10 P0.10S
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P0.10S
Student skills
Practical support 10 Concentrations and
dilutions
Concentrations
The concentration of a solution is the amount of the dissolved substance (solute) per
unit volume of solvent liquid. The SI unit for concentration is mol m–3; however, solutions
are usually given as either molar concentrations, i.e. mol dm–3 (this is the same as mol l–1,
mol per litre and M), or mass concentrations, i.e. g dm–3 or g l–1.
Making a mol dm–3 solution
To make a 1 mol dm–3 (1 M) solution of, for example, potassium hydroxide (KOH), you
first need to work out the molar mass of potassium hydroxide. For this you need the
formula of potassium hydroxide and the relative atomic masses of potassium, oxygen
and hydrogen:
relative atomic mass of K is
39.10
relative atomic mass of O is
16.00
relative atomic mass of H is
1.01
The molar mass of KOH is therefore 39.10 16.00 1.01 56.11 g mol–1.
You need to dissolve 56.11 g of KOH in distilled water to give a solution of 1 mol dm–3.
The correct way to do this is to put at least half a litre of cold distilled water into a large beaker
and stand the beaker in a sink, in case of breakage. Then, wearing eye protection, add the solid
potassium hydroxide a little at a time whilst stirring with a long glass rod. Wait for the solid
to completely dissolve before adding the next amount. When it has all dissolved transfer the solution
to a 1 dm-3 volumetric flask. Then put some more distilled water in the beaker, swirl it round
and pour it into the volumetric flask. This rinses out any potassium hydroxide remaining in the beaker.
Repeat the rinsing process twice more, not forgetting to rinse the glass rod used for stirring. Then
fill the volumetric flask to just short of the 1 dm-3 mark before carefully making the solution to
the correct volume using a dropping pipette. Note the 1 M solution of potassium hydroxide
will be CORROSIVE.
To work out the mass of potassium hydroxide needed to prepare 100 cm3 of 2 mol dm–3
KOH, first work out:
amount of substance in moles volume of solution required (dm3)
required concentration (mol dm–3)
0.1* dm3 2 mol dm–3
0.2 moles
Then convert the number of moles to the mass needed in grams:
mass of substance (g) amount of substance in moles
molar mass (g mol–1)
0.2 moles 56.11 g mol–1
11.22 g
*Volume of solution required in cm3 divided by 1000 (1000 cm3 1 dm3), in this case
100 cm3 1000 0.1 dm3.
Salters-Nuffield Advanced Biology, Harcourt Education Ltd 2005. ©University of York Science Education Group.
This sheet may have been altered from the original.
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10 P0.10S
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P0.10S
Practical support 10 Concentrations and
dilutions
Student skills
Q1 Work out the mass of potassium chloride needed to prepare 200 cm3 of 0.5 mol dm3 KCl.
(Note the relative atomic masses are potassium 39.10 and chlorine 35.45.)
Q2 Fill in the gaps in the instructions below for making 1 litre (1 dm3) of 0.1 mol dm3 sodium
hydroxide. (The relative atomic mass of sodium is 22.99.)
The formula for sodium hydroxide is NaOH. The molar mass is
solution weigh out
g and make this up to
. To make a 0.1 mol dm3
dm3 using distilled water.
Making a w/v solution
W/v means weight to volume. To make a 10% w/v aqueous sodium chloride solution
you dissolve 10 g of sodium chloride in 100 cm3 of water or 20 g in 200 cm3, etc. If only
50 cm3 of the solution were required you would only require 5 g of sodium chloride in
50 cm3 of water. To make a litre of 20% w/v aqueous sodium chloride solution would
require 200 g in 1000 cm3.
Q3 Work out the mass of sodium chloride required to make 200 cm3 of 15% w/v sodium
chloride.
Q4 Work out the mass of potassium chloride needed to make 100 cm3 of a 1% w/v solution.
Dilutions
It is often necessary to make dilutions of a stock solution for an experiment. For
example, suppose you are supplied with a bottle of 1 mol dm3 hydrochloric acid but
the experiment requires 100 cm3 of 0.1 mol dm3 hydrochloric acid. You will need to
dilute the 1 mol dm–3 hydrochloric acid. What precisely do you do? You can use the
formula below to work out the volume of stock solution that needs to be diluted.
volume of stock solution
to be diluted
final volume of diluted solution
concentration of diluted solution
concentration of stock solution
For the example above:
100 cm3 0.1 mol dm3
volume of 1 mol dm3 hydrochloric
3
acid to be diluted (cm )
1 mol dm3
10 cm3
Therefore 10 cm3 of the 1 mol dm–3 hydrochloric acid is diluted with water to give
100 cm3 of 0.1 mol dm–3 hydrochloric acid.
