Molarity in Chemical Reactions: Practical Examples

Understanding Molarity in Chemical Reactions

Molarity is a crucial concept in chemistry that measures the concentration of a solute in a solution. It is expressed as moles of solute per liter of solution (mol/L). Understanding molarity allows chemists to predict how substances will react and to carry out reactions accurately. Below are three practical examples of using molarity in various chemical reactions.

Example 1: Preparing a Sodium Chloride Solution

In a laboratory setting, preparing a sodium chloride (NaCl) solution is a common task. Molarity is used to ensure that the concentration of the solution is correct for subsequent experiments or reactions.

To prepare a 0.5 M NaCl solution, you first need to know the molar mass of NaCl, which is approximately 58.44 g/mol. To find out how many grams are needed to create 1 liter of solution:

• Calculate the mass required: 0.5 mol/L × 58.44 g/mol = 29.22 g
• Weigh out 29.22 grams of NaCl.
• Dissolve the NaCl in enough water to make the total volume 1 liter.

This solution can be utilized in various experiments, such as electrolysis or reaction kinetics studies.

Notes:

• If you need a different volume, the mass can be scaled accordingly using the formula: mass (g) = molarity (mol/L) × volume (L) × molar mass (g/mol).

Example 2: Titration of Acetic Acid with Sodium Hydroxide

Titration is an analytical technique used to determine the concentration of an unknown solution. In this example, we will titrate acetic acid (CH₃COOH) with sodium hydroxide (NaOH) to find out the molarity of the acetic acid solution.

1. Prepare the NaOH solution: Assume you have a 0.1 M NaOH solution ready for titration.
2. Setup: Fill a burette with the 0.1 M NaOH solution and use a pipette to measure 25 mL of the acetic acid solution into a flask.
3. Conduct the titration: Slowly add the NaOH from the burette to the acetic acid solution while continuously stirring until the endpoint is reached, indicated by a color change from phenolphthalein.

4. Calculate the molarity of acetic acid: If 20 mL of NaOH is required to reach the endpoint, use the titration formula:

   M₁V₁ = M₂V₂

   Where:

   • M₁ = molarity of acetic acid
   • V₁ = volume of acetic acid (0.025 L)
   • M₂ = molarity of NaOH (0.1 M)
   • V₂ = volume of NaOH used (0.020 L)

   Rearranging gives:
   M₁ = (M₂V₂) / V₁ = (0.1 mol/L × 0.020 L) / 0.025 L = 0.08 M

This indicates that the molarity of the acetic acid solution is 0.08 M.

Notes:

• The indicator used can affect the accuracy, so it’s essential to choose one suitable for the pH range of the titration.

Example 3: Reaction of Ammonium Hydroxide and Hydrochloric Acid

In this example, we will investigate the reaction between ammonium hydroxide (NH₄OH) and hydrochloric acid (HCl) using molarity to determine how much of each reactant is needed.

1. Define the reaction: The balanced chemical equation is:
   NH₄OH + HCl → NH₄Cl + H₂O
   This reaction shows a 1:1 molar ratio between NH₄OH and HCl.

2. Determine the required amounts: Suppose you want to react 0.1 M NH₄OH with HCl. If you plan to use 50 mL of NH₄OH:

   • Calculate moles of NH₄OH:
     Moles = Molarity × Volume = 0.1 mol/L × 0.050 L = 0.005 moles

   • Since the reaction ratio is 1:1, you will need 0.005 moles of HCl as well.

   • Calculate the volume of HCl needed if it is 0.1 M:
     Volume = Moles / Molarity = 0.005 moles / 0.1 mol/L = 0.050 L or 50 mL.

Mix the solutions to carry out the reaction safely and observe the formation of ammonium chloride (NH₄Cl) and water.

Notes:

• Ensure to conduct this reaction in a controlled environment, as both reactants can be hazardous.

These examples illustrate the practical applications of molarity in various chemical reactions and how it aids in precise calculations and experimental setups.