Henry-Law Gas Solubility Problems

Introduction

Increasing the partial pressure of a gas above a liquid often increases how much of that gas dissolves at equilibrium. Henry's law describes this proportionality for a dilute dissolved gas at fixed temperature. Numerical problems become error-prone because textbooks use more than one definition of a “Henry constant.” Write the supplied equation and inspect the constant's units before multiplying or dividing.

Core explanation

In one common solubility convention, c = kp, where c is dissolved gas concentration in mol L⁻¹, p is that gas's partial pressure, and k has units mol L⁻¹ atm⁻¹ if p is measured in atm. A larger k then means greater solubility at a fixed pressure. Some sources use a volatility convention p = Kc or p = Kx, where x is dissolved-gas mole fraction; their K has different units and generally larger K means less solubility. The symbol k H by itself is therefore ambiguous. IUPAC recommends specifying which Henry-law solubility or volatility constant is meant, and the problem's equation and units settle the calculation.

With c = kp and k = 1.50 × 10⁻³ mol L⁻¹ atm⁻¹ at a stated temperature, a gas partial pressure of 0.400 atm gives c = 6.00 × 10⁻⁴ mol L⁻¹. In 2.00 L of solution, the equilibrium dissolved amount is 1.20 × 10⁻³ mol if solution volume is 2.00 L. If partial pressure doubles to 0.800 atm while temperature and solution behaviour are unchanged, the predicted concentration doubles to 1.20 × 10⁻³ M. The constant is for that gas–solvent pair at that temperature; it is not a universal constant.

Use the gas's partial pressure rather than the total pressure of a mixture. If a headspace has total pressure 2.00 atm and gas A has mole fraction 0.25 under an ideal-gas-mixture model, p A = 0.500 atm. Applying c = kp with 2.00 atm would overpredict A's dissolved concentration by a factor of four. A carrier gas can change total pressure without changing p A if A's amount, headspace volume and temperature remain fixed in a suitable idealised setting.

The proportionality describes an equilibrium condition. A newly pressurised bottle may need time to approach it, while opening a bottle lowers gas partial pressure and can cause dissolved gas to escape. Temperature can change k substantially, so a value tabulated at one temperature cannot generally be reused at another. Gases that react chemically with the solvent may have an apparent total uptake that is not described by the simple physical-dissolution relation alone. Oxygen in water is often a useful simple example; carbon dioxide in water can require care because dissolved species interconvert.

The IUPAC Gold Book definition for a concentration-over-pressure constant gives this convention explicitly, while its general Henry-constant entry explains the multiple solubility and volatility forms. The arithmetic here follows whichever form and units a question states.

Step-by-step reasoning

1. Identify the gas, solvent, temperature and explicit Henry-law equation supplied. 2. Determine the target gas's partial pressure from its gas-phase mixture if necessary. 3. Convert pressure to the unit expected by the stated constant. 4. Multiply or divide according to that equation, keeping units visible. 5. Convert concentration to total dissolved moles only if solution volume is given and the equilibrium assumption applies.

Visual explanation

Draw a gas box above a liquid box. Mark only A's contribution to headspace pressure as p A, even when other gas dots are present. An arrow from p A into the liquid is labelled c A = kp A. Sketch a straight line through the origin on a graph of c A versus p A for a fixed temperature and dilute concentration range.

Real-world analogy

A warehouse accepts parcels in proportion to the number of delivery trucks assigned to it, not all trucks in the city. The gas's own partial pressure is the relevant driving condition for its dissolution. Total pressure includes unrelated gas contributions that need not represent additional molecules of the gas being studied.

Real-world example

In a closed fizzy drink, elevated carbon-dioxide partial pressure supports substantial dissolved CO₂. When the container opens, the pressure above the liquid falls, and the old dissolved amount is no longer the new equilibrium amount; bubbles can form. Real drinks also involve carbon dioxide chemistry, so a precise speciation model may be richer than the basic Henry-law picture.

Why?

Why does doubling p double c in the stated model? At fixed temperature and in a dilute regime, k is held constant. The direct proportionality c = kp therefore doubles its right-hand side when p doubles. It does not predict that every gas remains linear at arbitrarily high pressure.

Common misconception

“Any printed Henry constant should be multiplied by pressure.” Some definitions use p/c or p/x, requiring division or a different composition measure. Inspect units: mol L⁻¹ atm⁻¹ naturally multiplies atm to produce mol L⁻¹, while atm L mol⁻¹ would multiply concentration to produce pressure.

Worked example

At a fixed temperature, a gas in water follows c = kp with k = 2.00 × 10⁻³ mol L⁻¹ atm⁻¹. Above 1.50 L of solution is a gas mixture at 1.20 atm total pressure; the target gas has gas-phase mole fraction 0.30. Its partial pressure is p = 0.30 × 1.20 = 0.360 atm. Thus c = (2.00 × 10⁻³)(0.360) = 7.20 × 10⁻⁴ mol L⁻¹. Dissolved moles at equilibrium are cV = (7.20 × 10⁻⁴)(1.50) = 1.08 × 10⁻³ mol. Using 1.20 atm instead would count all gas species as though they were the target.

Quick check

1. For c = kp at fixed temperature, k = 0.0040 M atm⁻¹ and p = 0.25 atm. What is c? Answer: c = (0.0040 M atm⁻¹)(0.25 atm) = 0.0010 M.

Exam focus

Write the supplied Henry relation, constant units, and target-gas partial pressure before computing. Keep temperature fixed for pressure-ratio shortcuts. Distinguish equilibrium dissolved concentration from a measured amount during a transient process.

Advanced insight

When a finite closed headspace shares gas with a finite liquid volume, gas dissolution lowers headspace moles and thus its partial pressure. An exact equilibrium calculation may need a gas-phase material balance combined with c = kp and PV = nRT. Treating headspace pressure as fixed is an additional reservoir assumption.

Summary

Henry-law solubility in the stated c = kp convention scales with the target gas's partial pressure at fixed temperature. Other Henry-constant conventions invert or change the composition measure, so equation and units are decisive. Equilibrium, temperature and possible chemical reaction limit the simple relation.

Practice questions

1. A gas has p = 0.60 atm and k = 1.0 × 10⁻³ M atm⁻¹ in c = kp. Find c. Answer: c = 6.0 × 10⁻⁴ M. 2. If total gas pressure is 2.0 atm and the target mole fraction is 0.10, what p enters Henry's law? Answer: p = 0.10 × 2.0 = 0.20 atm. 3. If a problem defines K = p/c instead of c/p, how do you obtain c from p and K? Answer: Rearrange p = Kc to c = p/K; do not multiply p by K.