and the concentration of solutions
See also 14.3
dilution of solutions calculations
the terms solubility, concentration, strength and molarity
Why are the terms 'concentration', 'strength' and 'molarity' important?
Quite a lot of analytical
procedures in chemistry involve the use of solutions of accurately known
concentration e.g. standard solutions for various analytical purposes
If you want to
analyse an acid solution you need to titrate it with a standard solution of
alkali of accurately known concentration e.g. an accurately known
molarity (concentration usually expressed in mol/dm3,
lots more on this on the rest of this page!).
The solubility of a
substance is the maximum amount
of solute that dissolves in a given volume of solvent.
It is important
to know the solubility of substance in various liquids, quite often quoted
maximum solubility of salts in water, but often quoted, not as molarity, but
in g salt /100 g of water and plotted in graphs known as solubility
the maximum concentration possible for a given solute and solvent.
For more on solubility see
of compounds, salt solubility and water of crystallisation
There are differences in
using the words ‘concentration’ and ‘strength’ in science
language concentration is a specifically defined term e.g.
of solute per
unit volume of solvent e.g. g/dm3,
g/cm3 (g dm-3, g cm-3),
(ii) moles per unit volume of solvent e.g.
mol/dm3 (mol dm-3),
in terms of solution concentration is not a scientifically defined term and tends to be
used in everyday language to 'crudely' indicate a concentration e.g.
'great/high strength' indicating a very concentrated solution, and
conversely, 'low/weak strength' to indicate a low concentration solution.
Unfortunately this every use of the term is widespread, so take care,
because it does not apply to the science of chemistry!
In chemistry, for solutions, the
word 'strength' in chemistry is applied to e.g. an acid to indicate
how much it ionises
in aqueous solution - is it a weak or strong acid or alkali (soluble base).
A low strength
solution of an acid indicates it is a weak acid and only ionises a few %
to give hydrogen ions.
The equilibrium is about 2% to the right,
indicating a very weak acid.
An acid of high
strength indicates that it ionises to a very percent to form hydrogen ions
i.e a strong acid.
HCl(g) + aq ===> H+(aq) + Cl–(aq)
When you dissolve hydrogen chloride in
water (aq) you get virtually 100% ionisation into the hydrogen ion and chloride
ion indicating a very strong acid.
both cases the term
applies to the concentration of the original acid molecules:
For more details
More on acid-base theory and weak and strong
acids and their properties
Revise section 7. moles and
mass before proceeding in this section 11 and eventually you may need to be
familiar with the use of the apparatus illustrated above, some of which give
great accuracy when dealing with solutions and some do not.
It is very useful to be know exactly how much of a dissolved
substance is present in a solution of particular concentration or volume of a
- So we need a standard way of comparing the concentrations of
solutions in some standard units.
- The more you dissolve in a given volume of
solvent, or the smaller the volume you dissolve a given amount of solute in,
the more concentrated the solution.
Note: A standard solution is one whose precise concentration is known
Reminders: The dissolved substance is
called the solute and the liquid dissolving it is the
solvent and the result is a solution.
- The more of the substance you dissolve in the same
volume of liquid, the
more concentrated the solution, the solute particles on average are closer
- The diagrams represent two substances dissolved in the
solvent, the right-hand diagram represents a more concentrated solution e.g.
a mixture of two salts in water.
- The pictures do not mean the right hand one is a
- Unfortunately, in everyday language, it would be
described as such, but this is science and the correct use of scientific
language is essential!
TOP OF PAGE
(b) Measures of concentration and simple calculations of
(b)(i) Concentration in terms of mass of
solute per unit volume of solution
A summary of how to do basic
calculations and rearrangement of the solution concentration formula
- We will look at moles in (b)(ii)
- The simplest measure of concentration is mass of
per unit volume of solvent e.g.
- concentration = mass of solute / volume of
- Take 5.0 g of salt dissolved in 500 cm3
- The concentration can be expressed in several ways.
- concentration = 5.0/500 = 0.01 g/cm3
- 1 dm3 = 1000 cm3, so 500 cm3
= 500/1000 = 0.50 dm3
- concentration = 5.0/0.50 =
- For interconversion: g/cm3 x 1000 = g/dm3
AND g/dm3/1000 = g/dm3
- Sometimes the general formula c = m/v is used
- c = concentration, m = mass, v = volume
- rearrangements: m = c x v and
Example questions (not using moles)
What is the concentration in g/dm3 if 6.0 g of salt is dissolved
in 150 cm3 of water?
