Volumetric
Analysis: Acids &
Bases
Quantitative vs
Qualitative
Quantitative


Qualitative
‘Quantity’

‘Quality’
- Finding the amounts of
materials present
-
Identifying substances
-
Deals with types of
reaction
Volumetric Analysis
•
Liquids
•
Involves reacting volumes
of solutions with each other.
Solute + Solvent = Solution
Solute –
A substance that dissolves in a
solvent.
 Solvent –
Substance in which a solute dissolves
in.
 Solution –
Completely perfect mixture of a
solute and solvent.

 Copper Sulfate + Water → Copper Sulfate Solution
 Solute + Solvent → Solution
Homogeneous
•
Can no longer distinguish
between ions of salt and
molecules of water.
•
Both are in the same phase –
i.e. homogeneous.

Concentrated Solution – large amount of solute relative to quantity of
solvent.

Dilute Solution – small amount of solute relative to quantity of solvent.

The concentration of a solution is the amount of solute dissolved in a
given volume of solution.
Concentration: g/L, mol/L, M (molarity)

Expressing
Percentage of
Solute
•
3 Ways of Expressing:
1.
Weight per weight (w/w)
2.
Weight per volume (w/v)
3.
Volume per volume (v/v)

Q. 65 g of a solution contains 8 g of solute. Express the
concentration of this solution as % w/w.
65 g of a solution contains 8 g solute.
Therefore, % w/w =
8
65
𝑥 100
= 12.31%𝑤/𝑤
Parts Per Million

Only used for very dilute solutions

1ppm = 1 mg/L

Always bring to g/L and then multiply by 1000

Q. On analysing a sample of water, a chemist finds that 100 𝑐𝑚3 of
it contains 0.005 g of calcium carbonate. What is the concentration
of calcium carbonate in ppm?
0.005 g per 100 𝑐𝑚3 = 0.005 x 10 g/L
= 0.05 g/L
= 0.05 x 1000 mg/L
= 50 mg/L
= 50 ppm
1mg/L = 1ppm
Moles of Solute Per Litre of Solution
(Molarity)

The molarity of a solution is the number of moles of solute per
litre of solution.

Unit: M or mol/L

A 1M (one molar) solution is one that contains one mole of the
solute per litre of solution.
Calculation Involving Solutions
a.) Converting Moles/L to Grams/L
b.) Converting Grams/L to Moles/L (Molarity)
- as the volume gets smaller the concentration increases.
c.) Calculating the Number of Moles Given the Volume and Molarity
of a Solution.
(a) Converting Moles/L to Grams/L

How many grams of NaCl per litre are present in a solution marked 0.25M
NaCl?
Ans: Mr of NaCl = 23 + 35.5 = 58.5
1M solution of NaCl contains 58.5 g/L
=> 0.25M NaCl solution contains 58.5 x 0.25 = 14.625 g/L

Calculate the concentration in grams per litre of bench dilute sulfuric acid
whose concentration is 1.5 mol/L
(b) Converting Grams/L to Moles/L

What is the molarity of a solution that contains 3.68 g of NaOH per litre of
solution?
Ans: Mr of NaOH = 23 + 16 + 1 = 40
40 g in 1L solution → 1M NaOH solution
1 g in 1L solution → 1/40 M NaOH solution
3.68 g in 1L solution → 3.68/40 M NaOH solution
= 0.092 M NaOH solution
(b) Converting Grams/L to Moles/L

Calculate the concentration in moles per litre of a solution containing 46 g of
sulfuric acid (H2SO4) per 250 𝒄𝒎𝟑 of solution.
Ans: Mr of H2SO4 = 2(1) + 32 + 4(16) = 98
98 g H2SO4 in 1L → 1M H2SO4 solution
1 g H2SO4 in 1L → 1/98 M solution
46 g H2SO4 in 1L → 46/98 M solution
we have a 250 𝑐𝑚3 solution → 4 x 46/98 M solution
= 1.84 M

Since the sulfuric acid is actually dissolved in
only 250 𝑐𝑚3 of solution, the concentration
of the solution is 4 times higher.
(c) Calculating the Number of Moles Given the
Volume and the Molarity of a Solution
𝑣𝑜𝑙𝑢𝑚𝑒 𝑐𝑚3 𝑥 𝑚𝑜𝑙𝑎𝑟𝑖𝑡𝑦
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 =
1000
(c) Calculating the Number of Moles Given the
Volume and the Molarity of a Solution

