DILUTION CALCULATIONS
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14.
Other GCSE chemical calculations  e.g. working out dilutions needed or the
resulting concentration from diluting a stock solution
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Keywords: 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
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See also
14.1
% purity of a product and assay calculations
14.2a
% reaction yield and theoretical yield calculations
and why you can't actually get 100% yield in practice
14.2b atom economy calculations
*
14.4
water of crystallisation
calculations
14.5
how
much of a reactant is needed? calculation of quantities required, limiting
reactant quantities
Chemical &
Pharmaceutical Industry Economics & Sustainability, Life Cycle
Assessment, Recycling
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.

Dilution calculation Example 14.3.1

A purchased standard solution of sodium
hydroxide had a concentration of 1.0 mol/dm^{3}.

(a) 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,

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.

(b)
Given a standard concentrated solution of hydrochloric acid of concentration
2.0 mol/dm^{3}, how much of this solution in cm^{3}, is
needed to make up 250 cm^{3} of a more dilute standard solution of
concentration 0.20 mol/dm^{3}? Also, briefly describe the
procedure to make this solution.

molarity = moles / volume, mol = molarity
x volume, volume = moles / molarity

1st consider the solution to be made up:

2nd consider the original standard
solution:

volume = moles / molarity = 0.05 /
2.0 = 0.025 dm^{3}

Therefore, volume of original
standard solution needed is 0.025 x 1000 =
25.0 cm^{3}.

This would be measured out with a 50 ml
burette or better, a calibrated 25.0 cm^{3} pipette, and carefully
transferred into a 250 cm^{3} calibrated flask and topped up with
deionised water to the calibration mark.

For more details see steps
and
in section 11(d)
How
to make up a standard solution



Dilution calculation 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/dm^{3} 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

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!

For picture details see 11(d)
How to make up a
standard solution



Dilution calculation
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

Dilution calculation 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?

(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.



Dilution calculation 14.3.5

Oleic has the structural formula CH_{3}(CH_{2})_{7}CH=CH(CH_{2})_{7}COOH,
molecular formula C_{18}H_{34}O_{2}

It is not very soluble in water, and
solutions of it are often prepared using an organic solvent like petroleum
ether.

(a) Calculate the molecular mass of oleic
acid (Relative atomic masses C = 12, H = 1, O =16)

(b) A stock solution of oleic acid in a
solvent has a molarity of 4.0 x 10^{4} mol dm^{3}.

2 cm^{3} of this solution is
added to 8 cm^{3} of the solvent to make a more dilute solution.

1 cm^{3} of this diluted solution
is further diluted and mixed with 9.0 cm^{3} of the same solvent.

What is the final molarity of the doubly
diluted solution?

The first dilution is 2 ==> 10 (or 1 ==>
5 by ratio)

so concentration = 4.0 x 10^{4 }
/ 5 = 0.8 x 10^{4} = 8.0 x 10^{5} mol dm^{3}

The 2nd dilution is 1 ==> 10 by ratio

so concentration = 8.0 x 10^{5}
/ 10 = 8.0 x 10^{6}
mol dm^{3} (final molarity)



(c) What mass of oleic acid would be in
5.0 cm^{3} of this final diluted solution?

Molarity is mol per dm3 (litre) or mol
per 1000 cm^{3} (ml)

Therefore in each cm^{3} of the
final dilution there are 8.0 x 10^{6} / 1000 = 8.0 x 10^{9}
mol

Therefore in 5.0 cm3 there are 5 x 8.0 x
10^{9} = 40.0 x 10^{9} = 4.0 x 10^{8} mol

mol = mass (g) / M_{r},
so mass = mol x M_{r},

Therefore mass of oleic acid in the 5.0
cm^{3} of solution
See also
14.1
% purity of a product and assay calculations
14.2a
% reaction yield and theoretical yield calculations
and why you can't actually get 100% yield in practice
14.2b atom economy calculations
*
14.4
water of crystallisation
calculations
14.5
how
much of a reactant is needed? calculation of quantities required, limiting
reactant quantities
Chemical &
Pharmaceutical Industry Economics & Sustainability, Life Cycle
Assessment, Recycling
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