Question

The solubility of nitrogen in water is \(8.21 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\) at \(0^{\circ} \mathrm{C}\) when the \(\mathrm{N}_{2}\) pressure above water is \(0.790 \mathrm{~atm} .\) Calculate the Henry's law constant for \(\mathrm{N}_{2}\) in units of \(\mathrm{mol} / \mathrm{L} \cdot\) atm for Henry's law in the form \(C=k P\), where \(C\) is the gas concentration in mol/L. Calculate the solubility of \(\mathrm{N}_{2}\) in water when the partial pressure of nitrogen above water is \(1.10 \mathrm{~atm}\) at \(0{ }^{\circ} \mathrm{C}\).

Short answer

The Henry's Law constant for N2 is approximately \(1.0392 \times 10^{-3}\,\mathrm{mol/L\cdot atm}\), and the solubility of N2 in water when the partial pressure is 1.10 atm at 0°C is approximately \(1.1431 \times 10^{-3}\,\mathrm{mol/L}\).

Step-by-step solution

Understand Henry's Law and the given variables

Henry's Law can be written as \(C = kP\), where \(C\) is the gas concentration in mol/L, P is the pressure in atm, and k is the Henry's Law constant. We are given the solubility of N2 in water at 0°C when the N2 pressure above water is 0.790 atm, which is \(8.21 \times 10^{-4}\) mol/L.

Calculate Henry's Law constant for N2

We can find the Henry's Law constant by rearranging the equation: \[k = \frac{C}{P}\] Use the given values for solubility and pressure to calculate k: \[k = \frac{8.21 \times 10^{-4}\,\mathrm{mol/L}}{0.790\,\mathrm{atm}}\]

Simplify the expression

Now, divide \(8.21 \times 10^{-4}\) by 0.790: \[k \approx 1.0392 \times 10^{-3}\, \mathrm{mol/L\cdot atm}\] So, the Henry's Law constant for N2 is approximately \(1.0392 \times 10^{-3}\,\mathrm{mol/L\cdot atm}\).

Use Henry's Law constant to calculate the solubility of N2 for the given pressure

We are asked to find the solubility of N2 when the partial pressure above water is 1.10 atm at 0°C. Using Henry's Law: \[C = kP\] Substitute the values of k and P: \[C \approx (1.0392 \times 10^{-3}\,\mathrm{mol/L\cdot atm})(1.10\,\mathrm{atm})\]

Compute the solubility of N2 for 1.10 atm pressure

Now, multiply \(1.0392 \times 10^{-3}\) by 1.10: \[C \approx 1.1431 \times 10^{-3}\,\mathrm{mol/L}\] So, the solubility of N2 in water when the partial pressure is 1.10 atm at 0°C is approximately \(1.1431 \times 10^{-3}\,\mathrm{mol/L}\).

Key concepts

Solubility

Solubility is a measure of how much of a particular solute, in this case, nitrogen (\(\mathrm{N}_{2}\), can dissolve in a solvent like water to form a solution at a given state of temperature and pressure.
It is typically expressed in terms of concentration, such as mol/L.
The solubility of gases in liquids is influenced by various factors, including temperature and pressure. Generally, gases become less soluble in liquids as the temperature increases.
In our problem, the solubility of nitrogen in water is given under the condition of a certain pressure and temperature.

Partial Pressure

Partial pressure is a term used to describe the pressure exerted by a single type of gas in a mixture of gases.
In any mixture of gases, each gas has its own partial pressure, which adds up to the total pressure of the mixture.
hen considering the dissolution of a gas into a liquid, the partial pressure directly above the solvent plays a crucial role.
In this exercise, nitrogen's partial pressure was initially given as 0.790 atm, and then changed to 1.10 atm to calculate how it affects solubility. Henry's Law beautifully illustrates how the gas' partial pressure relates to its concentration or solubility within the liquid.

Gas Concentration

Gas concentration in this context refers to the amount of nitrogen that has dissolved in the water, measured in moles per liter (mol/L).
According to Henry's Law, this concentration is directly proportional to the partial pressure of the gas just above the liquid surface.
This relationship can be expressed with the formula \(C = kP\), where:

• \(C\) is the gas concentration.
• \k\(k\) is the Henry's Law constant.
• \(P\) is the partial pressure.

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By knowing the partial pressure and the Henry's Law constant for nitrogen, one can easily calculate how much nitrogen dissolves in water.
This is critical in applications like carbonated beverages or understanding how gases dissolve in biological systems.