Question

Using henry's law, calculate the molar concentration of o2 in the surface water of a mountain lake saturated with air at 20 ∘c and an atmospheric pressure of 685 torr .

Answer 1

Answer: The molar concentration of oxygen gas in water is $1.43\times 10^{-7} mol/L$.

Explanation:

Partial pressure of the $O_2$gas = 685 torr = 0.8905 bar

1 torr = 0.0013 bar

According Henry's law:

$p_{o_2}=K_H\times\chi_{O_2}$

Value of Henry's constant of oxygen gas at 20 °C in water = 34860 bar

$0.0013=34860 bar\times \chi_{O_2}$

$\chi_{O_2}=\frac{0.8905 bar}{34680 bar}=2.56\times 10^{-5}$

Let the number of moles of $O_2$ gas in 1 liter water be n.

1 Liter water = 1000 g of water

Moles of water in 1 L $n_w=\frac{1000 g}{18 g/mol}=55.55 mol$

$\chi_{O_2}=\frac{n}{n+n_w}$

$2.56\times 10^{-5}=\frac{n}{n+55.55}$

$n=1.43\times 10^{-7} moles$

$Molarity=\frac{\text{Moles of}O_2}{Volume}$

Molar concentration of oxygen gas in 1 L of water:

$=\frac{1.43\times 10^{-7} moles}{1 L}=1.43\times 10^{-7} mol/L$

The molar concentration of oxygen gas in water is $1.43\times 10^{-7} mol/L$.