Question

The Henry's constant for oxygen dissolved in water is $4.34\times {10}^4$ atm at ${25}^{o}C$ . If the partial pressure of oxygen in air is $0.2atm$ , under atmospheric conditions, calculate the concentration (in moles per litre ) of dissolved oxygen in water in equilibrium with air at ${25}^{o}C$.

Answer 1

According to Henry's law, $P=K_H\times x$

$\therefore x_{O_2}=\frac{P_{O_2}}{K_H}=\frac{0.2}{4.34\times {10}^4}=4.6\times {10}^{-6}$

Moles of water= $\frac{1000}{18}=55.5mol$

$\therefore x_{O_2}=\frac{n_{O_2}}{n_{H_{2}O}+n_{O_2}}\approx \frac{n_{O_2}}{n_{H_{2}O}}$

$\Rightarrow n_{O_2}=4.6\times {10}^{-6}\times 55.5=2.55\times {10}^{-4}$ mole

$\therefore$ Molarity $=$ $2.55\times {10}^{-4}M$