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Open a cold bottle of soda and it fizzes gently; open a warm one and it can spray. Both are Henry's law at work. The law is short, but students lose marks on it through unit slips and by mixing up which way the constant runs. This post explains the law, works three problems with full numbers, and lists the traps.
What does Henry's law say?
The NCERT Class 12 Solutions unit states it this way: at constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The form NCERT uses most is
p = KH x
where p is the partial pressure of the gas, x is its mole fraction in the solution and KH is the Henry's law constant. NCERT points out two consequences:
A higher KH means lower solubility at a given pressure, because x = p / KH.
KH increases with temperature for gases such as N₂ and O₂, so gases become less soluble as water warms. NCERT links this to aquatic species being more comfortable in cold water.
Where is Henry's law in the syllabus?
It appears in all the major syllabi, though not always in the same unit. The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list Henry's law under Equilibrium ("equilibria involving physical processes"). The JEE (Advanced) 2026 syllabus lists it under Solutions. NCERT teaches it in the Solutions unit, which is where most questions draw their data.
Which KH values should you know how to read?
NCERT's Table 1.2 gives KH in kbar for gases in water. Some values:
He at 293 K: 144.97 kbar
N₂ at 293 K: 76.48 kbar; at 303 K: 88.84 kbar
O₂ at 293 K: 34.86 kbar; at 303 K: 46.82 kbar
CO₂ at 298 K: 1.67 kbar
CO₂ has a far smaller KH than N₂ or O₂, so it is far more soluble, which is why soda can hold so much of it.
Worked problem 1: nitrogen in water
This is NCERT's Example 1.4. N₂ at a partial pressure of 0.987 bar is bubbled through water at 293 K. How much dissolves in 1 litre?
Convert KH to bar: 76.48 kbar = 76,480 bar.
x = p / KH = 0.987 / 76,480 = 1.29 × 10⁻⁵.
1 litre of water is about 55.5 mol. Since n is tiny, x ≈ n / 55.5, so n = 1.29 × 10⁻⁵ × 55.5 = 7.16 × 10⁻⁴ mol = 0.716 mmol.
Worked problem 2: CO₂ in soda water
This is NCERT intext question 1.7. KH for CO₂ is 1.67 × 10⁸ Pa at 298 K. How much CO₂ is in 500 mL of soda water packed under 2.5 atm of CO₂?
Convert pressure to Pa: 2.5 × 101,325 = 253,312.5 Pa.
x = 253,312.5 / (1.67 × 10⁸) = 1.517 × 10⁻³.
Take 500 mL of water as 500 g: 500 / 18 = 27.78 mol.
n(CO₂) ≈ x × 27.78 = 1.517 × 10⁻³ × 27.78 = 0.0421 mol.
Mass = 0.0421 × 44 g mol⁻¹ ≈ 1.85 g.
So about 1.85 g of CO₂ is dissolved. Neglecting n in the denominator changes the answer by about 0.15%, which is fine.
Worked problem 3: oxygen in warm water
How much does warming water from 293 K to 303 K reduce dissolved oxygen? Take an O₂ partial pressure of 0.21 bar (a sample value close to that in air at sea level).
At 293 K: x = 0.21 / 34,860 = 6.02 × 10⁻⁶; n = 6.02 × 10⁻⁶ × 55.5 = 3.34 × 10⁻⁴ mol per litre, about 10.7 mg.
At 303 K: x = 0.21 / 46,820 = 4.49 × 10⁻⁶; n = 2.49 × 10⁻⁴ mol per litre, about 8.0 mg.
The ratio is 34.86 / 46.82 = 0.745, so a 10 K rise cuts dissolved oxygen by about a quarter. You can get this ratio without any of the other steps, which is a quick check in an MCQ.
Which mistakes cost the most marks?
Units of KH. NCERT's table uses kbar; problems may give Pa, atm or mm Hg. Convert pressure and KH to the same unit before dividing.
Total pressure instead of partial pressure. For a gas in a mixture, use its own partial pressure (problem 1.39 in NCERT gives air at 10 atm with oxygen and nitrogen fractions).
Reading KH backwards. A large KH means low solubility in the p = KH x form. Some books write the law as concentration = kH × p, where a large constant means high solubility. A 2015 compilation of Henry's law constants gathered 17,350 values from 689 references and had to convert them into one uniform format, so always check which form a question uses.
Forgetting the mole fraction step. Convert x to moles using the moles of solvent.
Henry's law also links to Raoult's law: NCERT shows that Raoult's law is a special case of Henry's law where KH equals the vapour pressure of the pure component. For the related topic of how dissolved particles change boiling and freezing points, see our post on colligative properties.
A practice routine
Redo the three problems above without looking, then check each step.
Solve NCERT intext question 1.6 and exercises 1.35 and 1.39, writing units on every line.
Make a one-line note: "x = p / KH; big KH, low solubility; KH up with T."
Mix two Henry's law problems into each chemistry set for the next two weeks.
If a step in a solution does not make sense, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Take NCERT exercise 1.39 tonight and solve it with units written on every line.
