Osmotic Pressure

Key Concepts

• Osmotic pressure arises when two solutions of different concentrations, or a pure solvent and a solution, are separated by a semipermeable membrane.
  Molecules such as solvent molecules that can pass through the membrane will migrate from the side of lower solute concentration to the side of higher solute concentration in a process known as osmosis.
  So the dilute solution becomes more concentrated over time, and at the same time, the concentrated solution becomes more dilute.
• The pressure required to stop osmosis is called the osmotic pressure.
• In dilute solutions, osmotic pressure (Π) is directly proportional to the molarity of the solution (c) and its temperature in Kelvin (T).

  Π ∝ cT

• van't Hoff Equation: Π = cRT

  Π = osmotic pressure
  c = molarity = moles ÷ volume (L)
  R = ideal gas constant
  T = temperature (K)

• Solvent can be removed from a solution using a pressure greater than the osmotic pressure.
  This is known as reverse osmosis.

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Example: Osmotic Pressure Calculation for a Nonelectrolyte Solution

Osmotic Pressure Problem:

Calculate the osmotic pressure exhibited by a 0.10 mol L^-1 sucrose solution at 20^oC.

Osmotic Pressure Problem Solution:

1. For pressure in atmospheres (atm)

   Π = cRT
   c = 0.10 mol L^-1
   R = 0.0821 L atm K^-1mol^-1 (from data sheet)
   T = 20^oC = 20 + 273 = 293 K

   Π = 0.10 x 0.0821 × 293 = 2.4 atm

2. For pressure in kilopascals (kPa)

   Π = cRT
   c = 0.10 mol L^-1
   R = 8.314 J K^-1mol^-1 (from data sheet)
   T = 20^oC = 20 + 273 = 293 K

   Π = 0.10 × 8.314 x 293 = 243.6 kPa

Example: Osmotic Pressure Calculation for an Electrolyte Solution

Osmotic Pressure Problem:

Calculate the osmotic pressure exhibited by a 0.42 mol L^-1 KOH solution at 30^oC.

Osmotic Pressure Problem Solution:

1. For pressure in atmospheres (atm):
   Π = cRT
   Since KOH → K⁺_(aq) + OH^-_(aq)
   c(K⁺_(aq)) = 0.42 mol L^-1
   c(OH^-_(aq)) = 0.42 mol L^-1
   c(solute) = 0.42 + 0.42 = 0.84 mol L^-1
   R = 0.0821 L atm K^-1mol^-1 (from data sheet)
   T = 30^oC = 30 + 273 = 303 K

   Π = 0.84 × 0.0821 × 303 = 20.9 atm

2. For pressure in kilopascals (kPa)
   Π = cRT
   Since KOH → K⁺_(aq) + OH^-_(aq)
   c(K⁺_(aq)) = 0.42 mol L^-1
   c(OH^-_(aq)) = 0.42 mol L^-1
   c(solute) = 0.42 + 0.42 = 0.84 mol L^-1
   R = 8.314 J K^-1mol^-1 (from data sheet)
   T = 30^oC = 30 + 273 = 303 K

   Π = 0.84 × 8.314 × 303 = 2116 kPa

Example: Calculation of Molar Mass Using Osmotic Pressure

Osmotic Pressure Problem:

0.500 g hemoglobin was dissolved in enough water to make 100.0 mL of solution.
At 25^oC the osmotic pressure was found to be 1.78 × 10^-3 atm.
Calculate the molar mass of the hemoglobin.

Osmotic Pressure Problem Solution:

1. Calculate the molarity, c, of the solution:
   c = Π ÷ RT
   Π = 1.78 × 10^-3 atm
   R = 0.0821 L atm K^-1 mol^-1 (from data sheet)
   T = 25^oC = 25 + 273 = 298 K
   c = 1.78 × 10^-3 ÷ (0.0821 × 298) = 7.28 × 10^-5 mol L^-1
2. Calculate the moles, n, of hemoglobin present in solution:
   n = c × V
   c = 7.28 × 10^-5 mol L^-1
   V = 100.0 mL = 100.0 × 10^-3 L
   n = 7.28 × 10^-5 × 100.0 × 10^-3 = 7.28 × 10^-6 mol
3. Calculate the molar mass of the hemoglobin:
   molar mass = mass ÷ moles
   mass = 0.500 g
   moles = 7.28 × 10^-6 mol
   molar mass = 0.500 ÷ (7.28 × 10^-6) = 68 681 g mol^-1