Question

What is the concentration of $O_{2}$ in a freshwater stream in equilibrium with air at $30^{\circ}C$ and 2.0 bar? Given, $K_{H}$ (Henry’s law constant) of $O_{2}=2.0\times 10^{-3} mol/kg$ bar at $30^{\circ}C$

• A. $8.736 \times 10^{-3} g/kg$
• B. $1.344 \times 10^{-3} g/kg$
• C. $24.04 \,g/kg$
• D. $114.5 \,g/kg$

Answer 1

Answer: (B)

In this case, $K_{H}$ has been given in the unit of mole/kg bar
Thus, solubility of $O_{2}$ (or concentration) $=K_{H}\,P_{gas}$
$\chi_{o_2}$ (in air) = 0.21
$\therefore p_{o_2}=p_{\text{air}} \chi_{o_2}=0.21 \,bar$
$\therefore $ Solubility $=\frac{2.0\times10^{-3}mol}{kg \,bar}\times0.21 $bar
$=4.2\times10^{-4}\times32 g/ kg$
$=1.344\times10^{-3} g/ kg$