What is Van't Hoff Factor?

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 at the solvent determines the intensity of the effect or the joint ownership (tonoscopy, ebullioscopy, cryoscopy, osmoscopy). 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 suffers the phenomenon of dissociation (release of ions in the middle) or ionization (production of ions in the medium), increasing the number of particles.

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

To represent this

factor, Van't Hoff used the letter i, which starts a mathematical expression that takes into account the degree of dissociation (α) and the number of moles of each ion released on dissolution in water (q):

i = 1 + α .(q – 1)

Note: As α is provided as a percentage, whenever we use it in the expression of the Van't Hoff factor, we must divide it by 100 before.

After calculating the Van't Hoff correction factor, we can use it in the following practical situations:

• To correct the number of particles of a solute, obtained from its mass;

• To correct the colligative effect of osmoscopy, that is, the osmotic pressure of a solution:

π = M.R.T.i

In this case, we have the osmotic pressure (π) of the solution, the molar concentration (M), the general gas constant (R) and the solution temperature (T).

• To correct the colligative effect of tonometry, that is, correct the lowering of the maximum vapor pressure of the solvent in the solution:

?P = kt. W.i
 P₂

For this, we consider the absolute lowering (?p) of the maximum vapor pressure, the maximum vapor pressure of the solvent (p₂), the tonometric constant (Kt) and the molality (W).

• To correct the colligative effect of cryometry, that is, to correct the lowering of the freezing temperature of the solvent in the solution:

?θ = kc. W.i

In this case, we have the lowering of the freezing temperature of the solvent (? a), the cryometric constant (Kt) and the molality (W).

• To correct the colligative effect of ebulliometrics, that is, to correct the rise in the boiling temperature of the solvent in the solution:

?te = ke. W.i

For this, we have the increase in the boiling temperature of the solvent (?te), the ebulliometric constant (Ke) and the molality (W).

Follow now examples of calculation and application of the Van't Hoff factor:

1st Example: What is the iron chloride III (FeCl) correction factor value₃), knowing that its degree of dissociation is 67%?

Exercise data:

• i =?

• α = 67% or 0.67 (after dividing by 100)

• Formula of salt = FeCl₃

1st Step: Determine the number of moles (q) of ions released.

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.

2nd Step: Use the data in the formula of the Van't Hoff factor:

i = 1 + α .(q – 1)

i = 1 + 0.67.(4 - 1)

i = 1 + 0.67.(3)

i = 1 + 2.01

i = 3.01

2nd Example: What is the number of particles present in water when 196 grams of phosphoric acid (H₃DUST₄), whose degree of ionization is 40%, are they added to it?

Exercise data:

• i =?

• α = 40% or 0.4 (after dividing by 100)

• Acid formula = H₃DUST₄

1st Step: Calculate the molar mass of the acid.

To do this, we must multiply the atomic mass of the element by the atomic index and then add the results:

Molar mass = 3.1 + 1.31 + 4.16

Molar mass = 3 + 31 + 64

Molar mass = 64 g/mol

2nd Step: Calculate the number of particles present in 196 grams of H₃DUST₄.

This calculation is performed from a rule of three and uses the molar mass and the mass provided by the exercise, but always assuming that in a 1 mol there are 6.02.10²³ particles:

1mol of H₃DUST₄98 grams6.02.10²³ particles

196 gramsx

98.x = 196. 6,02.10²³

98.x = 1179.92.10²³

x = 1179,92.10²³
98

x = 12.04.10²³ particles

3rd Step: Determine the number of moles (q) of ions released.

Analyzing the formula for salt, we have index 3 in H and index 1 in PO4, so the number of moles of ions will be equal to 4.

Step 4: Use the data in the formula of the Vant’ Hoff factor:

i = 1 + α .(q – 1)

i = 1 + 0.4.(4 - 1)

i = 1 + 0.4.(3)

i = 1 + 1.2

i = 2.2

5th Step: Calculate the actual number of particles in the solution.

To do this, just multiply the number of particles found in the second step by the correction factor:

Number of particles = x.i

Number of particles = 12.04.10²³.2,2

Number of particles = 26,488.10²³ particles.

3rd Example: An aqueous solution of sodium chloride has a concentration equal to 0.5 molal. What is the value of the rise in the boiling point suffered by water, in ^OÇ? Data: Water Ke: 0.52^OC/molal; α of NaCl: 100%.

Exercise data:

• i =?

• α = 100% or 1 (after dividing by 100)

• Molality (W) = 0.5 molal

• Formula of salt = NaCl

• Ke = 0.52^OWith molal

1st Step: Determine the number of moles (q) of ions released.

Analyzing the formula for salt, we have index 1 in Na and index 1 in Cl, so the number of moles of ions is equal to 2.

2nd Step: Use the data in the formula of the Van't Hoff factor:

i = 1 + α .(q – 1)

i = 1 + 1.(2 - 1)

i = 1 + 1.(1)

i = 1 + 1

i = 2

3rd Step: Calculate the boiling point elevation suffered by the water, using the data provided, the Van't Hoff factor calculated in the second step, in the formula below:

?te = ke. W.i

?te = 0.52.0.5.2

?te = 0.52 ^OÇ

* Image credit: Boris 15/ shutterstock.com

By Me. Diogo Lopes Dias

Source: Brazil School - https://brasilescola.uol.com.br/o-que-e/quimica/o-que-e-fator-vant-hoff.htm