- 150 / 1000 = 0.15 dm3
concentration = mass / volume = 6.0 / 0.15 =
Given a salt solution of concentration 16 g/dm3, what mass of
salt is in 40 cm3 of the solution?
- 1 dm3 = 1000 cm3
- c = m / v = 16 / 1000 = 0.16 g/cm3
(note this a way of converting g/dm3 to g/cm3)
of salt: m = c x v = 0.16 x 40 =
6.4 g of
Given 5.0 g of a salt, what volume of water in cm3, should it be
dissolved in to give a solution of concentration of 12.5 g/dm3?
- c = m / v, rearranging gives v = m / c
- v = 5.0 / 12.5 = 0.40 dm3
of water needed = 1000 x 0.40 =
- Its also good to be able to do dilution'
in section 14.3
dilution of solutions
TOP OF PAGE
Concentration in terms of moles of solute per unit volume of solution
A summary of how to do basic molarity
calculations and rearrangement of the molarity formula
most analytical and calculation purposes the concentration of an aqueous solution is usually
expressed in terms of moles of dissolved substance per cubic decimetre of
(reminder mole formula triangle on the right). 1 cubic decimetre (dm3)
= 1 litre (l) in old money!
- concentration =
molarity = moles of solute /
volume of solvent in dm3 (litres)
- Make sure you know how to calculate moles, see
the triangle on the right!
- Using concentration units of mol
dm-3 (or mol/dm3), the concentration is called molarity, sometimes denoted in
shorthand as M (old money again, take care!) and the word molar
is used too.
- Note: 1dm3 = 1 litre = 1000ml = 1000 cm3, so dividing
cm3/1000 gives dm3, which is handy to know since
most volumetric laboratory apparatus is calibrated in cm3 (or ml),
but solution concentrations are usually quoted in molarity, that is mol/dm3
- Concentration is also expressed in a
'non-molar' format of mass per volume e.g. g/dm3
- You need to know all about moles to proceed
further on this page and get into 'molarity' ...
- ... so read
section 7. on moles and
mass - essential pre-reading for section 11 ...
- AND, if you can't understand molarity, you cannot
do titration calculations either!
- Equal volumes of solution of the
same molar concentration contain the same number of moles of solute i.e.
the same number of particles as given by the chemical formula you use in
defining a specific molarity.
- A note on solutions of ionic compounds e.g.
- A 1.0 molar solution of magnesium chloride MgCl2(aq),
contains 1.0 mol/dm3 of magnesium ions (Mg2+), BUT 2.0
mol/dm3 in terms of the chloride ion (Cl-)
- You need to be able to calculate
- the number of moles or mass of substance in an aqueous
solution of given volume and concentration
- the concentration of an aqueous solution given the
amount of substance and volume of water, for this you use the equation
.... (reminder molarity formula triangle on the right), so, for a
substance Z ...
molarity (concentration) of Z
= moles of Z / volume in dm3
- This is sometimes referred to as the molar concentration
- and you need to be able to rearrange
this equation ... therefore ...
- (1b) moles =
molarity (concentration) x volume in dm3 and
volume in dm3
= moles / molarity (concentration)
- You can use the triangle on the right to help you
rearrange the equation for the basic definition of molarity, BUT it
is much better to know how to rearrange the equation:
molarity = moles ÷ volume(dm3)
- You may also need to know that ...
molarity x formula mass of
solute = solute concentration in g/dm3
- This is sometimes referred to as the mass-concentration,
- and dividing this by 1000 gives
the concentration in g/cm3, and
concentration in g/dm3
/ formula mass = molarity in mol/dm3
- both equations (2) and (3)
result from equations (1) and (4), work it out for yourself.
- and to sum up, by now you should
- (4) moles Z
= mass Z / formula mass of Z
- (5) 1 mole = formula mass in
- (6) molarity = moles/dm3
Molarity calculation Example 11.1
If 5.00g of sodium chloride is
dissolved in exactly 250 cm3 of water in a calibrated volumetric flask,
(a) what is the concentration in g/dm3?
(b) What is the
molarity of the solution?
