GCSE Edexcel Chemistry Key Concepts in Chemistry Complete Revision Summary

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Mendleev’s Table Shortcomings

• Argon atoms have a greater relative mass than potassium which will place Argon in the group of sodium and lithium and potassium in the group of noble gases.
• Many other elements were found not fitting this pattern and were swapped by Mendleev’s to maintain the periodicity.

Present Periodic Table

• Organise the elements in the order of increasing atomic number
• All the shortcoming due to atomic weights were solved by organising the elements in the order of increasing atomic number.
• Heavy atoms are due to the presence of different isotopes of the elements.

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Groups in the Periodic Table

Group 1: Alkali Metals

GROUP 1: Alkali Metals Physical Properties

• They are highly reactive
• Reactivity increases down the group – because tendency to loose one electron increases down the group due to increase increase in size and decrease in nuclear charge
• They loose one electron and form +1 Ions.
• They are stored in kerosene or oil to prevent them reacting from air and water
• They are soft, silvery and shinny.
• They look dull in air as they react with oxygen and form oxide which coats their surface
• Lithium is less reactive and francium is highly reactive.
• They have low melting and boiling point and the melting and boiling point decreases down the group.

Group 1: Alkali Metals Chemical Properties

REACTION WITH WATER

Reacts with water to metal hydroxide

2M + H₂O                  2MOH + H₂

Metal hydroxide are alkali therefore the pH increases. Reactivity increases down the group so potassium reacts violently

Fizzing is produced due to the formation of hydrogen.

Eg – 2Li(s) + H₂O(l)                   2LiOH(aq) + H₂(g)

Reaction with Oxygen

Reacts with oxygen to form a metal oxide

2M + O₂                        M₂O

Metals go dull in air due to this reaction

4Li(s) + 4O₂(g)                  2Li₂O(s)

Reaction with Halogens

React with halogens to form metal Halides

2M +X₂                   2MX [X= F, Cl, Br, I]

Metal Halides are while solids but dissolve in water to form colourless solutions.

2Li(s) + F₂(g)                  2LiF(s)

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WHY REACTIVITY OF GROUP 1 INCREASES DOWN THE GROUP ?

The Reactivity of Group 1 increases down the group as the tendency to loose an electron increases down the group. – React by loosing an electron

FACTORS AFFECTING TENDENCY TO LOOSE AN ELECTRON

• To loose an electron small nuclear charge greater size of atom and greater shielding is required
• Nuclear charge – Great the size of the atom, the outer electron becomes further away from the nucleus decreasing the nuclear charge
• Shielding – More the number of inner electrons due to increases in number of shell greater will be the shielding of the outer electron from the nuclear charge
• Size of the atoms – Greater the size of the atom, the outer electron will become further away from the nucleus resulting in decreases in nuclear charge
• Down the group the atom size increases due to increase in number of electron shells. This results in the outer electron being further away from the nucleus.
• As the outer electron becomes further away from the nucleus the nuclear charge decreases. Increase in number of shells also increases the shielding and shields the outer electron from the nuclear charge.
• Therefore, the tendency of atom to loose an electron increases down the grou resulting in increase in reactivity down the group.

GROUP 7 : Halogens ( Salt Forming)

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GROUP 7 : Halogens Physical Properties

• They are non metals
• They gain an electron to form -1 ions.
• They have low melting and boiling points
• Their melting point increases down the group due to increases in intermolecular forces.
• They are found in pairs and exist as diatomic molecules (X₂)
• They are poisonous and smelly
• Their reactivity increase down the group
• Their density increases down the group.
• They are poor conductors of heat and electricity

HALOGEN REACTION X[F,Cl,Br,I]

REACTION WITH HYDROGEN

They react with hydrogen to form hydrogen halides.

X₂(s) + H₂(g)                  2HX(g)

Reactivity decreases down the group so fluorine and chlorine reacts explosively and bromine and iodine reacts at higher temperature in the presence of catalyst.

