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Dilution Calculations
Previously in this lesson, the concentration calculations that we have done essentially
involved preparing a solution from scratch. We started with separate solvent and solute
and figured out how much of each you would need to use.
Quite often, however, solutions are prepared by diluting a more concentrated solution.
For example, if you needed a one molar solution you could start with a six molar solution
and dilute it. Consequently, you also need to be familiar with the calculations that are
associated with dilutions.
| There is an element of simplicity in calculations of these types and the
element of simplicity is that the number of moles of solute stays the same, as shown here.
The number of moles of solute in the concentrated solution (indicated by the subscripted
molescon) is equal to the number of moles in the dilute solution. You
have simply increased the amount of solvent in the solution. |
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| Of course you know that the number of moles of solute in the concentrated
solution is equal to the molarity of the concentrated solution times the volume of the
concentrated solution. Also, the number of moles of solute in the dilute solution is equal
to the molarity of the dilute solution times the volume of the dilute solution. Since we
are really interested in the molarities and volumes we can substitute and use the equation
shown in the second line (or third line, depending on your preference for how to show
multiplication). Let's use this equation in a few examples. |
|
| Mcon x volcon |
= |
Mdil x voldil |
| (Mconc) (Vconc) |
= |
(Mdil) (Vdil) |
|
|
Examples (Ex. 6)
Calculating New Concentration (Ex. 6a)
| In this example you are asked what the concentration of a solution would
be if it were made by diluting 50.0 ml of 0.40 M NaCl solution to 1000. ml. As a
general procedure, I recommend that you first write down the equation that is given in the
first line: the molarity of the concentrated solution times the volume of the concentrated
solution is equal to the molarity of the dilute solution times the volume of the dilute
solution. Next, substitute the known values into the equation as shown in the next line.
Then rearrange the equation to solve for the unknown value. In this case that is done by
dividing both sides of the equation by 1000. ml. Last, carry out the necessary calculations
to get, in this case, .020 M. |
| A chemist starts with 50.0 mL of a 0.40 M NaCl solution and
dilutes it to 1000. mL. What is the concentration of NaCl in the new solution? |
(Mcon) (Vcon) = (Mdil)
(Vdil) |
(0.40 M) (50.0 mL) = (Mdil)
(1000. mL) |
(0.40 M) (50.0 mL)
(1000. mL) |
= (Mdil) |
|
0.020 M = Mdil |
|
Calculating Initial Volume (Ex. 6b)
| The question in this example is different only in that you are asked to
determine a volume instead of a concentration. The process used to answer the question is
the same as in the previous example. Write down the algebraic equation that represents the
relationship. Rearrange the equation to solve for the unknown quantity. Substitute numbers
(and units) for the known values. (The second and third steps can be reversed if you
wish.) Calculate the unknown value (which now becomes known). |
| A chemist wants to make 500. mL of 0.050 M HCl by diluting a 6.0 M
HCl solution. How much of that solution should be used? |
(Mcon) (Vcon) = (Mdil)
(Vdil) |
(Vcon) = |
(Mdil) (Vdil)
--------------
(Mcon) |
|
(Vcon) = |
(0.050 M)(500.mL)
----------------------
(6.0 M) |
|
Vcon = 4.2 mL |
|
If you have any questions about how to do these kinds of problems, please ask the
instructor.
Practice (Ex. 7)
Now you should practice working on dilution problems by answering the following
questions (from exercise 7 in your workbook). Do those now and check your answers before
you continue.
Dilution Calculations: Practice
 | How much 2.0 M NaCl solution would you need to make 250 mL of 0.15 M NaCl
solution? |
| What would be the concentration of a solution made by diluting 45.0 mL of 4.2 M
KOH to 250 mL? |
| What would be the concentration of a solution made by adding 250 mL of water to 45.0 mL
of 4.2 M KOH? |
| How much 0.20 M glucose solution can be made from 50. mL of 0.50 M glucose
solution? |
Answers (Ex.7)
Here are the answers to the questions in exercise 7.
| How much 2.0 M NaCl solution would you need to make 250 mL of 0.15 M NaCl
solution? 19 mL (2 s.d. from 18.75 mL) |
| What would be the concentration of a solution made by diluting 45.0 mL of 4.2 M
KOH to 250 mL? 0.76 M (2 s.d.) |
| What would be the concentration of a solution made by adding 250 mL of water to 45.0 mL
of 4.2 M KOH? 0.64 M |
| How much 0.20 M glucose solution can be made from 50. mL of 0.50 M glucose
solution? 130 mL (2 s.d. from 125 mL) |
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Eden Francis
Clackamas Community College
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