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Solutions · TOPIC LESSON

Gas Solubility and Henry Law

Read the constant’s definition before deciding whether a larger value means greater solubility.

JEE Main · NEET UG38 min study guide9 indexed sections

BY THE END, YOU CAN

  • Use gas partial pressure rather than total pressure in Henry-law calculations.
  • Distinguish concentration-form and pressure-over-mole-fraction constants.
  • Convert gas solubility into moles and mass with consistent units.
  • Apply a closed-headspace material balance when the gas reservoir is finite.
  • Separate physical dissolution from chemical speciation and recognize temperature and nonideality limits.

BEFORE YOU START

  • Ideal-gas partial pressures and mole fractions.
  • Molar concentration, density and mole conversion.
  • Dynamic equilibrium and simple algebra.

Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.

CONNECT THE CONCEPT TO THE EXAM

Selected exam questions & worked solutions

These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.

No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.

01 / UNDERSTAND & APPLY

1. A dissolved gas remains in exchange with the gas phase

At a gas–liquid interface, molecules enter and leave the liquid. At equilibrium their macroscopic transfer rates balance, giving a reproducible dissolved concentration for specified temperature, solvent and gas-phase chemical potential. The gas has not stopped crossing the interface. A larger gas partial pressure generally supports a larger dissolved amount in the dilute physical-solubility regime.

Henry’s law describes the limiting proportionality between the activity or concentration of a dissolved dilute gas species and its gas-phase partial pressure or fugacity. In an elementary ideal treatment, the relation is linear at fixed temperature. It is most useful when the dissolved species is dilute and the gas does not undergo substantial unaccounted chemical reaction.

Gas solubility is not the same as the speed of dissolution. Stirring, bubble size and interface area can accelerate mass transfer without changing the final equilibrium concentration at fixed conditions. A solution can remain temporarily supersaturated after pressure is lowered because bubble nucleation and escape take time.

Name the dissolved species carefully. Physically dissolved CO₂ is not identical to the sum of all dissolved inorganic carbon forms, and dissolved NH₃ is not identical to NH₃ plus NH₄⁺. Henry’s law for one neutral molecular species can remain useful while additional acid–base equilibria determine the total analytical amount.

02 / UNDERSTAND & APPLY

2. Two common Henry constants point in opposite numerical directions

One common form is C=k_cp p, where C is dissolved amount concentration and p is that gas’s partial pressure. The coefficient k_cp has concentration-per-pressure units, such as mol L⁻¹ bar⁻¹. With this definition, a larger coefficient means more dissolved gas at a given pressure, and therefore greater solubility under the same conditions.

Another common classroom form is p=K_H x, where x is the dissolved gas mole fraction. K_H has pressure units. With this definition, x=p/K_H, so a larger K_H means a smaller dissolved mole fraction at the same pressure. These statements are not contradictory: the two symbols refer to differently defined quantities.

IUPAC terminology distinguishes solubility-type constants from reciprocal volatility-type constants and specifies the concentration measure used. The safest practice is to write the defining equation and unit next to every numerical value. The bare phrase “Henry constant” is insufficient to determine whether it should multiply or divide pressure.

Constants based on molarity, molality and mole fraction are not simple reciprocals of one another unless their concentration conversion is also included. For a very dilute gas in a solvent with molar concentration C_solvent, x≈C_gas/C_solvent. Combining the two elementary forms then gives k_cp≈C_solvent/K_H with compatible units.

C=k_cp p; p=K_H x

Different constant definitions: k_cp has concentration/pressure units, K_H has pressure units.

At very low dissolved fraction: k_cp≈C_solvent/K_H

Requires a solvent molar concentration and compatible concentration units.

ORIGINAL WORKED EXAMPLE

Read the supplied constant correctly

A hypothetical gas has K_H=4.00×10⁴ bar in the definition p=K_Hx. Its partial pressure is 2.00 bar. Find x.

  1. Use the stated pressure-over-mole-fraction form, not C=K_Hp.
  2. x=p/K_H=2.00/(4.00×10⁴)=5.00×10⁻⁵.
  3. The pressure units cancel, leaving a dimensionless mole fraction.
  4. Multiplying by K_H would give an impossible large number and incorrect units.

Answer x=5.00×10⁻⁵.

03 / UNDERSTAND & APPLY

3. A gas mixture contributes one partial pressure per species

For an ideal gas mixture, p_i=y_iP_total. Henry’s law for dissolved species i uses p_i, not the total pressure of all gases. If a gas mixture is 20% A by mole at 5.00 bar total pressure, the A partial pressure is 1.00 bar. Doubling total pressure at fixed gas composition doubles every partial pressure and doubles each dilute dissolved concentration in the elementary model.