REACTION WITH METALS

They react with metals to form ionic compounds. In Ionic compounds, halogens gain one electron from the metals to form -1 ions and attain noble gas electronic configurations.

2Na(s) + Cl₂(g)                  NaCl(s)

Mg(s) + Cl₂(g)                  MgCl₂(s)

DISPLACEMENT REACTION

The more reactive halogen displaces the less reactive halogen from its salt

As the reactivity decreases down the group, the halogen at the top can take the position of the halogen at the bottom in its compounds and will displace the less reactive halogen.

Cl₂    +               2NaBr                               2NaCl + Br₂(yellow solution formed)

(more               (salt of less reactive        (chlorine being more reactive has taken the reactive)     halogen than Chlorine)     position of less reactive bromine in its

compounds)

Cl₂ +2NaF                    No reaction

F2 – can displace all halogens

Cl₂ – can displace all Halogen except Fluorine

Br₂ – can displace only Iodine

F          Cl         Br           I

decreasing reactivity

WHY REACTIVITY OF GROUP 7 DECREASES DOWN THE GROUP ?

The Reactivity of Group 7 decreases down the group as the electron affinity or tendency to gain the electron decreases down the qroup. – React by gaining electron

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FACTORS AFFECTING TENDENCY TO GAIN AN ELECTRON

• To gain an electron, smaller nuclear charge smaller size and less shielding is required
• Nuclear Charge – Smaller the size of the atom, greater will be the force of the nucleus as the electron will be closer to the nucleus.
• Shielding – Less electrons and shells, smaller will *be the shielding which will in turn increa the nuclear charge.
• Size of Atom – Greater the size of the atom, the outer electron will become further away from th nucleus resulting in decreases in nuclear charge
• Down the group the atom size increases due to increase in number of electron shells. As a result the nuclear charge decreases.
• The size of the atom also increases down the group which makes the nuclear charge weaker
• The electron shells also increases which decreases the effective nuclear charge on the incoming electron.
• Due to all these factors, the nuclear charge decreases which decreases the tendency of gaining electrons down the group of halogen making them less reactive.

COMPARISON BETWEEN GROUP 1 and GROUP 7

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TRANSITION METALS

• Found between group 2 and group 3
• They are hard
• They are strong
• They are malleable and ductile
• They have higher densities than group 1 and group 2 hence they are used in construction purpose like iron.
• They show variable oxidation states
• They are used commercially as catalyst
• They form coloured compounds.
• They show the reaction with oxygen, water and halogen like group 1 but they react much slowly than alkali metals.

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BASIC TERMS OF PERIODIC TABLE

Periodic Table – A table that shows arrangement of all the known elements in the order of increasing atomic number. The table is organised into periods and groups.

Metals- Elements found to the left of the periodic table which are soft, shinny, conductors malleable and ductile eg Group 1; group 2 and group 3 elements

Non Metals — Elements found to the right of the periodic table which are dull, insulators Group 4,516 and 7 are non metals.

Halogens Group 7 elements are halogens as they are salt forming.

Alkali Metals Group 1 elements which react with water to form alkali

Noble Gases — Group 0 elements which are stable and do not react as they have complete outer shell

Elements found between group 2 and group 3 which have high densities, show variable oxidation states

Transition Metals and form coloured compounds

Periods- Horizontal rows of the periodic table

Groups – Vertical columns of the periodic table

Group Number – Indicates the number of electrons in the outermost shell.

