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 $m$₁v₁=m₂v₂, where m₁ and v₁ are the molarity and volume before dilution, and m₂ and v₂ are after dilution. Dilution decreases molarity, transitioning from a concentrated to a dilute solution.

Dilutions

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 solvent here, which is actually a 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. That's showing us that it's not as concentrated as it was before. So here, this represents our diluted solution. Just remember, when we're talking about dilutions, we're just talking about adding water to our original solution to make it less concentrated.

Dilutions Example 1

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 molar. 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 molar. 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 molar. So arranging it from lowest molarity to highest molarity, we're going to say the order would be b, then c, and then finally, a would have the highest molarity.

Dilutions

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:

M1∙V1=M2∙V2

Here, M1 and V1 represent the molarity and volume before dilution, while M2 and V2 are after the dilution. We are 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

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 one compound. And with that one compound, to be talking about dilution, we tend to deal with two molarities. So, the fact that we're dealing with just hydrobromic acid and we have two molarities associated with it is a strong indication that we're dealing with dilution. This means we're going to use M1V1=M2V2. Now, remember M1 is larger than M2 because M1 represents your concentrated solution before you've begun dilution.

Because of this, 5.2 molar has to be our M1. It's the larger molarity. 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 between two numbers, means multiply. We're going to say here that this is the smaller molarity, so this has to be M2, that's our diluted molarity. We're multiplying it with 3.5 liters, so based on the dilution equation, 3.5 liters has to be V2. We're going to isolate V1. So, divide both sides by 5.2 molar. The molarities cancel out and look, we'll have V1 but here it'll be in liters.

That comes out to be 1.8173 liters. We want the answer in milliliters so just do a quick metric prefix conversion. Liters on the bottom, milliliters on top. One milli is 10 to the negative third power, so liters cancel out, and that comes out to be 1817.3 milliliters. Here, 5.2, 3.5, and 2.7 all have two significant figures. So, if we wanted two significant figures here, we would just write this as 1800 milliliters. So just remember, when we're dealing with one compound and we have different molarities, that's a strong indication that we're dealing with a 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?

• A. 72 mL
• B. 289 mL
• C. 330 mL
• D. 930 mL
• E. 1400 mL

If 880 mL of water is added to 125.0 mL of a 0.770 M HBrO₄ solution what is the resulting molarity?

• A. 0.096 M
• B. 0.136 M
• C. 0.257 M
• D. 0.892 M
• E. 1.76 M

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).

• A. 0.0500 M
• B. 0.025 M
• C. 0.005 M
• D. 0.500 M
• E. 0.0100 M

Here’s what students ask on this topic:

What is a stock solution in chemistry?

A stock solution in chemistry is a concentrated solution that is prepared for the purpose of being diluted to a lower concentration for various laboratory uses. By having a stock solution, you can easily prepare solutions of different concentrations by adding a specific amount of solvent, usually water. This is particularly useful in experiments where precise concentrations are required. The process of dilution is governed 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 know the initial concentration and volume, and the final volume, you can find the final concentration using $m_2=\frac{m_1}{v_1}{v}_2$.

What is the purpose of diluting a solution in a laboratory setting?

The purpose of diluting a solution in a laboratory setting is to achieve a desired concentration that is suitable for a specific experiment or application. Dilution allows scientists to work with solutions that are less concentrated, which can be necessary for accurate measurements, reactions, or analyses. By starting with a stock solution and diluting it, you can prepare multiple solutions of varying concentrations efficiently. This is particularly important in experiments where precise control over the concentration of reactants is required to ensure reproducibility and accuracy.

What is the equation used to describe the dilution process?

The equation used to describe the dilution process is $m_1{v}_1=m_2{v}_2$. In this equation, $m_1$ represents the initial molarity (concentration) of the solution, $v_1$ is the initial volume, $m_2$ is the final molarity after dilution, and $v_2$ is the final volume after dilution. This equation helps in calculating the new concentration or volume when a solution is diluted by adding more solvent.

How does dilution affect the molarity of a solution?

Dilution affects the molarity of a solution by decreasing it. When you add more solvent to a solution, the number of moles of solute remains the same, but the total volume of the solution increases. This results in a lower concentration of solute per unit volume. The relationship between the initial and final concentrations and volumes is given by the equation $m_1{v}_1=m_2{v}_2$. Here, $m_1$ and $v_1$ are the molarity and volume before dilution, and $m_2$ and $v_2$ are after dilution. As the volume increases, the molarity decreases proportionally.