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.
Dilutions - Video Tutorials & Practice Problems
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 HBrO4 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?
Your GOB Chemistry tutor
- The Environmental Protection Agency has set the limit for arsenic in drinking water at 0.010 ppm. To what volu...
- Determine the final volume, in milliliters, of each of the following: b. a 2.0% (m/v) LiCl solution prepared ...
- You need 500. mL of a 5.0% (m/v) glucose solution. If you have a 25% (m/v) glucose solution on hand, how many ...
- An aqueous solution that contains 285 ppm of potassium nitrate (KNO₃) is being used to feed plants in a garden...
- What is the concentration of a NaCl solution, in (m/v)%, prepared by diluting 65 mL of a saturated solution, w...
- Select the container that represents the dilution of a 4% (m/v) KCl solution to give each of the following: (9...
- In a laboratory experiment, a 10.0-mL sample of NaCl solution is poured into an evaporating dish with a mass o...