Question

The Henry's law constant \(\left(k_{H}\right)\) for \(\mathrm{O}_{2}\) in water at \(20^{\circ} \mathrm{C}\) is \(1.28 \times 10^{-3} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{atm}\) (a) How many grams of \(\mathrm{O}_{2}\) will dissolve in \(2.50 \mathrm{~L}\) of \(\mathrm{H}_{2} \mathrm{O}\) that is in contact with pure \(\mathrm{O}_{2}\) at \(1.00 \mathrm{~atm} ?\) (b) How many grams of \(\mathrm{O}_{2}\) will dissolve in \(2.50 \mathrm{~L}\) of \(\mathrm{H}_{2} \mathrm{O}\) that is in contact with air, where the partial pressure of \(\mathrm{O}_{2}\) is \(0.209 \mathrm{~atm} ?\)

Short answer

For (a), about 0.1024 grams, and for (b), about 0.021 grams of \( \textrm{O}_2 \) will dissolve.

Step-by-step solution

Understanding Henry's Law

Henry's Law states that the amount of gas dissolved in a liquid (at a constant temperature) is directly proportional to the partial pressure of the gas in contact with the liquid. The formula is given by \[ C = k_H \times P \]where: -\( C \) is the concentration of the gas in the liquid (in mol/L)-\( k_H \) is the Henry's law constant (in mol/L·atm)-\( P \) is the partial pressure of the gas (in atm).

Calculate the concentration of dissolved \(\textrm{O}_{2}\) for Part (a)

Given that \( k_H = 1.28 \times 10^{-3} \) mol/L·atm, \( P = 1.00 \) atm, and the volume of water is 2.50 L. First, find the concentration \( C \) of \( \textrm{O}_2 \) dissolved in the water:\[ C = k_H \times P \]\[ C = (1.28 \times 10^{-3} \text{ mol/L·atm}) \times (1.00 \text{ atm}) \]\[ C = 1.28 \times 10^{-3} \text{ mol/L}\]

Convert moles of \(\textrm{O}_{2}\) to grams for Part (a)

Now, calculate the number of moles of \( \textrm{O}_2 \) that dissolve in 2.50 L of water:\[ \text{moles of } O_2 = C \times V \]\[ \text{moles of } O_2 = (1.28 \times 10^{-3} \text{ mol/L}) \times 2.50 \text{ L}\]\[ \text{moles of } O_2 = 3.2 \times 10^{-3} \text{ mol}\]Next, convert moles to grams using the molar mass of \( \textrm{O}_2 \) (32 g/mol):\[ \text{mass } = \text{moles} \times \text{molar mass} \]\[ \text{mass } = (3.2 \times 10^{-3} \text{ mol}) \times 32 \text{ g/mol} \]\[ \text{mass } = 0.1024 \text{ grams}\]

Calculate concentration of dissolved O2 for Part (b)

Now, for part (b), using the partial pressure of \( \textrm{O}_2 \) in air, which is 0.209 atm:\[ C = k_H \times P \]\[ C = (1.28 \times 10^{-3} \text{ mol/L·atm}) \times (0.209 \text{ atm}) \]\[ C = 2.6752 \times 10^{-4} \text{ mol/L}\]

Convert moles of \(\textrm{O}_{2}\) to grams for Part (b)

Calculate the number of moles of \( \textrm{O}_2 \) that dissolve in 2.50 L of water:\[ \text{moles of } O_2 = C \times V \]\[ \text{moles of } O_2 = (2.6752 \times 10^{-4} \text{ mol/L}) \times 2.50 \text{ L} \]\[ \text{moles of } O_2 = 6.688 \times 10^{-4} \text{ mol}\]Convert moles to grams using the molar mass of \( \textrm{O}_2 \) (32 g/mol):\[ \text{mass } = \text{moles} \times \text{molar mass} \]\[ \text{mass } = (6.688 \times 10^{-4} \text{ mol}) \times 32 \text{ g/mol} \]\[ \text{mass } = 0.021 \text{ grams}\]

Key concepts

Gas Solubility

Gas solubility is a crucial concept in chemistry. It refers to the amount of a gas that can dissolve in a liquid at a given temperature and pressure. This property is heavily influenced by the nature of the gas and the solvent, as well as the conditions of the environment, particularly temperature and pressure.

According to Henry's Law, the solubility of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid. As the partial pressure increases, more gas molecules collide with the liquid surface and dissolve. On the other hand, lower partial pressures mean fewer collisions, leading to lower solubility.

Partial Pressure

Partial pressure is an essential concept in understanding Henry's Law. It is defined as the pressure exerted by a single type of gas in a mixture of gases. For instance, in our atmosphere, gases like nitrogen and oxygen collectively contribute to the total atmospheric pressure, but each gas also exerts its own partial pressure.

The partial pressure of a gas is denoted as 'P' in Henry's Law equation: \[ C = k_H \times P \] Here, 'C' is the concentration of the gas, 'k_H' is the Henry's law constant, and 'P' is the partial pressure. To solve Henry's law problems, you need to know the Henry's law constant and the partial pressure of the gas.

In our exercise, pure oxygen has a partial pressure of 1 atm, while oxygen in air has a partial pressure of 0.209 atm. Using these partial pressures, we can determine the concentration of dissolved oxygen in water.

Molar Concentration

Molar concentration, or molarity, is a measure of the concentration of a solute in a solution. It is expressed in moles of solute per liter of solution (mol/L).

To find the molar concentration of oxygen in water using Henry's Law, you would use the formula: \[ C = k_H \times P \] Let's consider part (a) of the exercise, where the Henry's law constant (\