This document defines concentration and discusses two common units - molarity and g/dm3 concentration. It provides equations to convert between the units and calculate various parameters like moles of solute given volume and molarity. Standard solutions are solutions of known concentration prepared using a volumetric flask. Dilution methods are described to calculate new concentration when an existing solution is diluted with solvent. The pH of acids and alkalis depends on degree of dissociation and concentration - higher values lower pH for acids and raise it for alkalis.

1. Concentration of Acids
and Alkalis

2.  Definition:
- Quantity of solute in a given volume of solution which
is usually 1 dm3 (text book)
- The mass (in grams) or the number of moles of solute
dissolved in a solvent to form 1.0 dm3 (1ooo cm3 ) of
solution (reference book)

3.  Hence, the concentration of a solution can be defined
in two ways:
Concentration = Mass of solute dissolved (g)
(g/dm3) Volume of solution (dm3)
Concentration = Number of moles of solute (mol)
(mol/dm3) Volume of solution (dm3)

4.  Number of moles of solute that are present in 1 dm3
of solution
Molarity = Concentration (g/ dm3)
(mol/dm3) Molar mass (g / mol)
= Number of moles of solute (mol)
Volume of solution (dm³)

5.  However, the concentration unit that is widely used by
chemists is molarity (mol/dm³) or molar
concentration (M)
 The two units of concentration can be inter-converted
as below:
Molarity
(mol/dm³)
Concnetration
(g/dm³)
Relationship between molarity and concentration
X Molar mass
÷ Molar mass

6.  Molarity = Number of moles of solute (mol)
Volume of solution (dm³)
 From the above equation, we can calculate the number of
moles of solute in a given volume of solution.
 Number of moles of solute = Molarity x volume of
solution
 In short,
 n= Number of moles of solute (mol)
 M= Molarity of solution (mol dm¯³)
 V= Volume of solution (dm³)
n = MV

7.  5 g of copper (II) sulphate dissolved in water to form
500 cm³ solution. Calculate the concentration of
copper (II) sulphate solution in g/dm³.
 What is the mass of sodium carbonate required to
dissolve in water to prepare a 200 cm³ solution and
contains 50 g dm¯³.
 Calculate the number of moles of ammonia in 150 cm³
of 2 mol dm¯³ aqueous ammonia.
 A 250 cm³ solution contains 0.4 moles of nitric acid.
Calculate the molarity of the nitric acid.

8.  4.0 g of sodium carbonate powder, Na₂CO₃, is
dissolved in water and made up to 250 cm³. what is the
molarity of the sodium carbonate solution?
[r.a.m.: C=12, O=16, Na=23]
 Dilute hydrochloric acid used in school laboratories
usually has a concentration of 2.0 mol dm¯³. calculate
the mass of hydrogen chloride in 250 cm³ of the
hydrochloric acid?
[r.a.m.: H=1, Cl=35.5]

9.  Standard solution – a solution in which its
concentration is accurately known (tx book) / a
solution with a known concentration (ref book)
 Volumetric flask (standard flask) – apparatus with a
known volume such as 100 cm³, 200 cm³, 250 cm³, 500
cm³ and 1000 cm³.
- Used to prepare standard solution.
- Can measure the volume of a liquid accurately, up to
one decimal point.

10.  Steps involved in the preparation of a standard
solution:
a) Calculate the mass (m g) of the chemical required to
prepare v cm³ of solution where v is the volume of the
volumetric flask.
b) Weight out the exact mass (m g) of the chemical
accurately in a weighting bottle using an electronic
balance.
c) Dissolve m g of the chemical in a small amount of
distilled water.
d) Transfer the dissolved chemical into the volumetric
flask.
e) Add enough water until the graduation mark.

11.  Dilution method
Dilution – process of diluting a concentrated solution
by adding a solvent such as water to obtain a more
dilute solution.
When a solution is diluted, the volume of solvent
increases but the number of moles of solute remain
s constant.
Hence, the concentration of the solution decreases.

12. If a solution with volume of V₁ cm³ and molarity of M₁
mol dm¯³ is diluted to become V₂ cm³, the new
concentration of the diluted solution, M₂ mol dm¯³
can be determined by:
Number of moles of = M₁V₁
solute before dilution 1000
Number of moles of = M₂V₂
solute after dilution 1000

13. However, the number of moles of solute before
dilution is the same as the number of moles of solute
after dilution.
M₁V₁ = M₂V₂
------- ------- OR M₁V₁ =M₂V₂
1000 1000
M₁ - molarity of solution before water is added
V₁ - volume of solution before water is added
M₂ - molarity of solution after water is added
V₂ - volume of solution after water is added

15.  pH value of an acid or an alkali depends on 2 factors:
a) degree of dissociation
b) molarity or concentration

16.  At the same concentration, the pH value of an acid or
an alkali depends on the degree of dissociation.
a) Degree of dissociation of an acid
increases, the pH value of the acid is
decreases
b) Degree of dissociation of an alkali
increases, the pH value of the alkali is
increases

17.  For an acid or alkali, its pH value depends on the
molarity of the solution.
a) Molarity of an acid increases, the pH
value of the acid is decreases
b) Molarity of an alkali increases, the pH
value of the alkali is increases