Question

Derive the relationship between Vant Hoff factor and degree of dissociation

Answer 1

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 MgSO4is 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% dissociated.

Solution:

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

α = 0.60 
n = 3

i = 1 + 0.60(3 - 1)

i = 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


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