$H_2S$,a toxic gas with rotten egg like smell,is used for the qualitative analysis. If the solubility of $H_2S$ in water at $STP$ is $0.195 \, m$,calculate Henry's law constant.
(N/A) Given,molality $(m) = 0.195 \, mol \, kg^{-1}$.
This means $0.195 \, mol$ of $H_2S$ is dissolved in $1000 \, g$ $(1 \, kg)$ of water.
Moles of water $= \frac{1000 \, g}{18 \, g \, mol^{-1}} = 55.56 \, mol$.
Mole fraction of $H_2S$ $(x) = \frac{n_{H_2S}}{n_{H_2S} + n_{H_2O}} = \frac{0.195}{0.195 + 55.56} = \frac{0.195}{55.755} \approx 0.0035$.
At $STP$,pressure $(p) = 0.987 \, bar$.
According to Henry's law,$p = K_H \cdot x$.
$K_H = \frac{p}{x} = \frac{0.987 \, bar}{0.0035} \approx 282 \, bar$.
This means $0.195 \, mol$ of $H_2S$ is dissolved in $1000 \, g$ $(1 \, kg)$ of water.
Moles of water $= \frac{1000 \, g}{18 \, g \, mol^{-1}} = 55.56 \, mol$.
Mole fraction of $H_2S$ $(x) = \frac{n_{H_2S}}{n_{H_2S} + n_{H_2O}} = \frac{0.195}{0.195 + 55.56} = \frac{0.195}{55.755} \approx 0.0035$.
At $STP$,pressure $(p) = 0.987 \, bar$.
According to Henry's law,$p = K_H \cdot x$.
$K_H = \frac{p}{x} = \frac{0.987 \, bar}{0.0035} \approx 282 \, bar$.
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The $K_{H}$ values for some gases at the same temperature '$T$' are given below:
Gas
$K_{H} / \text{kbar}$
$Ar$
$40.3$
$CO_{2}$
$1.67$
$HCHO$
$1.83 \times 10^{-5}$
$CH_{4}$
$0.413$
where $K_{H}$ is Henry's Law constant in water. The order of their solubility in water is:
| Gas | $K_{H} / \text{kbar}$ |
|---|---|
| $Ar$ | $40.3$ |
| $CO_{2}$ | $1.67$ |
| $HCHO$ | $1.83 \times 10^{-5}$ |
| $CH_{4}$ | $0.413$ |
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