Alkali Bases that are soluble in water

Displacement Reaction — When a more reactive element displaces the less reactive element from its salt

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Q1 Look at the periodic table and give two examples of each

Metal — Li, Na

Non Metal – O₂, F₂

Alkali Metal – K, Li

Halogens – F, Cl

Noble Gas – He, Ar

Semi metal or metalloid – Si

Metals that form +1 ions – Li, K

Non metal that form -1 ions – F, Cl

Metal that form +2 ions – Mg, Ca

Transition metal – Fe, Cu

Q2 Write the name of most reactive halogen and most reactive alkali metals

Halogen –F Alkali Metal – Fr

Q3 Why the alkali metals gets more reactive down the group

Down the group the atom size increases due to increase in number of electron shells. This results in the outer electron being further away from the nucleus. As the outer electron becomes further away from the nucleus the nuclear charge decreases. Increase in number of shells also increases the shielding and shields the outer electron from the nuclear charge. Therefore, the tendency of atom to loose an electron increases down the group resulting in increase in reactivity down the group.

Q4 Why halogens get less reactive down the group

Down the group the atom size increases due to increase in number of electron shells. As a result the nuclear charge decreases. The size of the atom also increases down the group which makes the nuclear charge weaker. The electron shells also increases which decreases the effective nuclear charge on the incoming electron. Due to all these factors, the nuclear charge decreases which decreeases the tendency of gaining electrons down the group of halogen making them less reactive.

Q5 Write the balanced chemical equation with state symbols of

1. a) Potassium with water

2K(s) + 2H₂O(g)                    2KOH(aq) +H₂(g)

1. b) Lithium with oxygen

4Li(s) + O₂(g)                    2Li₂O (s)

1. c) Sodium with bromine

2Na(s) + Br₂(l)                    2NaBr (s)

1. d) Chlorine with hydrogen

Cl₂ (s) + H₂(g)                    2HCl (g)

Q6 Explain displacement reaction of halogens with examples.

F₂ + 2NaCl                   2NaF + Cl₂

Cl₂ +2NaF                   No reaction

COVALENT BONDING

• It is between two non metals
• It involves the sharing of electrons between two non metals.
• More than one electron pair can also be shared resulting in the formation of single double and triple bonds.

 Properties of Covalent Compounds

Simple Molecule

• Eg – O₂, CH₄
• They have weak intermolecular forces in them so have a lower melting and a boiling points
• The intermolecular forces increases with increase in size as the surface area between the molecules increases.
• Therefore, polymers which have covalent bonding between them have high melting and boiling point due to increase in chain length.

Giant Covalent

• Diamond
• Graphite
• Silicon Dioxide

GIANT COVALENT STRUCTURES

Substances which have huge network of atoms joined together by covalent bonds form giant covalent structures.

PROPERTIES OF GRAPHITE

Q1 Why graphite is soft and slippery?
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In graphite, Carbon atoms are bonded in the form of layers in the form of hexagons. No covalent bonding between the layers so they can slide past each other. The layers have only weak intermolecular forces between them. By applying a little pressure then layers can easily slide past each other making Graphite soft and slippery.

Q2 Why graphite conduct electricity ?

In graphite, Carbon atoms are bonded in the form of layer in the form of hexagons. No covalent bonding between the layers so they can slide past. Each carbon atom is bonded with three other carbon leaving the fourth electron has delocalized. These delocalized electrons are mobiles electrons which can move and conduct electricity.

FULLERENE AND GRAPHENE

Fullerene: Hollow shaped molecule having hexagonal rings like a bucky ball.

• Also known as bucky ball or buckminsterfullerene.
• Carbon can be in the form of pentagon or hexagon rings
• Used as catalyst, drug delivery and treating cancer.
• Graphene: Layer of interlocking hexagonal rings like single sheet of graphite.

• It is a better conductor than graphite, light and have low density.
• Used in making computer chips and flexible electronic displays.

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CARBON NANOTUBES

• Cylinderical fullerene with the length greater than the diameter.
• High tensile strength – Used in making reinforced composite materials
• High electronic conductivity – used in electronic industry

METALLIC BONDING

• It is between two metals.
• There are fixed positive ions present in the sea of delocalised electrons.
• There is strong electrostatic force of attraction between fixed positive ions and delocalized electrons resulting in metallic bonding.

