A standard, sometimes referred to as a stock solution, is a concentrated solution that will be diluted for some laboratory use later on. We're going to say dilution is just the addition of more solvent, usually water, to a solution in order to create a lower concentration. So if we take a look here, we have our purple solution. It's pretty dark purple, meaning that it's concentrated. And what we're doing here is we're slowly adding more water to dilute it. As a result of this, it goes from being a dark purple to a lighter type of fuchsia or purple, showing us that it's not as concentrated as it was before. This represents our diluted solution. So just remember, when we're talking about dilutions, we're just talking about adding water to our original solution to make it less concentrated.

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# Dilutions - Online Tutor, Practice Problems & Exam Prep

A stock solution is a concentrated solution that can be diluted by adding more solvent, typically water, to achieve a lower concentration. This process is represented by the equation $m$_{1}_{1}_{2}_{2}, where _{1} and _{1} are the molarity and volume before dilution, and _{2} and _{2} are after dilution. Dilution decreases molarity, making it essential for various laboratory applications.

In Dilutions, a solvent (usually water) is added to a concentrated solution.

## Concentrated & Diluted Solutions

### Dilutions

#### Video transcript

### Dilutions Example 1

#### Video transcript

In this example question, it says, if each sphere represents a mole of solute from the images provided below, arrange the solutions from least concentrated to most concentrated. Alright. So least concentrated is the same thing as saying the lowest molarity. Most concentrated means we have the highest molarity. Now remember, molarity itself represents moles of solute divided by liters of solution. So if we take a look here for a, a has in it 1, 2, 3, 4, 5 spheres. So that would be 5 moles of solute divided by 1 liter of solution, so that'd be 5 M. For b, b is 1, 2, 3 spheres, so that's 3 moles of solute over 2 liters of solution, so that'd be 1.5 M. And then finally, c we have 1, 2, 3, 4, 5, 6 spheres, so 6 moles divided by 3 liters of solution, so that's 2 M. So arranging it from lowest molarity to highest molarity, we're gonna say the order would be b then c, and then finally, a would have the highest Molarity.

### Dilutions

#### Video transcript

So at this point, we know that a dilution makes our solutions less concentrated. It takes us from a larger molarity value to a smaller molarity value. We're going to say dilution can be expressed by the following equation:

M1V1 = M2V2Here, M1 and V1 represent the molarity and volume before dilution, while M2 and V2 are after the dilution. We're going to say here M1 is before a solvent is added. So M1, which is the more concentrated solution, is always larger than M2, which will be the diluted solution. Now V2 represents your final volume, and how exactly did we get to V2? Well, we started out with an initial volume and we added water to it. So V2 = V1 + the volume of solvent added.

### Dilutions Example 2

#### Video transcript

In this example question, it asks what volume in milliliters of 5.2 molar hydrobromic acid must be used to prepare 3.5 liters of 2.7 molar hydrobromic acid. Now, how do we know this is a dilution question? Well, typically in a dilution question, we're only talking about 1 compound. And with that one compound, to be talking about dilution, we tend to deal with 2 molarities. So the fact that we're dealing with just hydrobromic acid and have 2 molarities associated with it is a strong indication that we're dealing with the dilution. That means we're going to use the formula M1V1=M2V2. Now remember, M1 is larger than M2 because it represents the concentrated solution before you've begun dilution. Since 5.2 molar is the larger molarity, it must be M1. Associated with M1 is V1. We don't see any number around it, so V1 is what we're looking for. Now remember also that the word "of", when it's in between two numbers, means multiply. We're going to say here the 2.7 molar is M2, our diluted molarity. We're multiplying it with 3.5 liters, so based on the dilution equation, 3.5 liters must be V2. We will isolate V1. So, divide both sides by 5.2 molar. The molarities cancel out and look, we'll have V1, but it will be in liters. It comes out to be 1.8173 liters. We want the answer in milliliters, so just do a quick metric prefix conversion. Liters are on the bottom, milliliters on top. One milliliter is 10 to the negative 3 liters, so liters cancel out, and that comes out to be 1817.3 milliliters.

