Problem 83 - Chemistry and Chemical Reactivity

Question

Nitrous oxide, \(\mathrm{N}_{2} \mathrm{O},\) laughing gas, is used as an anesthetic. Its Henry's law constant is \(2.4 \times 10^{-2} \mathrm{mol} /\) kg \cdotbar. Determine the mass of \(\mathrm{N}_{2} \mathrm{O}\) that will dissolve in \(500 .\) mL of water, under an \(\mathrm{N}_{2} \mathrm{O}\) pressure of 1.00 bar. What is the concentration of \(\mathrm{N}_{2} \mathrm{O}\) in this solution, expressed in ppm \(\left(d\left(\mathrm{H}_{2} \mathrm{O}\right)=1.00 \mathrm{g} / \mathrm{mL}\right) ?\)

Step-by-Step Solution

Answer

The mass of dissolved \( \mathrm{N}_{2} \mathrm{O} \) is 0.528 g, and the concentration is 1056 ppm.

Step 1: Understanding Henry's Law

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. The law can be expressed as \( C = k_H \cdot P \), where \( C \) is the concentration of the dissolved gas, \( k_H \) is the Henry's Law constant, and \( P \) is the partial pressure of the gas.

Step 2: Using Henry's Law to Find Solubility

Given that \( k_H = 2.4 \times 10^{-2} \text{ mol/kg} \cdot \text{bar} \) and \( P = 1.00 \text{ bar} \), we use the formula \( C = k_H \cdot P \) to find the concentration of \( \mathrm{N}_{2} \mathrm{O} \) in water. Therefore, \( C = 2.4 \times 10^{-2} \text{ mol/kg} \).

Step 3: Calculating the Mass of Dissolved \( \, \mathrm{N}_{2} \mathrm{O} \)

To find the mass of \( \mathrm{N}_{2} \mathrm{O} \) dissolved, note that the concentration found is in moles per kg of water. We'll convert \( C = 2.4 \times 10^{-2} \text{ mol/kg} \) to grams. The molar mass of \( \mathrm{N}_{2} \mathrm{O} \) is approximately 44 g/mol. Thus, mass \( = 2.4 \times 10^{-2} \text{ mol/kg} \times 44 \text{ g/mol} = 1.056 \text{ g/kg} \).

Step 4: Determining the Mass in 500 mL of Water

Since 500 mL of water is equivalent to 0.5 kg (considering the density of water is 1.00 g/mL), the mass of \( \mathrm{N}_{2} \mathrm{O} \) that will dissolve is \( 1.056 \text{ g/kg} \times 0.5 \text{ kg} = 0.528 \text{ g} \).

Step 5: Calculating the Concentration in ppm

Parts per million (ppm) is defined as the grams of solute per million grams of solution. In this case, the solution is approximately 500 grams in total (assuming the mass of dissolved gas is negligible). Thus, \( \text{ppm} = \left( \frac{0.528 \text{ g}}{500 \text{ g}} \right) \times 10^6 = 1056 \text{ ppm} \).

Key Concepts

Solubility

Solubility is a measure of how much of a given substance can be dissolved in a solvent to form a homogeneous solution at a specified temperature and pressure. When discussing gases like nitrous oxide \(\mathrm{N}_{2} \mathrm{O}\), solubility is influenced by factors like pressure and temperature. According to Henry's Law, which is key in understanding the solubility of gases in liquids, the solubility is directly proportional to the partial pressure of the gas above the liquid. This means, as the pressure increases, more gas will dissolve in the liquid.

For instance, in our exercise, we've considered the solubility of \(\mathrm{N}_{2} \mathrm{O}\) in water at a specific pressure (1.00 bar). The constant, known as Henry's Law constant \(k_H\), plays a significant role and varies with different substances and conditions. To calculate how much \(\mathrm{N}_{2} \mathrm{O}\) can dissolve in water, we used the formula \(C = k_H \cdot P\). Here, the concentration \(C\) of dissolved gas is a result of the given constant and the applied pressure.

Partial Pressure

In the context of Henry's Law, partial pressure is a crucial factor. Partial pressure refers to the pressure exerted by a single type of gas in a mixture of gases. When a gas, like \(\mathrm{N}_{2} \mathrm{O}\), is dissolved in a liquid, its behavior is influenced significantly by its partial pressure over the liquid.

In our exercise, the partial pressure of \(1.00 \, \text{bar}\) was used to determine how much of the gas dissolves in water. That pressure is applied directly in the Henry's Law equation \(C = k_H \cdot P\), showing how much concentration of the gas will be seen in the liquid phase. The higher the pressure over the liquid, the greater the solubility of the gas.

Concentration

Concentration refers to the amount of solute that is present in a given quantity of solvent or solution. It can be expressed in a variety of units, but in this exercise, we're using parts per million (ppm), which helps in understanding very dilute solutions.

Calculating the concentration of \(\mathrm{N}_{2} \mathrm{O}\) in the water involved several steps. First, we found the mass of gas dissolved per liter of water using the solubility formula. Then, this mass was converted to ppm, enabling us to express how much \(\mathrm{N}_{2} \mathrm{O}\) is present in a million parts of the solution. This method of expressing concentration is particularly useful in scenarios like this, where solutions contain very small amounts of solute compared to the solvent.