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Q.
$H_2S$ is a toxic gas used in qualitative analysis. If solubility of $H_2S$ in water at $STP$ is $0.195\, m$, what is the value of $K_H$ ?
Solutions
Solution:
No. of moles of $H_2S = 0.195$
No. of moles of $H_{2}O = \frac{1000}{18} = 55.55\,mol$
Mole fraction of $H_{2}S =\frac{0.195}{0.195 + 55.55} = 0.0035$
Pressure at $STP = 0.987$ bar
According to Henry’s law, $p = K_{H}x$
or $K_{H} = \frac{p_{H_2S}}{x_{H_2S}}= \frac{0.987}{0.0035}$
$= 282\,$bar