Question

Calculate the solubility of \(\mathrm{O}_{2}\) in water 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 \(1.3 \times 10^{-3} \mathrm{~mol} / \mathrm{L} \cdot\) atm for Henry's law in the form \(C=k P\), where \(C\) is the gas concentration \((\mathrm{mol} / \mathrm{L})\).

Short answer

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

Step-by-step solution

Pressure conversion formula

\(P_{atm} = \frac{P_{torr}}{760}\) #Step 2: Calculate pressure in atm# Now we'll plug in the given pressure value and calculate the pressure in atm.

Pressure in atm calculation

\(P_{atm} = \frac{120 \: torr}{760 \: torr} = 0.1579 \: atm\) #Step 3: Use Henry's law to find solubility# Now that we have the pressure value in atm, we'll use Henry's law formula to find the solubility of O₂ in water.

Henry's law formula

\(C = kP\) Here, the Henry's law constant, k, is given as \(1.3 \times 10^{-3}\: \mathrm{mol} / \mathrm{L} \cdot \mathrm{atm}\) #Step 4: Calculate solubility of O₂ in water at given pressure# Finally, we'll use the values of k and P to find the solubility of O₂ in water (C).

Solubility calculation

\(C = (1.3 \times 10^{-3}\: \mathrm{mol} / \mathrm{L} \cdot \mathrm{atm}) \times 0.1579 \: atm\) \(C = 2.05 \times 10^{-4}\: \mathrm{mol} / \mathrm{L}\) Hence, the solubility of O₂ in water at a partial pressure of 120 torr and \(25^{\circ}C\) is approximately \(2.05 \times 10^{-4}\: \mathrm{mol} / \mathrm{L}\).

Key concepts

Solubility Calculation

Understanding how to calculate the solubility of a gas in a liquid like water is crucial for many scientific applications. Solubility refers to the amount of gas that can dissolve in a given volume of liquid. This depends on factors such as temperature, pressure, and the nature of the gas and liquid. Henry's Law provides a handy way to calculate this. The formula used in Henry's Law is: \(C = kP\), where \(C\) is the concentration (or solubility) of the gas in the liquid, \(k\) is the Henry's Law constant, and \(P\) is the partial pressure of the gas.
In the example, \(k\) is given as \(1.3 \times 10^{-3}\) mol/L ⋅ atm, and \(P\) is 0.1579 atm, converted from 120 torr. Plugging these into the formula allows us to find \(C\), the solubility of oxygen in water. This step-by-step calculation helps in understanding how solubility can be determined from basic principles.

Partial Pressure Conversion

Converting partial pressure is a crucial step in solubility calculations when using Henry's Law. Gases often have their pressure mentioned in different units like torr or mmHg, whereas Henry's Law requires the pressure in atmospheres (atm). To convert from torr to atm, use the formula: \(P_{atm} = \frac{P_{torr}}{760}\).
In the problem, we have a given pressure of 120 torr. By dividing by 760 (since 1 atm equals 760 torr), we find the pressure in atm to be approximately 0.1579. This conversion is necessary to ensure consistency and accuracy in the solubility calculation using Henry's Law.

Gas Solubility in Water

The solubility of gases like oxygen in water is affected by several factors, including temperature and pressure. Higher pressure increases gas solubility, while higher temperatures typically decrease it. Henry's Law helps quantify this relationship by linking solubility to pressure.
In this scenario, oxygen's solubility is calculated at a specific temperature, \(25^{\circ}C\), and pressure, using the Henry's Law constant for oxygen. The final value shows the amount of oxygen that can dissolve in a liter of water under these conditions. This information is vital in fields like environmental science and engineering, where understanding gas exchange in water bodies is essential.