Changing total pressure by adding another gas is different from compressing the original mixture. At fixed temperature and volume, adding ideal inert gas does not change the existing species’ partial pressures n_iRT/V. Its dissolved equilibrium concentration therefore does not change merely because the total pressure rises, assuming no solvent or interaction change.

If a quoted gas composition is on a dry basis but the gas over water contains water vapor, account for the water-vapor pressure when calculating the dry-gas partial pressures. For a dry fraction y_i,dry, p_i=y_i,dry(P_total−p_water) under the ideal mixture assumption. A wet-basis fraction already includes water in its denominator and must not be corrected twice.

Pressure units and reference conventions must be consistent. A coefficient expressed per atmosphere is not directly multiplied by a pressure stated numerically in bar or pascals. Convert pressure first or convert the constant with its units; do not treat nearby atmospheric units as exactly equal.

ORIGINAL WORKED EXAMPLE

Use composition before Henry’s law

At a fixed temperature, k_cp for hypothetical gas A is 1.20×10⁻³ mol L⁻¹ bar⁻¹. A gas mixture has y_A=0.250 and total pressure 4.00 bar. Find dissolved C_A.

  1. p_A=y_AP=0.250×4.00=1.00 bar.
  2. C_A=k_cpp_A=(1.20×10⁻³)(1.00)=1.20×10⁻³ M.
  3. Using total pressure would overestimate C_A by a factor of four.

Answer C_A=1.20×10⁻³ mol L⁻¹.

04 / UNDERSTAND & APPLY

4. Convert the equilibrium concentration into the requested amount

Once the dissolved concentration is known, multiply by final liquid volume to obtain moles, then by gas molar mass to obtain dissolved mass. This is a solution-volume calculation, not the volume the gas would occupy as a separate gas phase. If a gas-volume equivalent is requested, use the ideal-gas equation at the specified reporting temperature and pressure after finding moles.

A mole-fraction result requires a different conversion. If x_gas=n_gas/(n_solvent+n_gas), then n_gas=x_gas n_solvent/(1−x_gas). At very low x, approximate n_gas≈x_gas n_solvent. Identify the approximation rather than silently replacing the exact denominator.

If solvent mass and molar mass are supplied, n_solvent follows directly. If only liquid volume is supplied, a density may be needed for a mole-fraction calculation. Do not assume every solvent has water’s density or molar mass. Conversely, a concentration-form constant and final solution volume can often avoid that conversion entirely.

Report the species basis. A dissolved oxygen mass means mass of O₂ molecules under that model, not mass of isolated oxygen atoms. For a chemically reacting gas, the total amount absorbed may differ from the amount of the neutral gas species predicted directly by Henry’s law.

ORIGINAL WORKED EXAMPLE

From partial pressure to dissolved mass

A gas with molar mass 32.0 g mol⁻¹ has k_cp=1.00×10⁻³ mol L⁻¹ bar⁻¹. It equilibrates at partial pressure 0.800 bar with 2.50 L of dilute solution. Find dissolved mass.

  1. C=(1.00×10⁻³)(0.800)=8.00×10⁻⁴ mol L⁻¹.
  2. n=CV=(8.00×10⁻⁴)(2.50)=2.00×10⁻³ mol.
  3. Mass=nM=2.00×10⁻³×32.0=0.0640 g.

Answer 64.0 mg of the specified gas.

05 / UNDERSTAND & APPLY

5. A sealed vessel may not maintain the original gas pressure

An effectively infinite gas reservoir can hold p constant while gas dissolves. A sealed finite headspace cannot generally do so: transfer into liquid reduces gas-phase moles and hence its partial pressure. The final pressure and dissolved amount must satisfy both Henry’s law and a total-mole balance.

For one physically dissolving ideal gas in headspace volume V_g above liquid volume V_l at fixed T, n_total=pV_g/(RT)+k_cp pV_l. Therefore p=n_total/[V_g/(RT)+k_cpV_l], using compatible pressure–volume units. This treats liquid volume and headspace volume as fixed and neglects solvent vapor or handles it separately.

If the liquid initially contains gas, include those moles in n_total. If other gases are present but noninteracting, write the component balance for the species of interest. The component’s partial pressure, not necessarily total headspace pressure, appears in both terms.

Check limiting cases. If k_cpV_l is negligible compared with V_g/(RT), little gas dissolves and the headspace pressure approaches the no-dissolution ideal-gas value. If the liquid capacity is large, more material resides dissolved and the gas pressure is smaller. Neither phase can contain more than the total supplied amount.

n_total=pV_g/(RT)+k_cp pV_l

Finite closed headspace, ideal gas, fixed volumes, dilute physical dissolution.

p=n_total/[V_g/(RT)+k_cpV_l]

Use one pressure unit consistently in R and k_cp.

