11.
Molarity, volumes
and the concentration of solutions
See also 14.3
dilution of solutions calculations
(a)
Explaining
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
including titrations.

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/dm^{3},
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
as the
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
curves.

This is
the maximum concentration possible for a given solute and solvent.


For more on solubility see
Important formulae
of compounds, salt solubility and water of crystallisation

Misconceptions

There are differences in
using the words ‘concentration’ and ‘strength’ in science
compared to
everyday language

In scientific
language concentration is a specifically defined term e.g.

(i) the
mass
of solute per
unit volume of solvent e.g. g/dm^{3},
g/cm^{3} (g dm^{3}, g cm^{3}),
OR,

(ii) moles per unit volume of solvent e.g.
mol/dm^{3} (mol dm^{3}),

'Strength'
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.

e.g.
ethanoic acid:
CH_{3}COOH_{(aq)}CH_{3}COO^{–}_{(aq)}
+ H^{+}_{(aq)}

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.

hydrochloric 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.

In
both cases the term
concentration
applies to the concentration of the original acid molecules:

For more details
see
More on acidbase 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
solution.
 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
together.

 The diagrams represent two substances dissolved in the
solvent, the righthand 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
stronger solution!
 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
and subindex
(b) Measures of concentration and simple calculations of
molarity
(b)(i) Concentration in terms of mass of
solute per unit volume of solution
A summary of how to do basic
concentration
calculations and rearrangement of the solution concentration formula
 We will look at moles in (b)(ii)
 The simplest measure of concentration is mass of
solute
per unit volume of solvent e.g.
 concentration = mass of solute / volume of
solvent
 Take 5.0 g of salt dissolved in 500 cm^{3}
of water.
 The concentration can be expressed in several ways.
 concentration = 5.0/500 = 0.01 g/cm^{3}
 1 dm^{3} = 1000 cm^{3}, so 500 cm^{3}
= 500/1000 = 0.50 dm^{3}
 concentration = 5.0/0.50 =
10.0 g/dm^{3}
 For interconversion: g/cm^{3} x 1000 = g/dm^{3}
AND g/dm^{3}/1000 = g/dm^{3}
 
 Sometimes the general formula c = m/v is used
 c = concentration, m = mass, v = volume
 rearrangements: m = c x v and
v
= m/c
 

Example questions (not using moles)
 Q1
What is the concentration in g/dm^{3} if 6.0 g of salt is dissolved
in 150 cm^{3} of water?
 150 / 1000 = 0.15 dm^{3}

concentration = mass / volume = 6.0 / 0.15 =
40.0 g/dm^{3}
 
 Q2
Given a salt solution of concentration 16 g/dm^{3}, what mass of
salt is in 40 cm^{3} of the solution?
 1 dm^{3} = 1000 cm^{3}
 c = m / v = 16 / 1000 = 0.16 g/cm^{3}
(note this a way of converting g/dm^{3} to g/cm^{3})
 Therefore
mass
of salt: m = c x v = 0.16 x 40 =
6.4 g of
salt
 
 Q3
Given 5.0 g of a salt, what volume of water in cm^{3}, should it be
dissolved in to give a solution of concentration of 12.5 g/dm^{3}?
 c = m / v, rearranging gives v = m / c
 v = 5.0 / 12.5 = 0.40 dm^{3}
 volume
of water needed = 1000 x 0.40 =
400 cm^{3}
 
 Its also good to be able to do dilution'
calculations
in section 14.3
dilution of solutions
TOP OF PAGE
and subindex
(b)(ii)
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

For
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
solution
(reminder mole formula triangle on the right). 1 cubic decimetre (dm^{3})
= 1 litre (l) in old money!
 concentration =
molarity = moles of solute /
volume of solvent in dm^{3} (litres)
 Make sure you know how to calculate moles, see
the triangle on the right!
 Using concentration units of mol
dm^{3} (or mol/dm^{3}), the concentration is called molarity, sometimes denoted in
shorthand as M (old money again, take care!) and the word molar
is used too.
 Note: 1dm^{3} = 1 litre = 1000ml = 1000 cm^{3}, so dividing
cm^{3}/1000 gives dm^{3}, which is handy to know since
most volumetric laboratory apparatus is calibrated in cm^{3} (or ml),
but solution concentrations are usually quoted in molarity, that is mol/dm^{3}
(mol/litre).
 Concentration is also expressed in a
'nonmolar' format of mass per volume e.g. g/dm^{3}
 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 prereading 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 MgCl_{2}(aq),
contains 1.0 mol/dm^{3} of magnesium ions (Mg^{2+}), BUT 2.0
mol/dm^{3} in terms of the chloride ion (Cl^{})
concentration.
 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 ...

