Dilution: Definition and Calculations

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To dilute a solution means to add more solvent without the addition of more solute. Of course, the resulting solution is thoroughly mixed so as to ensure that all parts of the solution are identical.

The fact that the solute amount stays constant allows us to develop calculation techniques.

First, we write:

moles solute before dilution = moles solute after dilution

From rearranging the equation that defines molarity, we know that the moles of solute equals the molarity times the volume. (Calculating the moles of solute from molarity times volume will be very useful in other areas of chemistry, particularly acid base. Remember that the volume must be in liters.)

So we can substitute MV (molarity times volume) into the above equation, like this:

M1V1 = M2V2

The "sub one" refers to the situation before dilution and the "sub two" refers to after dilution.

This equation does not have an official name like Boyle's Law, so we will just call it the dilution equation.


Example #1: 53.4 mL of a 1.50 M solution of NaCl is on hand, but you need some 0.800 M solution. How many mL of 0.800 M can you make?

Solution:

Using the dilution equation, we write:

(1.50 mol/L) (53.4 mL) = (0.800 mol/L) (x)

x = 100. mL

Notice that the volumes need not be converted to liters. Any old volume measurement is fine, just so long as the same one is used on each side. (However, as mentioned above, if you are calculating how many moles of solute are present, you need to have the volume in liters.)


Example #2: 100.0 mL of 2.500 M KBr solution is on hand. You need 0.5500 M. What is the final volume of solution which results?

Placing the proper values into the dilution equation gives:

(2.500 mol/L) (100.0 mL) = (0.5500 mol/L) (x)

x = 454.5454545 mL (oops, my fingers got stuck typing.) (Bad attempt at humor, really bad!)

x = 454.5 mL

Sometimes the problem might ask how much more water must be added. In this last case, the answer is 454.5 - 100.0 = 354.5 mL.

Go ahead and answer the question, if your teacher asks it, but it is bad technique in the lab to just measure out the "proper" amount of water to add and then add it. This is because the volumes (the soltion and the diluting water) are not necessarily additive. The only volume of importance is the final solution's volume. You add enough water to get to the final solution volume without caring how much the actual volume of water you added is.


Example #3 - 8:

3. A stock solution of 1.00 M NaCl is available. How many milliliters are needed to make 100.0 mL of 0.750 M

4. What volume of 0.250 M KCl is needed to make 100.0 mL of 0.100 M solution?

5. Concentrated H2SO4 is 18.0 M. What volume is needed to make 2.00 L of 1.00 M solution? Concentrated HCl is 12.0 M. What volume is needed to make 2.00 L of 1.00 M solution?

6. A 0.500 M solution is to be diluted to 500.0 mL of a 0.150 M solution. How many mL of the 0.500 M solution are required?

7. A stock solution of 10.0 M NaOH is prepared. From this solution, you need to make 250.0 mL of 0.375 M solution. How many mL will be required?

8. 2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 1.50 M in concentration. How many mL are required?

Go to Dilution answers #3-8


These next two are a bit harder and involve slightly more calculation than the discussion above. Here's a summary of the steps:

1) calculate total moles
2) calculate total volume
3) divide moles by volume to get molarity

You can also think of it this way:

M1aV1a + M1bV1b = M2V2

Where the '1a' refers to one starting solution, '1b' to the other and '2' refers to the mixed solution (hints: V2 is the total volume after mixing and M2 is the unknown).

Example #9: Calculate the final concentration if 2.00 L of 3.00 M NaCl and 4.00 L of 1.50 M NaCl are mixed. Assume there is no volume contraction upon mixing.

Example #10: Calculate the final concentration if 2.00 L of 3.00 M NaCl, 4.00 L of 1.50 M NaCl and 4.00 L of water are mixed. Assume there is no volume contraction upon mixing.

Go to answers for 9 & 10


Go to some dilution problems

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