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The solubility of H_2Sgas in water at STP.195m .Calculate its Henry's law constant at STP

Answer»

SOLUTION :SOLUBILITY of `H_2S` at STP = 0.195m,
i.e., 0.195 moles of `H_2S` is DISSOLVED in 1000g (1kg) of water
Number of moles in 1000g `H_2O` = 1000/18 = 55.55
`therefore` Mole fraction of `H_2S` inthe solution (x) =` 0.195/(0.195+55.55)` = `0.195/55.745`= 0.0035
STP pressure =1 BAR
Applying Henry.s law,`P_(H_2S)= K_H xx x_(H_2S) therefore K_H of H_2S = P_(H_2S)/x_(H2S)` = 1 bar/0.0035 = 285.7 bar


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