Content Reviewers:Rishi Desai, MD, MPH
Molarity is a way of quantifying the concentration of a solution.
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 this 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 are 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 want to make 500. milliliters of a 1.00 M 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 want to 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 grams per mole.
Since we need 0.500 moles of sodium hydroxide to make our solution, we can multiply 0.500 moles times 40.0 grams per mole.
The “moles” cancel out, and we end up with 20.0 grams of sodium hydroxide.
So to prepare 500. milliliters of a 1.00 molar solution of sodium hydroxide, we would dissolve 20.0 grams of sodium hydroxide in enough distilled water to get 500mL for our total solution.
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 grams 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 want to use distilled water because distilled water is 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.
Note that we want the bottom of the meniscus to just touch the calibration mark.
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 where we start out with a more concentrated stock 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 stock solution of copper sulfate to create 250. mL of a 0.100 molar solution of copper(II) sulfate.
When doing a dilution calculation, we want to 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 a 1.00 molar solution, which we plug in for “M” initial.
However, we do not know what volume of this solution to use, and so we leave “V” 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.