Question

The solubility of a gas in water is $0.001M$ at STP. Determine its Henry’s law constant.

Answer 1

Hint:The best way to approach the numerical is to think about the relation between given quantities under a given condition with the unknown quantity which needs to be determined. So think about Henry's law constant and its relation with solubility.
Formula Used:
$X = {p_{gas}}{K_H}$ Where,
$X$ Represents solubility of a gas in a liquid,
${p_{gas}}$ Represents partial pressure of the gas present above the surface of liquid or solution,
${K_H}$ Represents Henry’s law constant.

Complete step by step answer:
First of all, we will write the given quantities given in the question as follows:
 $X = 0.001M$ and ${p_{gas}} = 1atm\left( {STP} \right)$ (at STP ${p_{gas}} = 1atm$)
 $X$ Represents solubility of a gas in a water,
${p_{gas}}$ Represents partial pressure of the gas present above the surface of liquid or solution,
Now we will define the relation between solubility and partial pressure of gas using Henry’s law.
Henry’s law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution.
Henry’s law can be expressed as:
  $X \propto {p_{gas}}$, after removing proportionality constant we get,
 $ \Rightarrow X = {K_H}{p_{gas}}$
The next step is calculation. Now we will substitute all the given quantities in the expression of Henry’s law. The given quantities are as follows:
$X = 0.001M$ and ${p_{gas}} = 1atm\left( {STP} \right)$ (at STP ${p_{gas}} = 1atm$)
Now after substituting the given quantities we get,
$ \Rightarrow X = {K_H}{p_{gas}}$
$ \Rightarrow 0.001M = {K_H} \times 1atm$
$ \Rightarrow {K_H} = 0.001M/atm$

Final result: Henry's law constant is ${K_H} = 0.001M/atm$ when solubility of gas in water is $0.001M$ at STP.

Note:
The important point to remember about Henry's constant ${K_H}$ is that it is a function of the nature of the gas. This can be explained with the fact that different gases have different ${K_H}$ values at the same temperature.
It can also be concluded that mole fraction of gas in the solution is proportional to the partial pressure of the gas over the solution.