Ar(Na) = 23, Ar(Cl)
= 35.5, so Mr(NaCl) = 23 + 35.5 = 58.5
mole NaCl = 5.0/58.5 = 0.08547
volume = 250/1000 = 0.25 dm3
molarity = mol of solute / volume
Molarity = 0.08547/0.25 =
calculation Example 11.2
potassium bromide was dissolved in 400cm3 of water.
its molarity. [Ar's: K = 39, Br = 80]
moles = mass / formula
mass, (KBr = 39 + 80 = 119)
mol KBr = 5.95/119 = 0.050
400 cm3 =
400/1000 = 0.400 dm3
molarity = moles of
solute / volume of solution
molarity of KBr
solution = 0.050/0.400 =
(b) What is the concentration in grams
concentration = mass / volume, the volume
= 400 / 1000 = 0.4 dm3
concentration = 5.95 / 0.4 =
calculation Example 11.3
What mass of
sodium hydroxide (NaOH) is needed to make up 500 cm3 (0.500 dm3)
of a 0.500 mol dm-3
(0.5M) solution? [Ar's: Na = 23, O = 16, H = 1]
1 mole of NaOH = 23 + 16 + 1 = 40g
molarity = moles / volume,
so mol needed = molarity x volume in dm3
500 cm3 = 500/1000 =
mol NaOH needed = 0.500 x 0.500 =
0.250 mol NaOH
therefore mass = mol x formula
= 0.25 x 40 =
calculation Example 11.4
many moles of H2SO4 are there in 250 cm3 of
a 0.800 mol dm-3
(0.8M) sulphuric acid solution?
What mass of acid is in this solution?
H = 1, S = 32, O = 16]
(a) molarity = moles /
volume in dm3, rearranging equation for the sulfuric acid
(b) mass = moles x formula
11.5 This involves calculating concentration in other ways e.g. mass/volume units
is the concentration of sodium chloride (NaCl) in g/dm3 and g/cm3
in a 1.50 molar solution?
At. masses: Na = 23, Cl
= 35.5, formula mass NaCl = 23 + 35.5 = 58.5
since mass = mol x formula mass,
for 1 dm3
concentration = 1.5 x 58.5 = 87.8 g/dm3, and
87.75 / 1000 =
A solution of calcium
sulphate (CaSO4) contained 0.500g dissolved in 2.00 dm3 of water.
Calculate the concentration in (a) g/dm3, (b) g/cm3
and (c) mol/dm3.
(a) concentration = 0.500/2.00
0.250 g/dm3, then since 1dm3
= 1000 cm3
(b) concentration = 0.250/1000
0.00025 g/cm3 (or from 0.500/2000)
(c) At. masses:
Ca = 40, S = 32, O = 64, formula mass CaSO4 = 40 + 32 + (4 x 16) = 136
Molarity calculation Example
- Its also good to be able to do dilution'
in section 14.3
dilution of solutions
There are more questions
involving molarity in section 12. on
section 14.3 on
dilution calculations and
TOP OF PAGE
(c) APPENDIX 1 on SOLUBILITY
and concentration calculations
How do you find out how soluble
a substance is in water?
Reminder: solute + solvent ==> solution
i.e. the solute is what dissolves, the solvent is what dissolves it and the resulting homogeneous mixture is the
The solubility of a substance is the maximum
amount of it that will dissolve in a given volume of solvent e.g. water.
The resulting solution is known as a saturated
solution, because no more solute will dissolve in the solvent.
Solubility can be measured and expressed in with
different concentration units e.g. g/100cm3, g/dm3
and molarity (mol/dm3).
Solubility can also be expressed as mass of
solute per mass of water e.g. g/100g of water.
You can determine solubility by titration if the
solute reacts with a suitable reagent e.g. acid - alkali titration and it
is especially suitable for substances of quite low solubility in water e.g.
calcium hydroxide solution (alkaline limewater) can be titrated with standard
hydrochloric acid solution.
However, many substances like salts are very
soluble in water and a simple evaporation method will do which is described below
e.g. for a thermally stable salt like sodium chloride.
(1) A saturated solution is prepared by mixing
the salt with 25cm3 of water until no more dissolves at room
(2) The solution is filtered to make sure no
undissolved salt crystals contaminate the saturated solution.
(3) Next, an evaporating dish (basin) is
accurately weighed. Then, accurately pipette 10 cm3 of the saturated
salt solution into the basin and reweigh the dish and contents.
By using a pipette, its possible to express
the solubility in two different units.
(4) The basin and solution are carefully heated
to evaporate the water.
