SME
Physical Chemistry I: Thermodynamics & Kinetics · Problem type
Gas solubility (Henry’s law)
Use this for “Air holds N₂ at 0.78 atm. With kH = 6.1 × 10⁻⁴ mol/(L·atm) at 25 °C, find the N₂ dissolved in water.”
Chart
c(0.21) ≈ 0.000273
This is an example. Type your own values to start a new problem.
c (M)
=
kH (mol/(L·atm))
×
p (atm)
Formulas
c
=
kH
×
p
2.73 ×
10
-4
=
0.0013
×
0.21
Assumptions
•
Henry’s law: at one temperature, the dissolved concentration of a gas is proportional to its partial pressure above the liquid, c = kHp.
•
A dilute solution of a gas that does not react with the solvent; kH belongs to that gas, solvent and temperature.
•
The graph is the line c = kHp through the origin, with the point at this partial pressure.
Step-by-step
WE KNOW
kH = 0.0013 mol/(L·atm), p = 0.21 atm
FIND
dissolved concentration (c)
Step 1 · Find dissolved concentration (c)
Use c = kH × p
Multiply Henry’s constant by the partial pressure.
c
=
0.0013
×
0.21
c
=
2.73 ×
10
-4
M
Check
Put the numbers back in. Both sides should match.
2.73 ×
10
-4
=
0.0013
×
0.21
✓
Topic check
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