Question

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.

Answer 1

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 = ${\displaystyle \frac{\text{Mass of the substance}}{\text{Molar mass of the substance}}}$, ${\text{X}}_{\text{A}}={\displaystyle \frac{{\text{n}}_{\text{A}}}{{\text{n}}_{\text{A}}+{\text{n}}_{\text{B}}}}$ 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 = ${\displaystyle \frac{\text{Mass of the substance}}{\text{Molar mass of the substance}}}$
Using the above formula, we get
Number of moles of water (${\text{H}}_2\text{O}$), ${\text{n}}_{{\text{H}}_2\text{O}}$ = ${\displaystyle \frac{\text{Mass of }{\text{H}}_2\text{O}}{\text{Molar mass of }{\text{H}}_2\text{O}}}$
By substituting the known values in the above equation, we get
.. Number of moles of water (${\text{H}}_2\text{O}$), ${\text{n}}_{{\text{H}}_2\text{O}}$ = ${\displaystyle \frac{1000}{18}}=55.56$ 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
${\text{X}}_{\text{A}}={\displaystyle \frac{{\text{n}}_{\text{A}}}{{\text{n}}_{\text{A}}+{\text{n}}_{\text{B}}}}\to \text{(1)}$
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 ${\text{H}}_2\text{S}$, ${\text{X}}_{{\text{H}}_2\text{S}}={\displaystyle \frac{{\text{n}}_{{\text{H}}_2\text{S}}}{{\text{n}}_{{\text{H}}_2\text{S}}+{\text{n}}_{{\text{H}}_2\text{O}}}}$
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 ${\text{H}}_2\text{S}$, ${\text{X}}_{{\text{H}}_2\text{S}}={\displaystyle \frac{0.195}{0.195+55.56}}=0.003497$
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,
${\text{P}}_{\text{gas}}={\text{K}}_{\text{H}}\times {\text{X}}_{\text{gas}}\to (\text{2)}$
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
$$\begin{gathered} \text{P}={\text{K}}_{\text{H}}\times {\text{X}}_{{\text{H}}_2\text{S}} \\ \Rightarrow {\text{K}}_{\text{H}}={\displaystyle \frac{\text{P}}{{\text{X}}_{{\text{H}}_2\text{S}}}} \\ \Rightarrow {\text{K}}_{\text{H}}={\displaystyle \frac{\text{0}\text{.987}}{0.003497}}=282.24\text{ atm} \end{gathered}$$

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.