Chapter 10: Problem 50

Calculate the solubility of $\mathrm{O}_{2}$ in water at a partial pressure of $\mathrm{O}_{2}$ of 120 torr at $25^{\circ} \mathrm{C}$. The Henry's law constant for $\mathrm{O}_{2}$ is $1.3 \times 10^{-3} \mathrm{mol} / \mathrm{L} \cdot \mathrm{atm}$ for Henry's law in the form $C=k P$ where $C$ is the gas concentration (mol/L).

Short Answer

Expert verified

The solubility of O₂ in water at a partial pressure of 120 torr and a temperature of $25^{\circ} \mathrm{C}$ is $2.05 \times 10^{-4}\,\mathrm{mol} / \mathrm{L}$.

Step by step solution

List the given values

- Partial pressure of O₂, $P = 120$ torr - Temperature, $T = 25^{\circ} \mathrm{C}$ - Henry's Law constant for O₂, $k = 1.3 \times 10^{-3} \mathrm{mol} / \mathrm{L} \cdot \mathrm{atm}$

Convert the pressure to atm

We need to convert the partial pressure of O₂ from torr to atm because the Henry's law constant is given in atm. The conversion factor is $1 \mathrm{atm} = 760 \mathrm{torr}$. $$ P_{\text{atm}} = \frac{120 \mathrm{torr}}{760 \mathrm{torr} \cdot \mathrm{atm}^{-1}} = 0.1579 \mathrm{atm} $$

Apply Henry's Law

Now we can apply Henry's Law to calculate the concentration of O₂ in water. The formula for Henry's Law is: $$ C = k \cdot P $$ Plugging in the given values: $$ C = (1.3 \times 10^{-3} \mathrm{mol/L}\cdot\mathrm{atm}) \cdot (0.1579 \mathrm{atm}) $$