about 151 g mol⁻¹ instead of 122.
3. In water, the same benzoic acid gives i slightly above 1 rather than below it. Explain the difference and estimate i for a 0.100 mol kg⁻¹ aqueous solution.
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In water the solvent competes for hydrogen bonding, so the cyclic dimer is broken up and the acid behaves as a normal solute, partly ionised: C6H5COOH ⇌ H⁺ + C6H5COO⁻. Ionisation increases the particle count, so i = 1 + α is greater than 1, opposite to the association seen in benzene.
With Ka = 6.3 × 10⁻⁵ (benzoic acid) and c = 0.100 mol kg⁻¹: α ≈ √(6.3 × 10⁻⁵/0.100) = √(6.3 × 10⁻⁴) ≈ 0.025, so i ≈ 1.025 - a depression about 2.5% larger than that of a non-electrolyte of the same molality.
The key judgement is that the solvent decides which effect dominates: hydrogen-bond-accepting water favours dissociation into ions, while a non-polar solvent favours self-association, so the sign of (i − 1) can flip for the same compound.