DILUTION CALCULATIONS
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Brown's Chemistry  GCSE/IGCSE/GCE (basic A level)
O Level Online Chemical Calculations
14.
Other GCSE chemical calculations  SOLUTION DILUTION CALCULATIONS
Quantitative chemistry calculations Diluting solutions. Help for problem solving dilution calculations. How do you do solution dilution
calculations? Using dilution factors to solve concentration problems 
fully worked out example calculations for diluting solutions. Online
practice exam chemistry CALCULATIONS and solved problems for KS4 Science
GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level
AS/A2/IB courses. These revision notes and practice questions on
how to do solution dilution chemical calculations and worked examples
should prove useful for the new AQA, Edexcel and OCR GCSE (9–1)
chemistry science courses.
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14.
Other GCSE chemical calculations
14.1
% purity of a product * 14.2a
% reaction yield * 14.2b
atom economy * 14.3
dilution of solutions
14.4
water of crystallisation
calculation * 14.5
how
much of a reactant is needed?
14.3
Dilution of solutions calculations
calculating
dilutions  volumes involved etc.

In conjunction with this page it
be important to study

It is important to know how to
accurately dilute a more concentrated solution to a specified solution of
lower concentration. It involves a bit of logic using ratios of volumes. The
diagram above illustrates some of the apparatus that might be used when
dealing with solutions.

Example 14.3.1

A purchased standard solution of sodium
hydroxide had a concentration of 1.0 mol/dm^{3}. How would you prepare 100 cm^{3} of
a 0.1 mol/dm^{3} solution to do a titration of an acid?

The required
concentration is 1/10th of the original solution.

To make 1dm^{3} (1000 cm^{3}) of
the diluted solution you would take 100 cm^{3} of the original solution and mix
with 900 cm^{3} of water.

The total volume is 1dm^{3} but only 1/10th as much sodium
hydroxide in this diluted solution, so the concentration is 1/10th, 0.1
mol/dm^{3}.

To make only 100 cm^{3} of the diluted solution you would dilute 10cm^{3}
by mixing it with 90 cm^{3} of water.

How to do this in practice is described at
the end of Example 14.3.2 below and a variety of accurate/'less inaccurate' apparatus
is illustrated above.

Example 14.3.2

Given a stock solution of sodium chloride
of 2.0 mol/dm^{3}, how would you prepare 250cm^{3} of a 0.5 mol/dm^{3} solution?

The
required 0.5 mol/dm^{3} concentration is 1/4 of the original concentration of 2.0
mol/dm^{3}.

To make 1dm^{3} (1000 cm^{3}) of a 0.5 mol/dm3 solution you would take 250
cm^{3} of the stock solution and add 750 cm^{3} of water.

Therefore to make only 250
cm^{3} of solution you would mix 1/4 of the above quantities i.e. mix 62.5 cm^{3} of
the stock solution plus 187.5 cm^{3} of pure water.

This can be done, but rather
inaccurately, using measuring cylinders and stirring to mix the two liquids in
a beaker.

It can be done much more accurately by using a burette or a pipette
to measure out the stock solution directly into a 250 cm^{3} graduatedvolumetric
flask.

Topping up the flask to the calibration mark (meniscus should rest on
it). Then putting on the stopper and thoroughly mixing it by carefully shaking
the flask holding the stopper on at the same time!

Example
14.3.3

In
the analytical laboratory of a pharmaceutical company a laboratory assistant
was asked to make 250 cm^{3} of a 2.0 x 10^{2} mol dm^{3}
(0.02M)
solution of paracetamol (C_{8}H_{9}NO_{2}).

(a) How much
paracetamol should the laboratory assistant weigh out to make up the
solution?

