Solubility — Solids, Gases & Henry's Law

What Does "Soluble" Really Mean?

Drop sugar into water and it disappears; drop sand into water and it just sits there. Why? This section is about solubility — the maximum amount of a solute that dissolves in a given amount of solvent at a given temperature.

Definition: The solubility of a substance is the maximum amount of it that can be dissolved in a specified amount of solvent at a specified temperature.

Solubility depends on three things: the nature of solute and solvent, the temperature, and (for gases) the pressure.

The golden rule of dissolution is "like dissolves like":

• Polar solutes (NaCl, sugar, urea) dissolve in polar solvents (water).
• Non-polar solutes (naphthalene, fats, iodine) dissolve in non-polar solvents (benzene, CCl$_4$).

This is because dissolution requires the solute-solvent attractions to be comparable to the solute-solute and solvent-solvent attractions they replace. A polar solvent can pull apart a polar/ionic lattice; a non-polar solvent cannot.

Solubility of a Solid in a Liquid

When a solid is added to a liquid, two opposite processes begin:

• Dissolution — solute particles leave the solid and enter solution.
• Crystallisation — dissolved particles collide with the solid and re-deposit.

When the two rates become equal, the system reaches dynamic equilibrium:

$$\text{Solute (solid)} \rightleftharpoons \text{Solute (in solution)}$$

At this point the solution is saturated — it holds the maximum solute possible at that temperature. The concentration of a saturated solution is the solubility.

• Unsaturated: can still dissolve more solute.
• Saturated: at the dissolving limit (dynamic equilibrium).
• Supersaturated: holds more than the equilibrium amount (unstable; excess crystallises out on disturbance).

Effect of temperature

Apply Le Chatelier's principle to the dissolution equilibrium:

• If dissolution is endothermic ($\Delta_{sol}H > 0$, e.g. KNO$_3$, NH$_4$Cl), heating increases solubility.
• If dissolution is exothermic ($\Delta_{sol}H < 0$, e.g. anhydrous Na$_2$SO$_4$, Ce$_2$(SO$_4$)$_3$), heating decreases solubility.

[JEE Tip] Pressure has almost no effect on the solubility of solids and liquids (they are nearly incompressible). Pressure only matters for gases.

Solubility of a Gas in a Liquid — Henry's Law

Gases dissolve in liquids too: oxygen in water keeps fish alive; CO$_2$ dissolved under pressure makes soda fizzy. The amount that dissolves is governed by pressure and temperature.

Henry's law quantifies the pressure effect:

The partial pressure of a gas in the vapour phase ($p$) is proportional to the mole fraction of the gas ($x$) in the solution.

$$p = K_H \, x$$

where $K_H$ is the Henry's law constant (units of pressure, e.g. kbar or Pa).

Reading the constant correctly is everything:

• Higher $K_H$ → lower solubility of the gas (the gas "wants" to escape, so a high pressure is needed to keep even a small mole fraction dissolved).
• $K_H$ is characteristic of the gas-solvent pair and increases with temperature — so gases become less soluble as temperature rises (warm soda goes flat; warm water holds less dissolved O$_2$).

Applications of Henry's Law

This law shows up everywhere — and these applications are favourite exam questions:

• Soda water and soft drinks are bottled under high CO$_2$ pressure to force more gas into solution. Opening the bottle drops the pressure, so dissolved CO$_2$ escapes as fizz.
• Scuba diving and "the bends": at depth, high pressure dissolves more N$_2$ in a diver's blood. Rising too fast lets that N$_2$ bubble out in the tissues, causing the painful, dangerous decompression sickness. Divers use air diluted with helium (less soluble) to reduce this.
• High-altitude anoxia: low atmospheric pressure means low partial pressure of O$_2$, so less O$_2$ dissolves in the blood of climbers — leading to weakness and impaired thinking ("anoxia").
• Aquatic life in cold water: cold water holds more dissolved O$_2$ (lower temperature → gas more soluble), which is why cold mountain streams are oxygen-rich.

A note on limitations

Henry's law holds when: the pressure is not too high, the temperature is not too low, and the gas does not react with or dissociate in the solvent. (Gases like NH$_3$ and CO$_2$ react with water, so they deviate.)

[NEET Important] Raoult's law (next section) is actually a special case of Henry's law where the solute is volatile and $K_H = p^0$ (the vapour pressure of the pure component). Keep this link in mind — examiners test it directly.

Solved Examples

Example 1: Reading the Henry constant

Two gases A and B have Henry's law constants $K_H(A) = 6.0$ kbar and $K_H(B) = 70$ kbar at the same temperature. Which gas is more soluble in water at a given partial pressure?

