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Chemistry · Ch 9 — Solutions

Van't Hoff Factor

9.11.2

Van't Hoff Factor

To quantify exactly how much a solute's association or dissociation shifts the observed (abnormal) molar mass away from its true value, van't Hoff introduced a factor, denoted ii and now called the van't Hoff factor. It is defined as the ratio of the solute's true (normal) molar mass to its observed (abnormal) molar mass -- and, equivalently, as the ratio of the observed colligative property to the colligative property calculated assuming no association or dissociation:

i=Normal (actual) molar massObserved (abnormal) molar mass=Observed colligative propertyCalculated colligative propertyi = \frac{\text{Normal (actual) molar mass}}{\text{Observed (abnormal) molar mass}} = \frac{\text{Observed colligative property}}{\text{Calculated colligative property}}

For example, the estimated van't Hoff factor for an acetic acid solution in benzene is about i=0.5i=0.5 (consistent with the dimerisation of section 9.11.1, which roughly halves the true particle count), while for a sodium chloride solution in water it is about i=2i=2 (consistent with complete dissociation into two ions).

Relating ii to the degree of dissociation or association. If a solute dissociates into nn ions/species per formula unit, its degree of dissociation is

αdissociation=i−1n−1\alpha_{dissociation} = \frac{i-1}{n-1}

If instead nn solute molecules associate together into a single aggregate, the degree of association is

αassociation=(1−i) nn−1\alpha_{association} = \frac{(1-i)\,n}{n-1}

Incorporating ii into the four colligative-property equations. Once association or dissociation is accounted for, the equations of sections 9.9.1-9.9.4 generalise by simply inserting a factor of ii:

  • Relative lowering of vapour pressure: Psolvent∘−PsolutionPsolvent∘=i nsolutensolvent\dfrac{P^\circ_{solvent}-P_{solution}}{P^\circ_{solvent}} = i\,\dfrac{n_{solute}}{n_{solvent}}
  • Elevation of boiling point: ΔTb=i Kb m\Delta T_b = i\,K_b\,m
  • Depression of freezing point: ΔTf=i Kf m\Delta T_f = i\,K_f\,m
  • Osmotic pressure: π=i wsoluteMsolute⋅RTV\pi = i\,\dfrac{w_{solute}}{M_{solute}}\cdot\dfrac{RT}{V}

Interpreting the three regimes of ii:

  • i=1i=1: the solute neither dissociates nor associates; the molar mass calculated from the colligative property matches the true (actual) molar mass exactly.
  • i<1i<1: the solute associates in solution; the observed (abnormal) molar mass comes out greater than the true molar mass.
  • i>1i>1: the solute dissociates in solution; the observed (abnormal) molar mass comes out less than the true (normal) molar mass. …
Misc Example Problem 7Van't Hoff factor of NaCl from freezing point depression

Worked out. 1 g NaCl in 200 g water gives ΔTf=0.24\Delta T_f=0.24 K (Kf=1.86K_f=1.86 K kg mol−1^{-1}). The (abnormal) molar mass from the depression is M=1000×Kf×wNaClΔTf×wwater=1000×1.86×10.24×200=38.75 g mol−1M=\dfrac{1000\times K_f\times w_{NaCl}}{\Delta T_f\times w_{water}}=\dfrac{1000\times1.86\times1}{0.24\times200}=38.75\ \text{g mol}^{-1}. NaCl's true (theoretical) molar mass is 58.5 g mol−1^{-1}, so i=58.538.75=1.50i=\dfrac{58.5}{38.75}=1.50. …

Misc Evaluate Yourself 14Van't Hoff factor of KCl from its freezing point

Worked out. An in-text practice box: a 0.2 m aqueous solution of KCl freezes at −0.68∘-0.68^\circC; calculate the van't Hoff factor, given Kf=1.86K_f=1.86 K kg mol−1^{-1} for water. …