Van’t Hoff Factor – Definition and how to find it?

Van’t Hoff Factor is a mathematical correction code and was proposed by the Dutch physicist and chemist Jacobus Henricus Van’t Hoff (1852-1911) in order to correct the number of dispersed particles of a solute in a solvent.

This correction of the number of particles is important because the amount of solute in the solvent determines the effect or intensity of the colligative property. Thus, the greater the number of particles, the greater the effect.

The need to correct the number of particles is due to the fact that, when an ionic solute dissolves in water, it undergoes the phenomenon of dissociation (release of ions in the medium) or ionization (production of ions in the medium), increasing the number of particles.

The number of particles in a molecular solute, however, does not need to be corrected by the Van’t Hoff factor because this type of solute does not ionize or dissociate and, therefore, does not change its quantity.

Van’t Hoff introduced a factor “i” which is known as the Van’t Hoff factor and is defined as the ratio of the experimental value of a colligative property to the calculated value of that property.

i = Experimental value of any colligative property / Calculated value of the same colligative property

Or we may write,

i = (∆Tb )obs / ( ∆Tb )cal

i = (∆Tf )obs / ( ∆Tf )cal

i = (∆P/P° )obs / ( ∆P/P° )cal

i = πobs / πcal

i = Mobs / Mcal

Where M is the molar mass of solute and ∆Tb, ∆Tf, ∆P/P°, and π are the elevation in boiling point, depression in freezing point, relative lowering in vapor pressure, and the osmotic pressure of the solution respectively. The subscripts ‘obs’ and ‘cal’ refer to the experimental and calculated values of the colligative properties.

Suppose we have a solution and this solution has a single component which is made up of solvent and this solvent is volatile in nature. Therefore, it shows some vapor pressure.

Let us have a solution that is made up of again solvent but within the solvent, the solute is dissolved. This solution is consists of two components (solute + solvent) and this solute is non-volatile in nature. By the presence of non-volatile solute, the escaping tendency of solvent is reduced which results in a lowering in vapor pressure, and this property is called colligative property. Colligative property depends upon a number of solute particles rather than the nature of the solute.

Effects of electrolytes on colligative properties:

There are two types of electrolytes:

  1. Strong electrolytes
  2. Weak electrolytes

suppose we have a solvent containing a non-electrolyte solute, non-electrolyte are not completely dissociated so, they will represent the same molecules and the colligative property of non-electrolyte depends on the no of solute particles in a solution. We have another solvent containing an electrolyte solute like NaCl, it dissociates into Na+ and Cl-. Now in the case of electrolytes, the no of particles in solution is increased because of the complete dissociation of electrolytes. Thereby, the colligative property also increased. As compared to non-electrolyte, electrolytes will have more colligative properties which can be estimated by Van’t Hoff factor ”i”.

Van’t Hoff factor ”i” :

”Total no. of moles of solute after dissociation/association divided by total no. of moles of solute before dissociation/association”

i = Total no. of moles of solute after dissociation or association / Total no. of moles of solute before dissociation or association

Case 1:

Dissociation of electrolytes:

In dissociation, a substance break into two or more particles and the determined molecular mass is less than the calculated value, but the colligative property change is greater than expected.

Let electrolytes are 100% dissociate,

  • NaCl → Na+ + Cl

1 mole 1 mole 1 mole

i = 2/1

  • MgCl2 → Mg2+ + 2Cl

1 mole 1 mole 2 moles

i = 3/1

  • Al2 (SO4 )3 → 2Al3+ + 3SO4 2-

1 mole 2 moles 3 moles

i = 5/1

Note:

In case of dissociation, the value of i is always greater than 1.