What mass of sodium hydroxide is contained in 25𝒄𝒎𝟑 of a 0.75 M solution od
sodium hydroxide?
Number of moles =
Ans:
25 𝑥 0.75
1000
𝑣𝑜𝑙𝑢𝑚𝑒 𝑥 𝑚𝑜𝑙𝑎𝑟𝑖𝑡𝑦
1000
= 0.019 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻
Mr NaOH = 23 + 16 + 1 = 40
0.019 x 40g
= 0.76 g
𝒎𝒂𝒔𝒔 𝒚𝒐𝒖 𝒉𝒂𝒗𝒆
# 𝒎𝒐𝒍𝒆𝒔 =
𝑴𝒓

What volume of 0.15 M sodium hydroxide solution will contain 5 g of sodium
hydroxide?
# moles =
𝑚𝑎𝑠𝑠 𝑦𝑜𝑢 ℎ𝑎𝑣𝑒
𝑀𝑟
Ans: # moles =
# moles =
0.125 =
5
40
= 0.125 𝑚𝑜𝑙𝑒𝑠
𝑣𝑜𝑙𝑢𝑚𝑒 𝑥 𝑚𝑜𝑙𝑎𝑟𝑖𝑡𝑦
1000
𝑣𝑜𝑙𝑢𝑚𝑒 𝑥 0.15
1000
Volume =
0.125 𝑥 1000
0.15
= 833.33𝑐𝑚3
Reaction Between a Solution and a Solid

In a number of chemical reactions, solids react with solutions, i.e. many metals react with
acidic solutions

What mass of magnesium will react with 50 𝒄𝒎𝟑 of 0.5 M H2SO4
Mg + H2SO4 → MgSO4 + H2
Ans: No. of moles =
=
𝑣𝑜𝑙𝑢𝑚𝑒 𝑥 𝑚𝑜𝑙𝑎𝑟𝑖𝑡𝑦
1000
50 𝑥 0.5
1000
= 0.025 moles H2SO4
Magnesium : Sulfuric Acid as 1:1 => # moles Mg = 0.025
= 0.025 x 24 g
= 0.6 g Mg

Sodium carbonate, Na2CO3, reacts with dilute hydrochloric acid according to
the equation:
Na2CO3 + 2HCl → 2NaCl + CO2 +H2O
What volume of HCl of concentration 0.75M would be needed to neutralise 7.5g
of anhydrous sodium carbonate?
Describe the
concentration
of the beakers
Dilution of Solutions

Most lab experiments require reagents to be in their diluted state.

When a solution is diluted more solvent is added but the quantity of solute is unchanged.

Since the volume of the solution increases and the # moles are the same => the concentration of the
solution must decrease. (MiWadi)
𝑵𝒐. 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝒊𝒏 𝒅𝒊𝒍𝒖𝒕𝒆𝒅 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 = 𝑵𝒐. 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝒊𝒏 𝒄𝒐𝒏𝒄𝒆𝒏𝒕𝒓𝒂𝒕𝒆𝒅 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏
∗ 𝑽𝒐𝒍𝒅𝒊𝒍 𝒙 𝑴𝒅𝒊𝒍
∗ 𝑽𝒐𝒍𝒄𝒐𝒏𝒄 𝒙 𝑴𝒄𝒐𝒏𝒄
=
𝟏𝟎𝟎𝟎
𝟏𝟎𝟎𝟎
Volume must be in 𝒄𝒎𝟑 **

What volume of 12 M HCl solution is required to make up 500 𝒄𝒎𝟑 of 3 M HCl
solution?
Ans:
Voldil x Mdil
1000
=
Volconc x Mconc
1000
500 x 3 = Volconc x 12
Volconc =
500 𝑥 3
12
Volconc = 125 𝑐𝑚3

Q 13.8
What volume of 10M HCl would be required to make 500 𝒄𝒎𝟑 of 0.25M?

Q13.9
What volume of 18M H2SO4 would be required to prepare 10L of 1M H2SO4?
Standard Solution
A standard solution is a solution whose concentration is accurately known.
Chemists refer to a standard solution which minimises error as a primary standard
solution.
A primary standard is a substance that can be:
1.
obtained in a highly pure state,
2.
is stable in air,
3.
dissolved easily in water,
4.
should have a high relative molecular mass, so it can be weighed out,
5.
dissolves in water to give a known concentration
Examples
Sodium Carbonate,
Na2CO3 – soluble in
water.
Potassium Dichromate,
K2Cr2O7
Titrations

Titration – laboratory procedure where a measured volume of one solution is
added to a known volume of another solution until the reaction is complete.