ORIGINAL WORKED EXAMPLE

Solve a closed-headspace balance

A sealed model system contains 0.0100 mol gas,1.00 L gas space and 1.00 L liquid at 300 K. Use R=0.08314 L bar mol⁻¹ K⁻¹ and k_cp=0.0200 mol L⁻¹ bar⁻¹. Initially the liquid is gas-free. Find final partial pressure and dissolved moles.

  1. Gas capacity per bar is V_g/(RT)=1/(0.08314×300)≈0.04009 mol bar⁻¹.
  2. Liquid capacity per bar is k_cpV_l=0.0200 mol bar⁻¹.
  3. p=0.0100/(0.04009+0.0200)≈0.1664 bar.
  4. n_dissolved=0.0200×0.1664≈0.00333 mol; gas-phase remainder≈0.00667 mol.

Answer Final p≈0.166 bar; dissolved amount≈0.00333 mol.

06 / UNDERSTAND & APPLY

6. Temperature and the solvent change the constant

Henry coefficients are temperature dependent and specific to the gas–solvent pair. Many gases become less physically soluble in water as temperature rises over familiar ranges, reflecting exothermic dissolution in those cases. This is a common trend, not a universal rule for every gas, solvent and temperature interval.

The direction of the numerical constant change depends on its definition. If solubility decreases on warming, a concentration-per-pressure coefficient k_cp decreases, while a pressure-over-mole-fraction K_H generally increases when the solvent-basis conversion is treated consistently. A statement about a constant without its equation can therefore appear to reverse the physical trend.

Use constants measured at the relevant temperature. If a temperature-dependence parameter or dissolution enthalpy is supplied, a van ’t Hoff-type relation may estimate changes under its assumptions. Do not invent a universal ten-degree correction factor or use one gas’s coefficient for another gas.

Dissolved salts can alter gas solubility through changes in solvent interactions, often producing a salting-out effect for particular systems. Organic cosolvents can also change solubility markedly. These are changes in the medium; the pure-water Henry coefficient should not be reused unchanged unless the problem explicitly permits it.

ORIGINAL WORKED EXAMPLE

Interpret opposite-looking trends

A gas becomes less soluble on heating at fixed partial pressure. How do the two definitions C=k_cpp and p=K_Hx respond qualitatively?

  1. Lower solubility means lower C and lower x at the same p.
  2. In C=k_cpp, k_cp must decrease.
  3. In p=K_Hx, K_H must increase as x decreases, within the stated concentration convention.
  4. The apparent opposite trends arise from reciprocal-style definitions, not contradictory chemistry.

Answer k_cp decreases; K_H increases under the stated definitions.

07 / UNDERSTAND & APPLY

7. Henry’s law may describe one species while chemistry controls total absorption

If dissolved gas reacts, distinguish physical dissolution from subsequent chemical conversion. For a weakly acidic dissolved gas species G, proton-transfer reactions may convert it to ionic forms. Removing neutral G into those forms allows more gas to enter from the headspace, increasing total analytical dissolved material beyond the neutral-species concentration alone.

For a simplified monoprotic model G⇌H⁺+G⁻ with K_a=[H⁺][G⁻]/[G], and Henry concentration [G]=k_cpp, the total[G]_T=[G]+[G⁻]=k_cpp(1+K_a/[H⁺]) if pH is maintained and the model includes all relevant species. This illustrates coupling; real carbon dioxide, for example, involves multiple equilibria and convention choices.

For a basic gas such as ammonia, protonation to its conjugate acid can similarly enhance total uptake in acidic solution. A finite acid supply is consumed as absorption proceeds, so an initially low pH may not remain fixed. A complete closed-system calculation must include acid/base balances as well as gas and liquid inventories.

The gas’s Henry coefficient should be associated with the specified molecular species and solvent conditions, while an “effective” coefficient based on total analytical concentration can depend strongly on pH. Mixing these definitions can produce a false claim that the fundamental physical-solubility law failed.

ORIGINAL WORKED EXAMPLE

A simple coupled-equilibrium model

A hypothetical neutral dissolved gas G has [G]=2.00×10⁻⁴ M from Henry’s law. It behaves as a monoprotic acid with K_a=1.00×10⁻⁵ at maintained [H⁺]=1.00×10⁻⁶ M. Find total [G]+[G⁻], ignoring all other species.

  1. [G⁻]/[G]=K_a/[H⁺]=10.0.
  2. Total concentration=[G](1+10.0)=2.00×10⁻⁴×11.0.
  3. The buffer must maintain the stated proton concentration as gas is absorbed.
  4. The neutral G concentration remains the Henry-law value; the total includes its ionic product.

Answer Total concentration=2.20×10⁻³ M.