(1a)
molarity (concentration) of Z
= moles of Z / volume in dm^{3}
 This is sometimes referred to as the molar concentration
(moleconcentration),
 and you need to be able to rearrange
this equation ... therefore ...
 (1b) moles =
molarity (concentration) x volume in dm^{3} and
...
 (1c)
volume in dm^{3}
= 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(dm^{3})
 You may also need to know that ...
 (2)
molarity x formula mass of
solute = solute concentration in g/dm^{3}
 This is sometimes referred to as the massconcentration,
 and dividing this by 1000 gives
the concentration in g/cm^{3}, and

(3)
concentration in g/dm^{3}
/ formula mass = molarity in mol/dm^{3}
 both equations (2) and (3)
result from equations (1) and (4), work it out for yourself.
 and to sum up, by now you should
know:
 (4) moles Z
= mass Z / formula mass of Z
 (5) 1 mole = formula mass in
grams
 (6) molarity = moles/dm^{3}

Molarity calculation Example 11.1

If 5.00g of sodium chloride is
dissolved in exactly 250 cm^{3} of water in a calibrated volumetric flask,

(a) what is the concentration in g/dm^{3}?

(b) What is the
molarity of the solution?

A_{r}(Na) = 23, A_{r}(Cl)
= 35.5, so M_{r}(NaCl) = 23 + 35.5 = 58.5

mole NaCl = 5.0/58.5 = 0.08547

volume = 250/1000 = 0.25 dm^{3}

molarity = mol of solute / volume
of solvent

Molarity = 0.08547/0.25 =
0.342 mol/dm^{3}



Molarity
calculation Example 11.2

5.95g of
potassium bromide was dissolved in 400cm^{3} of water.

(a)
Calculate
its molarity. [A_{r}'s: K = 39, Br = 80]

moles = mass / formula
mass, (KBr = 39 + 80 = 119)

mol KBr = 5.95/119 = 0.050
mol

400 cm^{3} =
400/1000 = 0.400 dm^{3}

molarity = moles of
solute / volume of solution

molarity of KBr
solution = 0.050/0.400 =
0.125 mol/dm^{3}



(b) What is the concentration in grams
per dm^{3}?

concentration = mass / volume, the volume
= 400 / 1000 = 0.4 dm^{3}

concentration = 5.95 / 0.4 =
14.9 g/dm^{3}



Molarity
calculation Example 11.3

What mass of
sodium hydroxide (NaOH) is needed to make up 500 cm^{3} (0.500 dm^{3})
of a 0.500 mol dm^{3}
(0.5M) solution? [A_{r}'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 dm^{3}

500 cm^{3} = 500/1000 =
0.50 dm^{3}

mol NaOH needed = 0.500 x 0.500 =
0.250 mol NaOH

therefore mass = mol x formula
mass

= 0.25 x 40 =
10g NaOH
required



Molarity
calculation Example 11.4

(a) How
many moles of H_{2}SO_{4} are there in 250 cm^{3} of
a 0.800 mol dm^{3}
(0.8M) sulphuric acid solution?

(b)
What mass of acid is in this solution?
[A_{r}'s:
H = 1, S = 32, O = 16]

(a) molarity = moles /
volume in dm^{3}, rearranging equation for the sulfuric acid

(b) mass = moles x formula
mass

Molarity
calculation Example
11.5 This involves calculating concentration in other ways e.g. mass/volume units

What
is the concentration of sodium chloride (NaCl) in g/dm^{3} and g/cm^{3}
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 dm^{3}

concentration = 1.5 x 58.5 = 87.8 g/dm^{3}, and

concentration =
87.75 / 1000 =
0.0878 g/cm^{3}



Molarity
calculation
Example 11.6

A solution of calcium
sulphate (CaSO_{4}) contained 0.500g dissolved in 2.00 dm^{3} of water.

Calculate the concentration in (a) g/dm^{3}, (b) g/cm^{3}
and (c) mol/dm^{3}.

