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.
Do not stop now... There's more after the advertising ;)
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: 0.52OC/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 = 0.52OWith 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
