Solvent Extraction and Colligative Properties

Lessons

Senior 6 · Chemistry

Solvent Extraction and Colligative Properties

5 lessons

Lesson 5 of 5

Osmotic Pressure and Applications of Colligative Properties

25 min

Why a drip is not pure water

When a patient in a health centre needs fluid straight into a vein, the bag usually contains "normal saline", 0.9 g of sodium chloride in every 100 cm³ of water. If pure water were run into the vein instead, red blood cells would take in water, swell and burst. And when a child has diarrhoea, the oral rehydration salts (ORS) mixed at home must be made up with exactly the volume of water stated on the packet: too little water and the drink pulls water out of the body into the gut. Both rules come from the fourth colligative property, osmotic pressure.

Definition

Semipermeable membrane: a membrane that lets solvent molecules pass but not solute particles. Examples: cellophane, a cell membrane (approximately), the polymer film in a water purifier.

Osmosis: the net movement of solvent through a semipermeable membrane from a less concentrated solution (or pure solvent) into a more concentrated solution.

Osmotic pressure, $π \pi$ : the pressure that must be applied to a solution to stop solvent flowing into it through a semipermeable membrane from the pure solvent. Unit: Pa.

Isotonic solutions have the same osmotic pressure; a hypotonic solution has a lower one and a hypertonic solution a higher one than the solution it is compared with. Example: 0.9 % NaCl is isotonic with blood plasma.

Reverse osmosis: forcing solvent out of a solution through a semipermeable membrane by applying a pressure greater than the osmotic pressure.

Osmosis and osmotic pressure

Solvent molecules strike the membrane from both sides. On the solution side some of the surface is occupied by solute particles, so fewer solvent molecules pass through per second than from the pure-solvent side: there is a net flow of solvent into the solution. It is the same lowering of the solvent's "escaping tendency" that lowers the vapour pressure in Lesson 4.

In the osmometer of Figure 5.1, water enters the funnel and the solution rises up the tube. The column of liquid pushes back, and when its pressure is large enough the flow stops: at equilibrium the hydrostatic pressure $\rho g h$ equals the osmotic pressure.

In 1887 van 't Hoff showed that for dilute solutions the osmotic pressure obeys an equation with the same form as the ideal gas equation:

$\pi V = nRT \qquad\Longleftrightarrow\qquad \pi = cRT$

where $c$ is the concentration of solute particles in $\mathrm{mol\,m^{-3}}$ (1 mol dm⁻³ = 1000 mol m⁻³), $R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}$ and $T$ is in kelvin; $π \pi$ is then in pascals. For electrolytes, $\pi = icRT$ . Writing $n = \frac{m}{M}$ gives the molar mass of the solute:

$M = \frac{mRT}{\pi V}$

Figure 5.1 A simple osmometer. Water passes up through the membrane into the sugar solution until the column of height $h$ exerts a pressure $\rho g h$ equal to the osmotic pressure.

The best method for giant molecules

Osmotic pressures are large even for dilute solutions. A protein solution of only 10 g dm⁻³ (Example 1) has an osmotic pressure that supports a water column about 4 cm high, easy to measure, while its freezing point is lowered by less than 0.0003 K, which no ordinary thermometer can detect. So osmotic pressure is the method of choice for the molar masses of polymers and proteins, and it is measured at room temperature, where the molecules are not damaged.

Table 5.1: Choosing a colligative method

Method	Measured	Best for
lowering of vapour pressure	$\Delta p$ , very small	rarely used; it is the basis of the other three
elevation of boiling point	$\Delta T_b$ , small	solutes stable at the solvent's boiling point
depression of freezing point	$\Delta T_f$ , larger	small molecules (M up to a few hundred); camphor or cyclohexane for larger effects
osmotic pressure	$π \pi$ , large	polymers and proteins (M of thousands to millions); heat-sensitive solutes

Worked example

Problem. 0.500 g of a protein is dissolved in water to make 50.0 cm³ of solution. At 27 °C its osmotic pressure is 378 Pa. Calculate the molar mass of the protein and the height of the water column in an osmometer ( $\rho = 1000\,\mathrm{kg\,m^{-3}}$ , $g = 9.81\,\mathrm{m\,s^{-2}}$ ).

