Question

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 \(1.3 \mathrm{X}\) \(10^{-3} \mathrm{~mol} / \mathrm{L} \cdot\) 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 \(\mathrm{O}_2\) in water at a partial pressure of 120 torr at 25°C is \(2.05 \times 10^{-4}\ \mathrm{mol}/\mathrm{L}\).

Step-by-step solution

Convert the partial pressure of \(\mathrm{O}_2\) into atmospheres.

First, we need to convert the given partial pressure from torr to atmospheres. The conversion factor is: 1 atm = 760 torr So, the partial pressure in atm can be found as follows: Partial pressure of \(\mathrm{O}_2\) (in atm) = (Partial Pressure in torr) / 760

Calculate the solubility of \(\mathrm{O}_2\).

Now, we can use Henry's law to calculate the solubility of \(\mathrm{O}_2\) in water: C = kP Where C is the solubility of the gas (concentration in \(\mathrm{mol}/\mathrm{L}\)), k is Henry's law constant for the gas (in this case, \(\mathrm{O}_2\)), and P is the partial pressure of the gas in atm. Substitute the given values: C = (1.3 x 10^{-3} \frac{\mathrm{mol}}{\mathrm{L} \cdot \mathrm{atm}}) x \frac{120 \ \mathrm{torr}}{760 \ \mathrm{torr \cdot \mathrm{atm}^{-1}}}

Solve for the solubility of \(\mathrm{O}_2\).

Finally, calculate the solubility of \(\mathrm{O}_2\) in water: C = 1.3 x 10^{-3} \frac{\mathrm{mol}}{\mathrm{L} \cdot \mathrm{atm}} \cdot \frac{120}{760} \ \mathrm{atm} C = 1.3 x 10^{-3} \frac{\mathrm{mol}}{\mathrm{L}} \cdot 0.1579 C = 2.05 x 10^{-4} \ \mathrm{mol}/\mathrm{L} The solubility of \(\mathrm{O}_2\) in water at a partial pressure of 120 torr at 25°C is \(2.05 \times 10^{-4}\ \mathrm{mol}/\mathrm{L}\).

Key concepts

Solubility of Gases in Liquids

When we try to understand how gases dissolve in liquids, we are exploring a process that is common in both nature and technology. The solubility of gases in liquids is a phenomenon that plays a crucial role in aquatic life, soda carbonation, and even in our own blood as it transports oxygen. To comprehend this concept, it's important to note that gases dissolve in liquids based on their nature and the conditions present, such as temperature and pressure.

Solubility is affected by Henry's law, which states that at a constant temperature, the amount of gas that dissolves in a liquid is directly proportional to the partial pressure of that gas above the liquid. In other words, higher pressure above the liquid leads to more gas being dissolved until an equilibrium is reached where the rates of dissolution and release are equal. This is why a soda fizzles when opened; the pressure is released, so the gas escapes until a new balance is attained.

In our exercise, the solubility of oxygen in water is influenced by the partial pressure of oxygen above the water. If we increase the pressure of oxygen, say by compressing the air above the water, we could dissolve more oxygen. However, each gas has a unique constant in Henry's law, known as the Henry's law constant, which you use to calculate exactly how much gas will dissolve under certain conditions. It's essential to know this constant for accurate calculations.

Partial Pressure in Atmospheres

To understand the partial pressure in atmospheres, we first need to know that air is a mixture of gases, each exerting its own pressure. These pressures all add up to the total atmospheric pressure. The pressure contributed by an individual gas in a mixture is called its partial pressure. For example, in our atmosphere at sea level, the total pressure is approximately 760 torr, which equals 1 atmosphere (atm).

When calculating solubility using Henry's law, we must express this partial pressure in atmospheres to be used with the Henry's law constant. In the exercise, we converted the pressure of oxygen from torr to atm by dividing by 760, since the Henry's law constant provided is in units of mol/L⋅atm. This accurate conversion ensures that our units are consistent for the final calculation, preventing any discrepancies in our solubility result.

The process of converting partial pressures is a fundamental skill in chemistry because pressure plays such a crucial role in the behavior of gases. Getting the units right is essential for accurate predictions in not just solubility, but also in calculations involving gas laws and reactions under various atmospheric conditions.

Chemical Concentration Calculation

The concentration of a chemical in a solution represents the amount of solute that has been dissolved in a given volume of solvent. It's a measure of how 'crowded' the solution is with solute particles, and it's crucial for predicting how chemicals will react with one another in the solution. Concentrations can be expressed in different ways, but one common unit is moles per liter (mol/L), often referred to as molar concentration.

In the context of Henry's law and our exercise, concentration calculation is done after converting the partial pressure of the gas to atmospheres. Using the formula from Henry's law (\(C=kP\)), where 'k' is the Henry's law constant and 'P' is the partial pressure of the gas, we can find the solubility (C), or the concentration of the dissolved gas. Simply multiply the constant by the pressure after making sure both are in the correct units, and the result gives us a precise molar concentration.

Understanding how to calculate concentrations is a foundational skill in chemistry. It allows you to predict the yield of reactions, the strength of acids or bases, and the behavior of solutions under various conditions. Furthermore, knowing concentration is vital for creating solutions with precise properties, necessary for experiments, industrial processes, and even cooking recipes.