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Properties of Metals

Metals are malleable

• Malleable means that the metals can be hammered into any shape.
• Metals have layered structure and layers can slide past each other by hammering giving metals different shapes.

Metals are ductile

• Ductile means that the metals can be drawn into thin wires.
• Metals have layered structure and layers can slide past each other by hammering giving metals a wire shape.

Metallic Bonding

• Atoms in a metal are arranged in a regular manner and vibrate about fixed positions.
• The outermost electrons move freely, forming a ‘sea of electrons’ enveloping the positive metal ions.

Metals are good conductors of electricity

• Metals have delocalised electrons.
• They are mobile and conduct electricity.
• These mobile electrons or delocalised electrons conduct heat and electricity.

Metals have high melting and boiling points

• There is strong electrostatic force of attraction between fixed positive ions and delocalized electrons.
• Large amount of energy is required to overcome strong electrostatic force of attraction.

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ALLOYS

• Alloys are the mixture of metals with another metal or a non metal which make the metal stronger.
• In metals the particles are arrranged in layers. There is a regular arrangement of fixed positive ions which can slide past each by applying pressure.
• In alloys there is a mixture of metals with another metal or a non metals. Another metal being different in shape and size distort the regular arrangement of the metal lattice.
• As a result the layers of the metal can no longer slide past each other making it strong

Example: Steel is the alloy of iron which is more strong and resistant to corrossion.

NANOPARTICLES

Nanoparticles are the particles that deals with the paricles of size 1 to 100 nm.

KIL- Killing

MET – Metal

MIL -Milo

MIC – Mickey

NAN – Nano

PIC – Pictures

Convert 10 nm to

1. Metre = 10/10⁹m = 10^-8m
2. Micrometer = 10/10⁶m = 10^-5m

SURFACE AREA TO VOLUME RATIO

As the size decreases the surface area to volume ratio Increases.

Therefore Nano particles being very small in size have large surface area to volume ratio making them very useful in Science and Medicine.

Surface area = 6 x side x side m²

= 6 x 1000 x 1000

= 6 10⁶ m²

Volume = side x side x side

= 10⁹m³

SA: Volume = 6 x 10⁶/10⁹ = 6 x 10³ m

Surface area = 6 x side x side m²

= 6 x 1000 x 1000

= 6 10⁴ m²

Volume = side x side x side

= 10⁶m³

SA: Volume = 6 x 10² m
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APPLICATIONS OF NANOPARTICLES

MEDICINES

• To kill cancer and tuomour cells
• For drug delievery

CATALYST

• They have large surface area to volume ratio.
• Used in small quantities so highly effective

COSMETICS

• Used in Sunscreen to block sunlight

HOUSEHOLD

• Self cleaning window panes
• Nano particles breaks dirty in the presence of sunshine which is washed away by water while raining.

RISKS OF NANOPARTICLES

• Due to small size can cause difficulty in breathing
• They can accumulate in the envrionment and cause air pollution
• Due to their large surface area a small spark can result in violent explosion making them risky to use.
• They are toxic and cause breathing and respiratory problems.
• Due to their small size they can also cause water pollution and risk the aquation life.

STATES OF MATTER

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KEY TERMS !!!!

Ions – charged atoms with unequal number of protons and electrons

Ionic Bonding — bond formed between a metal and a non metal which involves complete transfer of electrons from metal to a non metal

Dot and Cross — diagram that show transfer of electron in an ionic bond or sharing of electrons in a covalent bond.

Covalent Bonding— bonding between two non metals which involves sharing of electrons.

Metallic Bonding— bonding in metals which involves strong electrostatic forces of attraction between fixed positive ions and delocalised electrons.

Intermolecular Forces — The forces between the molecules which determines the melting or a boiling point.

Giant Covalent Molecules — Covalently bonded molecules which forms large giant structure

Polymers – Molecules which are made up of many repeating units

Delocalised electrons — Mobile electrons that are free to move as they are not associated with a bond or an atom.

