Osmosis, osmotic pressure and reverse osmosis
Solvent moves through a semipermeable membrane into the stronger solution: osmosis. The pressure that just stops it is π = CRT; pushing harder reverses the flow.
Osmosis and osmotic pressure
A semipermeable membrane (SPM) lets solvent molecules through but not solute molecules. Animal bladder, parchment, cellophane and cell membranes behave like this. When an SPM separates a solvent from a solution, the solvent flows from the pure solvent (or dilute solution) into the more concentrated solution. This flow is osmosis. The extra pressure that must be applied on the solution to just stop this flow is its osmotic pressure, π. For dilute solutions, π = CRT, where C is the molar concentration.
How to draw the osmosis set-up
- Draw a beaker of pure water.
- Dip into it an inverted thistle funnel whose mouth is tied over with a semipermeable membrane, filled with sugar solution.
- Show the solution level rising in the stem, with arrows for water moving inward through the membrane.
- For osmotic pressure, draw a piston on the solution side, pressing just enough to stop the rise. Label: solvent, solution, SPM, piston, π.
- For reverse osmosis, draw the same set-up with a bigger arrow on the piston labelled 'pressure > π', and water moving out of the solution through the membrane.
Why osmotic pressure suits big molecules
Osmotic pressure is measured at room temperature. Its value is large even for very dilute solutions, while the change in boiling or freezing point would be too small to measure. So it is the method of choice for the molar mass of proteins and polymers: biomolecules break down on heating, and polymers dissolve only a little. It uses molarity: π = (n₂ ÷ V) RT, so M₂ = (w₂RT) ÷ (πV).
Worked example: molar mass of a protein
2.0 g of a protein is dissolved in water to make 100 mL of solution. Its osmotic pressure at 300 K is 8.3 × 10⁻³ bar. Find the molar mass of the protein (R = 0.083 L bar mol⁻¹ K⁻¹).
- Given: w₂ = 2.0 g, V = 100 mL = 0.100 L, T = 300 K, π = 8.3 × 10⁻³ bar.
- Formula: M₂ = (w₂ × R × T) ÷ (π × V).
- Top: 2.0 × 0.083 × 300 = 49.8 (g L bar mol⁻¹).
- Bottom: 8.3 × 10⁻³ × 0.100 = 8.3 × 10⁻⁴ (bar L).
- M₂ = 49.8 ÷ (8.3 × 10⁻⁴) = 6.0 × 10⁴ g mol⁻¹.
- Compare: the same solution would lower the freezing point by only about 6 × 10⁻⁴ K, far too small to measure.
M₂ = 6.0 × 10⁴ g mol⁻¹
Isotonic, hypertonic and hypotonic
| Solution around a cell | Compared with the fluid inside the cell | What happens to the cell |
|---|---|---|
| Isotonic, such as 0.9% NaCl (normal saline) | Same osmotic pressure | No net flow of water; the cell keeps its shape |
| Hypertonic, such as 2% NaCl | Higher osmotic pressure | Water flows out; the cell shrinks |
| Hypotonic, such as 0.5% NaCl or pure water | Lower osmotic pressure | Water flows in; the cell swells and may burst |
Reverse osmosis and water purification
If a pressure larger than the osmotic pressure is applied on the solution side, pure solvent flows out of the solution through the membrane. This is reverse osmosis (RO). It is used to desalinate sea water. A suitable membrane is a film of cellulose acetate, which lets water through but holds back ions and impurities. Coastal cities such as Chennai get part of their drinking water from RO desalination plants, and home RO purifiers work the same way.
Raw mango pieces shrivel in brine while pickling, and raisins swell when soaked overnight: osmosis in both directions. A hospital drip uses 0.9% saline because it is isotonic with blood cells.
2 marks: “What is reverse osmosis? Give one use.” When a pressure greater than the osmotic pressure is applied on the solution side, pure solvent flows from the solution through the semipermeable membrane. It is used to desalinate sea water.
2 marks: “Why is osmotic pressure preferred over other colligative properties for finding the molar mass of proteins?” It is measured at room temperature, so proteins are not damaged by heat, and its value is large enough to measure even for very dilute solutions, where ΔTb or ΔTf would be tiny.
3 marks: “Calculate the osmotic pressure at 300 K of a solution made by dissolving 3.42 g of sucrose (M = 342 g mol⁻¹) in water to make 500 mL of solution (R = 0.083 L bar mol⁻¹ K⁻¹).” n = 3.42 ÷ 342 = 0.01 mol; C = 0.01 ÷ 0.500 = 0.02 mol L⁻¹. π = CRT = 0.02 × 0.083 × 300 = 0.498 bar.
Water walks towards salt. Osmosis: dilute to concentrated. Push harder than π and the water walks back out: reverse osmosis.