Question

$H_{2}S$ 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.

Answer 1

GivenL Molarity $=0.195m$

Molarity $=\frac{no.ofmolesofsolute}{massofsolvent\left(inkg\right)}$

Let us consider mass of water $=1kg$

$\therefore$ moles of solute $=$ molarity $\times$mass of solvent

$=0.195\times 1=0.195mol$

Molar mass of $H_{2}O=2\times 1+16=18$

No. of moles of water $=\frac{mass}{molarmass}=\frac{1000}{18}=55.56mol$ ($\because$ $1kg-1000g$)

mole fraction of $H_{2}S=\frac{0.195}{(0.195+55.56)}=0.0035$

At STP,

$P=0.987atm$

According to Henry's law

$P=K_H\times x$

$K_H=\frac{p}{x}=0.987/0.0035=282$ atm