Solution: From $p = K_H x$, at fixed $p$ we have $x = p/K_H$. The smaller $K_H$ gives the larger mole fraction. Since $K_H(A) < K_H(B)$, gas A is more soluble.

Takeaway: Higher Henry constant → lower solubility. Memorise this inverse relationship.

Example 2: Effect of temperature on dissolved gas

Explain why a warm bottle of soda goes flat faster than a cold one.

Solution: $K_H$ increases with temperature, so gas solubility decreases as temperature rises. In a warm bottle, less CO$_2$ stays dissolved and more escapes as gas — the drink loses its fizz faster. This is also why fish struggle in warm, oxygen-poor water.

Example 3: Henry's law — finding solubility

If $N_2$ gas is bubbled through water at 293 K at a partial pressure of 0.987 bar, and $K_H$ for $N_2$ at 293 K is $76.48$ kbar, find the mole fraction of $N_2$ in water.

Solution:

1. Formula: $p = K_H x \Rightarrow x = p/K_H$.
2. Convert units: $K_H = 76.48$ kbar $= 76.48\times10^3$ bar.
3. Compute: $x = \dfrac{0.987}{76.48\times10^3}=1.29\times10^{-5}$.

Answer: $x_{N_2} = 1.29\times10^{-5}$ (extremely small — N$_2$ is barely soluble).

Example 4: Henry constant from data

H$_2$S, a toxic gas, has a solubility (mole fraction) in water of $0.195\times10^{-3}$ at 0.10 bar partial pressure and 298 K. Calculate $K_H$.

Solution:

1. Formula: $K_H = \dfrac{p}{x}$.
2. Compute: $K_H = \dfrac{0.10}{0.195\times10^{-3}} = 512.8$ bar.

Answer: $K_H = 512.8$ bar $\approx 0.0513$ kbar.

Example 5: 'Like dissolves like'

Predict whether naphthalene is more soluble in water or in benzene, and justify.

Solution: Naphthalene is a non-polar aromatic solid. By "like dissolves like", it dissolves in non-polar solvents. Benzene is non-polar; water is polar. Answer: Naphthalene is far more soluble in benzene.

Example 6: Endothermic vs exothermic dissolution

The solubility of KNO$_3$ rises sharply on heating, while that of anhydrous Na$_2$SO$_4$ falls. What does this tell you about the sign of $\Delta_{sol}H$ for each?

Solution: By Le Chatelier's principle:

• KNO$_3$ solubility increases with temperature → dissolution is endothermic ($\Delta_{sol}H > 0$); heat favours dissolving.
• Na$_2$SO$_4$ (anhydrous) solubility decreases with temperature → dissolution is exothermic ($\Delta_{sol}H < 0$).

Example 7: Pressure and solid solubility

A student claims that increasing pressure greatly increases the amount of sugar that dissolves in water. Is this correct?

Solution: No. Solids and liquids are nearly incompressible, so pressure has a negligible effect on their solubility. Pressure significantly affects only gas solubility (Henry's law). The claim is incorrect.

Example 8: Why divers use helium-diluted air

Explain, using Henry's law, why deep-sea divers breathe air in which nitrogen is partly replaced by helium.

Solution: At depth, the high pressure raises the partial pressure of breathing gases, dissolving more of them in blood (Henry's law). Dissolved N$_2$ can bubble out during ascent, causing the bends. Helium has a lower solubility (higher $K_H$) in blood than nitrogen, so replacing some N$_2$ with He reduces the amount of dissolved gas and lowers the risk.

Example 9: Calculating dissolved gas mass via Henry's law

The partial pressure of CO$_2$ over a soft drink is 2.5 atm at 298 K. If $K_H$ for CO$_2$ in water is $1.67\times10^3$ atm (mole-fraction basis), find the mole fraction of CO$_2$ dissolved.

Solution:

1. $x = \dfrac{p}{K_H} = \dfrac{2.5}{1.67\times10^3} = 1.50\times10^{-3}$.

Answer: $x_{CO_2} = 1.5\times10^{-3}$.

Takeaway: The high CO$_2$ pressure inside a sealed bottle keeps this (already small) mole fraction dissolved; release the pressure and most of it escapes.

Example 10: High-altitude anoxia

Why do mountaineers often feel weak and unable to think clearly at very high altitudes?

Solution: At high altitude the atmospheric pressure — and therefore the partial pressure of oxygen — is low. By Henry's law, less O$_2$ dissolves in the blood. Low blood-oxygen ("anoxia") causes weakness and impaired thinking.