Freezing-point data from three colourless solutions
In a Class XII practical, three 0.100 m aqueous solutions are prepared from glucose, sodium chloride and acetic acid. Freezing points are read with a digital thermometer whose stated accuracy is ±0.1 K: glucose −0.186 °C, sodium chloride −0.350 °C and acetic acid −0.191 °C. The teacher reminds the group that Kf for water is 1.86 K kg mol−1 and that the thermometer was not recalibrated that morning. A faded label is found on a fourth bottle; 1.00 g of the solid, known to be a non-electrolyte, is dissolved in 50.0 g of water and the solution freezes at −0.620 °C. The group must decide which data can be trusted, estimate the molar mass of the unknown, and explain the behaviour of acetic acid without simply quoting ‘weak acid’.
1. Calculate the van’t Hoff factor i for each of the three 0.100 m solutions and state what each value implies about dissociation. Account for the sodium chloride value being below 2.
Read the worked answer
Use (i=\Delta T_f/(K_f m)), with (K_f m=1.86\times0.100=0.186\ \mathrm{K}).
- Glucose: (i=0.186/0.186=1.00). No dissociation; one particle per formula unit.
- NaCl: (i=0.350/0.186=1.88). Close to 2, so it is essentially fully dissociated into Na+ and Cl−. The small shortfall from 2 can arise from ion pairing and from the thermometer uncertainty; the difference corresponds to only 0.022 K, well inside the stated ±0.1 K.
- Acetic acid: (i=0.191/0.186=1.03). Slightly greater than 1, so only a small fraction of molecules has ionised; it is not fully dissociated like NaCl.
The comparison is meaningful because all three solutions have the same molality and the same solvent. The NaCl result should not be over-interpreted as ‘incomplete dissociation’ without first checking the thermometer, because a 0.1 K error changes i by about 0.5.
2. For acetic acid, estimate the degree of ionisation α from the freezing data and calculate an approximate Ka. Is the result chemically reasonable? What does the thermometer uncertainty do to your conclusion?
Read the worked answer
For a monobasic acid that ionises as HA ⇌ H+ + A−, the van’t Hoff factor is (i=1+\alpha) if no association occurs. Thus:
[ \alpha=i-1=1.027-1=0.027 ]
Using (c=0.100\ \mathrm{mol\ kg^{-1}}):
[ K_a=\frac{\alpha^2 c}{1-\alpha} =\frac{(0.027)^2(0.100)}{0.973} \approx 7.5\times10^{-5} ]
The order of magnitude is reasonable for a weak acid, but the value is larger than the accepted Ka of acetic acid (about (1.8\times10^{-5})). The freezing-point depression is small, so the thermometer uncertainty dominates: a ±0.1 K error in ΔT changes α by about ±0.54, which is larger than α itself. Therefore the data support partial ionisation qualitatively, but they cannot give a reliable Ka. A more precise method, such as pH measurement or conductometry, would be needed.
3. Calculate the molar mass of the unknown and identify the most plausible compound among urea (60), glucose (180) and acetamide (59). State the assumptions in your calculation and one reason the result could be low.
Read the worked answer
For the unknown:
[ \Delta T_f=0.620\ \mathrm{K},\quad m=\frac{0.620}{1.86}=0.333\ \mathrm{mol\ kg^{-1}} ]
Mass of solvent = (50.0\ \mathrm{g}=0.0500\ \mathrm{kg}), so:
[ n=0.333\times0.0500=0.0167\ \mathrm{mol} ]
[ M=\frac{1.00\ \mathrm{g}}{0.0167\ \mathrm{mol}}\approx60\ \mathrm{g\ mol^{-1}} ]
The value matches urea (60) and acetamide (59) closely; glucose (180) is ruled out. A nitrogen test or melting point would be needed to distinguish urea from acetamide. The calculation assumes the solute is a non-electrolyte, does not associate, and does not volatilise. If association occurred, i would be less than 1, so the calculated molar mass would be lower than the true value.