### The relationship between the van 't Hoff factor and the degree of dissociation

Following Wikipedia's van 't Hoff factor discussion, the van 't Hoff factor can be computed from the degree of ionization as follows:

i = αn + (1 - α)

where α is the degree of dissociation and n equals the number of ions formed from one formula unit of the substance. The formula above is often rearranged as follows:

i = 1 + α(n - 1)

The form just above is what I will use in the solutions below.

Example #1: What is the expected van 't Hoff factor for a substance (such as glucose) that does not ionize at all in solution.

Solution:

α = 0
n = 1

i = 1 + 0(1 - 1)

i = 1

Example #2: What is the expected van 't Hoff factor for a substance (such as NaCl) that ionizes into two ions per formula unit.

α = 1
n = 2

i = 1 + 1(2 - 1)

i = 2

Example #3: What is the osmotic pressure of a 0.30 M solution of MgSO4 if the MgSO4 is 80% dissociated at 20.0 °C?

Solution #1:

1) Calculate the van 't Hoff factor from the degree of dissociation:

α = 0.80
n = 2

i = 1 + 0.80(2 - 1)

i = 1.80

2) Solve for the osmotic pressure:

π = iMRT = (1.80)(0.30 mol/L) (0.08206 L-atm/mol-K) (293 K)

π = 12.98 atm

to two sig figs, 13 atm

Solution #2:

1) Magnesium sulfate ionizes as follows:

MgSO4 ---> Mg2+ + SO42-

2) Determine the concentration of all particles in solution:

For 80% ionization, [Mg2+] = 0.30 M x 0.8 = 0.24 = [SO42-]
[MgSO4] unionized = 0.3 M x 0.2 = 0.06 M
Total concentration of all species = 0.24 + 0.24 + 0.06 = 0.54 M

3) Solve for the osmotic pressure:

π = MRT = (0.54 mol/L) (0.08206 L-atm/mol-K) (293 K)

Note the lack of an explicit van 't Hoff factor. It is implicit in the development of the 0.54 M value.

π = 12.98 atm

to two sig figs, 13 atm

Example #4: 2.00 mols of Ba(ClO4)2 were placed in 1.00 L of solution at 45.0 °C. 15% of the salt was dissociated at equilibrium. Calculate the osmotic pressure of the solution.

Solution #1:

1) Calculate the van 't Hoff factor from the degree of dissociation:

α = 0.15
n = 3

i = 1 + 0.15(3 - 1)

i = 1.30

2) Solve for the osmotic pressure:

π = iMRT = (1.30) (2.00 mol/L) (0.08206 L-atm/mol-K) (318 K)

π = 67.85 atm

to three sig figs, 67.8 atm

Solution #2:

Ba(ClO4)2 ---> Ba2+ + 2ClO4¯

[Ba2+] = 2 M times (1 x 0.15) = 0.3 M
[ClO4¯] = 2 M (2 x 0.15) = 0.6 M

[Ba(ClO4)2] = 2 M x 0.85 = 1.7 M (this is the undissociated Ba(ClO4)2

Total molarity of all ions and undissociated salt = 1.7 M + 0.3 M + 0.6 M = 2.6 M

π = MRT = (2.6) (0.08206) (318) = 67.8 atm

Example #5: Find the osmotic pressure of an aqueous solution of BaCl2 at 288 K containing 0.390 g per 60.0 mL of solution. The salt is 60.0% dissociated.

Solution:

1) Calculate the van 't Hoff factor from the degree of dissociation:

Method One:
α = 0.60
n = 3

i = 1 + 0.60(3 - 1)

i = 2.20

Method Two:

In solution, we have this situation:

40.0% --> that's the undissociated BaCl2, call it 1 unit . . .
60.0% --> that's 1 Ba + 2 Cl, call it 3 units . . .

Therefore:

i = (0.400 x 1) + (0.600 x 3) = 2.20

2) Calculate the molarity of the barium chloride solution:

MV = mass / molar mass

(x) (0.0600 L) = 0.390 g / 208.236 g/mol

x = 0.0312146 M (keep some guard digits)

3) Solve for the osmotic pressure:

π = iMRT = (2.20) (0.0312146 mol/L) (0.08206 L-atm/mol-K) (318 K)

π = 1.623 atm

to three sig figs, 1.62 atm

Example #6: 3.58 g of NaCl was dissolved in 120.0 mL of solution at 77.0 °C. The osmotic pressure is 26.31 atm. Calculate the degree of dissociation of NaCl.

Solution #1:

1) Calculate the molarity of the NaCl:

MV = mass / molar mass

(x) (0.1200 L) = 3.58 g / 58.443 g/mol

x = 0.510469 M

2) Calculate the van 't Hoff factor:

26.31 atm = (i) (0.510469 mol/L) (0.08206 L atm / mol K) (350 K)

i = 1.7945

3) Use the van 't Hoff factor to determine the percent dissociation:

α = x
n = 2

1.7945 = 1 + x(2 - 1)

x = 0.7945 = 79.45% dissociated

Solution #2:

1) Calculate the molarity of all particles in solution:

NaCl(aq) ---> Na+ + Cl¯

When 'x' amount of NaCl ionizes, the [NaCl] goes down by 'x' and both [Na+] and [Cl¯] go up by 'x.' Therefore, when all dissociation at equilibrium, we have this in solution:

[NaCl] = 0.510469 - x
[Na+] = x
[Cl¯] = x

and the total molarity of everything is solution is this:

0.510469 - x + x + x = 0.510469 + x

2) Let us solve for x:

26.31 = (0.510469 + x) (0.08206) (350)

26.31 = 14.6612 + 28.721x

x = 0.405585 M <--- this is the concentration of the NaCl that ionized

3) Calculate the percent dissociation:

0.405585 M / 0.510469 M = 0.7945 = 79.45%