Solution. $V = 50.0\,\mathrm{cm^3} = 5.00 \times 10^{-5}\,\mathrm{m^3}$ ; $T = 300\,\mathrm{K}$ .

$M = \frac{mRT}{\pi V} = \frac{0.500 \times 8.314 \times 300}{378 \times 5.00 \times 10^{-5}} = 6.60 \times 10^{4}\,\mathrm{g\,mol^{-1}}$

$h = \frac{\pi}{\rho g} = \frac{378}{1000 \times 9.81} = 0.0385\,\mathrm{m} = 3.85\,\mathrm{cm}$

Answer: $6.60 \times 10^{4}\,\mathrm{g\,mol^{-1}}$ , close to that of haemoglobin, and a 3.85 cm column. (The same solution would freeze only $2.8 \times 10^{-4}$ K below 0 °C.)

Osmosis in the body, in food and in water treatment

Blood and drips. Red blood cells are surrounded by a membrane that water crosses easily. In an isotonic solution they keep their shape; in a hypotonic one they swell and may burst (haemolysis); in a hypertonic one they shrink (Figure 5.2). Fluids for injection are therefore made isotonic with blood: 0.9 % NaCl or 5 % glucose.

Oral rehydration salts. Since 2004 the WHO has recommended a "reduced-osmolarity" ORS that contains, per litre, 75 mmol of glucose and 75 mmol of sodium, together with potassium, chloride and citrate: 245 mmol of particles in all, slightly below the osmolarity of blood plasma. Glucose helps the gut absorb sodium, and water follows by osmosis. A drink made too concentrated would be hypertonic and draw water into the gut.

Plants and food. Water enters root cells by osmosis and keeps cells firm (turgid). Soil made salty by poor irrigation, or too much fertiliser near the roots, reverses the flow and the plant wilts. Salting fish and meat, and preserving fruit as jam, work because bacteria and moulds lose water by osmosis in the concentrated solution.

Reverse osmosis. If a pressure greater than $π \pi$ is applied to salt water on one side of a membrane, pure water is forced out to the other side. This is how desalination plants turn sea water into drinking water, and how many household and bottled-water purifiers remove dissolved salts.

Other colligative applications (Lesson 4): ethane-1,2-diol antifreeze in radiators; salt or calcium chloride to melt ice on roads; salt and ice mixtures to freeze ice cream; and measuring the molar masses of new compounds.

Figure 5.2 Red blood cells in hypotonic, isotonic and hypertonic solutions. Blue arrows show the net movement of water.

Worked example

Problem. (a) Calculate the osmotic pressure at body temperature (37 °C) of WHO reduced-osmolarity ORS, which contains 245 mmol of particles per litre. (b) Normal saline contains 9.0 g of NaCl per dm³. What mass of glucose ( $M = 180.2$ ) per dm³ gives a solution isotonic with it? (Assume NaCl gives 2 particles.)

Solution. (a) $c = 245\,\mathrm{mol\,m^{-3}}$ , $T = 310\,\mathrm{K}$ :

$\pi = cRT = 245 \times 8.314 \times 310 = 6.32 \times 10^{5}\,\mathrm{Pa} = 632\,\mathrm{kPa}$

(b) Particles in saline: $2 \times \frac{9.0}{58.44} = 0.308\,\mathrm{mol\,dm^{-3}}$ . Glucose gives one particle per molecule, so it needs 0.308 mol dm⁻³:

$m = 0.308 \times 180.2 = 55\,\mathrm{g\,dm^{-3}}$

Answer: (a) about 630 kPa, more than six times atmospheric pressure; (b) about 55 g dm⁻³, which is why 5 % glucose (50 g dm⁻³) is used as a near-isotonic drip.

Worked example

Problem. Sea water behaves roughly like 0.55 mol dm⁻³ NaCl. Estimate the minimum pressure needed to desalinate it by reverse osmosis at 25 °C.