Fullerene— Allotrope of carbon which forms a cage like structure like bucky ball.

Graphene— Allotrope of carbon which is equivalent to single layer of graphite

Alloys— Mixture of metals with another metal or a non metal.

Nanoparticles- particles which are of the size of 1 nm to 100 nm_

Nanoscience—lt is the branch of science that deals with nanoparticles

State Of Matter-Different forms that a matter can take They are solids, liquids and gas

Solids — States of matter with fixed shape and volume.

Liquids— States of matter without fixed shape but fixed volume.

Gases— States of matter with fixed shape and volume.

TEST YOURSELF

Q1 Name the type of bonding in the following compounds :

1. a) Sodium Chloride – Ionic
2. b) Magnesium – Metallic
3. c) Nitrogen – Covalent
4. d) Carbon Dioxide – Covalent
5. e) Water – Covalent
6. f) Ammonia – Covalent

Q2 Draw dot and cross diagram to represent bonding in the following

1. a) Sodium chloride

1. b) Water
2. c) Magnesium

Q3 Differentiate Between Diamond and Graphite

Q4 Why Ionic compounds do not conduct electricity in solids ?

In solids, the ions are held together by strong electrostatic force of attraction in the giant ionic lattice. In molten state the ions are free to move therefore conduct electricity

Q5 Why Alloys are stronger than metals

Alloys are the mixture of metals which distors the regular arrangement of metal as a result of which layers are not able to slide past each other making alloys stronger than metals.

Q6 Why alumunium has a stronger melting point than sodium

Aluminium has a greater charge. Due to greater charge of aluminium there is a stronger electrostatic forces of attraction between fixed positive ions and delocalised elecctrons. As a result aluminium has a greater melting point than sodium.

Q7 What are nanoparticles? Write the properties and applications of nanoparticles

Nanoparticles are the particles between the size of 1 to 100 nm_ Due to smaller size they have large surface area to volume ratio making them highly useful in medicine, catalysts, cosmetics and electronic industry.

RELATIVE FORMUALE MASS

It is the sum of relative atomic mass of all the atoms present in a compound.

The ratio of the average mass of one atom compared to 1/12th of the mass of C-12

Eg – CO₂ = Mass of C + 2(mass of O)

= 12+ (2× 16)

= 44g

CaCO₃ = Mass of Ca + Mass of C + 3(mass of O)

= 40+12+ (3× 16)

= 100g

Ca(PO₄)₃ = Mass of C + 3 × mass of P+ 12(mass of O)

= 310g

MOLES  – Amount of substance.

Examples

Q1 Calculate the moles of following

1. 22g of CO₂ Moles = Mass/Mr = 22/44 = 0.5 moles

1. 17g of NH₃ Mass/Mr = 24/17 =2 moles

1. 48dm³ of O₂ Volume/24 = 48/24 = 2 moles

1. 24000cm³ of CO₂ 24000/1000dm³ = 24dm³ = 24/24 = 1 mole

1. 20g of NaOH dissolved in 50cm³ of Solution Moles = 20/40 = 0.5 moles

Q2 Calculate the mass of :-

1. a) 2 moles of calcium carbonate Mass = Moles × Mr =Mr CaCO₃ =100 = 2 × 100g= 200g
2. b) 0.1 moles of hydrochloric acid Mr of Hcl = 36.5 = 36.5×0.1 = 0.365g

Q3 Calculate the concentration of the following

1. a) 2 moles of NaOH dissolved in 10 dm³ of solution = Concentration = Moles/V(dm)³ = 2/10 = 0.2 mol dm^-3
2. b) 20 g of NaOH dissolved in 50 cm³ of solution = Moles of NaOH = 20/40 = 0.5 moles

C = 0.5/0.05 moldm^-3 = 10moldm^-3

V= 50/1000 = 0.05dm³

Avogadros Constant

1 mole = 6.02 x 10²³ atoms

Q1 Calculate the number of molecules in the following :

