Van’t Hoff Factor

The van’t Hoff factor (i) measures a solute's effect on the solute's colligative properties, such as boiling point elevation, freezing point depression, vapour pressure lowering, and osmotic pressure. 

1.0Introduction 

It has been observed that the difference between the observed and calculated molecular masses of a solute is due to the association or dissociation of solute molecules in the solution. This phenomenon leads to a change in the number of particles present in the solution.

In 1880, Van’t Hoff introduced the Van’t Hoff factor (i) to account for this difference. The factor i is defined as:

The ratio of the actual number of particles in solution to the expected number of formula units dissolved.

This factor quantifies the degree to which a substance undergoes association or dissociation in a solution.

For instance:

• For a non-electrolyte (a substance that doesn't dissociate into ions, such as sugar), i = 1, meaning it does not change the number of particles when dissolved.
• For an ionic compound like NaCl, i equals the total number of ions produced when the compound dissociates. For example, CaCl₂ dissociates into one Ca²⁺ ion and two Cl⁻ ions, so ideally, i = 3. However, ion pairing or association in the solution may reduce the total number of particles and thus lower the observed Van’t Hoff factor.

The Van’t Hoff factor is named after Jacobus Henricus Van’t Hoff, a Dutch chemist who first received the Nobel Prize in Chemistry. It is important to note that the measured value of i in electrolytic solutions is typically lower than the predicted value due to the pairing of ions, and the deviation becomes more significant when the ions carry a higher charge.

Association

Association refers to the process in which particles combine to form larger entities, while dissociation involves splitting molecules into smaller particles or ions.

When particles aggregate into larger units, the number of dissolved particles decreases. For example:

• Carboxylic acids may dimerize in benzene, meaning two molecules join to form a pair.
• Effects:
• • The observed molar mass will be greater than the predicted value.
  • The Van’t Hoff factor (i) will be less than 1.
  • Due to fewer particles in the solution, the colligative properties (such as boiling point depression and freezing point depression) will be lower than expected.

Dissociation

When molecules separate into ions, the number of dissolved particles increases. For example:

• (NaCl) Sodium chloride dissociates into Na⁺ and Cl⁻ ions in water.
• Effects:
• • The observed molar mass will be less than expected.
  • The Van’t Hoff factor (i) will be greater than 1.
  • Due to the increased number of particles, the colligative properties (like osmotic pressure and boiling point elevation) will be higher than expected.

2.0Abnormal Molar Masses

Sometimes, the calculated molar mass from colligative properties doesn't match the expected value. These discrepancies are known as abnormal molar masses.

• Dissociation increases the number of particles in solution, enhancing colligative properties (e.g., freezing point depression). The calculated molar mass will be lower than the actual mass because the formula only accounts for the original solute molecules, not the dissociated particles.
• On the other hand, if a substance is associated with a solution, the total number of particles in the solution decreases. This leads to a weaker effect on colligative properties, and the calculated molar mass will appear higher than expected.

For example:

• If NaCl is dissolved in water, it dissociates into 2 moles of ions (Na⁺ and Cl⁻) per mole of NaCl. However, we typically consider only 1 mole of NaCl when calculating molar mass based on colligative properties.

The overall abnormality arises from the number of solute particles influencing the colligative properties and the molar mass. Therefore:

• Dissociation results in more particles and lower molar mass.
• Association results in fewer particles and higher molar mass than expected

3.0Formula of Van’t Hoff Factor

 It is defined mathematically as:

$i=\frac{\text{Total number of moles of particles after association/dissociation}}{\text{Number of moles of solute before association/dissociation}}$

In terms of measurable colligative properties, the factor can also be expressed as:

$i=\frac{\text{Observed colligative property}}{\text{Calculated colligative property (assuming no dissociation/association)}}$

$i=\frac{\text{Normal molar mass}}{\text{Abnormal molar mass}}$

Since most ionic compounds dissociate or associate in solution, the expressions for colligative properties must be corrected using the Van’t Hoff factor.