(5) When you seem to have dry salt crystals, you
let the basin cool and reweigh it.
(6) The basin is then gently heated again and
then cooled and weighed again.
This is repeated until the weight of the dish and
salt is constant, proving that all the water is evaporated
By subtracting the original weight of the
dish from the final weight you get the mass of salt dissolved in the volume
or mass of saturated salt solution you started with.
You can repeat the experiment to obtain a
more accurate and reliable result.
By using a pipette it is possible to
calculate the solubility in two ways, expressed as two quite different
Suppose the dish weighed 95.6g.
With the 10.0 cm3 of salt
solution in weighed 107.7g
After evaporation of the water the dish
Mass of 10.0 cm3 salt solution = 107.7 - 95.6 =
Mass of salt in 10 cm3
of salt solution = 96.5 - 95.6 = 0.9g
Mass of water evaporated = 107.7 -
96.5 = 11.2g
(a) Expressing the solubility in grams salt
per 100 g of water
From the mass data above 0.9g of salt
dissolved in 11.2g of water
Therefore X g of salt dissolves in 100g of
water, X = 100 x 0.9 / 11.2 = 8.0
Therefore the solubility of
the salt = 8.0g/100g water
You can scale this up to 80.0g/1000g H2O,
or calculate how much salt would dissolve in any given mass of water.
You can also express the solubility as g
salt/100g of solution.
0.9g salt is dissolved in 12.1g of
solution, X g in 100g of solution
Therefore X = 100 x 0.9 / 12.1 = 7.4, so
solubility = 7.4g/100g solution
These calculations do not require
the original salt solution to be pipetted. You can just measure
out approximately 10cm3 of the salt solution with
10cm3 measuring cylinder, and do the experiment and
these calculations in the exactly the same way.
(b) However, if you know the exact volume of
salt solution and the mass dissolved in it, then you can calculate the
concentration in g/dm3, and if you know the formula mass of the
salt, you can calculate the molarity of the solution.
From part (a) we have 0.9g of salt in
Therefore X g will dissolve in 1000cm3
solution, X = 1000 x 0.9 / 10 = 90g/1000 cm3
Solubility of salt = 90g/dm3
Suppose the formula mass of the salt was
200, calculate the molarity of the saturated solution.
moles salt = mass / formula mass = 90/200
= 0.45 moles
Therefore solubility of saturated
salt solution in terms of molarity = 0.45 mol/dm3
NOTE Solubility varies with temperature,
Gas and salt solubility
in water and solubility curves, and it usually (but not always)
increases with increase in temperature. So, in the experiment described
above, the temperature of the saturated solution should be noted, or perhaps
controlled to be saturated at 20oC or 25oC.
TOP OF PAGE
2 - How to make up a standard solution - a solution of precisely known
The method and procedure of how to make up a
standard solution of a soluble solid e.g. a salt, is fully described.
making up a standard solution of known molarity
The method and procedure of how to make up a
standard solution of a soluble solid e.g. a salt, is fully described
Suppose you want to make up 250 cm3 of a salt
solution of concentration 20g/dm3 (20g/litre, 20g/1000cm3,
c = m / v, m = c x v, m = 20 x 250 / 1000 = 5g
so 5g of the salt is needed to be made up into an
aqueous solution of exactly 250.0 cm3.
described in detail after example 2. below.
To prepare a solution of known molarity, you need to work backwards from the
volume required and the molarity to see how much solid you need.
Suppose you want to make up 250cm3 of a sodium chloride solution of concentration
formula (on the right): moles = molarity (mol/dm3) x
and volume in cm3 / 1000 = dm3
moles NaCl needed = 0.20 x 250/1000 = 0.20 x 0.25 = 0.05 mol NaCl
Atomic masses: Na =23 and Cl = 35.5, so molar mass of NaCl = 23 + 35.5 = 58.5
From basic mole formula: mass of NaCl needed = mol NaCl x formula mass NaCl
mass of NaCl needed = 0.05 x 58.5 = 2.925 g (which is ok if you
have a 3 decimal place balance!), so
2.295g of pure NaCl salt is needed to made up 250.0 cm3 of
solution with a precise concentration of 0.20 mol/dm3.
