$H_{2} S$ is 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 .

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Solution

1 Kg of solvent contains =
$\frac{1000}{18} = 55.55 m o l e s$
$HencethemolefractionofH2Sgasinsolution=$
$χ_{H_{2} S} = \frac{n_{H_{2} S}}{n_{H_{2} S} + n_{H_{2} O}} = \frac{0.195}{0.195 + 55.5} = \frac{0.195}{55.745} = 0.0035$
$P r e s s u r e a t S . T . P (b e c a u s e n o r m a l l y p r e s s u r e i s t a k e n a t S t a n d a r d c o n d i t i o n) = 1 a t m = 0.987 b a r$
$H e n c e t h e f o r m u l a f o r H e n r y^{'} s c o n s t a n t i s$
$p_{H_{2} S} = K_{H} \times χ_{H_{2} S}$
$H e n c e K_{H} = \frac{0.987}{0.0035} = 282 b a r$
$H e n c e t h e h e n r y^{'} s c o n s t a n t i s 282 b a r$

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