Question

$H_{2}S$ is a toxic gas with rotten egg smell is used for the qualitative analysis. If the solubility of $H_{2}S$ in water at STP is $0.2m$, calculate Henry's constant.

• A. $208atm$
• B. $243bar$
• C. $310bar$
• D. $288bar$

Answer 1

The correct option is D $288bar$

$\text{The solubility of}{H}_{2}S\text{gas}=0.2m$
i.e.,
$0.2$ moles of $H_{2}S$ is dissolved in $1kg$ of water (solvent).
$\text{1 kg of water contains}=\frac{1000}{18}=55.55\text{moles of water}$
Mole fraction of $H_{2}S$ gas in solution is,
$X_{H_{2}S}=\frac{n_{H_{2}S}}{n_{H_{2}S}+n_{H_{2}O}}=\frac{0.2}{0.2+55.55}=\frac{0.2}{55.75}=0.0035\text{Pressure at STP =}1.01bar=1atm\text{By Henry law,}{P}_{H_{2}S}=K_H\times x_{H_{2}S}{K}_H=\frac{1.01bar}{0.0035}=288.6\approx 288bar$
Thus,
$\text{Henry's constant}=288bar$