Solutions · Lesson 10 of 11
Association, Dissociation and Abnormal Molar Mass
“Count particles after chemical equilibria and use the van ’t Hoff factor to interpret colligative measurements.”
• Explain why association and dissociation alter apparent molar mass. • Derive particle-count formulas for the van ’t Hoff factor. • Apply corrected colligative equations and distinguish analytical from effective concentration. • Infer weak-acid dissociation and an approximate dissociation constant from freezing data.
Formula units and independent particles are different counts
One mole of glucose normally contributes approximately one mole of independent dissolved molecules. One mole of KCl can supply about one mole K⁺ plus one mole Cl⁻ in dilute water. By contrast, two carboxylic acid molecules can associate into one hydrogen-bonded dimer in a suitable non-polar solvent. Colligative measurements respond to independently acting particles, so the analytical amount calculated from weighed solute is not always the effective particle amount.
If KCl supplies approximately twice as many particles as an equal analytical amount of glucose, its thermal or osmotic effect is approximately twice as large in the ideal dilute comparison. If a molar-mass formula assuming one particle per KCl formula unit is applied, it infers too many “molecules” and a molar mass about half the actual formula molar mass. Association has the opposite effect: fewer particles give a smaller measured effect and an apparent molar mass larger than the monomer value.
In an ideal particle-count model, i is the ratio of independent particle amount after the change to initial solute amount. It is greater than one for dissociation and less than one for association. In real electrolyte solutions, a measured i is an effective factor that also reflects ionic interactions. A non-integer i does not by itself prove that a strong electrolyte remains partly undissociated as neutral molecules.
Derive the dissociation expression by counting
Start with one mole of a solute that forms ν ions per formula unit. If a fraction α dissociates, 1−α moles of formula units remain and να moles of ions form. Total particle amount is (1−α)+να=1+(ν−1)α. Dividing by the original one mole gives i. This elementary model neglects interactions and any other equilibria.
| Salt | i at 0.1 mol kg⁻¹ | i at 0.01 mol kg⁻¹ | i at 0.001 mol kg⁻¹ | Ideal dilute complete-dissociation limit |
|---|---|---|---|---|
| NaCl | 1.87 | 1.94 | 1.97 | 2 |
| KCl | 1.85 | 1.94 | 1.98 | 2 |
| MgSO₄ | 1.21 | 1.53 | 1.82 | 2 |
| K₂SO₄ | 2.32 | 2.70 | 2.84 | 3 |
These effective factors tend toward the stoichiometric limits on dilution as ionic non-ideal effects diminish. MgSO₄ can exhibit particularly strong ion interactions and ion pairing. It is misleading to interpret every shortfall from an integer solely as incomplete dissociation; the ideal particle-count formula is a simplified model, while the measured property also contains non-ideality.
Association reduces the number of independent particles
Suppose a fraction α of initial monomer units forms n-mers. From one initial mole of monomers, 1−α mole remains free and α/n mole of aggregates forms. The aggregate amount is α/n because each aggregate consumes n monomers. Add the two independent particle amounts to obtain i=1−α+α/n. For complete dimerisation i=1/2, while the apparent molar mass is twice the monomer mass.
Carboxylic acids can form cyclic dimers in benzene through two hydrogen bonds between their carboxyl groups. The carbon skeleton and covalent O–H bonds remain present; the dotted intermolecular contacts are hydrogen bonds. A dimer counts as one associated particle in this simplified model. In a different solvent, hydrogen bonding with solvent and acid ionisation can favour different behaviour.
Problem
A 2.00 g benzoic acid sample, monomer molar mass 122 g mol⁻¹, in 25 g benzene gives ΔT_f=1.62 K. Use the problem’s K_f=4.9. Assume only dimerisation.
- 1.Ignoring association initially, apparent molar mass=1000×4.9×2/(1.62×25)=241.9753 g mol⁻¹.
- 2.i=122/241.9753=0.5041837. Fewer effective particles than monomers are indicated.
- 3.For dimers i=1−α/2, so α=2(1−i)=0.9916327, or 99.1633%≈99.2%.
- 4.The degree is below one and i is slightly above 0.5, consistent with nearly complete dimerisation. Use 4.9 throughout rather than substituting a different table value midway.
Weak-acid dissociation links a colligative result to equilibrium
For a monoprotic acid in water, write HA + H₂O ⇌ H₃O⁺ + A⁻. One acid unit creates two solute ions, while solvent water is not counted as a new solute species. If α is the dissociated fraction, the ideal result is i=1+α. At analytical molarity C, [HA]=C(1−α) and [H₃O⁺]=[A⁻]=Cα, neglecting water autoionisation. The concentration-form equilibrium expression is K_a≈Cα²/(1−α). Thermodynamic K_a is dimensionless when activities are referenced to their standard states; this concentration approximation is used for dilute calculations.
Problem
A 0.600 mL sample of acetic acid of density 1.06 g mL⁻¹ is added to 1.00 L water. Observed ΔT_f=0.0205 K. Use acid 60 g mol⁻¹, K_f=1.86 and water density 1 g mL⁻¹. Estimate i, α and K_a.
