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31/12/2021 4:53 pm
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H2S, a toxic gas with a rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.195 m, calculate Henry's law constant.
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31/12/2021 4:59 pm
It is given that the solubility of H2S in water at STP is 0.195 m, i.e., 0.195 mol of H2S is dissolved in 1000 g of water.
Moles of water = \(\frac{1000g}{18\;g\;mol^{-1}}\)
= 55.56 mol
∴Mole fraction of H2S, x = \(\frac{Moles\;of\;H_2S}{Moles\;of\;H_2S + Moles\;of\;water}\)
= \(\frac{0.195}{0.195 + 55.56}\)
= 0.0035
At STP, pressure (p) = 0.987 bar
According to Henry's law
p = KHx
⇒ KH = \(\frac{p}{x}\)
= \(\frac{0.987}{0.0035}\)bar
= 282 bar