Henry's Law Calculations: Videos & Practice Problems

Henry's law describes the relationship between the solubility of a gas and its partial pressure. The law is expressed through the formula:

$S_{\mathrm{gas}}=k_H\cdot P_{\mathrm{gas}}$

where Sgas is the solubility in molarity, kH is Henry's law constant (molarity/pressure), and Pgas is the partial pressure of the gas. The constant varies with temperature and solvent. Units for pressure can be in atmospheres, torrs, or millimeters of mercury, but it's essential to maintain consistency in units. When comparing two sets of conditions, the formula:

$\frac{S_1}{P_1}=\frac{S_2}{P_2}$

is used, where S1 and S2 are the initial and final solubilities, and P1 and P2 are the initial and final pressures. This demonstrates how changes in pressure can affect gas solubility, with solubility being directly proportional to the gas's partial pressure.

Henry's Law Calculations

The concentration of a dissolved gas, often referred to as its solubility, can be calculated using Henry's Law, which relates the solubility of a gas to its partial pressure. The key variable in this relationship is Henry's law constant, denoted as \( k_h \). This constant indicates the solubility of a gas at a specific temperature in a given solvent, expressed in molarity (M). It's important to note that the terms concentration and molarity are often used interchangeably in academic settings.

Henry's Law can be expressed with the formula:

\( S_{\text{gas}} = k_h \times P_{\text{gas}} \)

In this equation, \( S_{\text{gas}} \) represents the solubility of the gas in molarity, \( k_h \) is the Henry's law constant (with units typically in molarity per atmosphere), and \( P_{\text{gas}} \) is the partial pressure of the gas, measured in atmospheres.

While the standard unit for pressure in this context is atmospheres, it is also possible to encounter other units such as torr or millimeters of mercury (mmHg). When using these alternative units, it is advisable to convert them to atmospheres to maintain consistency with the units of \( k_h \). This ensures accurate calculations when applying Henry's Law to determine the solubility of gases in various solvents.

CO2 Solubility Example

To determine the solubility of carbon dioxide gas (CO₂) in a solution, we can utilize Henry's Law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid. The formula for calculating the solubility (S) is given by:\[S = k_H \times P\]where:- \( S \) is the solubility of the gas in molarity (M),- \( k_H \) is Henry's law constant (in molarity per atmosphere),- \( P \) is the pressure of the gas (in atmospheres).In this case, the Henry's law constant for CO₂ is \( 8.20 \times 10^2 \, \text{M/atm} \), and the pressure of CO₂ is \( 3.29 \, \text{atm} \). Plugging these values into the formula, we have:\[S = (8.20 \times 10^2 \, \text{M/atm}) \times (3.29 \, \text{atm})\]When we perform the multiplication, the units of atmospheres cancel out, leaving us with molarity:\[S = 2.70 \times 10^3 \, \text{M}\]Thus, the concentration of carbon dioxide in this scenario is \( 2.70 \times 10^3 \, \text{M} \). This calculation illustrates how gas solubility can be influenced by pressure, highlighting the practical application of Henry's Law in understanding gas behavior in solutions.

Henry's Law Two-point Formula

Henry's Law describes the relationship between the solubility of a gas in a liquid and the pressure of that gas above the liquid. Specifically, it states that the solubility of a gas is directly proportional to the pressure exerted by that gas. This principle is crucial in understanding how gases dissolve in liquids under varying conditions.

In the context of the 2.4 method, the formula can be expressed as:

\(\frac{S_1}{P_1} = \frac{S_2}{P_2}\)

Here, \(S_1\) represents the initial solubility of the gas, \(P_1\) is the initial pressure, \(S_2\) is the final solubility, and \(P_2\) is the final pressure. This version of Henry's Law is particularly useful when comparing two different solubilities at two distinct pressures for a given gas.

The solubility can be measured in various units, including molarity (moles per liter) or other mass per volume units, allowing for flexibility in application across different scenarios. Understanding this relationship is essential for fields such as chemistry, environmental science, and engineering, where gas solubility plays a critical role in processes like carbonation, respiration, and pollutant behavior in aquatic systems.

Henry's Law Two-point Formula Example

To determine the new pressure of dichloromethane (CH₂Cl₂) when its solubility changes, we can use the relationship between solubility and pressure, which is described by Henry's Law. According to this law, the ratio of solubility to pressure remains constant for a given solute at a specific temperature. This can be expressed mathematically as:

\[ \frac{s_1}{p_1} = \frac{s_2}{p_2} \]

In this scenario, we have:

• Initial pressure, \( p_1 = 2.88 \) atm
• Initial solubility, \( s_1 = 0.384 \) mg/L
• New solubility, \( s_2 = 0.225 \) mg/L
• New pressure, \( p_2 \) (unknown)

Substituting the known values into the equation gives:

\[ \frac{0.384 \, \text{mg/L}}{2.88 \, \text{atm}} = \frac{0.225 \, \text{mg/L}}{p_2} \]

Cross-multiplying to solve for \( p_2 \) results in:

\[ 0.384 \, \text{mg/L} \times p_2 = 0.225 \, \text{mg/L} \times 2.88 \, \text{atm} \]

Calculating the right side:

\[ 0.225 \, \text{mg/L} \times 2.88 \, \text{atm} = 0.648 \, \text{mg atm/L} \]

Now, dividing both sides by \( 0.384 \, \text{mg/L} \) to isolate \( p_2 \):

\[ p_2 = \frac{0.648 \, \text{mg atm/L}}{0.384 \, \text{mg/L}} \approx 1.69 \, \text{atm} \]

Thus, the new pressure \( p_2 \) is approximately 1.69 atmospheres. This demonstrates how solubility decreases with a reduction in pressure, consistent with Henry's Law.

Henry's Law Constant for nitrogen in water is 1.67 × 10^-4 M • atm^–1. If a closed canister contains 0.103 M nitrogen, what would be its pressure in atm?

• A. 617 atm
• B. 1.72 × 10^–5 atm
• C. 1.62 × 10^–3 atm
• D. 778 atm

Answer: (A)

At 0°C and 1.00 atm, as much as 0.84 g of O₂ can dissolve in 1.0 L of water. At 0°C and 4.00 atm, how many grams of O₂ dissolve in 1.0 L of water?

• A. 0.105 g
• B. 3.36 g
• C. 6.72 g
• D. 4.68 g

Answer: (B)

The atmospheric pressure in a lab is calculated as 1.3 atm. If oxygen gas contributes 62% of this atmospheric pressure, determine its mass (in g) dissolved at room temperature in 25 L of water. The Henry's Law Constant for oxygen in water at this temperature is 5.3 × 10^–5 M/atm.

• A. 1.4 × 10^–3 g
• B. 6.9 × 10^–5 g
• C. 0.055 g
• D. 0.034 g

Answer: (D)

Here’s what students ask on this topic:

How do you calculate Henry's law constant?

Henry's Law constant can be calculated by determining the ratio of the concentration of a gas in a liquid to the partial pressure of the gas above the liquid when the system is at equilibrium. The formula to calculate Henry's Law constant (_kH) is:

$k_H=\frac{C}{P}$

where:

• C is the concentration of the gas in the liquid (usually expressed in mol/L or M),
• P is the partial pressure of the gas above the liquid (usually expressed in atmospheres, atm, or Pascals, Pa).

The units of Henry's Law constant will depend on the units used for concentration and pressure. It's important to maintain consistency in the units when applying the formula.

To experimentally determine _kH, you would measure the concentration of the gas in the liquid at a known partial pressure at a constant temperature. This is because Henry's Law constant is temperature-dependent and will change with varying temperatures.