H2S is a toxic gas with a rotten egg-like smell and is used for qualitative analysis. If the solubility of H2S in water at STP is 0.195M, calculate Henry’s law constant.
By BYJU'S Exam Prep
Updated on: September 13th, 2023
We know that
Henry’s law states that the amount of gas dissolved in a liquid is directly proportional to the partial pressure of that gas above the liquid when the temperature is kept constant. The constant of proportionality for this relationship is called Henry’s law constant (usually denoted by ‘kH’).
P ∝ C (or) P = kH.C
Where,
- ‘P’ is the partial pressure of the gas in the atmosphere above the liquid.
- ‘C’ is the concentration of the dissolved gas.
- ‘kH’ is Henry’s law constant of the gas.
Solvent mass = 1 kg
Molality = 0.195m
Moles of solute = 0.195 mol
So the number of moles of water = 1000/18 = 55.56 mol
Molecular fraction of H2S = 0.195/ (0.195 + 55.5)
x = 0.035
As Henry’s constant KH = P/x
Substitute the values
KH = 0.987/0.0035
KH = 282 atm
Therefore, Henry’s law constant is 282 atm.
Summary:
H2S is a toxic gas with a rotten egg-like smell and is used for qualitative analysis. If the solubility of H2S in water at STP is 0.195M, calculate Henry’s law constant.
H2S is a toxic gas with a rotten egg-like smell and is used for qualitative analysis. If the solubility of H2S in water at STP is 0.195M, Henry’s law constant is 282 atm.
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