Question

\(1 M \mathrm{HCl}\) and \(2 \mathrm{M} \mathrm{HCl}\) are mixed in volume ratio of \(4: 1 .\) What is the final molarity of \(\mathrm{HCl}\) solution? (a) \(1.5\) (b) 1 (c) \(1.2\) (d) \(1.8\)

Short answer

The final molarity of the HCl solution is 1.2 M.

Step-by-step solution

Understand the concept of molarity

Molarity is defined as the number of moles of a solute divided by the volume of the solution in liters. When two solutions of different molarity are mixed, the final molarity can be calculated by using the weighted average formula based on the molarities and volumes of the solutions mixed.

Set up the molarity equation for the mixture

Let V1 be the volume of the 1 M HCl solution and V2 be the volume of the 2 M HCl solution. The final molarity (Mf) is given by the equation: Mf = (M1*V1 + M2*V2) / (V1 + V2), where M1 and M2 are the molarities of the first and second solutions, respectively.

Calculate the final molarity

Given the volume ratio of 4:1, let's assume V1 = 4x and V2 = x for easier calculation. Now, substituting the known values into the equation: Mf = (1M * 4x + 2M * x) / (4x + x) = (4 + 2) / 5 = 6 / 5. Therefore, the final molarity is 1.2 M.

Key concepts

Understanding Chemical Solutions

When we talk about chemical solutions, we are referring to a homogeneous mixture composed of two or more substances. In the context of chemistry, a solution typically consists of a solute and a solvent. The solute is the substance that is dissolved, and the solvent is the medium in which the solute is dissolved.

One of the most common ways of expressing the concentration of a solution is through molarity, which is a measure of the number of moles of solute per liter of solution. Molarity is an essential concept when mixing different solutions to achieve a specific concentration, which is exactly the scenario students encounter in our exercise.

Understanding the molarity of a solution is vital in many aspects of chemistry, including reaction stoichiometry and the preparation of different chemical reagents. Making sure that students can perform molarity calculations confidently is a key step in their chemical education.

Calculating Mixture Volumes

When combining two or more chemical solutions, it's essential to understand how their mixture volumes impact the resulting solution's concentration. Volume is a measure of the space that a substance occupies, and when solutions with different volumes are mixed, the volume of the resulting solution is the sum of the individual volumes.

In our exercise, the focus is on the effect of mixing two solutions with different molarities in a specified volume ratio. This concept teaches students not only to be careful with measurements but also the significance of proportionality in chemistry. The resulting molarity after mixing depends on both the initial concentrations and the volumes of the solutions used.

Being able to calculate mixture volumes accurately is a foundational skill, necessary for experiments, and often used in laboratory settings, pharmaceutical applications, and any scenario where precise chemical composition is critical.

Applying the Weighted Average Formula

The weighted average formula is a mathematical technique used to find the average value of a set of numbers, with each number having a different weight that reflects its importance. This formula holds much significance in the context of mixing solutions.

For chemical solutions, the weighted average takes into account both the concentration of the solutions and their respective volumes. The formula as applied to our exercise is:

Mf = (M1*V1 + M2*V2) / (V1 + V2).

Here, M1 and M2 are the molarities of the individual solutions and V1 and V2 are their respective volumes. The final molarity, Mf, reflects the weighted importance of the two different concentrations of the solutions being mixed.

This approach to calculating concentration after mixing solutions is crucial for students to learn as it provides a way to predict the outcome of real-world applications such as diluting concentrated solutions to prepare them for use in a variety of settings.