Solution. With $i = 2$ , $c = 2 \times 550 = 1100\,\mathrm{mol\,m^{-3}}$ :

$\pi = icRT = 1100 \times 8.314 \times 298 = 2.73 \times 10^{6}\,\mathrm{Pa}$

Answer: at least 2.7 MPa (about 27 atmospheres) must be applied to the sea-water side before any fresh water passes through. In practice a higher pressure is used, because the remaining brine becomes more concentrated as water is removed.

Warning

Common mistakes. (1) Using $c$ in mol dm⁻³ with $R = 8.314$ : convert to mol m⁻³ (× 1000), or the answer is 1000 times too small. (2) Using °C instead of kelvin. (3) Forgetting $i$ for salts. (4) Saying water moves "from high concentration to low concentration": the solvent moves towards the more concentrated solution.

Health. Follow the packet: dissolve one sachet of ORS in exactly the volume of clean (boiled and cooled) water stated, usually one litre, and do not add sugar or salt. A stronger mixture is hypertonic and can make dehydration worse. Seek medical help for a child with diarrhoea who cannot drink, is very sleepy or has blood in the stool.

Tip

$\pi = cRT$ is $p V = n R T$ in disguise: the solute particles behave like a gas filling the volume of the solution. Units as for gases: Pa, m³, mol, K.

The Organic Chemistry Tutor: Molar mass from osmotic pressure. Watch how π = icRT is rearranged to find a molar mass, and how the van 't Hoff factor is included for ionic solutes.

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Practice

Define osmosis and osmotic pressure.
Calculate the osmotic pressure of 0.0500 mol dm⁻³ glucose at 25 °C.
At the same temperature, which has the higher osmotic pressure, 0.10 mol dm⁻³ NaCl or 0.15 mol dm⁻³ glucose? Explain.
2.00 g of a polymer is dissolved to make 100 cm³ of solution. At 25 °C the osmotic pressure is 620 Pa. Calculate the molar mass of the polymer.
Explain why osmotic pressure, rather than freezing-point depression, is used to find the molar mass of a polymer.
Describe and explain what happens to red blood cells placed in pure water, and why a drip contains 0.9 % NaCl.
Calculate the minimum pressure needed to purify brackish water, roughly 0.050 mol dm⁻³ NaCl, by reverse osmosis at 25 °C.
A parent dissolves a sachet of ORS meant for 1 litre in half a litre of water "to make it stronger". Evaluate this decision.

Show answersHide answers

Osmosis: the net movement of solvent through a semipermeable membrane into a more concentrated solution. Osmotic pressure: the pressure that must be applied to the solution to stop this flow from pure solvent.
$\pi = 50.0 \times 8.314 \times 298 = 1.24 \times 10^{5}\,\mathrm{Pa}$ (124 kPa).
NaCl: it gives about 0.20 mol dm⁻³ of particles ( $i = 2$ ), more than the 0.15 mol dm⁻³ of glucose molecules, and $π \pi$ depends on the concentration of particles.
$M = \frac{2.00 \times 8.314 \times 298}{620 \times 1.00 \times 10^{-4}} = 8.0 \times 10^{4}\,\mathrm{g\,mol^{-1}}$ .
For a large molar mass, the number of particles in a practical solution is tiny: $\Delta T_f$ would be far too small to measure, but $π \pi$ is large enough to measure accurately, and at room temperature.
Pure water is hypotonic: water enters the cells by osmosis, they swell and burst (haemolysis). 0.9 % NaCl is isotonic with blood plasma, so there is no net flow of water and the cells keep their shape.
$\pi = 2 \times 50 \times 8.314 \times 298 = 2.5 \times 10^{5}\,\mathrm{Pa}$ (about 250 kPa).
Harmful. Doubling the concentration (about 490 mmol of particles per litre) makes the drink hypertonic to body fluids, so water moves from the body into the gut by osmosis, which worsens diarrhoea and dehydration and can raise blood sodium. The packet's volume must be followed exactly.

1Why a drip is not pure water

2Osmosis and osmotic pressure

3The best method for giant molecules

4Osmosis in the body, in food and in water treatment

Why a drip is not pure water

Osmosis and osmotic pressure

The best method for giant molecules

Osmosis in the body, in food and in water treatment