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

Answer 1

Answer: (D)

No. of moles of $H_{2}S=0.195$
No. of moles of $H_{2}O=\frac{1000}{18}=55.55mol$
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 $K_H=\frac{p_{H_{2}S}}{x_{H_{2}S}}=\frac{0.987}{0.0035}$
$=282$bar