1. 49 g of sulphuric acid

Mr of H₂SO₄ = 98g

=98/49 =0.5 moles

= 0.5 x 6.02 x 10²³ atoms

= 3.01 x 10²³ atoms

1. 8 g of oxygen gas

Mr of O₂ = 32g

=8/32 =0.25 moles

= 0.25 x 6.02 x 10²³ atoms

= 1.50 x 10²³ atoms

1. 48 dm³ of Nitrogen

Mr of N₂ =48/24 = 2 moles

= 2 x 6.02 x 10²³ atoms

= 1.20 x 10²⁴ atoms

1. d) 50 cm³ of 0.1 mol dm³ of sodium hydroxide

Moles = C x V = 0.1 x 50/1000 = 5 x 10^-3 moles

=5 x 10^-3 x 6.02 x 10²³ atoms

= 3.01 x 10²¹ atoms

Balanced Chemical Equations

C + O₂                        CO₂

One mole of carbon is reacting with one mole of oxygen to form one mole of carbon dioxide.

2Al + 3O₂                       2Al₂O₃

2 moles aluminium reacts with 3 moles of oxygen to form 2 moles of aluminium oxide.

Mg + 2HCl                       MgCl₂ + H₂

One mole of magnesium is reacting with 2 moles of hydrochloric acid to form one moles of magnesium chloride and one mole of hydrogen gas.

 
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CALCULATIONS FROM EQUATIONS

Calculate the mass of Magnesium oxide produced from 6 g of Magnesium when burned complete y in air ?

STEP 1: Write the balanced chemical equation.

2Mg + O₂                        2MgO

STEP2: Write known to the left and unknown to the right

Mg = MgO

STEP3: Write the moles relationship from the balanced chemical equation

2moles = 2 moles

STEP4: Convert moles into mass

48g = 80g

STEP5: Do the Maths.

48/48 = 1g                                                 80/48 = 1.66g

1g× 6 = 6g                                                 1.66g × 6 = 10g

Calculate the mass of alumium needs to produce 306 g of Aluminium Oxide ?

STEP 1: Write the balanced chemical equation.

4Al + 3O₂                        2Al₂O₃

STEP2: Write known to the left and unknown to the right

2Al₂O₃                         Al

STEP3: Write the moles relationship from the balanced chemical equation

2moles = 4 moles

STEP4: Convert moles into mass

204g = 108

STEP5: Do the Maths.

306g                       108/204 x 306 = 162g
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LIMITING REAGENTS

To work out the limiting reagent work out moles of all the reagents and then work from the balanced chemical equation which reagent is in excess and which is in limiting

The reactant that get all used up completely is the limiting reagent. The other reagent which is present in greater quantity than required is in excess.

13.5 gm Aluminium reacts which 32 g of Oxygen

1. a) Work out the moles of Aluminium and Oxygen
2. b) Which reagent is limiting and which is in excess
3. c) Calculation the mass of aluminium oxide produced ?

4Al + 3O₂                        2Al₂O₃

13. Moles of Al = 13.5/27 = 0.5 moles

Moles of O₂ = 32/32 = 1 moles

1. 4 moles of Alumunium = 3 moles of oxygen

1 moles of alumnum = ¾ O₂

0.5 mol of aluminium = ¾ x 0.5 = 0.375

So 0.375 moles oxygen is required hence oxygen is in excess and alumunium is limiting

1. Al = A₂O₃

4 moles = 2 moles

• 2/4

0.5 – 2/4 x 0.5

=0.25 x 102

=25.5g
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PERCENTAGE YIELD

% yield = observed mass/expected mass x 100

= Actual yield/Theoretical yield x 100

When 28 g of nitrogen combined with hydrogen 30 g of ammonia is made ? Calculate the percentage yield ?

