Henry's Law Calculations — Formula, Units and Worked Examples
Open a cold bottle of soda and it hisses; leave the same bottle open and it goes flat. Both observations are Henry's law at work. The law tells you how much of a gas dissolves in a liquid, and it is the one part of the Solutions chapter where students routinely lose marks not because the physics is hard but because two different constants share the same name. This guide fixes that first, then works through three calculations in full.
The law
At a constant temperature, the amount of a gas dissolved in a given volume of liquid is proportional to the partial pressure of that gas above the liquid. There are two standard ways to write it, and both are correct.
Form 2 (concentration form): C = kH · p
| Form 1: p = KH x | Form 2: C = kH p | |
|---|---|---|
| What is on the left | partial pressure of the gas | concentration of the dissolved gas |
| Solubility measure | mole fraction x of the gas in solution | molarity C in mol/L |
| Unit of the constant | pressure (bar, atm, kbar) | mol L⁻¹ atm⁻¹ or mol L⁻¹ bar⁻¹ |
| A large constant means | low solubility | high solubility |
Read that last row twice. The two constants are reciprocals of one another in spirit, so the direction of the trend flips. In NCERT's form, KH(He) is far larger than KH(CO₂) precisely because helium is far less soluble. If a question gives a constant in kbar, you are in Form 1; if it gives mol L⁻¹ atm⁻¹, you are in Form 2.
What Henry's law assumes
- The pressure is not very high and the solution is dilute.
- The gas does not react chemically with the solvent. HCl and NH₃ in water react or ionise heavily, so they do not follow the law well. CO₂ follows it approximately, because only a small part of the dissolved CO₂ actually forms carbonic acid.
- The gas does not associate or dissociate in solution.
- It is the partial pressure of that one gas that matters, not the total pressure of the gas mixture above the liquid.
Worked example 1 — nitrogen dissolving in water
Problem: Nitrogen is bubbled through water at 293 K. If the partial pressure of N₂ is 0.987 bar and KH for N₂ in water at this temperature is 76.48 kbar, how many millimoles of N₂ dissolve in 1 litre of water?
Step 1 — put the constant into the same unit as the pressure.
KH = 76.48 kbar = 76.48 × 1000 = 76 480 bar
Step 2 — mole fraction of dissolved N₂.
x = p ÷ KH = 0.987 ÷ 76 480 = 1.2905 × 10⁻⁵
Step 3 — moles of water in 1 litre. 1 L of water has a mass of about
1000 g and M(H₂O) = 18.02 g/mol:
n(H₂O) = 1000 ÷ 18.02 = 55.494 mol
Step 4 — the dilute approximation. Because x is tiny,
x = n(N₂) ÷ [n(N₂) + n(H₂O)] ≈ n(N₂) ÷ n(H₂O)
n(N₂) = 1.2905 × 10⁻⁵ × 55.494 = 7.162 × 10⁻⁴ mol =
0.716 mmol
Under 1 millimole of nitrogen in a whole litre of water. Gases really are very poorly soluble in water, which is why fish need water that is constantly re-aerated.
Worked example 2 — why a soft drink fizzes
Problem: Take kH for CO₂ in water at 25 °C as 3.4 × 10⁻² mol L⁻¹ atm⁻¹. Compare the dissolved CO₂ in (a) water open to the air, where the partial pressure of CO₂ is about 4.0 × 10⁻⁴ atm, and (b) a sealed carbonated drink at a CO₂ pressure of 2.5 atm.
(a) Open to the air:
C = kH p = (3.4 × 10⁻²) × (4.0 × 10⁻⁴) = 1.36 × 10⁻⁵ mol/L
(b) Sealed bottle:
C = (3.4 × 10⁻²) × 2.5 = 8.5 × 10⁻² mol/L = 0.085 mol/L
Ratio: 0.085 ÷ (1.36 × 10⁻⁵) = 6250 times more
As a mass: M(CO₂) = 12.011 + 2 × 15.999 = 44.009 g/mol, so the sealed drink holds 0.085 × 44.009 = 3.74 g of CO₂ per litre.
When you open the cap, the CO₂ pressure above the liquid crashes from 2.5 atm to 0.0004 atm. Henry's law says the solubility must fall by the same factor, so almost all of that 3.74 g has to leave — and it leaves as bubbles.