Here, 5.2, 3.5, and 2.7 all have 2 significant figures. So if we wanted 2 significant figures here, we would just write this as 1800 milliliters. Oh, it's 1800 milliliters as our final answer. Just remember, when we're dealing with 1 compound and we have different molarities, that's a strong indication that we're dealing with the dilution equation. So use the dilution formula and solve for the missing variable.

To what final volume would 100 mL of 5.0 M KCl have to be diluted in order to make a solution that is 0.54 M KCl?

If 880 mL of water is added to 125.0 mL of a 0.770 M HBrO_{4} solution what is the resulting molarity?

A student prepared a stock solution by dissolving 25.00 g of NaOH in enough water to make 150.0 mL solution. The student took 20.0 mL of the stock solution and diluted it with enough water to make 250.0 mL solution. Finally taking 75.0 mL of that solution and dissolving it in water to make 500 mL solution. What is the concentration of NaOH for this final solution? (MW of NaOH:40.00 g/mol).

0.0500 M

0.025 M

0.005 M

0.500 M

0.0100 M

## Do you want more practice?

### Here’s what students ask on this topic:

What is a stock solution and how is it used in dilutions?

A stock solution is a concentrated solution that is prepared for the purpose of being diluted to a lower concentration for various laboratory uses. It allows for the convenient preparation of solutions of varying concentrations by simply adding a solvent, typically water. The process of dilution involves adding more solvent to the stock solution, which decreases its molarity. This is represented by the equation ${m}_{1}{v}_{1}={m}_{2}{v}_{2}$, where ${m}_{1}$ and ${v}_{1}$ are the molarity and volume before dilution, and ${m}_{2}$ and ${v}_{2}$ are after dilution.

How do you calculate the final concentration after dilution?

To calculate the final concentration after dilution, you can use the equation ${m}_{1}{v}_{1}={m}_{2}{v}_{2}$. Here, ${m}_{1}$ is the initial molarity, ${v}_{1}$ is the initial volume, ${m}_{2}$ is the final molarity, and ${v}_{2}$ is the final volume. Rearrange the equation to solve for the unknown. For example, if you need to find the final molarity ${m}_{2}$, the equation becomes ${m}_{2}=\frac{{m}_{1}{v}_{1}}{{v}_{2}}$.

What is the purpose of performing dilutions in a laboratory setting?

Performing dilutions in a laboratory setting serves several purposes. It allows scientists to achieve desired concentrations for experiments, ensuring that reactions occur under optimal conditions. Dilutions are also essential for preparing solutions that are safe to handle and use, as highly concentrated solutions can be hazardous. Additionally, dilutions help in achieving accurate and reproducible results by providing precise control over the concentration of reactants. This is crucial in quantitative analyses, such as titrations and spectrophotometry, where the concentration of a solution directly affects the outcome of the experiment.

How does the equation m1v1 = m2v2 apply to dilutions?

The equation ${m}_{1}{v}_{1}={m}_{2}{v}_{2}$ is fundamental to understanding dilutions. It states that the product of the initial molarity (${m}_{1}$) and initial volume (${v}_{1}$) of a solution is equal to the product of the final molarity (${m}_{2}$) and final volume (${v}_{2}$) after dilution. This relationship helps in calculating the new concentration or volume needed when a solution is diluted. It ensures that the number of moles of solute remains constant before and after dilution, as the added solvent only changes the volume and concentration, not the amount of solute.

What are some common mistakes to avoid when performing dilutions?

When performing dilutions, common mistakes to avoid include: 1) Incorrectly measuring volumes, which can lead to inaccurate concentrations. Always use calibrated equipment like pipettes and volumetric flasks. 2) Not mixing the solution thoroughly after adding the solvent, which can result in an uneven concentration. 3) Misinterpreting the dilution equation ${m}_{1}{v}_{1}={m}_{2}{v}_{2}$, especially in rearranging it to solve for the unknown variable. 4) Forgetting to account for the final volume, which includes both the initial volume and the added solvent. 5) Using contaminated or impure solvents, which can affect the outcome of the dilution.

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