N₂ + 3H₂                        2NH₃

N₂ = NH₃

1mole = 2moles of ammonia

28g = 32g of ammonia

Expected = 32g

Observed = 30g

% yield = 30/32 x 100 = 93.75%

When 80 g of iron oxide reacts with carbon, 20 g of iron is produced. Calculate the % yield ?

2Fe₂O₃ + 3C               Fe + 3CO₂

Fe₂O₃ = Fe

2 moles = 4 moles

320g = 224 g

1g =224/320

80g = 224/320 x 80 = 56g

20/56 x 100 = 35.71%

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WHY PERCENTAGE YEILD IS NOT 100 %

The percentage yield can never by greater than 100%. It is very difficult to get 100% percentage yield but scientist always look for that route that gives maximum percentage yield.

• Reason for not getting 100% yield
• The reaction does not go to completion so complete products are not formed.
• Some of the reaction can start moving to the reverse direction if they are reversible
• Some of the reaction can go and form alterative or unwanted product.
• During the reaction some of the reactants can get lost or stick to the reaction vessel so do not react.
• If the reaction involves gaseous reactants they can escape.
• Some of the products can also get lost in the reaction vessel. If the reaction involves gaseous products they can also escape.
• The reagent might not be pure therefore did not react completely to give the desired yield.

Atom Economy

Atom economy = Mr of desired product/Total Mr of all the Products x 100

Calcium Oxide is produced using the following reaction

CaCO₃                   CaO + CO₂

Calculate the atom economy

Desired Product = CaO =56g

Mass of reactants = CaCO₃ = 100g

Atom Economy = 56/100 x 100 = 56%

Iron oxide is reduced by carbon to form iron and carbon monoxide

Fe₂0₃ + 3C                   2Fe +3CO

Calculate the atom economy

Desired Product = 2Fe =56 x 2 = 112g

Mass of reactants = Mass of Fe₂CO₃ + 3(Mass of C) = 196g

% Yield = 112/196 x 100 = 57%

REACTIONS WITH 100 % ATOM ECONOMY

• Addition reaction
• Reactions with only one product have 100 atom economy
• How to increase atom economy?
• Chemist should look for reaction that produce single product
• If the by products are produced they should look for recycling the byproducts or use them in some other reactions to increase the atom economy.

Titration

It is the technique used to determine the exact volume of acid and base to carry out a neutralization reaction.

TITRATION PROCEDURE

• Known Concentration solution either acid or base is measured by the pipette and is added into the conical flask. For this solution both concentration and volume is known.
• The indicator is also added in the conical flask. When the indicator changes colour the end point is reached i.e the solution gets completely neutralised.
• The unknown concentration solution is added into the burette. The starting volume is noted from the burette. The tap is then opened and the unknown solution is added dropwise into the conical flask with regular mixing.
• As soon as the indicator changes colour, the tap is closed and the final reading from the burette is noted.
• The entire process is repeated three times and the values are noted in the following format.
• The concordont reading are taken. The anamolous results are not taken into account.
• The means of the concordont readings are noted and used in the calculation

Titration Example

10 cm³ of 0.5 mol dm³ of sulphuric acid is titrated with sodium hydroxide. The results are given in the table.

2NaOH + H₂SO₄                  Na₂SO₄ + H₂0

Calculate the concentration of sodium hydroxide required to completely neutralize the acid ?

Steps

1. Write the balanced chemical equation

2NaOH + H₂SO₄                  2NaOH + H₂0

1. Underneath each equation write the numerical value given for each

Moles = 5 x 10^-3 x 2 = 10^-2 moles

1. The quantity that has two value use the concentration triangle to find the moles

V = 10cm³

= 0.01dm³

C = 0.5 mol dm^-3

Moles = C x V

0.01 x 0.5=5 x 10^-3 moles

1. Use the molar ratio to find the moles of unknown quantity
2. Use titration volume and find the concentration

Mean Titre = 25.24+25.34+ 25.29/3

= 25.29cm3

Q1 In a Titration 50 cm³ of 0.1 mol dm³ of potasisum hydroxide is neutralized by 20 cm3 of sulphuric Acid.

