2.6    H_{2}S, a toxic gas with rotten egg-like smell, is used for the qualitative analysis. If the solubility of H_{2}S,  in water at STP is 0.195\; m. calculate Henry’s law constant.

Answers (1)

For finding Henry's constant we need to know about the mole fraction of H2S.

Solubility of H2S in water is given to be 0.195 m .

i.e.,  0.195 moles in 1 Kg of water.

                                                             Moles\: of\: water :=\frac{1000}{18} = 55.55\ moles

So          x_{H_2S} = Mole\ fraction\ of\ H_2S 

                                              = \frac{0.195}{0.195+55.55} = 0.0035

At STP conditions, pressure = 1 atm or 0.987 bar

Equation is :                            p_{H_2S} = K_h\times x_{H_2S}

 So we get :

                                                  K_h =\frac{0.987}{0.0035} = 282\ bar

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