Dilutions - Online Tutor, Practice Problems & Exam Prep

In the process of dilution, a solvent, typically water, is added to a concentrated solution, known as a standard or stock solution, to decrease its concentration. This results in a solution of lighter color, indicating a lower concentration of solute. The relationship between the concentration and volume of the original solution and the diluted solution can be described by the equation:

$M_1{V}_1=M_2{V}_2$

Here, M1 and V1 represent the molarity and volume before dilution, while M2 and V2 represent the molarity and volume after dilution. The final volume (V2) is the sum of the initial volume (V1) and the volume of solvent added. The molarity after dilution (M2) is always less than the molarity before dilution (M1), reflecting the reduced concentration of the solution. Understanding this concept is crucial for accurate preparation of solutions in laboratory settings, ensuring precision in experimental procedures.

Dilutions

A standard sometimes referred to as a stock solution is a concentrated solution that will be diluted for some laboratory use later on. And 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 purple solution. Actually, it's pretty dark purple, meaning that it's concentrated. And what we're doing here is we're slowly adding more water, so we're adding water to it 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. 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.

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. All right. 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 12345 Sears, so that would be 5 moles of solute divided by one liter of solution, so that'd be 5 molar. For BB is 123 sphere, so that's three moles of solute over 2 liters of solution, so that'd be 1.5 molar.

And then finally C we have 123456 spheres so 6 moles divided by 3 liters of solution so that's two 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 solution 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. So it's M1V1=M2V2.

Here 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 equals V1 plus the volume of solvent added.

Dilutions Example 2

In this example question, it says what volume and milliliters, a 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 a dilution.

So that 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 in between 2 numbers, that means multiply. We're going to say here that this is the smaller molarity, so this has to be M2. So 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 so that comes out to be 1.8173 liters. We want the answer milliliters, so just do a quick metric prefix conversion. Leaders on the bottom, milliliters on top. One milli is 10 to the -3, so leaders cancel out and that comes out to be 1817.3 milliliters.

Here 5235 and 27 all have two sig figs. So if we wanted two significant figures here, we just write this as 1800 milliliters. Oops, 1800 milliliters as our final answer. 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 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 dilution?

Dilution in a scientific context typically refers to the process of reducing the concentration of a solute in a solution, usually by adding more solvent. For instance, if you have a strong salt solution and you add more water to it, you are diluting the solution. The amount of salt remains the same, but the volume of the solution increases, which means the salt concentration decreases.

In a broader context, dilution can also refer to the process of weakening any mixture or substance. For example, in business and finance, dilution can occur when a company issues more shares of stock, which reduces the ownership percentage of the existing shareholders. Similarly, in biology, dilution can refer to the process of reducing the concentration of cells, bacteria, or other particles in a sample.

How do you calculate dilution?

To calculate dilution, you need to understand the concept of concentration, which is the amount of solute divided by the volume of the solution. When you dilute a solution, you are adding more solvent without adding more solute, thereby decreasing the solution's concentration.

The dilution equation is:

$C_1{V}_1=C_2{V}_2$

Where:

• C1 is the initial concentration of the solution.
• V1 is the initial volume of the solution.
• C2 is the final concentration of the solution after dilution.
• V2 is the final volume of the solution after dilution.

To perform the calculation, you need to know three of these values to solve for the fourth. For example, if you have 1 liter of a 0.5 M solution and you want to dilute it to 0.1 M, you would solve for V2:

$\left(0.5\text{M}\right)\left(1\text{L}\right)=\left(0.1\text{M}\right)\left(V_2\right)$$V_2=\frac{\left(0.5\text{M}\right)\left(1\text{L}\right)}{\left(0.1\text{M}\right)}=5\text{L}$

This means you would need to add enough solvent to the original 1 liter of solution to make a total volume of 5 liters to achieve the desired concentration of 0.1 M.

How can I find the dilution factor?

To find the dilution factor, you'll need to know the volume of the original concentration and the final volume after dilution. The dilution factor is the ratio of the final volume to the initial volume. Here's the formula you can use:

$\text{Dilution Factor (DF) =}\frac{\text{Total Volume}}{\text{Solute Volume}}$

For example, if you have 1 mL of a stock solution and you add 9 mL of water to it, your total volume is now 10 mL. The dilution factor would be:

$\text{DF =}\frac{10\text{mL (total volume)}}{1\text{mL (solute volume)}}=10$

This means the concentration has been diluted by a factor of 10. If you're working with serial dilutions, where you dilute a solution multiple times, the overall dilution factor is the product of all individual dilution factors in the series. Remember, the dilution factor is a unitless number, representing how much the original solution has been diluted.

How to do dilution calculations?

Dilution calculations are used to determine the concentration of a solution after it has been diluted with a solvent. To perform a dilution calculation, you can use the formula:

$C_1{V}_1=C_2{V}_2$

Where:

• C1 is the concentration of the original solution before dilution.
• V1 is the volume of the original solution before dilution.
• C2 is the concentration of the solution after dilution.
• V2 is the final volume of the solution after dilution.

Here's how you do it:

1. Identify the concentration and volume of the original solution (C1 and V1).
2. Determine the final volume of the diluted solution (V2). This includes the volume of the original solution plus the volume of the solvent added.
3. Decide the concentration you want in the final solution (C2).
4. Plug the known values into the equation and solve for the unknown variable.

For example, if you have 10 mL of a 5 M solution and want to dilute it to 1 M, you would solve for V2:

$\left(5\text{M}\right)\left(10\text{mL}\right)=\left(1\text{M}\right)\left(V_2\right)$$V_2=\frac{\left(5\text{M}\right)\left(10\text{mL}\right)}{\left(1\text{M}\right)}=50\text{mL}$

This means you would need to add 40 mL of solvent to your original 10 mL solution.

What is the dilution equation?

The dilution equation is a formula used in chemistry to calculate the concentration of a solution after it has been diluted with a solvent. The most common form of the dilution equation is:

$C_1{V}_1=C_2{V}_2$

Here's what each variable represents:

• C1 is the initial concentration of the solution before dilution.
• V1 is the volume of the solution before dilution.
• C2 is the final concentration of the solution after dilution.
• V2 is the final volume of the solution after dilution.

This equation is based on the principle of conservation of mass, which states that the amount of solute remains the same before and after dilution. When you dilute a solution, you're adding more solvent, which increases the volume (V2) and decreases the concentration (C2) of the solution. By using this equation, you can calculate any one of the four variables if you know the other three. This is particularly useful in laboratory settings for preparing solutions of desired concentrations.