Calculate the concentration of sulphuric acid.

1. a) Write the balanced chemical equation

2KOH + H₂SO₄                        K₂SO₄ + 2H₂O

1. b) Underneath each equation write the numerical value given for each
2. c) The quantity that has two value use the concentration triangle to find the moles
3. d) Use the molar ratio to find the moles of unknown quantity
4. e) Use titration volume and find the concentration

V = 0.05dm³ V = 0.02 dm³

C = 0.1 mol dm^-3

= 0.005 moles

m = 0.0025 mole

C = 0.125 mol dm^-3

Key Terms

• Relative atomic mass — It is the ratio of the average mass of an atom compared to one twelfth of the mass of carbon-12
• Relative formula mass — It is the sum of relative atomic masses of all the atoms present in a formulae
• Moles – It is the amount of substance that has the same number of particles found in 12 g of carbon-12.
• Avogadros Constant —Number of particles present in one mole of the substance.
• 1mole = 6.02 x 10²³ atoms
• Limiting Reagent – It is the reagent that is completely used up in the reaction
• Yield— The mass of desired product obtained in a chemical reaction
• Percentage yield = Actual yeild /Theoretical Yield x 100
• Atom Economy = Mass of the desired product/total mass of all the reactants x 100

Concentration = Mass of Solute(g)/Volume of Solution(dm³)

Titration — It is the technique used to determine the exact volume of acids and alkali required to carry out complete neutralization.

Neutralization Reaction Reaction in which acid and base react to form salt and water.

Acids—Substance that has pH less than T

Alkali — Soluble bases that has pH greater than 7

Pipette – It is a glass tube with a bulge in the middle that is used to take out the exact volume of known concentration solutiom

Burette— It is a long tube with the.tap at the bottom that is used to measure the titre

Concordant – Values which are in the range of difference 0.1 to 0.2 cm³

TEST YOURSELF

Q1 Calculate the number of moles in the following

1. 10 g of calcium carbonate

Mr = CaCO₃ = 100g = 10/100 = 0.1 moles

1. 98 g of sulphuric acid

Mr = H₂SO₄ =98/98 = 1 mole

1. 18 g glucose

C₆H₁₂0₆

Mr = 180

18/180 = 0.1 moles

1. 90 cm3 of oxygen gas

Firstly convert 90cm³ in to dm³

90/1000 = 0.09 dm³

0.09/24 = 0.00375 moles

1. 10 cm3 of 0.2 mol dm3 sodium hydroxide solution

0.2 x 10/1000 = 0.002 moles

Q2 What is the mass of magnesium required to form 160 g of magnesium oxide ?

MgO    =       Mg

Mg + O₂                   2MgO

MgO    =       Mg

Moles     1                  1

Mass 4Og               24g

160g              96g

= 96g

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Q3 Hydrogen react with chlorine to form hydrogen chloride gas. When 71 g of chlorine reacts 70 g of hydrogen chloride is obtained. Calculate the percentage yield.

H₂ + Cl₂                             2HCl

Cl₂                            HCl

Cl₂    =       HCl

Moles     1                  2

Mass   71g               73g

160g              96g

% yield = 70/73 x 100 = 95.8%

Q4 Calculate the atom economy of both of these reactions:-

N₂ + 3H₂                             2NH₃

34/28+6 x 100 = 100%

CaCO₃                             CaO + CO₂

40+16/100 x100 = 56%

Q5 50 cm3 of 0.2 mol dm3 of HCI reacts with 10 cm3 of NaOH. Determine the concentration of NaOH ?

NaOH + HCl                             NaCl + H₂O

V = 50/1000 = 0.05dm³

C = 0.2

Moles = 0.1

NaOH V = 10/1000 = 0.01dm³

Moles = 0.1 moles

C = 10 moldm^-3

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