Question

What is the concentration of $O_2$ in a fresh water stream in equilibrium with air at ${25}^{o}C$ and $1.0bar$ ?
Given that , at $1bar$ oxygen in air is $21\%\text{by volume}$
$K_H^{\prime }$ (Henry's law constant) for $O_2=1.3\times {10}^{-3}mol/kgbar$ at ${25}^{o}C.$

• A. $8.736\times {10}^{-3}g/kg$
• B. $4.16\times {10}^{-2}g/kg$
• C. $24.04g/kg$
• D. $114.5g/kg$

Answer 1

The correct option is A $8.736\times {10}^{-3}g/kg$

In this case, $K_H$ has been given in the unit of mol/kg bar.
By Henry's law,
Solubility of $O_2$ $=K_H{P}_{gas}$
${\chi }_{O_2}\text{(in air)}$ $=\frac{21}{100}=0.21$
$\therefore p_{O_2}=P_{air}\times {\chi }_{O_2}=0.21bar$
$\therefore \text{Solubility}=\frac{1.3\times {10}^{-3}mol}{\text{kg bar}}\times 0.21bar$
$\text{Solubility}=2.73\times {10}^{-4}mol/kg=2.73\times {10}^{-4}\times 32g/kg$
$\text{Solubility}=8.736\times {10}^{-3}g/kg$