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.
Solution:
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 $=\frac{1000 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}$
= 55.56 mol
$\therefore$ Mole fraction of $\mathrm{H}_{2} \mathrm{~S}, x=\frac{\text { Moles of } \mathrm{H}_{2} \mathrm{~S}}{\text { Moles of } \mathrm{H}_{2} \mathrm{~S}+\text { Moles of water }}$
$=\frac{0.195}{0.195+55.56}$
= 0.0035
At STP, pressure (p) = 0.987 bar
According to Henry’s law:
p = KHx
$\Rightarrow \mathrm{K}_{\mathrm{H}}=\frac{p}{x}$
$=\frac{0.987}{0.0035}$ bar
= 282 bar