Molarity and dilutions

Definitions & Key takeaways

Molarity and dilutions are important concepts in chemistry, particularly when it comes to concentrations of solutions. Molarity is a unit of concentration that is expressed as moles per liter (M). It describes the number of moles of a solute that are present in one liter of solution. Dilution is the process of reducing the concentration of a solution by adding more solvent. This can be done in order to make the solution weaker, or to create different concentrations for various purposes.

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Molarity is a way of quantifying the concentration of a solution and dilution is a way of decreasing the concentration of a solution.
Both molarity and dilution are essential concepts for correctly performing chemical experiments in a laboratory. Let's say that we have two glasses of water in one glass, we put a small amount of sodium chloride which is table salt.
We can assume that all of the sodium chloride dissolves into the water to form an aqueous solution. In our other glass, we put a much larger amount of sodium chloride which we will also assume completely dissolves in the water because the first glass has less sodium chloride in the solution.
We would say that this glass contains a more dilute solution of sodium chloride because the second glass has more sodium chloride.
We would say that this glass has a more concentrated solution of sodium chloride. The words dilute and concentrated allow us to compare our two solutions of sodium chloride.
However, they don't quantify exactly how much more concentrated one solution is compared to another. When we're quantifying the concentration of a solution, we need to use the concept of molarity.
The definition of molarity is moles of solute divided by the total volume of the solution in liters. Let's use the definition of molarity to make a solution of sodium hydroxide.
Suppose we wanna make 500 mL of a 1.00 molar solution of sodium hydroxide. So we know that the definition of molarity is moles of solute divided by liters of solution.
We plug 1.00 molar into our equation for the molarity. We also plug in our desired quantity of the final solution which is 500 mL.
But because the definition of molarity is moles per liter, we need to convert 500 mL to 0.500 L. So the equation for molarity is that X divided by 0.500 L equals 1.00 molar.
Where X is the number of moles of sodium hydroxide. We want in our solution solving for X, we can see that we wanna use 0.500 moles of sodium hydroxide to make our solution.
But how many grams of sodium hydroxide does this correspond to? We now need to convert moles to grams to do this with any substance.
We always need to use the molar mass. The molar mass of sodium hydroxide is 40.0 g per mole.
Since we need 0.500 moles of sodium hydroxide to make our solution, we can multiply 0.500 moles times 40.0 g per mole, the moles cancel out and we end up with 20.0 g of sodium hydroxide.
So to prepare 500 mL of a 1.00 molar solution of sodium hydroxide, we would dissolve 20.0 g of sodium hydroxide in enough distilled water to get 500 mL for a total solution.
Now, let's talk about how to make 500 mL of our 1.00 molar solution of sodium hydroxide in the laboratory, our first step is to mass out 20.0 g of sodium hydroxide using a balance.
Then we add the sodium hydroxide to a 500 mL volumetric flask. Then we add some distilled water.
When we make an aqueous solution, we wanna use distilled water because it's free from any impurities that might affect the making of the solution.
After adding our solvent, we swirl to dissolve the sodium hydroxide. Once the sodium hydroxide has completely dissolved, we add distilled water until the total volume of the solution reaches the 500 mL calibration mark on the neck of the volumetric flask.
Then we swirl a final time to mix the solution. Next, let's do an example of how to make a dilution.
A dilution is when we start out with a more concentrated stalk solution, which is a solution with a high molarity. And we try to create a solution with a lower molarity.
Let's say we want to use a 1.00 molar stalk solution of copper sulfate to create 250 mL of a 0.100 molar solution of copper two sulfate.
When doing a deletion calculation, we wanna use the formula M initial times V initial is equal to M final times V final.
In this formula M stands for molarity and V stands for volume. So looking at the left hand side, we start with the 1.00 molar solution which we plug in for m initial.
However, we don't know what volume of the solution to use. And so we leave the initial as an unknown.
Now, looking at the right hand side of our equation, we know that we want our final molarity to be 0.100 molar and that our desired volume of the solution is 250 mL.
And so we plug these numbers in for M final and V final. Now that we've plugged in some numbers, we can simplify things a bit.
The only unknown left in this equation is the initial. And so we can solve for this number using algebra.
The initial equals 250 mL times 0.100 molar, divided by 1.00 molar. The units of molar cancel out leaving us with the initial equal to 25.0 mL.
This is how much of our starting high concentration solution that we wanna use to make our new solution. Now, let's do the deletion in the laboratory.
Our first step would be to draw up 25.0 mL of our 1.00 molar stalk solution in a pipette. Then we add the 25.0 mL to a 250 mL volumetric flask.
Next, we add enough distilled water to reach the 250 mL calibration mark on the volumetric flask. This corresponds to the addition of 225 mL of our solvent distilled water to our 25 mL of the more concentrated solution we can then shake or stir our flask to ensure that our solution is mixed evenly.
We have now created 250 mL of a 0.100 molar solution of copper. Two sulfate copper two sulfate solutions are unique because they have a rich blue color comparing our starting solution and our final dilute solution.
We noticed that the initial solution appears dark blue. While our final solution is light blue.
This is a nice check for our dilution. Our final solution appears paler because it has less copper two sulfate in it.
Now, let's do another dilution problem this time. Let's assume that we start with a concentrated solution of 12.0 molar hydrochloric acid.
We wanna use this concentrated acid to prepare 200 mL of a 3.00 molar solution of HCL. Once again, we use our equation M initial times V initial is equal to M final times V final, we plug in 12.0 molar for M initial and we plug in 3.00 molar for m final and 200 mL for V final.
Now we saw for our unknown variable, the initial which is equal to 200 mL times 3.00 molar, divided by 12.0 molar equals 50.0 mL.
So 50.0 mL is the amount of concentrated hydrochloric acid. We need to perform our dilution for this problem.
Let's consider in detail exactly how we would do our dilution in the laboratory. Let's start with a 200 mL volumetric flask.
We know that we're eventually going to add in 50.0 mL of HCL for a total volume of 200 mL. So we start by adding in about 148 mL of distilled water to the volumetric flask.
Next, we can measure out 50.0 mL of concentrated hydrochloric acid in a graduated cylinder and add the acid to the volumetric flask.
We should just be under the calibration mark on the volumetric flask. The next step is to use a pipette to add in just enough distilled water for the bottom of the meniscus to just touch the calibration mark.
Finally, we can stir the solution to ensure even mixing, we now have 200 mL of a 3.00 molar solution of HCL. You may have noticed that we added the acid to the water instead of putting the acid in first and then diluting with water.
The reason why you always add concentrated acid to water is because mixing a concentrated acid and water generates a lot of heat.
Water is able to absorb the heat and prevents the solution from getting dangerously hot. So by adding the water first and then adding the acid, we're able to prepare our solution more safely.
Adding the small amount of distilled water at the end to reach the final volume is safe because the acidic solution is now much more dilute and we're only adding a small amount of water.
All right, as a quick recap, molarity is a measure of how concentrated a solution is. The definition of molarity is moles of solute divided by liters of solution.
If we want to make an aqueous solution of a specific molarity, we can add the exact amount of solute necessary to a volumetric flask and then add enough distilled water to reach the calibration mark on the flask.
Sometimes we can make an aqueous solution by diluting a more concentrated solution with distilled water. However, if we start with a concentrated acid, we always add acid to water instead of adding water to acid.