and the concentration of solutions
How to make up a
standard solution is on a separate page
See also 14.3
dilution of solutions calculations
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
- 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.
The concentration of an aqueous solution is usually
expressed in terms of moles of dissolved substance per cubic decimetre
(reminder mole formula triangle on the right),
- with units mol
dm-3 (or mol/dm3)and this is called molarity, sometimes denoted in
shorthand as M.
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
- 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 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),
molarity (concentration) of Z
= moles of Z / volume in dm3
- This is sometimes referred to as the
- 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 may also need to know that ...
molarity x formula mass of
solute = solute concentration in g/dm3
- This is sometimes referred to as the
- 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
calculation Example 11.1
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 = 10g NaOH
calculation Example 11.2
many moles of H2SO4 are there in 250cm3 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
calculation Example 11.3
potassium bromide was dissolved in 400cm3 of water. Calculate
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 = 0.125M
11.4 This involves calculating concentration in other ways e.g.
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 = 0.0878 g/cm3
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
(c) At. masses:
Ca = 40, S = 32, O = 64, f. mass CaSO4 = 40 + 32 + (4 x 16) = 136
Molarity calculation Example 11.6
If 5.00g of sodium chloride is
dissolved in exactly 250 cm3 of water in a calibrated volumetric flask, 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 = 0.08547/0.25 =
Molarity calculation Example
There are more questions
involving molarity in section 12. on
section 14.3 on
dilution calculations and
How to make up a
standard solution is on a separate page
APPENDIX on SOLUBILITY
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 = 12.1g
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 Xg 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, Xg 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 Xg 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.
[msc] type in answer
Advanced level GCE-AS-A2
acid-alkali titration calculation questions
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
definition of molarity
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