Now we talk about weak electrolytes that are not 100% dissociate, Degree of dissociation means no. of moles dissociate per mole. Let NaCl is not 100 % dissociate,

  • NaCl → Na+ + Cl

1 0 0

1-α α α

i = α + α + 1-α / 1

i = 1 + α

if 100 % dissociation mean α = 1, putt the value of α = 1, we get;

i = 2

  • MgCl2 → Mg2+ + 2Cl

1 0 0

1-α α 2α

i = α + 2α + 1-α / 1

i = 1 + 2α

if 100 % dissociation mean α = 1, putt the value of α = 1, we get;

i = 3

Note:

The value of I depends upon the type of electrolyte.

Case 2:

Association of electrolytes:

Solute molecules in solution may get associated to produce bigger molecules. In such cases, the number of solute molecules decreases and a colligative property shows a proportional decrease.

Let acetic acid is 100 % associated,

  • 2CH3 COOH → (CH3 COOH)2

2 moles 1 mole

i = 1/2

  • 2C6H5 COOH → (C6H5 COOH)2

2 moles 1 mole

i = 1/2

Note:

Carboxylic acid in benzene solution forms Dimerization (dimer, trimer, tetramer etc.). In the case of dissociation, the value of i is always smaller than 1.

If acetic acid is not 100 % associate,

  • 2CH3 COOH → (CH3 COOH)2

1 0

1- α α/2

i = 1- α + α/2 / 1

i = 1- α/2

if 100 % association mean α = 1, putt the value of α = 1, we get;

i = 1/2

How to estimate Van’t Hoff factor ”i”:

we can also write ”i”,

i = observed colligative property/calculated colligative property

i = no of moles of solute after dissociation or association / no of moles of solute before dissociation or association

  • In case of dissociation i > 1.
  • In case of association i < 1.

Proof that i = obs C.P / cal C.P:

Obs C.P ∝ no of moles of solute after dissociation/association ——- (1)

Cal C.P ∝ no of moles of solute before dissociation/association ——-(2)

Divided by (1) by (2)

Obs C.P / Cal C.P = no of moles of solute after dissociation or association / no of moles of solute before dissociation or association

Obs C.P / Cal C.P = i

Colligative properties of electrolytes:

  1. Lowering of vapor pressure PA° – PA / PA° = Xsolute × i
  2. Elevation in boiling point ∆Tb = Kb × m × i
  3. Depression in freezing point ∆Tf = Kf × m × i
  4. Osmotic pressure π = MRT × i

Note:

  • All these formulas is apply for electrolyte and non-electrolyte.
  • For non-electrolytes like urea, glucose i=1.

Problems related to Van’t Hoff factor ”i”:

Q 1: At a given temperature, which of the following will have maximum osmotic pressure,

  • 1 M NaCl
  • 1 M Na2 SO4
  • 1 M Al2 (SO4 )3 having 100% dissociate.

As, π = MRT

Temp is same (at given temperature)

Molar mass same (1M)

R general gas constant (constant)

but NaCl, Na2 SO4 and Al2 (SO4 )3 is electrolyte so, π = MRT is not valid.

Formula used: π = MRT × i

M, R, T is same so, π = i mean osmotic pressure depends on i.

  • NaCl → Na+ + Cl

i = 2

  • Na2 SO4 → 2Na+ + SO4 2-

i = 3

  • Al2 (SO4 )3 → 2Al3+ + 3SO4 2-

i = 5

Osmotic pressure of NaCl = π = i = 2

Osmotic pressure of Na2 SO4 = π = i = 3

Osmotic pressure of Al2 (SO4 )3 = π = i = 5

Order:

NaCl < Na2 SO4 < Al2 (SO4 )3

Q 2: Which of the following have maximum osmotic pressure at a given temperature,

  • 1 M NaCl
  • 1 M urea
  • 1 M C6H5COOH having 100% dissociation.

