Question

Which of the following is true regarding Henry's Law constant for different aqueous solutions of $C{O}_2\left(g\right)$ and $O_2\left(g\right)$ at a given temperature?
(Assume both the gases have same partial pressure)

• A. $K_H\left(C{O}_2\right)=K_H\left(O_2\right)$
• B. Cannot be predicted
• C. $K_H\left(C{O}_2\right)>K_H\left(O_2\right)$
• D. $K_H\left(C{O}_2\right)<K_H\left(O_2\right)$

Answer 1

The correct option is D $K_H\left(C{O}_2\right)<K_H\left(O_2\right)$

$C{O}_2$ is more soluble in water as compared to $O_2$ because the two $C=O$ bonds in $C{O}_2$ are more polarized whereas in $O_2$ bonds are not polarized.
Hence more solubility is there for $C{O}_2$ in polar water molecule
By Henry's Law:
$p=K_H\times \chi$
Here $p$ is the partial pressure of gas over the solution.
$K_H$ is Henry’s law constant
$\chi$ is mole fraction of the gas in the solution.
Since ${\chi }_{C{O}_2}>{\chi }_{O_2}$
So, for p to be same
$K_H\left(C{O}_2\right)<K_H\left(O_2\right)$