- 1.Acid mass=0.600×1.06=0.636 g; amount=0.636/60=0.0106 mol. Solvent mass≈1.00 kg, giving analytical m=0.0106 mol kg⁻¹.
- 2.Undissociated prediction K_fm=1.86×0.0106=0.019716 K. Therefore i=0.0205/0.019716=1.0397647.
- 3.α=i−1=0.0397647, about 3.98%. Take final solution volume as approximately 1.00 L, so C≈0.0106 M.
- 4.K_a≈0.0106×α²/(1−α)=1.74551×10⁻⁵. This is a conditional estimate using ideal concentration behaviour and the supplied near-freezing data.
- 5.Rounding 0.019716 to 0.0197 before calculating i gives a noticeably different α and K_a because α is the small difference i−1. Preserve guard digits and avoid claiming more accuracy than the measurement supports.
Problem
Dissolve 10.0 g CH₃CH₂CHClCOOH, molar mass 122.5 g mol⁻¹, in 250 g water. Given K_a=1.4×10⁻³, K_f=1.86, estimate depression using C≈m numerically.
- 1.Amount=10/122.5=0.08163265 mol. Molality=0.08163265/0.250=0.3265306 mol kg⁻¹; assume C≈0.3265306 M.
- 2.From K_a=Cα²/(1−α), obtain Cα²+K_aα−K_a=0. Take its positive root: α=[−K_a+√(K_a²+4CK_a)]/(2C)=0.0633703.
- 3.Then i=1+α=1.0633703 and ΔT_f=iK_fm≈0.645835 K.
- 4.This uses an estimated molarity because no final volume or density was supplied. A more exact result requires additional composition-volume and activity information.
Electron-withdrawing substituents can stabilise a carboxylate conjugate base and increase acid dissociation. For equal analytical amounts in equal water masses under comparable conditions, acetic acid generally gives a smaller freezing depression than trichloroacetic acid, which in turn gives a smaller effect than trifluoroacetic acid because the effective ion count rises with stronger dissociation. Do not infer an unlimited depression: for a simple monoprotic acid model i approaches a maximum of two.
If a derived α lies outside zero to one, check units, concentration definitions, numerical precision and whether the assumed association or dissociation process is adequate. Strong electrolyte interactions, multiple equilibria and measurement error can invalidate a simple conversion from i to a molecular dissociation fraction.
Quick Quiz
Quiz
Dissociation of one formula unit into three independent ions has ideal limiting i equal to:
Explanation: Complete dissociation gives three independent particles per original formula unit.
Complete dimerisation gives which apparent molar mass if a one-particle formula is used?
Explanation: The particle number halves, so the inferred molar mass doubles and i=0.5.
For a monoprotic weak acid with dissociation fraction 0.10 in the ideal model, i is:
Explanation: i=1+α because each dissociated acid produces two solute ions instead of one acid molecule.
What does i=1.85 for aqueous KCl prove by itself?
Explanation: Measured i also reflects interionic non-ideality. A precise dissociation fraction requires a justified model.
Which concentration belongs in ΔT_f=iK_fm?
Explanation: m is calculated from weighed solute formula units; multiplying by i once accounts for the effective particle factor.
Practice Problems
- 19.5 g fluoroacetic acid, molar mass 78 g mol⁻¹, is dissolved in 500 g water. ΔT_f=1.00 K and K_f=1.86. Find i, α and K_a assuming C≈m.
- What mass of CaCl₂ gives Π=0.75 atm at 300 K in a final volume approximated as 2.5 L, if i=2.47? Use molar mass 111 and R=0.082057 L atm mol⁻¹ K⁻¹.
- Find Π for 25 mg K₂SO₄ in approximately 2 L solution at 25 °C, assuming complete ideal dissociation. Use molar mass 174 and R=0.082057.
- A solute forms trimers; 60% of initial monomer units associate. Find i and the ratio of apparent to monomer molar mass.
- A salt produces four ions per formula unit in the complete ideal limit. A simplified particle-count model gives i=2.5. Find α and state the limitation.
- Explain why acetic acid may show i<1 in benzene but i>1 in water.
Amount=19.5/78=0.250 mol; m=0.250/0.500=0.500 mol kg⁻¹. i=1.00/(1.86×0.500)=1.0752688. For one acid producing two solute ions, α=0.0752688. With C≈0.500 M, K_a≈0.500×α²/(1−α)=0.00306327, or about 3.06×10⁻³. This last step assumes density near unity and negligible volume change.
Key Takeaways
• Colligative effects depend on effective independent particles, not simply formula-unit amount. • Dissociation commonly gives i>1 and an apparent molar mass below the normal value. • Association commonly gives i<1 and an apparent molar mass above the monomer value. • For a simple dissociation model, i=1+(ν−1)α. • For association into n-mers, i=1−α(1−1/n). • Use analytical concentration and apply the particle factor once. • Real electrolyte factors can reflect ionic interactions as well as chemical equilibria. • Weak-acid constants inferred from colligative data require concentration, activity and rounding assumptions.