Question

H₂S, a toxic gas with 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.

Answer 1

Solubility of H₂S gas = 0.195 m

∴ Moles of H₂S = 0.195 mol

Mass of water = 1000 g

No. of moles of water = ${\displaystyle \frac{1000 \text{g}}{18 {\text{g mol}}^{-1}}}$

= 55.55 mol

∴ Mole fraction of H₂S gas in the solution (x) = ${\displaystyle \frac{0.195}{0.195+55.56}}$

= ${\displaystyle \frac{0.195}{55.745}}$

= 0.0035

Pressure at STP = 1 bar

According to Henry’s law,

${\displaystyle {\text{p}}_{{\text{H}}_2\text{S}}={\text{K}}_{\text{H}}\times {\text{x}}_{{\text{H}}_2\text{S}}}$

or, K_H = ${\displaystyle \frac{{\text{p}}_{{\text{H}}_2\text{S}}}{{\text{x}}_{{\text{H}}_2\text{S}}}}$

= ${\displaystyle \frac{1 \text{bar}}{0.0035}}$

= 285.7 bar