Question

$H_{2}S$ is a toxic gas used in qualitative analysis, if solubility of $H_{2}S$ in water at STP is $0.195m$, what is the value of $K_H$?

• A. $0.0263$ bar
• B. $69.16$ bar
• C. $192$ bar
• D. $282$ bar

Answer 1

The correct option is D $282$ bar

No of moles of $H_{2}S$ = $0.195mol$

No of moles of $H_{2}O$ = $\frac{1000}{18}=55.55mol$

Hence mole fraction of $H_{2}S$ = $\frac{0.195}{0.195+55.55}=0.0035$

Pressure at $STP$ = $0.987bar$

As per Henry's law,

$p=K_H\times x$

$K_H=\frac{p}{x}$

$=\frac{0.987}{0.0035}=282bar$

So, the correct option is $D$