Question

Calculate the proportion of ${{O}_{2}}$ and ${{N}_{2}}$ dissolved in water at 298 K. When air containing 20% ${{O}_{2}}$and 80% ${{N}_{2}}$ by volume is in equilibrium with water at 1 atm pressure. Henry’s law constants for two gases are ${{K}_{H}}({{O}_{2}})=4.6\times {{10}^{4}}atm$ and ${{K}_{H}}({{N}_{2}})=8.5\times {{10}^{4}}atm$.

Answer 1

Hint: Henry’s law states a relationship between partial pressure and mole fraction of a gas. The partial pressure can be calculated by total pressure multiplied with the mole fraction in the vapor phase. The mole fraction in vapor phase can be calculated by the percentage of the respective gases in the air. Find the particular mole fractions of gases as their proportion in the solution.

Complete answer:
We have been given two gases${{O}_{2}}$ and ${{N}_{2}}$ having 20% and 80% volume in air at equilibrium with water at 1 atm pressure. We have to find the proportion of these gases dissolved in water at 298 K. the proportion means we have to determine the mole fractions, as mole fractions give the proportion of the gases.
According to Henry’s law the partial pressure of a gas is directly proportional to the mole fraction of the gas. The relation is $p={{K}_{H}}\times \chi $ where ${{K}_{H}}$ is Henry’s constant and$\chi $ is the mole fraction.
Now, we have to calculate the partial pressure of both the gases, as given value of Henry’s constant are ${{K}_{H}}({{O}_{2}})=4.6\times {{10}^{4}}atm$ and ${{K}_{H}}({{N}_{2}})=8.5\times {{10}^{4}}atm$.
Suppose that the volume of air is V, so the volume of ${{O}_{2}}$ and ${{N}_{2}}$ will be 0.2V and 0.8V respectively, now the partial pressure will be calculated as $p={{p}_{total}}\times y$ , where y is the mole fraction in vapor phase. So,
${{p}_{{{O}_{2}}}}=1\times \dfrac{0.2V}{V}=0.2\,atm$ and
${{p}_{{{N}_{2}}}}=1\times \dfrac{0.8V}{V}=0.8\,atm$
Putting these values and Henry’s constant values in $p={{K}_{H}}\times \chi $for which solving for mole fraction$\chi $, we have
${{\chi }_{{{O}_{2}}}}=\dfrac{{{p}_{{{O}_{2}}}}}{{{K}_{H}}({{O}_{2}})}$ and ${{\chi }_{{{N}_{2}}}}=\dfrac{{{p}_{{{N}_{2}}}}}{{{K}_{H}}({{N}_{2}})}$
Substituting the values,
${{\chi }_{{{O}_{2}}}}=\dfrac{0.2}{4.6\times {{10}^{4}}atm}$ and${{\chi }_{{{N}_{2}}}}=\dfrac{0.8}{8.5\times {{10}^{4}}atm}$
${{\chi }_{{{O}_{2}}}}=4.3\times {{10}^{-2}}$ and${{\chi }_{{{N}_{2}}}}=9.4\times {{10}^{-2}}$
Hence, the proportion of ${{O}_{2}}$ and ${{N}_{2}}$ dissolved in water at 298 K${{\chi }_{{{O}_{2}}}}=4.3\times {{10}^{-2}}$ and ${{\chi }_{{{N}_{2}}}}=9.4\times {{10}^{-2}}$.

Note:
The proportion is taken in terms of mole fraction. Mole fraction is defined as the number of moles of a particular component (here of oxygen or nitrogen gases) divided by total number of moles of all components. Mole fraction is a fraction so it does not have any unit. Mole fraction is calculated in the solution, while vapor phase has mole fraction of vapor phase as y.