Why Deep-Sea Divers Risk the Bends When They Surface Too Fast

Sugar in tea, oxygen in river water, carbon dioxide in soda and the zinc in a brass vessel are all solutions — homogeneous mixtures of two or more substances. To work with them, chemists need precise ways to state how much solute is present, and laws that predict vapour pressure and gas solubility.

This lesson covers concentration terms, Raoult's law for ideal and non-ideal solutions, and Henry's law for gases.

Molarity is moles of solute per litre of solution, molality is moles of solute per kilogram of solvent, mole fraction is the moles of one component divided by the total moles, and normality is gram equivalents of solute per litre of solution.

$$M=\frac{n_{\text{solute}}}{V_{\text{solution}}\text{ in L}}m=\frac{n_{\text{solute}}}{\text{mass of solvent in kg}}{x}_A=\frac{n_A}{n_A+n_B}$$

- Mole fractions of all components add up to 1
- Equivalent mass = molar mass divided by the n-factor, and normality = molarity × n-factor

Worked example 1 — molarity and normality. 4.9 g of sulphuric acid ($M=98$ g mol${}^{-1}$) is dissolved to make 500 mL of solution:

$$n=\frac{4.9}{98}=0.050\text{ mol}\text{Molarity}=\frac{0.050}{0.500}=0.10\text{ M}$$

Sulphuric acid releases two ${\mathrm{H}}^{+}$ ions per molecule, so its n-factor is 2 and its normality is $0.10\times 2=0.20$ N.

Worked example 2 — molality and mole fraction. 9.0 g of glucose ($M=180$ g mol${}^{-1}$) is dissolved in 100 g of water:

$$m=\frac{0.050}{0.100}=0.50{\text{ mol kg}}^{-1}{x}_{\text{glucose}}=\frac{0.050}{0.050+5.56}=0.0089$$

where 5.56 mol is the amount of water in 100 g.

An everyday example. A hospital saline drip contains 0.9 g of sodium chloride in every 100 mL, or 9 g per litre — about 0.15 M.

The substance. Molality does not change with temperature, but molarity does — heating expands the volume of a solution, while the mass of solvent stays fixed.

Raoult's law states that the partial vapour pressure of each volatile component equals its mole fraction multiplied by the vapour pressure of the pure component; ideal solutions obey it at every concentration, while non-ideal solutions show positive or negative deviations.

$$p_A=x_A{p}_A^{\circ }{p}_{\text{total}}=x_A{p}_A^{\circ }+x_B{p}_B^{\circ }$$

Worked example. Liquids A and B have pure vapour pressures of 100 and 40 mm Hg. For a solution with $x_A=0.40$:

$$p_A=0.40\times 100=40\text{ mm Hg}{p}_B=0.60\times 40=24\text{ mm Hg}{p}_{\text{total}}=64\text{ mm Hg}$$

The mole fraction of A in the vapour is $\frac{40}{64}=0.625$, so the vapour is richer in the more volatile liquid.

Ideal solutions:

- Obey Raoult's law at all concentrations, with ${\Delta }_{\text{mix}}H=0$ and ${\Delta }_{\text{mix}}V=0$
- A-B attractions equal A-A and B-B attractions, as in benzene with toluene

Positive deviation:

- A-B attractions are weaker, so molecules escape more easily and the vapour pressure is higher than predicted
- ${\Delta }_{\text{mix}}H>0$ and ${\Delta }_{\text{mix}}V>0$, as for ethanol with water
- These mixtures can form minimum boiling azeotropes

Negative deviation:

- A-B attractions are stronger, so the vapour pressure is lower than predicted
- ${\Delta }_{\text{mix}}H<0$ and ${\Delta }_{\text{mix}}V<0$, as for acetone with chloroform, which form a hydrogen bond, and nitric acid with water
- These mixtures can form maximum boiling azeotropes

Non-volatile solutes. Only the solvent contributes, so $p=x_{\text{solvent}}{p}^{\circ }$, and vapour pressure falls as solute is added.

An everyday example. Surgical spirit is an ethanol-water mixture, and simple distillation cannot purify ethanol beyond about 95 per cent because the mixture forms an azeotrope.

The substance. Two liquids that warm up when mixed show negative deviation — the stronger new attractions release heat and also make the mixture harder to vaporise.

**Henry's law states that the partial pressure of a gas above a liquid is proportional to its mole fraction in the solution, $p=K_{H}x$, so a gas dissolves more at higher pressure and less at higher temperature.**

$$p=K_{H}x$$

Here $K_H$ is Henry's law constant; a larger $K_H$ means a lower solubility.

Worked example. For nitrogen in water at 293 K, $K_H$ is about $7.6\times 10^4$ bar. With nitrogen in air at a partial pressure of 0.78 bar:

$$x=\frac{p}{K_H}=\frac{0.78}{7.6\times 10^4}=1.03\times 10^{-5}$$

Factors affecting gas solubility:

- Pressure — solubility rises as partial pressure rises
- Temperature — solubility falls as temperature rises, because dissolving a gas is usually exothermic, so $K_H$ increases
- Reaction with water — gases such as ammonia and hydrogen chloride dissolve far more and do not obey Henry's law

Applications:

- Soft drinks are sealed under high carbon dioxide pressure to dissolve more gas
- Scuba divers breathe compressed air, so more nitrogen dissolves in the blood; surfacing too fast releases it as bubbles, causing the painful bends, so diving tanks may use air diluted with helium
- At high altitudes, the low partial pressure of oxygen reduces the oxygen dissolved in blood, causing weakness and confusion

An everyday example. A bottle of soda water left open in the sun goes flat faster than one kept in the fridge, because gas solubility falls as temperature rises.

The substance. Henry's law fails when a gas reacts with the solvent — ammonia dissolves far beyond the prediction because it forms ammonium and hydroxide ions.