Solve the following

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