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Question

H2S is a toxic gas with rotten egg smell is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.2 m, calculate Henry's constant.

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Solution

The solubility of H2S gas =0.2 m
i.e.,
0.2 moles of H2S is dissolved in 1 kg of water (solvent).
1 kg of water contains =100018=55.55 moles of water
Mole fraction of H2S gas in solution is,
XH2S=nH2SnH2S+nH2O=0.20.2+55.55=0.255.75=0.0035Pressure at STP = 1.01 bar=1 atmBy Henry law,PH2S=KH×xH2SKH=1.01 bar0.0035=288.6288 bar
Thus,
Henry's constant =288 bar

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