Calculate the concentration of \(\mathrm{CO}_{2}\) in a soft drink bottle after the bottle is opened and sits at \(25^{\circ} \mathrm{C}\) under a \(\mathrm{CO}_{2}\) partial pressure of \(3.0 \times 10^{-4} \mathrm{~atm}\). Henry's law constant for \(\mathrm{CO}_{2}\) in water is \(3.1 \times 10^{-2}\) mol/litre-atm at this temperature.

Short Answer

Expert verified
The concentration of \( \mathrm{CO}_2 \) in the soft drink after opening is \( 9.3 \times 10^{-6} \) mol/litre.

Step by step solution

01

Understanding Henry's Law

Henry's Law states that the solubility of a gas in a liquid at a given temperature is directly proportional to the partial pressure of the gas above the liquid. The mathematical representation of Henry's Law is given by: \( C = kP \), where \( C \) is the concentration of the gas in the liquid, \( k \) is Henry's Law constant, and \( P \) is the partial pressure of the gas.
02

Given Values

The partial pressure of \( \mathrm{CO}_2 \) is given as \( P = 3.0 \times 10^{-4} \) atm, and Henry's law constant \( k \) for \( \mathrm{CO}_2 \) in water is \( 3.1 \times 10^{-2} \) mol/litre-atm. We need to find the concentration \( C \).
03

Applying Henry's Law

By substituting the given values into Henry's Law formula \( C = kP \), we get \( C = (3.1 \times 10^{-2}) \times (3.0 \times 10^{-4}) \).
04

Calculating the Concentration

To find the concentration, perform the multiplication: \( C = (3.1 \times 10^{-2}) \times (3.0 \times 10^{-4}) \). This gives \( C = 9.3 \times 10^{-6} \) mol/litre.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gas Solubility
Gas solubility refers to the ability of a gas to dissolve in a liquid. This attribute is of significant importance not just in chemistry, but also in fields like environmental science and food technology. To understand how solubility works, picture a bottle of your favorite carbonated drink. When sealed, the gas (carbon dioxide, in this case) is under high pressure, staying dissolved in the liquid. However, as soon as you open the bottle, the pressure above the liquid drops, and the gas begins to escape, forming bubbles.

One key factor that determines the solubility of a gas in a liquid is temperature. Typically, gases become less soluble as the temperature increases. This is why your drink seems to lose its fizz faster on a hot day than on a cold one. Additionally, the nature of the gas and the liquid play roles in how well the gas will dissolve. Different gases have varying solubilities in various liquids based on their physical and chemical properties. For a refined understanding, we look at Henry's Law, which specifically quantifies the relationship between pressure and solubility for gases in liquids at a constant temperature.
Partial Pressure
Partial pressure is the pressure that a gas in a mixture of gases would exert if it alone occupied the entire volume. Imagine being at a concert with a crowd of people. The overall noise is loud, but if only one person were to sing, that voice would have its own distinct 'pressure' or sound level. In the same manner, each gas in a mixture contributes to the total pressure of the system, according to its proportion in the mixture.

When it comes to solving problems involving Henry's Law, knowing the partial pressure of the gas in question is crucial. It's the value you're going to use to determine how much of the gas will dissolve in a particular liquid at a given temperature. In our soft drink example, the carbon dioxide in the bottle has a partial pressure, and it's this pressure that determines how much CO2 remains in solution, giving the drink its characteristic fizziness.
Concentration Calculation
Concentration calculation involves determining the amount of a substance within a particular volume. Just as bakers must measure out ingredients to ensure a recipe turns out correctly, chemists need to calculate concentration to predict and understand the behavior of solutions. Concentration is a measure of how 'crowded' a solution is with a particular solute. In the context of our fizzy beverage, after opening the bottle, the concentration of CO2 in the liquid will change as the gas escapes.

To calculate the concentration from Henry's Law, you multiply the gas's partial pressure by its solubility constant at the given temperature. Using the formula from the exercise solution, \( C = kP \), we determined the concentration of CO2 in the soft drink. The subtleties of this calculation lie in the precise use of units for pressure (atmospheres, in this case) and the concentration (moles per liter). It's these details that translate scientific laws into practical tools for understanding and designing everyday systems like carbonated beverages.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the number \((n)\) of atoms contained within (a) cubic cell, (b) a body centred cubic cell, (c) a face centred cubic cell.

For the reaction; \(\mathrm{N}_{2 \mathrm{k} \mathrm{g}}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}\) At \(400 \mathrm{~K}, K_{\mathrm{p}}=41 \mathrm{~atm}^{-2}\). Find the value of \(K_{\mathrm{p}}\) for each of the following reactions at the same temperature: (i) \(2 \mathrm{NH}_{3(\mathrm{~g})} \Longrightarrow \mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})}\) (ii) \({ }_{2} \mathrm{~N}_{2(\mathrm{~g})}+\frac{3}{2} \mathrm{H}_{2(g)} \Longrightarrow \mathrm{NH}_{3(\mathrm{~g})}\); (iii) \(2 \mathrm{~N}_{2(\mathrm{~g})}+6 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 4 \mathrm{NH}_{3(\mathrm{~g})}\)

Nitric oxide reacts with bromine and gives nitrosyl-bromide as per reaction given below: $$ 2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{Br}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NOBr}_{(\mathrm{g})} $$ When \(0.087\) mole of \(\mathrm{NO}\) and \(0.0437\) mole of \(\mathrm{Br}_{2}\) are mixed is a closed container at constant temperature, \(0.0518\) mole of NOBr is obtained at equilibrium. Calculate equilibrium amount of nitric oxide and bromine.

Which of the following reactions will get affected by increase of pressure? Also mention, whether change will cause the reaction to go into the right or left direction? (i) \(\mathrm{CH}_{4(\mathrm{~g})}+2 \mathrm{~S}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CS}_{2(\mathrm{~g})}+2 \mathrm{H}_{2} \mathrm{~S}_{(\mathrm{g})}\) (ii) \(\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{C}_{(\mathrm{s})} \rightleftharpoons 2 \mathrm{CO}_{(\mathrm{g})}\) (iii) \(4 \mathrm{NH}_{3(\mathrm{~g})}+5 \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 4 \mathrm{NO}_{(\mathrm{g})}+6 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\) (iv) \(\mathrm{C}_{2} \mathrm{H}_{4(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6(\mathrm{~g})}\)

Sodium metal crystallizes in a body centred cubic lattice with the cell edge \(a=4.29 \AA\). What is the radius of sodium atom?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free