There are certain pieces of apparatus needed in order to perform a titration:
1.
Graduated cylinder
2.
Volumetric flask
3.
Pipette
4.
Burette
5.
Conical flask
Graduated Cylinder

Measures approximate volumes

Wash with deionized water.
Volumetric Flask

Used to prepare standard solutions

Measure definite volumes at particular temps

Wash with deionized water

Fill to 1 cm below meniscus and bring to mark
with a dropper

Invert 20 times – makes a homogeneous
solution i.e. both in same phase.
Pipette

Delivers accurate fixed volumes

Wash with deionized water and then the solution it will
contain

Fill with pipette filler. Allow the solution to drop to the
meniscus and tip of the side to the beaker

Release the solution into the conical flask; touch the tip
off the inside of the conical flask

Do not blow out the remaining drops
Burette

Delivers accurate variable volumes

Wash with deionized water and the solution it’ll contain

Use a funnel, fill above the ‘0’ ml/𝑐𝑚3 mark

Remove funnel and allow solution to run into waste
beaker

Adjust the level to below the meniscus

Tap the side of the burette to get rid of air bubbles

Clamp using a retort stand and clamp.
Conical Flask

Normally the base is put in the conical flask

Wash with deionized water
During Titration:

Swirl conical flask

Wash the sides with deionized water
Note: 𝟐𝟓 𝒄𝒎𝟑 of sodium carbonate contains a certain
number of moles, the water will not affect the number of
moles only the molarity
Solving Volumetric Problems
Calculating the concentration of a solution of unknown concentration from titration
data

Usually given a chemical equation
𝑽𝒂 𝒙 𝑴𝒂
𝑽𝒃 𝒙 𝑴𝒃
=
𝒏𝒂
𝒏𝒃
Va = Vol of acid (𝑐𝑚3 )
Vb = Vol of base (𝑐𝑚3 )
Ma = Molarity of acid
Mb = Molarity of base
Na = no. of moles of acid
Nb = no. of moles of base

Hydrochloric acid and sodium hydroxide react according to the equation:
HCl + NaOH → NaCl + H2O
25 𝑐𝑚3 of a sodium hydroxide solution was titrated against a 0.2M HCl solution. The average
titration figure was 23.5 𝑐𝑚3 . Calculate the concentration of the sodium hydroxide solution
in (a) moles per litre and (b) grams per litre.
Ethanoic Acid (Vinegar) Titration

A 20 𝑐𝑚3 sample of vinegar was diluted to 100 𝑐𝑚3 in a volumetric flask. This diluted
solution was then titrated against 20𝑐𝑚3 of 0.12M NaOH solution. The balanced equation for
the reaction is:
CH3COOH + NaOH → CH3COONa + H2O
The average titration figure was 14.2 𝑐𝑚3 . Calculate the concentration of ethanoic acid in the
original vinegar in (a) mol/L (b) g/L (c) % w/v
Va =
Ma =
Na =
Vb =
Mb =
Nb =

Q13.14
A 25𝑐𝑚3 sample of vinegar was diluted 250 𝑐𝑚3 in a volumetric flask. This diluted solution
was then titrated against 25 𝑐𝑚3 of 0.09M NaOH solution. The balanced equation for the
reaction is:
CH3COOH + NaOH → CH3COONa + H2O
The average titration figure was 26.55 𝑐𝑚3 . Calculate the concentration of ethanoic acid in
the original vinegar in (a) mol/L (b) g/L (c) % w/v
Calculating Water of Crystallisation &
Mr of a Compound from Titration Data

In this case water of crystallisation is water that is combined with the ions inside the
crystals of washing soda.

Crystals that contain water of crystallisation are said to
be hydrated.

Washing soda consists of hydrated crystals of sodium
carbonate, Na2CO3.xH2O.
To determine the amount of water of
crystallisation in hydrated sodium carbonate

Balanced Equation: Na2CO3 + 2HCl → NaCl + H2O + CO2

Indicator: methyl orange

Justification of Indicator: Titrating a strong acid against a weak
base.

A student is asked to analyse a sample of washing soda, Na2CO3.xH2O, in order to find the number of
moles of water per mole of sodium carbonate. He weighs out 8.61g of the washing soda, dissolves
this amount in water and makes the solution up to 250 𝑐𝑚3 in a volumetric flask. 25 𝑐𝑚3 of this
solution required an average of 28.7 𝑐𝑚3 of 0.21M HCl for neutralisation.
Calculate (i) the % of water of crystallisation in the washing soda crystals and (ii) the value of x in the
above formula. The balanced equation for the reaction is:
Na2CO3 + 2HCl → 2NaCl + H2O + CO2

A student is asked to analyse a sample of metal carbonate whose formula is M2CO3. She does this by
dissolving 2.11g of the carbonate in water and diluting it to 250 𝑐𝑚3 in a volumetric flask. 25 𝑐𝑚3
samples of this carbonate are titrated against 0.12M HCl solution. The average titration figure is 25.4
𝑐𝑚3 . The balanced equation for the reaction is M2CO3 + 2HCl → 2MCl +H2O +CO2.
Calculate the relative molecular mass of the carbonate and hence identify the metal M.