Question

The solubility of O2 in water is 0.590 g/L at an oxygen pressure of around 15.5 atm. What is the Henry's law constant for O2?

Answer 1

Answer: The Henry's constant for oxygen gas is $3.81\times 10^{-2}g/L.atm$

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{O_2}=K_H\times p_{O_2}$

where,

$K_H$ = Henry's constant = ?

$C_{O_2}$ = molar solubility of oxygen gas = $0.590g/L$

$p_{O_2}$ = partial pressure of oxygen gas = 15.5 atm

Putting values in above equation, we get:

$0.590g/L=K_H\times 15.5atm\\\\K_H=\frac{0.590g/L}{15.5atm}=3.81\times 10^{-2}g/L.atm$

Hence, the Henry's constant for oxygen gas is $3.81\times 10^{-2}g/L.atm$