4.0Modified Colligative Property Equations

Relative lowering of vapour pressure:

$PAo=iXB\frac{P_A^o-P_A}{P_A^o}=i$

Where:

$P_A^o$ = vapour pressure of the pure solvent

$P_A$ = vapour pressure of the solvent in the solution

$X_B$ = mole fraction of solute

Elevation in boiling point:

$\Delta T_b=i{K}_b$
Where:

ΔTb​ = elevation in boiling point

$K_b$ = molal elevation constant

m = molality

Depression in freezing point:

$\Delta T_f=i{K}_{f}m$

Where:

$\Delta T_f=$ = depression in freezing point

$K_f$ = molal depression constant

Osmotic pressure:

$\Pi =iCRT$

Where:

$\Pi$ = osmotic pressure

C= molar concentration of the solution

R = universal gas constant

T = absolute temperature

5.0Applications of the Van’t Hoff Factor

(a) Calculation of the Degree of Dissociation (α)

When a solute dissociates, it splits into multiple particles. Suppose a molecule Aₙ dissociates into nA in solution:

Total number of moles in solution:

$(1-\alpha )+n\alpha =1+(n-1)\alpha$

The Van’t Hoff factor is:

$i=\frac{\text{Moles of solute actually present in the solution}}{\text{Moles of solute originally dissolved}}$

$i=(1-\alpha )+n\alpha 1i=\frac{(1-\alpha )+n\alpha }{1}$

Solving for the degree of dissociation (α):

$\alpha =\frac{i-1}{n-1}$

(b) Calculation of the Degree of Association (α)

When a solute associates, multiple molecules combine into one larger unit, reducing the number of solute particles. Suppose n molecules of A associate to form Aₙ:

$nA\rightleftharpoons A_n$

Total number of moles in solution:

$(1-\alpha )+\frac{\alpha }{n}$

6.0Solved Examples

1. A solution of $BaCl2BaC{l}_{2}BaCl2$ has an observed osmotic pressure of 8 atm, while the calculated osmotic pressure (assuming no dissociation) is 5 atm. Find the degree of dissociation (α) of $BaCl2BaC{l}_{2}BaCl2.$

Solution:

Van’t Hoff factor formula:

$i=\frac{\text{Observed Colligative Property}}{\text{Calculated Colligative Property}}$

$i=\frac{8}{5}=1.6$

Dissociation equation for BaCl₂​:

$BaC{l}_2\rightleftharpoons B{a}^{2+}+2C{l}^{-}$

Total number of particles after dissociation:

$(1-\alpha )+\alpha +2\alpha =1+2\alpha$

Van’t Hoff factor relation: $i=1+2\alpha$

Substituting $1.6=1+2\alpha$ ,

2α = 0.62

α=0.3

Answer: Degree of dissociation (α) = 0.3 (or 30%)

2. A 0.5 m glucose solution (a non-electrolyte) is prepared in water. What will be its boiling point? Given Kb=0.52 K⋅kg⋅mol⁻¹   

Solution:

Since glucose is a non-electrolyte, it does not dissociate.

So, the Van’t Hoff factor (i) = 10

The formula for boiling point elevation:

ΔTb​=iKb​m

Substituting the values:

ΔT_b​=1×0.52×0.5 

ΔT_b=0.26K

Boiling point of pure water = 100°C

T_b=100+0.26=100.26°CT_b = 100 + 0.26 = 100.26°CTb​=100+0.26=100.26°C

Answer: 100.26°C

3. A solution of Al₂(SO₄)₃ in water shows 80% ionization. Find the Van’t Hoff factor (iii).

Solution:

Dissociation equation:

Al₂(SO₄)₃⇌2Al³⁺+3SO₄²⁻

The compound dissociates into five ions per formula unit:

• 2Al³⁺ (2 moles)
• 3SO₄²⁻ (3 moles)

Degree of ionization (α) = 80% = 0.80

The formula for Van’t a Hoff factor:

$i=1+(\text{total number of particles after dissociation}-1)\times \alpha$

Here, the total number of particles after dissociation = 5

i=1+(5−1)×0.80 

i=1+4×0.80

i=1+3.2=4.2

Answer: i=4.2