Worked example 3 — the same gas at two pressures
When the temperature is unchanged, KH is unchanged, so you can compare two states directly without ever using the constant.
Problem: The mole fraction of dissolved oxygen in water is 3.0 × 10⁻⁵ when the partial pressure of O₂ is 1.0 bar. What is it at 5.0 bar, and what mass of O₂ does 1 litre of water then hold?
Step 1 — new mole fraction.
x₂ = 3.0 × 10⁻⁵ × (5.0 ÷ 1.0) = 1.5 × 10⁻⁴
Step 2 — moles of O₂ per litre of water.
n(O₂) = 1.5 × 10⁻⁴ × 55.494 = 8.324 × 10⁻³ mol
Step 3 — mass. M(O₂) = 2 × 15.999 = 32.00 g/mol
mass = 8.324 × 10⁻³ × 32.00 = 0.266 g per litre
Five times the pressure gives exactly five times the dissolved amount. That linear relationship is the entire content of Henry's law.
Temperature: why warm water holds less gas
Dissolving a gas is normally exothermic — the gas molecules give up their freedom and settle into the liquid. By Le Chatelier's principle, raising the temperature pushes that equilibrium backwards, so the gas comes out. In the language of the constants, KH in Form 1 increases with temperature (and kH in Form 2 decreases), and solubility falls either way.
Two everyday consequences worth remembering for a viva:
- Small bubbles appear on the inside of a pan of water long before it boils. That is dissolved air escaping as the water warms.
- Warm water released by an industrial plant into a river holds less dissolved oxygen, so fish are stressed. This is the chemistry behind thermal pollution.
Two more applications you can be asked to explain
- Decompression sickness in divers. Under water the total pressure is high, so the partial pressure of nitrogen in the breathing mixture is high and more nitrogen dissolves in the blood. Surfacing too fast drops that pressure quickly and the nitrogen comes out as bubbles inside the body. Diving cylinders are often filled with helium diluted air because helium is much less soluble.
- Breathing at high altitude. Atmospheric pressure is lower, so the partial pressure of oxygen is lower, so less oxygen dissolves in the blood plasma. This is why climbers feel breathless — a direct pressure effect, not a change in the percentage of oxygen in the air, which stays close to 21% all the way up.
Common mistakes that cost marks
- Assuming a large constant always means high solubility. It depends on which form you are in. Check the unit of the constant before you interpret it.
- Forgetting the k in kbar. 76.48 kbar is 76 480 bar. Dropping the factor of 1000 makes the answer 1000 times too big.
- Using total pressure instead of partial pressure. If air at 1 atm is above the water, the nitrogen partial pressure is about 0.78 atm, not 1 atm.
- Stopping at the mole fraction. Most questions want millimoles, grams or mol/L. Converting mole fraction to moles needs the moles of solvent, which is mass ÷ molar mass.
- Applying the law to a reacting gas. HCl, NH₃ and SO₂ in water do not obey it well because they ionise or react.
- Saying solubility rises with temperature. That is true for most solids, and false for gases.
- Mixing bar and atm in one calculation. 1 atm = 1.013 bar; they are close but not equal, and a mixed calculation is marked wrong.
Where Henry's law appears in exams
| Exam | Typical question |
|---|---|
| CBSE/ICSE Class 12 | Millimoles of a gas dissolved at a given partial pressure; explain fizzing, divers or altitude |
| JEE/NEET | Comparing KH values to rank solubility; combined Henry and Raoult problems |
| IIT-JAM / CUET-PG | Deviations from Henry's law; Henry's law as the dilute-solution limit |
| GATE / CSIR-NET | Gas–liquid equilibria, absorption columns, activity of dissolved gases |
The step that trips people up is the conversion. Henry's law gives you a mole fraction; the marks are for millimoles or grams. The Mass ↔ Mole tool does that conversion both ways for any formula, so you can turn 7.162 × 10⁻⁴ mol of N₂ into grams — or a mass of dissolved CO₂ back into moles — without a slip.
Open the Mass ↔ Mole Calculator →Solutions is a short chapter with a high mark-to-effort ratio if the units are drilled properly. ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre and online classes across India: abcchemistry.in.