Procedure to make the standard solution i.e. one of known
concentration of solid (in this case)
accurate one pan electronic balanced is set to zero (preferably with an accuracy
of two decimal places). A beaker is placed on the balance and the reading noted
(ignore the figures on the diagram).
carefully, with a spatula (not shown), salt crystals are added to the beaker
until it weighs exactly 2.925 grams more than the beaker. This can be a very fiddly
procedure if you want exactly 2.925g of salt.
water (distilled/deionised) is then added to the beaker to completely dissolve the salt and use of a
stirring rod helps to speed up the process.
The amount of water you add to the
beaker should be much less than 250cm3 to allow for the transfer and
rinsing of the solution into the standard volumetric flask using a
'squeezy' wash bottle!
a clear solution of the salt should be seen, there should be no residual salt
crystals at the bottom of the beaker or on the sides of the beaker.
You can use
the wash bottle to rinse down any crystals on the side of the beaker, but watch
the volume you use..
accurately calibrated 250cm3 volumetric flask should be washed out
and cleaned several times with pure water.
Then, the whole of the solution in
the beaker is transferred into the flask with the help of a funnel to avoid the
risk of spillage.
To make sure every drop of the salt solution ends up in the
flask, a wash bottle of pure water is used to rinse out the beaker several
times, AND rinse the stirring rod and the funnel too.
This is to ensure nothing
is lost in the transfer fro beaker to flask.
very carefully, the flask is topped up with pure water so the meniscus rests
exactly on the 250.0cm3 calibration mark, a teat pipette is useful
for the last few drops of water.
The stopper is placed on and the flask carefully
shaken quite a few times to ensure the salt solution is completely mixed up.
Finally, check the meniscus lies on the calibration mark, in case another
few drops are needed.
Either way, the last drops of water should be added most
carefully with a teat pipette.
on standard solutions of acids and alkalis
You can purchase standard solutions ready for use.
OR, a phial of concentrated acid or alkali, which you
dilute into a specified volume to give a specific molarity.
Apart from weighing out a solid, the procedure is the
ensuring every drop from the phial is rinsed down the funnel into the
calibrated volumetric flask.
See dilution' calculations
in section 14.3
dilution of solutions
TOP OF PAGE
(e) Self-assessment Quizzes
on molarity calculations:
type in answer
QUIZ on molarity or
QUIZ on molarity
type in titration answer
(good revision for A level students)
GCE-AS-A2 acid-alkali titration calculation questions
Above is typical periodic table used in GCSE science-chemistry specifications in
doing molarity calculations,
and I've 'usually' used these values in my exemplar calculations to cover most
e.g. OTHER CALCULATION PAGES
What is relative atomic mass?,
relative isotopic mass and calculating relative atomic mass
formula/molecular mass of a compound or element molecule
Law of Conservation of Mass and simple reacting mass calculations
Composition by percentage mass of elements
in a compound
Empirical formula and formula mass of a compound from reacting masses
(easy start, not using moles)
Reacting mass ratio calculations of reactants and products
moles) and brief mention of actual percent % yield and theoretical yield,
and formula mass determination
Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations
(relating reacting masses and formula
moles to calculate empirical formula and deduce molecular formula of a compound/molecule
(starting with reacting masses or % composition)
Moles and the molar volume of a gas, Avogadro's Law
Reacting gas volume
ratios, Avogadro's Law
and Gay-Lussac's Law (ratio of gaseous
Molarity, volumes and solution
concentrations (and diagrams of apparatus)
How to do acid-alkali
titration calculations, diagrams of apparatus, details of procedures
Electrolysis products calculations (negative cathode and positive anode products)
e.g. % purity, % percentage & theoretical yield, dilution of solutions
(and diagrams of apparatus), water of crystallisation, quantity of reactants
required, atom economy
Energy transfers in physical/chemical changes,
Gas calculations involving PVT relationships,
Boyle's and Charles Laws
Radioactivity & half-life calculations including
keywords and phrases:
revision study notes for AQA Edexcel OCR IGCSE/GCSE
chemistry science topics modules on explaining the terms
solubility, concentration, strength and molarity definition, explaining
different measures of
concentration and simple calculations of molarity of solutions of
substances, explaining the difference between
strength and concentration, explaining molarity, how to calculate
molarity units molar concentration of solutions practice questions on
molarity apparatus and method on how to make up a standard solution, how
to questions on molarity practice chemical calculations for IB chemistry
AQA Edexcel OCR Salters advanced A level chemistry question exercise on
calculating molarities molarity of solution
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