(a) concentration = 0.500/2.00
=
0.250 g/dm^{3}, then since 1dm^{3}
= 1000 cm^{3}

(b) concentration = 0.250/1000
=
0.00025 g/cm^{3} (or from 0.500/2000)

(c) At. masses:
Ca = 40, S = 32, O = 64, formula mass CaSO_{4} = 40 + 32 + (4 x 16) = 136

Molarity calculation Example
11.7
 Its also good to be able to do dilution'
calculations
in section 14.3
dilution of solutions
There are more questions
involving molarity in section 12. on
titrations
and
section 14.3 on
dilution calculations and
TOP OF PAGE
and subindex
(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
solution.
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/100cm^{3}, g/dm^{3}
and molarity (mol/dm^{3}).
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 25cm^{3} of water until no more dissolves at room
temperature.
(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 cm^{3} 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.
(7) Calculations
By using a pipette it is possible to
calculate the solubility in two ways, expressed as two quite different
units.
Suppose the dish weighed 95.6g.
With the 10.0 cm^{3} of salt
solution in weighed 107.7g
After evaporation of the water the dish
weighed 96.5g
Mass of 10.0 cm^{3} salt solution = 107.7  95.6 =
12.1g
Mass of salt in 10 cm^{3}
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 H_{2}O,
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 10cm^{3} of the salt solution with
10cm^{3} 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/dm^{3}, 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
10.0 cm^{3}
Therefore X g will dissolve in 1000cm^{3}
solution, X = 1000 x 0.9 / 10 = 90g/1000 cm^{3}
Solubility of salt = 90g/dm^{3}
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/dm^{3}
NOTE Solubility varies with temperature,
see
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 20^{o}C or 25^{o}C.
TOP OF PAGE
and subindex
(d) APPENDIX
2  How to make up a standard solution  a solution of precisely known
concentration
The method and procedure of how to make up a
standard solution of a soluble solid e.g. a salt, is fully described.
Procedure
for
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
Example 1.
Suppose you want to make up 250 cm^{3} of a salt
solution of concentration 20g/dm^{3} (20g/litre, 20g/1000cm^{3},
20g/1000ml).
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 cm^{3}.
The procedure
to
is
described in detail after example 2. below.
Example 2.
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 250cm^{3} of a sodium chloride solution of concentration
0.20 moldm^{3}
Preliminary calculation:
From molarity
formula (on the right): moles = molarity (mol/dm^{3}) x
volume (dm^{3})
and volume in cm^{3} / 1000 = dm^{3}
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 cm^{3} of
solution with a precise concentration of 0.20 mol/dm^{3}.
Procedure to make the standard solution i.e. one of known
concentration of solid (in this case)
An
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).
Very
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.
Pure
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 250cm^{3} to allow for the transfer and
rinsing of the solution into the standard volumetric flask using a
'squeezy' wash bottle!
Eventually
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..
An
accurately calibrated 250cm^{3} 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.
Then,
very carefully, the flask is topped up with pure water so the meniscus rests
exactly on the 250.0cm^{3} 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.
Job done!
Note
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
same as
and
,
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
and subindex
(e) Selfassessment Quizzes
on molarity calculations:
type in answer
QUIZ on molarity or
multiple
choice
QUIZ on molarity
type in titration answer
QUIZ or
multiple choice
titration
QUIZ
(good revision for A level students)
See also
Advanced level
GCEASA2 acidalkali titration calculation questions
Above is typical periodic table used in GCSE sciencechemistry specifications in
doing molarity calculations,
and I've 'usually' used these values in my exemplar calculations to cover most
syllabuses
WHAT NEXT?
e.g. OTHER CALCULATION PAGES

What is relative atomic mass?,
relative isotopic mass and calculating relative atomic mass

Calculating relative
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
from equations
(NOT using
moles) and brief mention of actual percent % yield and theoretical yield,
atom economy
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
mass)

Using
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 GayLussac's Law (ratio of gaseous
reactantsproducts)

Molarity, volumes and solution
concentrations (and diagrams of apparatus)
(this page)

How to do acidalkali
titration calculations, diagrams of apparatus, details of procedures

Electrolysis products calculations (negative cathode and positive anode products)

Other calculations
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,
exothermic/endothermic reactions

Gas calculations involving PVT relationships,
Boyle's and Charles Laws

Radioactivity & halflife calculations including
dating materials
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