Question

You mix \(225.0 \mathrm{~mL}\) of a \(2.5 \mathrm{M} \mathrm{HCl}\) solution with \(150.0 \mathrm{~mL}\) of a \(0.75 \mathrm{M} \mathrm{HCl}\) solution. What is the molarity of the final solution?

Short answer

Calculating the molarity of the final solution involves finding the total moles of HCl and total volume of the final solution. Moles of HCl (first solution) = \(2.5 M \times 0.225 L\); moles of HCl (second solution) = \(0.75 M \times 0.150 L\). Total moles of HCl and total volume (L) are obtained after adding the respective values of both solutions. Final molarity = Total moles of HCl ÷ Total volume (L).

Step-by-step solution

Calculate moles of HCl in each solution

We'll first find the amount of HCl in each solution in moles using the given molarities and volumes. Use the formula: Moles of HCl = Molarity × Volume For the first solution: Moles of HCl (first solution) = \(2.5 M \times 225.0 mL\) For the second solution: Moles of HCl (second solution) = \(0.75 M \times 150.0 mL\)

Convert volume to liters

We should convert the volume of the solutions from milliliters to liters, as molarity is usually expressed in moles per liter. To do this, divide the volume by 1000: For the first solution: Volume (first solution, L) = \(225.0 mL ÷ 1000 = 0.225 L\) For the second solution: Volume (second solution, L) = \(150.0 mL ÷ 1000 = 0.150 L\)

Calculate the total moles of HCl

Add the moles of HCl in both solutions to get the total moles after mixing: Total moles of HCl = Moles HCl (first solution) + Moles HCl (second solution)

Calculate the total volume of the final solution

Add the volumes of both solutions to get the total volume after mixing: Total volume (L) = Volume (first solution, L) + Volume (second solution, L)

Calculate the molarity of the final solution

Now that we have the total moles of HCl and the total volume of the final solution, we can calculate the molarity using the formula: Final molarity = Total moles of HCl ÷ Total volume (L) This will give us the molarity of the final HCl solution.

Key concepts

Solution Concentration

Understanding solution concentration is essential when working with chemical solutions, as it tells us how much solute is dissolved in a given amount of solvent. One of the most common units of concentration is molarity, denoted as 'M'. Molarity is defined as the number of moles of a solute divided by the volume of the solution in liters.
When we deal with exercises involving dilution or mixing solutions, as in our given problem, we calculate the individual molarities and then derive the molarity after mixing. Molarity can inform us about the strength and reactivity of the solution, which is crucial in various chemical processes and reactions.

• To calculate molarity, you need to know the moles of solute and the volume of solution in liters.
• Molarity is a convenient measure because it directly relates to the number of moles of reactant in a given volume, linking it to stoichiometric calculations for reactions.
• In dilution and mixing problems, it is the combined volume and moles of solutes that determine the final concentration.

When improving your exercise-solving skills, remember to check the units of volume and convert them to liters if necessary to ensure the correct calculation of molarity.

Moles and Volume Relationship

The relationship between moles and volume is central to solving problems in chemistry, particularly those involving solutions. We use the units of moles to express the amount of a substance, while volume is usually given in liters for solutions.
For instance, in acid-base reactions, it is often necessary to know the amount of an acid or base in moles to understand how it will react with other substances.

Key Points to Remember:

• Moles provide a way to quantify substances based on the number of particles, rather than mass or volume.
• The volume of a solution is essential because we usually express molar concentration in terms of liters.
• Conversion between milliliters and liters is often needed when working with laboratory volumes and molarity.

The steps to solve the mixed solution exercise implicate this relationship by first calculating the moles of solute in each solution separately and then combining them to find the total moles. Converting milliliters to liters before calculating the final concentration ensures accurate results, as molarity is expressed in moles per liter.

Acid-Base Chemistry

The exercise presented relates to acid-base chemistry, which involves the study of the properties of acids and bases, their reactions, and how they can be quantified. The acid in our example, hydrochloric acid (HCl), is a strong acid and fully dissociates in water.
To comprehend how acids and bases behave when mixed, you need to understand concepts like molarity and the moles-volume relationship within a solution.

Important Aspects:

• Strong acids like HCl dissociate completely into ions in a solution, which affects the pH and reactivity.
• Studying the molarity of acids and bases can help predict the outcomes of their reactions and the extent of neutralization.
• In the mixing problem, combining two solutions of HCl at different concentrations affects the overall concentration and therefore the acidity of the final solution.

In practicing acid-base problems, it is also crucial to be familiar with titration calculations, neutralization reactions, and pH calculations, all of which depend on the accurate determination of solution concentrations.