Atomic masses: C = 12, H = 1, N = 14, O = 16

method (i): M_{r}(paracetamol)
= (8 x 12) + (9 x 1) + (1 x 14) + (2 x 16) = 151

1000 cm^{3} of
1.0 molar solution needs 151g

1000 cm^{3} of
2.0 x 10^{2} molar solution needs 151 x 2.0 x 10^{2}/1 =
3.02g

(this is just scaling
down the ratio from 151g : 1.0 molar)

Therefore to make 250
cm^{3} of the solution you need 3.02 x 250/1000 = 0.755 g

method (ii): M_{r}(paracetamol)
= 151

moles = molarity x
volume in dm^{3}

mol paracetamol required
= 2.0 x 10^{2} x 250/1000 = 5.0 x 10^{3} (0.005)

mass = mol x M_{r}
= 5.0 x 10^{3} x 151 = 0.755 g

(b) Using the 2.0 x
10^{2} molar stock solution, what volume of it should be added to a
100cm^{3} volumetric flask to make 100 cm^{3} of a 5.0 x 10^{3}
mol dm^{3} (0.005M) solution?

The ratio of the two
molarities is stock/diluted = 2.0 x 10^{2}/5.0 x 10^{3} =
4.0 or a dilution factor of 1/4 (0.02/0.005).

Therefore 25 cm^{3}
(^{1}/_{4} of 100) of the 2.0 x 10^{2} molar
solution is added to the 100 cm^{3} volumetric flask prior to
making it up to 100 cm^{3} with pure water to give the 5.0 x 10^{3}
mol dm^{3} (0.005M) solution.

There are more questions
involving molarity in
section 7.
introducing molarity and
section 12. on
dilution

Example 14.3.4

You are given a stock
solution of concentrated ammonia with a concentration of 17.9 mol dm^{3}
(conc. ammonia! ~18M!)

(a) What volume of the conc.
ammonia is needed to make up 1dm^{3} of 1.0 molar ammonia solution?

Method (i) using simple ratio
argument.

The conc. ammonia must be
diluted by a factor of 1.0/17.9 to give a 1.0 molar solution.

Therefore you need (1.0/17.9) x
1000 cm^{3} = 55.9 cm^{3} of the conc.
ammonia.

If the 55.9 cm^{3} of
conc. ammonia is diluted to 1000 cm^{3} (1 dm^{3}) you will
have a 1.0 mol dm^{3} (1M) solution.

Method (ii) using molar
concentration equation  a much better method that suits any kind
of dilution calculation involving molarity.

molarity = mol / volume (dm^{3}),
therefore mol = molarity x volume in dm^{3}

Therefore you need 1.0 x 1.0
= 1.0 moles of ammonia to make 1 dm^{3} of 1.0M dilute ammonia.

Volume = mol / molarity

Volume of conc. ammonia needed =
1.0 / 17.9 = 0.0559 dm^{3} (55.9
cm^{3}) of the conc. ammonia is required,

and, if this is diluted to 1 dm^{3},
it will give you a 1.0 mol dm^{3} dilute ammonia solution.

(b) What volume of conc.
ammonia is needed to make 5 dm^{3} of a 1.5 molar solution?

molarity = mol / volume (dm^{3}),
therefore mol = molarity x volume in dm^{3}

Therefore you need 1.5 x 5 =
7.5 moles of ammonia to make 5 dm^{3} of 1.5M dilute ammonia.

Volume (of conc. ammonia
needed) = mol / molarity

Volume of conc. ammonia needed =
7.5 / 17.9 = 0.419 dm^{3} (419 cm^{3})
of the conc. ammonia is required,

and, if this is diluted to 5 dm^{3},
it will give you a 1.5 mol dm^{3} dilute ammonia solution.


Above is typical periodic table used in GCSE sciencechemistry specifications in
doing chemical calculations,
and I've 'usually' used these values in my exemplar calculations to cover most
syllabuses
OTHER CALCULATION PAGES

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moles) and brief mention of actual percent % yield and theoretical yield,
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Introducing moles: The connection between moles, mass and formula mass  the basis of reacting mole ratio calculations
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and GayLussac's Law (ratio of gaseous
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