Redox Reactions at a glance

Oxidation And Reduction reaction (Redox)

 Oxidation-Reduction(redox) reactions are two opposite and complementary reactions which occur simultaneously

Redox reactions have been defined in several ways before attaining a more general and simplified definition. These definitions are as follows

1.     In term of addition of oxygen:  Oxidation is defined as the addition of oxygen to a substance while reduction is defined as the removal of oxygen from a substance. E.g.

_Reduction              

CuO + C_(s) → Cu_(s) +  CO_(g)   

 ^{O.A        R.A}                       
   the example above, carbon (C) is oxidized to carbon (II) oxide (CO) while Cupper (II) oxide (CuO) is reduced to metallic Copper (Cu).

   Carbon is removing oxygen from CuO and so is the reducing agent because it causes CuO to become reduced to Cu. CuO  supplies the oxygen atom that causes carbon to become oxidized to CO and so CuO  is the oxidizing agent.

2       In terms of removal of hydrogen:  Oxidation is defined as the removal of hydrogen from a substance while reduction is defined as the addition of hydrogen to a substance                    

                   _{——————
                     ↓Reduction       ↓}
H₂S   +   Cl₂ → S_(s) + HCl_(aq)
 ^{R A            O.A}    
 ↑   ^{Oxidation          ↑
      ——————}

                     

          Similarly in this reaction, H₂S is oxidized to atomic S_(s) due to the removal of hydrogen as chlorine is reduced by gaining or addition of hydrogen. H₂S is action as the reducing agent while Cl₂ is the oxidizing agent.

3        In terms of change in the oxidation number of an element: Oxidation is defined as the increase in oxidation number of an element while reduction is the decrease in oxidation number of an element. 

4.  Definition in terms of electronegative elements: - Oxidation is the addition of an electronegative element to a substance or the removal of electropositive element from a substance while reduction is the removal of electronegative element from a substance or the addition of electropositive element to a substance 

4        In terms of electron transfer: Oxidation is defined as the loss of electrons while reduction is defined as the gain of electrons. 

  When an element loses an electron to become an ion; the O.N increase to a higher number while a gain of electrons by an element will lead to a decrease in O.N of an element. For example,

        ₂₀Ca   → ₂₀Ca²⁺+ 2e^–
        20 protons      20 protons
   20 electrons         18 electrons
    O.N 0 (zero)        O.N (+2)

        

        ₁₇Cl    +   e^–   → 17Cl^–
     17 protons            17 protons
     17 electrons         18 electrons
     O.N 0 (zero)         O.N( –1)

          In other words, as noted from the above examples, loss of electrons means a higher O.N while gain of electrons means a decrease in O.N of an element.

                            _Oxidation
               Mg + Cl   →MgCl₂       
               ^{R.A   O.A     reduction}

Example of redox reactions that occurs generally around us include

1. Photosynthesis

2. Rusting of iron

  Fe_(s) + nH₂O → Fe₂O₃.nH₂O_(s)

3.   Combustion

           _{–——————}
        ↓ _Oxidation         ↓
      C₄H₁₀ + O₂ → CO₂ + H₂O
                      ↑ ^reduction       ↑
                         ^{——————}

Redox reactions always involve the movement of electrons ( i.e loss and gain of electrons) For example 

    Pb° →Pb²⁺ + 2e^–  

   Pb²⁺→Pb⁴⁺ + 2e^–. 

O.N increased from 0 to +2 and then to +4 in Pb. i.e., oxidation involves loss of electrons which will lead to increase in O.N.

In contrast reduction involves the gains of electrons which will lead to a decrease in the O.N of the element for example                   S° + e → S^– + e^– → S^2–.

Some examples of redox recitations are

1        Fe_(s)+ S_(s)→ FeS

          ^{R.A        O.A}

          Fe losses electrons in the above reaction to become iron (II) ions (Fe²⁺),  its O.N from 0 to +2, it is oxidized, and so it is the Reducing Agent. Sulphur on the other hand, gains electrons from the iron, its O.N decreases from 0 to (–2) and so it is reduced and so is the Oxidizing Agent

2. Pb⁺²O^-2 + C⁺²O^-2 → Pb⁰+ C⁴⁺O^2-₂

     In the above example, the O.N of Pb decreased from +2 to 0, so PbO is reduced to Pb and so PbO is the oxidizing agent while  the O.N of C increased from +2 to +4, so CO is oxidized and so CO is the reducing agent.

3        H_2(g)+ O_2(g) → H₂O_(l) 
          ^{R.A           O.A}

          O.N of H increased from 0 to (+1) i.e. H is oxidized.

