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 = 1i = 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 = 2i = 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 = 2i = 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 = 3i = 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 = 3i = 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 = 21.7945 = 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¯] = xand 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%