Question

Air contains \[O_2~\] and \[N_2\] ​ in the ratio \[1:4~\] calculate the ratio of their solubility in terms of mole fraction at atmospheric pressure and room temperature will be (given Henry’s constant for \[{{O}_{2}}=3.3~\times {{10}^{7}}~torr\] for \[{{N}_{2}}=6.6~\times {{10}^{7}}~torr\])

Answer 1

Hint:The ratio of Henry's law constants of two different gases is equal to the ratio of their solubility.
-Henry’s law develops a relationship between the concentration of the species in aqueous phase and the partial pressure of the gas under consideration.

Complete answer:
Henry’s law could be expressed by a simple equation, which is given below,
$H=\dfrac{c}{p}$
Here is $c$ the concentration of a species in the aqueous solution, and $p$ is the partial pressure of that species in the gas phase under equilibrium conditions. And $H$ is the henry’s law constant of proportionality.
We need to calculate the ratio of their solubility in terms of their mole fraction. The values of the henry’s law constants of oxygen and nitrogen are given as
 \[{{O}_{2}}=3.3~\times {{10}^{7}}~torr\] and \[{{N}_{2}}=6.6~\times {{10}^{7}}~torr\]
We know that the pressure of both the gases are equal as well as the concentration, so in order to find the ratio of their mole fraction,
\[\dfrac{{{\chi }_{{{O}_{2}}}}}{{{\chi }_{{{N}_{2}}}}}=\dfrac{{{K}_{H}}({{O}_{2}})\times {p}{{{O}}_{{2}}}}{{{K}_{H}}({{N}_{2}})\times p{{N}_{2}}}\]
Where ${{\chi }_{{{O}_{2}}}}$ and ${{\chi }_{{{N}_{2}}}}$ represents the mole fraction of oxygen and nitrogen, \[{{K}_{H}}({{O}_{2}})\] represents the henry’s constant of oxygen and the \[{{K}_{H}}({{N}_{2}})\] represents the henry’s law constant of nitrogen. And the partial pressures of oxygen and nitrogen are represented by $p{{O}_{2}}$ and $p{{N}_{2}}$ respectively.
The ratio of partial pressure is also given to us as $1:4$.
Now we will substitute all these values in the above equation,
\[\dfrac{{{\chi }_{{{O}_{2}}}}}{{{\chi }_{{{N}_{2}}}}}=\dfrac{3.3\times {{10}^{7}}}{6.6\times {{10}^{7}}}\times \dfrac{1}{4}\]
\[\dfrac{{{\chi }_{{{O}_{2}}}}}{{{\chi }_{{{N}_{2}}}}}=\dfrac{1}{8}\]
So the ratio would be $1:8$.

Additional information:
 In the field of physical chemistry, Henry's law is a law related to the gas which states that the amount of dissolved gas in a liquid is directly proportional to its partial pressure above the liquid. The factor of proportionality is termed Henry's law constant. It was expressed by the English chemist William Henry, who researched about the following topic in the early 19th century.
A very common example where Henry's law is being applied, is in dissolution of nitrogen and oxygen, which is by the way depth-dependent, in the blood of underwater divers that changes during the process of decompression, leading to a sickness usually called bents. As when they come out of the water, the dissolved gases start to come out, which causes tremendous pain in the veins. Another very commonly observed example is given by one's experience with carbonated soft drinks, which contain some specific amount of dissolved carbon dioxide in it. Before opening, the gas above the drink in its container is almost a pure form of carbon dioxide, at a pressure which is much higher than normal atmospheric pressure. After the opening of the bottle, this gas which was present in the bottle comes out, which moves the partial pressure of carbon dioxide which is present above the liquid to be much lower, resulting in a frizzling effect, as the dissolved carbon dioxide comes out of solution, all at once.

Note:
There are a number of applications of henry’s law, Such as,
-The mole fraction of any substance in a mixture is the amount of that substance present in that mixture, in terms of moles.
-A partial pressure of a gas is the pressure exerted by a particular gas in a mixture of gases.