Question

H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.195 m, calculate Henry’s law constant.

Answer 1

Answer :

Given :-

Solubility of $H_{2}S$ gas = 0.195 mol/kg = 0.195 moles of $H_{2}S$ is disolved in 1 kg of water (solvent)

1 kg of solvent contains = $\frac{1000}{18}$ = 55.55 moles

Mole fraction of $H_{2}S$ gas in solution,

$χ_{H_{2}S}$ = $\frac{η_{H_{2}S}}{η_{H_{2}S + η_{H_{2}O}}}$

$χ_{H_{2}S}$ = $\frac{0.195}{0.195 + 55.55}$

$χ_{H_{2}S}$ = 0.0035

Pressure at STP = 0.987 bar= 1 atm

$ρ_{H_{2}S}$ = $K_{H}$ × $χ_{H_{2}S}$

$K_{H}$ = $\frac{0.987 bar}{0.0035}$

$K_{H}$ ⇒ 282 bar

Answer 2

Answer :

• Henry's law constant = 282 bar.

Step-by-step explanation :

Given that,

• Moles of $\rm{H_2SO_4}$ = 0.195.
• Mass of water (solvent) = 1 kg = 1000 g.

Also,

• Pressure at STP = 0.987 bar.

$\hrulefill$

Moles of water = $\rm{\dfrac{1000}{18}=55.55}$

Now, Mole fraction of $\rm{H_2S\:(x)=}$

$\implies\rm{\dfrac{0.195}{55.55+0.195}}$

$\implies\rm{0.0035}$

$\hrulefill$

We know, $\rm{Pressure,\:p_{H_2S}=K_Hx_{H_2S}}$

$\bold{OR}$

$\rm{K_H=\dfrac{p_{H_2S}}{x_{H_2S}}}$

Putting the values, we get,

$\rm{K_H=\dfrac{0.987\:bar}{0.0035}=}\:\bold{282\:bar.}$