Q5 Complete the following sentence:
cm3 of
To prepare 500 cm3 of 0.5 mol dm3 copper sulphate solution you take
–3
1 mol dm copper sulphate solution and dilute with water to give a final volume of 500 cm3.
Q6 What volume of 1.0 mol dm3 copper sulphate solution must be diluted to give 250 cm3 of
0.2 mol dm3 copper sulphate solution?
Salters-Nuffield Advanced Biology, Harcourt Education Ltd 2005. ©University of York Science Education Group.
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Practical support 10 Concentrations and
dilutions
Student skills
General notes on dilutions
1
First work out the relative amounts of stock solution and solvent that you need.
2
Measure the required amount of stock solution into a volumetric flask.
3
Add the required amount of solvent to bring the level up to the calibration mark.
(NB: It is best to stop before you get to the calibration mark and add the final
amounts with a pipette drop by drop. You can’t take excess solvent out again!)
4
Mix the solution thoroughly.
Alternatively, if volumetric flasks are not available, work out the quantities of solvent and
stock solution, measure them accurately and then transfer them to a test tube, beaker or
conical flask. Mix thoroughly.
Dilution series
Serial dilution is commonly used to give a range of concentrations, for example when
testing the effects of concentration of substrates or enzymes on rates of reaction, and in
microbiology. The following are common series. In each case it is assumed that you are
making up 10 cm3 of solution.
Linear dilution
In some cases you will want to have a linear series of concentrations running, for
example 1 M, 0.8 M, 0.6 M, 0.4 M and 0.2 M.
In this case, assuming you start with a 1 M solution and want to make 10 cm3 of solution
at each dilution, use the volumes shown in Table 1.
Table 1 Linear dilution of stock solution.
Desired concentration/M
Volume of stock solution/cm3
Volume of solvent/cm3
1
0.8
0.6
0.4
0.2
0
10
8
6
4
2
0
0
2
4
6
8
10
Serial dilutions
Doubling dilutions
A doubling dilution is a serial dilution that halves the concentration at each step in the
series (e.g. to get the sequence 1 M, 0.5 M, 0.25 M, 0.125 M).
To make your 1st dilution (half the strength of the stock solution) take 5 cm3 of the stock
solution and 5 cm3 of the solvent and mix them thoroughly.
To make your 2nd dilution (a quarter the strength of the stock solution) take 5 cm3 of
your 1st dilution and 5 cm3 of the solvent and mix them thoroughly.
To make your 3rd dilution (an eighth the strength of the stock solution) take 5 cm3 of
your 2nd dilution and the solvent and mix them thoroughly.
Repeat as many times as needed, each time using equal volumes of the last dilution and
the stock solution.
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This sheet may have been altered from the original.
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Practical support 10 Concentrations and
dilutions
1 cm3
1 cm3
1 cm3
1 cm3
1 cm3
Student skills
1 cm3
9 cm3 of distilled
water in each tube
undiluted
solution
1
10
102
103
104
105
106
Concentration relative to the undiluted solution
Figure 1 Making a logarithmic dilution. Using a clean pipette each time, 1 cm3 of solution is transferred to the
next tube and thoroughly mixed.
Logarithmic dilutions
These dilute by 10-fold with each step in the dilution series (Figure 1). They are
otherwise very like the doubling dilutions.
To make the 1st dilution (1/10 strength of the stock solution) mix 1 cm3 stock solution
and 9 cm3 solvent.
To make the 2nd dilution (1/100 strength of the stock solution) mix 1 cm3 of the 1st
dilution solution and 9 cm3 solvent.
To make the 3rd dilution (1/1000 strength of the stock solution) mix 1 cm3 of the 2nd
dilution solution and 9 cm3 solvent.
Safety note
When diluting acids, always add the acid to water – not the other way around.
Answers
Q1 7.46 g
Q2 40, 4.00 g, 1 dm3
Q3 30 g
Q4 1 g
Q5 250 cm3
Q6 50 cm3
Salters-Nuffield Advanced Biology, Harcourt Education Ltd 2005. ©University of York Science Education Group.
This sheet may have been altered from the original.
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