Question

$H_{2}S$, 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 $0.195m$, calculate Henry's law constant.

Answer 1

Given:
Solubility of $H_{2}S$ in water at STP$=0.195m$.
means, moles of $H_{2}S=0.195mol\text{in}1000$ of water.
Moles of $H_{2}O\text{in}1000g$ of water
$=\frac{1000}{18}$
$=55.55mol$

Mole Fraction of $H_{2}S$,

${\chi }_{H_{2}S}=\frac{n_{H_{2}S}}{n_{H_{2}S}+n_{H_{2}O}}$

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

Henry's Law constant

According to Henry's Law,

$p=K_H.{\chi }_{H_{2}S}\Rightarrow K_H=\frac{p}{{\chi }_{H_{2}S}}$

Given pressure(p) at STP $=0.987bar$
$\Rightarrow K_H=\frac{0.987}{0.0035}=282bar$

Final Answer:
The Henry's law constant is $282bar$.