A toxic gas with rotten egg-like smell, ${{\text{H}}_2}{\text{S}}$, is used for qualitative analysis. If the solubility of ${{\text{H}}_2}{\text{S}}$ in water at STP is 0.195 m. Calculate Henry’s law constant.

Hint- Here, we will proceed by firstly finding out the number of moles of water and then, with the help of it we will calculate the number of moles of ${{\text{H}}_2}{\text{S}}$ gas. Then, we will apply Henry’s Law in order to find Henry’s law constant.

Complete answer:
Formulas Used- Number of moles of any substance = $\dfrac{{{\text{Mass of the substance}}}}{{{\text{Molar mass of the substance}}}}$, ${{\text{X}}_{\text{A}}} = \dfrac{{{{\text{n}}_{\text{A}}}}}{{{{\text{n}}_{\text{A}}} + {{\text{n}}_{\text{B}}}}}{\text{ }}$ and ${{\text{P}}_{{\text{gas}}}} = {{\text{K}}_{\text{H}}} \times {{\text{X}}_{{\text{gas}}}}$.
Given, Solubility of ${{\text{H}}_2}{\text{S}}$ in water at STP = 0.195 m
This means that 0.195 mol of ${{\text{H}}_2}{\text{S}}$ is dissolved in 1000 g of water
So, Number of moles of ${{\text{H}}_2}{\text{S}}$, ${{\text{n}}_{{{\text{H}}_2}{\text{S}}}}$ = 0.195 mol
Mass of water (${{\text{H}}_2}{\text{O}}$) = 1000 g
Also, Molar mass of ${{\text{H}}_2}{\text{O}}$ = 2(Atomic mass of H) + Atomic mass of O = 2(1) + 16 = 18 ${\text{gmo}}{{\text{l}}^{ - 1}}$
As we know that
Number of moles of any substance = $\dfrac{{{\text{Mass of the substance}}}}{{{\text{Molar mass of the substance}}}}$
Using the above formula, we get
Number of moles of water (${{\text{H}}_2}{\text{O}}$), ${{\text{n}}_{{{\text{H}}_2}{\text{O}}}}$ = $\dfrac{{{\text{Mass of }}{{\text{H}}_2}{\text{O}}}}{{{\text{Molar mass of }}{{\text{H}}_2}{\text{O}}}}$
By substituting the known values in the above equation, we get
.. Number of moles of water (${{\text{H}}_2}{\text{O}}$), ${{\text{n}}_{{{\text{H}}_2}{\text{O}}}}$ = $\dfrac{{1000}}{{18}} = 55.56$ mol
If a mixture contains two substances A and B having a number of moles as ${{\text{n}}_{\text{A}}}$ and ${{\text{n}}_{\text{B}}}$ respectively. Then, mole fraction of substance A is given by
${{\text{X}}_{\text{A}}} = \dfrac{{{{\text{n}}_{\text{A}}}}}{{{{\text{n}}_{\text{A}}} + {{\text{n}}_{\text{B}}}}}{\text{ }} \to {\text{(1)}}$
Here, the mixture or solution consists of ${{\text{H}}_2}{\text{S}}$ and ${{\text{H}}_2}{\text{O}}$. Using the formula given by equation (1), we have
Mole fraction of ${{\text{H}}_2}{\text{S}}$, ${{\text{X}}_{{{\text{H}}_2}{\text{S}}}} = \dfrac{{{{\text{n}}_{{{\text{H}}_2}{\text{S}}}}}}{{{{\text{n}}_{{{\text{H}}_2}{\text{S}}}} + {{\text{n}}_{{{\text{H}}_2}{\text{O}}}}}}$
By substituting ${{\text{n}}_{{{\text{H}}_2}{\text{S}}}}$ = 0.195 mol and ${{\text{n}}_{{{\text{H}}_2}{\text{O}}}}$ = 55.56 mol in the above equation, we get
$ \Rightarrow $ Mole fraction of ${{\text{H}}_2}{\text{S}}$, ${{\text{X}}_{{{\text{H}}_2}{\text{S}}}} = \dfrac{{0.195}}{{0.195 + 55.56}} = 0.003497$
At standard temperature pressure condition (STP),
Temperature = 273 K and Pressure = 0.987 atm
So, Partial pressure of ${{\text{H}}_2}{\text{S}}$ under equilibrium conditions, P = 0.987 atm
According to Henry’s Law,
${{\text{P}}_{{\text{gas}}}} = {{\text{K}}_{\text{H}}} \times {{\text{X}}_{{\text{gas}}}}{\text{ }} \to ({\text{2)}}$
where ${{\text{P}}_{{\text{gas}}}}$ denotes the partial pressure of a gas under equilibrium conditions, ${{\text{K}}_{\text{H}}}$ denotes henry’s law constant and ${{\text{X}}_{{\text{gas}}}}$ denotes the mole fraction of the gas
Using the formula given by equation (2) for ${{\text{H}}_2}{\text{S}}$, we get
$
  {\text{P}} = {{\text{K}}_{\text{H}}} \times {{\text{X}}_{{{\text{H}}_2}{\text{S}}}} \\
   \Rightarrow {{\text{K}}_{\text{H}}} = \dfrac{{\text{P}}}{{{{\text{X}}_{{{\text{H}}_2}{\text{S}}}}}} \\
   \Rightarrow {{\text{K}}_{\text{H}}} = \dfrac{{{\text{0}}{\text{.987}}}}{{0.003497}} = 282.24{\text{ atm}} \\
 $

Therefore, Henry's law constant for the given problem is equal to 282.24 atm (atmospheric pressure).

Note- Since, solubility refers to the maximum amount of solute (in moles) that can dissolve in a known quantity of solvent (in gram) at a certain temperature. In this particular problem, taking ${{\text{H}}_2}{\text{S}}$ as solute and water as solvent. From here, we have said that solubility of ${{\text{H}}_2}{\text{S}}$ in water at STP is 0.195 m means that 0.195 mol of ${{\text{H}}_2}{\text{S}}$ is dissolved in 1000 g of water.