08 / UNDERSTAND & APPLY

8. Pressure release changes equilibrium before bubbles necessarily appear

When a pressurized gas–liquid system is opened to a lower gas partial pressure, its previous dissolved concentration may exceed the new equilibrium value. Gas then tends to leave the liquid. Bubble formation, available nucleation sites, mixing and transport determine how quickly the visible release occurs. Henry’s law predicts the new equilibrium concentration, not the exact bubbling history.

The law also clarifies why a gas mixture’s composition matters in environmental or industrial equilibrium calculations. Oxygen solubility responds to oxygen partial pressure, while carbon dioxide responds to its own partial pressure and coupled aqueous chemistry. Total atmospheric or vessel pressure alone is insufficient without composition.

At high pressures, gas fugacity replaces ideal partial pressure for accurate work. At appreciable dissolved concentration, liquid activities and pressure effects on the solvent can matter. Strong chemical reaction, condensation or multiple liquid phases can require a different model. State the dilute physical-equilibrium range rather than present the linear relation as universal.

For an exam calculation, the best audit is compact: identify the constant definition, find the correct species partial pressure, match temperature and units, account for finite inventory if sealed, and distinguish free molecular species from total analytical gas-derived material. Each step corresponds to a real physical assumption.

09 / UNDERSTAND & APPLY

9. Original Henry-law practice

These original problems use ideal gas phases and dilute solutions with the supplied constant definitions. Hypothetical values are calculation data, not tabulated solubilities. No medical or engineering operation is prescribed by these conceptual examples.

Always report the equation used beside the answer. This makes a larger-versus-smaller constant argument checkable and prevents pressure-unit mistakes from being hidden behind the same symbol K_H.

ORIGINAL PRACTICE · 1

For C=k_cpp with k_cp=2.00×10⁻³ M bar⁻¹ and p=0.500 bar, find C.

Reveal answer & reasoning

1.00×10⁻³ M.

Multiply the concentration-per-pressure coefficient by the species partial pressure; bar cancels.

ORIGINAL PRACTICE · 2

A gas is 10.0% of an ideal mixture at 8.00 bar. What pressure enters its Henry equation?

Reveal answer & reasoning

0.800 bar.

Its partial pressure is yP=0.100×8.00. Total pressure includes other gases and is not the correct species driving pressure.

ORIGINAL PRACTICE · 3

For p=K_Hx, K_H doubles while p stays fixed. What happens to x?

Reveal answer & reasoning

It halves.

x=p/K_H. This volatility-style constant is inversely related to solubility at fixed pressure.

ORIGINAL PRACTICE · 4

A 0.00200 M dissolved gas solution occupies 0.750 L. How many moles of that species are dissolved?

Reveal answer & reasoning

0.00150 mol.

n=CV=0.00200×0.750. This is a liquid solution-volume calculation.

ORIGINAL PRACTICE · 5

Does stirring an already equilibrated gas–liquid system at unchanged temperature and partial pressure raise its equilibrium concentration?

Reveal answer & reasoning

No in the stated equilibrium model.

Stirring changes mass-transfer speed, not the Henry equilibrium relation when the chemical conditions are unchanged.

ORIGINAL PRACTICE · 6

Why can using total dissolved carbon in a Henry equation for neutral CO₂ be incorrect?

Reveal answer & reasoning

Total carbon can include bicarbonate and carbonate formed by subsequent equilibria.

The Henry relation applies to its specified dissolved molecular species or explicitly defined convention; total analytical concentration requires the coupled speciation model.

ORIGINAL PRACTICE · 7

An inert gas is added to a fixed-volume ideal headspace at fixed temperature without changing the amount of gas A. Does p_A rise?

Reveal answer & reasoning

No.

p_A=n_ART/V is unchanged. The total pressure rises, but Henry’s law for A responds to its unchanged partial pressure under the model.

ORIGINAL PRACTICE · 8

Why might a sealed finite headspace give less dissolved gas than a calculation using its initial pressure throughout?

Reveal answer & reasoning

Dissolution removes gas-phase moles and lowers the final partial pressure.

The final pressure must satisfy the combined gas-plus-liquid mole balance. Treating the initial pressure as a fixed reservoir can overestimate dissolution.

KEEP THE BIG IDEAS

Quick revision

  • Write the defining Henry equation and units before using a constant.
  • Gas solubility depends on the species partial pressure, not automatically total pressure.
  • A concentration-form constant and pressure-over-mole-fraction constant imply opposite numerical solubility trends.
  • A finite sealed headspace requires a material balance, not a fixed-pressure assumption.
  • Temperature, solvent and dissolved salts can change the coefficient.
  • Chemical conversion changes total uptake while the neutral-species Henry relation remains a separate constraint.
  • Equilibrium solubility and the rate of gas transfer are different questions.
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The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.

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