As, π = MRT × i

Temperature is the same (at a given temperature)

Molar mass same (1M)

R general gas constant (constant)

so, π = i

  • NaCl → Na+ + Cl

i = 2

  • Urea is neither dissociated nor associate because it is non-electrolyte so, i=0.
  • C6H5COOH is associated and forms a dimer in benzene solution 2C6H5 COOH → (C6H5 COOH)2 so, i=1/2

Osmotic pressure of NaCl = i = 2

Osmotic pressure of urea = i = 1

Osmotic pressure of C6H5COOH = i = 1/2

Order:

NaCl > Urea > C6H5COOH

Q 3: A 2 molal NaCl solution in water has an elevation in the boiling point of 1.88 K. if Kb for water is 0.52 k kg/mol. calculate ”i” and D.O.D ”α” for NaCl? (D.O.D)≠1

Given,

∆Tb = 1.88 k

m = 2 molal

Kb = 0.52

Calculate ”i” mean NaCl is not 100% dissociate,

  • NaCl → Na+ + Cl

1 0 0

1-α α α

i = α + α + 1-α / 1

i = 1 + α ——-(1)

As, ∆Tb = Kb × m × i

1.88 = 0.52 × 2 × i

i = 1.88 / 0.52 × 2

i = 1.807

put i = 1.807 in eq. 1

1.807 = 1 + α

α = 0.807 (80% dissociate)

Q 4: A solution ”A” is tetramerises in water up to 80%. if 2.5 g of ”A” in 100 g of water produces depression in freezing point 0.3° (Kf for water is 1.86 k kg/mol). calculate molar mass of ”A”.

Given,

∆Tf = 0.3°

Kf = 1.86

α = 80% = 0.8

m = 2.5 g

mass of solvent = 100 g

MA = ?

As, ∆Tf = Kf × m × i

”A” is tetramer 4A → A4

initially, 1 0

finally, 1-α α/4

i = 1-α + α/4 /1

i = 1 – 3α/4

as, α = 0.8

i = 1 – 3(0.8) / 4

i = 0.4

∆Tf = Kf × no of moles of solute / mass of solvent in kg × i

0.3 = 1.86 × 2.5/MA × 1000/100 × 0.4

0.3 = 1.86 × 2.5/MA × 10 × 0.4

MA = 1.86 × 2.5 × 10 × 0.4 / 0 .3

MA = 18.6 / 0.3

MA = 62 g

Q 5: What is the value of the iron chloride III (FeCl3) correction factor, knowing that its dissociation degree is 67%?

Given,

  • α = 67% or 0.67 (after dividing by 100)
  • Formula of salt = FeCl3
  • i =?

Step 1: 

Determine the number of moles (q) of released ions.

Analyzing the formula for salt, we have index 1 in Fe and index 3 in Cl, so the number of moles of ions is equal to 4.

Step 2:

Use the data in the Van’t Hoff factor formula :

i = 1 + α .(q – 1)

i = 1 + 0.67.(4 – 1)

i = 1 + 0.67.(3)

i = 1 + 2.01

i = 3.01

Abnormal molar mass:

Association:

  • 2CH3COOH (CH3COOH)2

M = 60 M = 120

(normal molar mass) (abnormal molar mass)

i = 1/2 = 60/120

i = normal molar mass/abnormal molar mass

Dissociation:

  • NaCl → Na+ + Cl

M =58.5 29.25 29.5

i = 2/1 = 58.5/29.25

i = normal molar mass/abnormal molar mass

Note:

  • In the case of dissociation, abnormal molar mass decreased.
  • In the case of an association, the abnormal molar mass increased.
  • If we don’t use ”i” and calculate molar mass using the colligative properties formula than, the molar mass comes out to be abnormal.

Proof that i = Normal molar mass / Abnormal molar mass:

i = no of moles of solute after dissociation or association / no of moles of solute before dissociation or association

i = obs colligative property/cal colligative property

Let any colligative property,

∆Tb = Kb × m

∆Tb = Kb × no of moles of solute/mass of solvent in kg

∆Tb = Kb × mass/molar mass /mass of solvent in kg

∆Tb ∝ 1/molar mass of solute

  • A colligative property is inversely proportional to the molar mass of the solute.
  • If colligative property is i = obs C.P/cal C.P so, molar mass is reverse.

i = cal molar mass/abnormal molar mass OR i = normal molar mass/abnormal molar mass

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