Q.Henry's law constant for $CO_2$ in water is $1.67 \times 10^8$ Pa at 298 K. Calculate the quantity of $CO_2$ in 500 mL of soda water when packed under 2.5 atm $CO_2$ pressure at 298 K.

The quantity of $CO_2$ dissolved in water is governed by Henry's Law, which relates the partial pressure of a gas above a liquid to its mole fraction in the solution.

1. First, convert the given $CO_2$ pressure from atmospheres to Pascals: $P_{CO_2} = 2.5 \text{ atm} \times 101325 \text{ Pa/atm} = 253312.5 \text{ Pa}$.
2. Next, use Henry's Law to find the mole fraction of $CO_2$ in the soda water: $x_{CO_2} = \frac{P_{CO_2}}{K_H} = \frac{253312.5 \text{ Pa}}{1.67 \times 10^8 \text{ Pa}} \approx 1.5168 \times 10^{-3}$.
3. Assume the density of soda water is approximately $1 \text{ g/mL}$. The moles of water in 500 mL are: $n_{H_2O} = \frac{500 \text{ mL} \times 1 \text{ g/mL}}{18.015 \text{ g/mol}} \approx 27.754 \text{ mol}$. …

Henry's Law describes the relationship between the partial pressure of a gas above a liquid and the concentration of that gas dissolved in the liquid. It's a fundamental concept for understanding how gases like $CO_2$ dissolve in beverages like soda water. The higher the partial pressure of the gas, the more of it will dissolve in the liquid, up to a certain point.

The law is expressed as:

In soda water, $CO_2$ gas is dissolved under pressure. When you open a bottle of soda, the pressure above the liquid decreases, causing the dissolved $CO_2$ to come out of solution as bubbles, which is why it fizzes. This problem asks us to calculate the amount of $CO_2$ dissolved when packed under a specific pressure.

Let's break down the calculation step-by-step:

1. Identify Given Values and Target:

   We are given:

   • Henry's Law constant for $CO_2$ ($K_H$) = $1.67 \times 10^8$ Pa
   • Volume of soda water = 500 mL
   • Partial pressure of $CO_2$ ($P_{CO_2}$) = 2.5 atm
   • Temperature = 298 K (This confirms the $K_H$ value is appropriate for the conditions, but isn't directly used in the calculation itself).

   Our goal is to find the "quantity" of $CO_2$, which typically means its mass in grams or moles.

2. Convert Units for Consistency:

   Henry's Law constant ($K_H$) is given in Pascals (Pa), but the pressure ($P_{CO_2}$) is in atmospheres (atm). To use the formula correctly, both pressure values must be in the same units. We'll convert atmospheres to Pascals.

   We know that $1 \text{ atm} = 101325 \text{ Pa}$.

   $P_{CO_2} = 2.5 \text{ atm} \times \frac{101325 \text{ Pa}}{1 \text{ atm}} = 253312.5 \text{ Pa}$

3. Apply Henry's Law to Find the Mole Fraction of $CO_2$:

   Now we can use Henry's Law to find the mole fraction ($x_{CO_2}$) of $CO_2$ dissolved in the water.

   $P_{CO_2} = K_H \cdot x_{CO_2}$

   Rearranging for $x_{CO_2}$:

   $x_{CO_2} = \frac{P_{CO_2}}{K_H}$

   $x_{CO_2} = \frac{253312.5 \text{ Pa}}{1.67 \times 10^8 \text{ Pa}}$

   $x_{CO_2} \approx 0.00151684$

   This mole fraction is a dimensionless quantity, representing the ratio of moles of $CO_2$ to the total moles in the solution.

4. Calculate Moles of Solvent (Water):

   Soda water is primarily water. We can assume that the 500 mL volume refers to the volume of water.

   The density of water is approximately $1 \text{ g/mL}$.

