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 MgSO_{4} if the MgSO_{4} 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:

MgSO_{4}---> Mg^{2+}+ SO_{4}^{2-}

2) Determine the concentration of all particles in solution:

For 80% ionization, [Mg^{2+}] = 0.30 M x 0.8 = 0.24 = [SO_{4}^{2-}]

[MgSO_{4}] 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(ClO_{4})_{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(ClO_{4})_{2}---> Ba^{2+}+ 2ClO_{4}¯[Ba

^{2+}] = 2 M times (1 x 0.15) = 0.3 M

[ClO_{4}¯] = 2 M (2 x 0.15) = 0.6 M[Ba(ClO

_{4})_{2}] = 2 M x 0.85 = 1.7 M (this is the undissociated Ba(ClO_{4})_{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 BaCl_{2} 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 BaCl

_{2}, 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%