Question

Arterial blood contains about 0.25 g of oxygen per liter at 37°C and standard atmospheric pressure. Under these conditions, the Henry's law constant is kH = 3.7 × 10^–2 mol/(L • atm), and the mole fraction of O2 in the atmosphere is 0.209Calculate the solubility (in M) of O2 in the blood of a climber on Mt. Everest, where Patm = 0.35 atm.________ M

Answer 1

Firstly we need to determine the partial pressure of O2:

$\begin{gathered} P_{O_2}=X\times P_T \\ P_{O_2}:partial\text{ }pressure \\ X:mole\text{ }fraction \\ P_T:total\text{ }pressure \\ \\ P_{O_2}=0.209\times0.35\text{ }atm \\ P_{O_2}=0.073\text{ }atm \end{gathered}$

We will now use the Henry's Law equation to determine the solubility of the gas:

$\begin{gathered} c=K_H\times P_{O_2} \\ c:solubility\text{ }or\text{ }concentration\text{ }of\text{ }the\text{ }gas(M) \\ K_H:Henry^{\prime}sLawconstant=3.7\times10^{-2}M\text{ }atm^{-1} \\ P_{O_2}:partial\text{ }pressure\text{ }of\text{ }the\text{ }gas=0.073atm \\ \\ c=3.7\times10^{-2}M\text{ }atm^{-1}\times0.073 \\ c=2.7\times10^{-3}M \end{gathered}$

Answer: Solubility is 2.7x10^-3 M