Question

Concentration of a gas is 0.75 M in water at 25C when the pressure of the gas over the solution is 3.5 atm. What is the Henry's law constant for this gas in water at that temperature?

Answer 1

the molarity of the gas in water is = 0.75 (M)

that means in 1000 mL gas-water solution, the number of moles of the gas dissolved=0.75

Now 1000 mL water = 1000 kg water = (1000/18) = 55.55 moles of H₂O.

now the mole fraction of the gas = 0.75/(55.55+0.75) = 0.0133

now Henry's law is

P= K_h*X₂ where P= pressure, K_h = Henry's law constant, X₂ = mole fraction of the gas

K_h = P/X₂

K_h =(3.5 atm)/0.0133

K_h = (3.5 *101325 Pascal)/0.0133 as 1 atm = 101325 Pascal

K_h = 8.64*10^6 Pascal

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