   Mass of water = Volume $\times$ Density

   Mass of water = $500 \text{ mL} \times 1 \text{ g/mL} = 500 \text{ g}$

   Next, we need the molar mass of water ($H_2O$).

   Molar mass of $H_2O = (2 \times 1.008 \text{ g/mol for H}) + (1 \times 16.00 \text{ g/mol for O}) = 18.016 \text{ g/mol}$.

   Moles of water ($n_{H_2O}$) = $\frac{\text{Mass of water}}{\text{Molar mass of water}}$

   $n_{H_2O} = \frac{500 \text{ g}}{18.016 \text{ g/mol}} \approx 27.753 \text{ mol}$

   Watch out

   We assume the density of soda water is approximately the same as pure water. This is a reasonable approximation because the amount of dissolved $CO_2$ is very small, as we will see.

5. Relate Mole Fraction to Moles of $CO_2$: …

Method: Henry’s Law Application (Mole–Mass Conversion)

Concept first:

Henry’s law states that at constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.

Mathematically:

where

• $p$ = partial pressure of the gas (in Pa)
• $K_H$ = Henry’s law constant (in Pa)
• $x$ = mole fraction of the gas in the solution

Steps

Step 1: Convert pressure to SI units

Given pressure = $2.5$ atm.

We know $1$ atm = $1.01325 \times 10^5$ Pa.

Step 2: Apply Henry’s law to find mole fraction

This means: in the solution, for every 1 mole of total particles, $1.517 \times 10^{-3}$ moles are $CO_2$.

Step 3: Relate mole fraction to actual moles

For dilute solutions (soda water is mostly water),

because $n_{CO_2} \ll n_{H_2O}$.

Step 4: Find moles of water in 500 mL …

Common Mistakes & How to Avoid Them — Henry’s Law Problem

Mistake 1: Confusing Henry’s Law Constant Units

The error:

Students often plug in $K_H$ without checking if pressure units match. Here, $K_H = 1.67 \times 10^8$ Pa, but the given pressure is 2.5 atm.

How to avoid:

Always convert all pressures to the same unit before using Henry’s Law.

• Convert 2.5 atm to Pa:

  $1 \text{ atm} = 1.01325 \times 10^5 \text{ Pa}$

  $P = 2.5 \times 1.01325 \times 10^5 = 2.533 \times 10^5 \text{ Pa}$

• Then apply Henry’s Law:

  $P = K_H \cdot x$

  where $x$ = mole fraction of $CO_2$ in water.

Mistake 2: Using the Wrong Form of Henry’s Law

The error:

Some students use $P = K_H \cdot C$ (concentration in mol/L) when the constant is given for mole fraction.

How to avoid:

Check the definition of $K_H$ in the problem. Here, $K_H$ is in Pa, which means it relates pressure to mole fraction (dimensionless).

• Correct formula: $x_{CO_2} = \dfrac{P}{K_H} = \dfrac{2.533 \times 10^5}{1.67 \times 10^8} = 1.517 \times 10^{-3}$

Mistake 3: Forgetting to Account for the Solvent (Water)

The error:

Students calculate moles of $CO_2$ directly from mole fraction without considering the moles of water.

How to avoid:

Mole fraction is:

$x_{CO_2} = \dfrac{n_{CO_2}}{n_{CO_2} + n_{H_2O}}$

• For dilute solutions, $n_{CO_2} \ll n_{H_2O}$, so:

  $x_{CO_2} \approx \dfrac{n_{CO_2}}{n_{H_2O}}$

• Calculate moles of water in 500 mL:

  Mass of water = 500 g (since density ≈ 1 g/mL)

  $n_{H_2O} = \dfrac{500}{18} = 27.78 \text{ mol}$

• Then:

  $n_{CO_2} = x_{CO_2} \times n_{H_2O} = 1.517 \times 10^{-3} \times 27.78 = 0.0421 \text{ mol}$

Mistake 4: Stopping at Moles Instead of Mass …