Table of contents  
Henry's Law  
Factors Affecting the Henry’s Law Constant  
Application of Henry's Law  
Limitations of Henry's Law 
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Henry’s Law gives a quantitative relation between pressure and gas solubility in a liquid. It states that:
The solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution.
The solubility is taken as the mass of the gas dissolved per unit volume of the liquid. Thus, if m is the mass of the gas dissolved per unit volume of the solvent and P is the pressure of the gas in equilibrium with the solution, then
m ∝ p
m = Kp
where K is the proportionality constant.
When P = 1, m = K, i. e., the solubility of the gas at unit pressure is equal to constant K.
Therefore, different gases have different Henry’s laws constant in different solvents, as illustrated graphically above.
This means at a given pressure higher the value of K_{H} (Henry’s Law constant), the lower is the solubility of the gas in a liquid and vice versa.
Example 1. If N_{2} gas is bubbled through water at 293 K, how many millimoles of N_{2} gas would dissolve in 1 litre of water? Assume that N_{2} exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for N_{2} at 293 K is 76.48 kbar.
Solution.
The solubility of gas is related to the mole fraction in aqueous solution.
The mole fraction of the gas in the solution is calculated by applying Henry’s law. Thus:
x (Nitrogen) = p(nitrogen)/K_{H} = 0.987 bar/76.480 = 1.29 × 10^{5}
As 1 litre of water contains 55.5 mol of it, therefore if n represents number of moles of N_{2} in solution,
x (Nitrogen) = n mol/n mol + 55.5 mol = n/55.5 = 1.29 × 10^{5}
(n in denominator is neglected as it is < < 55.5)
Thus n = 1.29 × 10^{–5} × 55.5 mol = 7.16 × 10^{–4} mol
Henry's law holds good if the following conditions are fulfilled:
Example.2. Calculate the concentration of CO_{2} in a soft drink that is bottled at partial pressure of CO_{2} of 4 atm' over the liquid at 25° C. The Henry’ Law constant for CO_{2} in water at 25 ˚C is 3.1 × 10^{2} mol/litreatm.
Solution.
According to Henry's Law:
S = KP
3.1 × 10^{2} × 4 = 0.12 mol litre^{1}
Try Yourself!
Q.1. At 20° C the solubility of nitrogen gas in water is 0.0150 g/litre when the partial pressure N_{2} is 580 torr. Find the solubility N_{2} in H_{2}O at 20°C when its partial pressure is 800 torr.
Ans. 0.0207 g/litre.
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