          The above reaction is a combustion reaction, and at this point it  is important to note that all combustion reactions are redox reactions with oxygen as the oxidizing agent.

   i.     Pb(NO3)2 + 2NaCl → PbCl2 + 2NaNO3

          The above reaction is a double decomposition reaction; it is not a redox reaction as there is no change in the O.N number of all the element s involved. Another non-redox reaction is a neutralization reaction, there is no change in the O.N of element involved in the reaction.

  ii  KOH_(aq) +HCl_(aq) → NaOH_(aq) + H₂O_(l)    (neutralization reaction)

          

Oxidizing and reducing agents (in summary)

OXIDIZING AGENT 1.      Supplies Oxygen 2.      Removes Hydrogen 3.      Decreases in oxidation number 4.      Gains electrons

REDUCING AGENT Supplies hydrogen Removes Oxygen Increases in oxidation number Loss electrons

Test for oxidizing agents

To common test or reactions that are used to test for an oxidizing agent involves the action on iron (II) chloride and hydrogen sulphide.

a)       Reaction with FeCl₂

          When an oxidizing agent is added to green iron (II) chloride; the green iron (II) ions become oxidized to yellow or brown Fe³⁺.

          Fe²⁺ →     Fe³⁺ + e^–

          green         yellow/brown

b)      Reaction with hydrogen sulphide

          When hydrogen sulphide is bubbled through a solution of an oxidizing agent, the sulphide ions S^2– becomes oxidized to elemental sulphur; and this is seen or observed as yellow deposits sulphur,   i.e. 

  S^2– → S(s) + 2e^–.

Test for reducing agents

Two commonest reagents that are used to test for a reducing agent are

1 Acidified potassium tetraoxomanganate(VI) (KMnO₄) and acidified potassium heptaoxodichromate(I) (K₂Cr₂O₇).

a)   Action of potassium hyptaoxodichromate (VI) (K₂Cr₂O₇)

  When acidified potassium heptaoxodichromate (VI) (K₂Cr₂O₇) is added to a sample of a reducing agent, its colour changes from orange to green, due to the reduction of the dichromate (VI) ion  (Cr⁶⁺) (orange) to chromium (III) (Cr³⁺) ion green

     Cr⁶⁺  +  3e^– → Cr^{3+
    Orange                green}

 b) Test using acidified potassium tetraoxomangane(VI) (KMnO₄)

  When acidified potassium tetraoxomanganate (VII) to a sample of reducing agent, the purple colour changes to colourless: due to the reduction of the manganate ion from (+7) which is purple to (+2) which is colourless and a more stable oxidation state.

MnO₄^- + 8H⁺ + 5e^– →Mn²⁺ + 4H₂O
 ^{purple              colourless}         

Mn⁷⁺ + 5e^– → Mn^{2
purple              colourless}

   This reaction is reversible as the purple colour is restored when an oxidizing agent is reintroduced into the mixture.

         Mn²⁺ + 5e →Mn^{7+   
       colourless          purple}

OBJECTIVE QUESTIONS.

1.How many electrons are removed from Cr^2- when it is oxidized to CrO₄^2- ?

a) 0

b) 2

c) 4

d) 8

2. Rusting of iron is an example of 

a) deliquescence

b)  decomposition 

c) displacement reaction

d) redox reaction

3.

THEORY QUESTIONS 

1(a) State what you will see 

i)  on bubbling SO₂ into acidified KMnO4 solution   [neco 2025]

ii). when hydrogen sulphide is bubbled into a solution of acidified potassium heptaoxodichromate VI

(b)(i). Write the ionic equation for the reaction between zinc powder and silver trioxonitrate (V) solution 

(ii). Which substance in bi above is I. Oxidized  II. Reduced 

WATER at a glance

 

WATER

CONTENT

·     Types, Uses and Structure of Water.

·      Laboratory Preparation of    Water.

·      Test for Water

·      Causes/ Removal of Hardness of Water.

·      Purification of Water for Municipal Supply.

 

WATER

Water is said to be a universal solvent , because it can dissolve almost all other substances.

SOURCES OF WATER

The following are the sources of water: the sources of water may be grouped into two

1.             Natural water: Rainwater, Well water, Spring water and Sea water, rivers and lakes

2.             Treated water: Distilled water, Pipe – borne water, deionized water and chlorinated water

STRUCTURE OF WATER

Because of the repulsion between the two lone pairs of electrons in the oxygen atom the two bonding pair are pushed towards each other resulting to a V-shape or angular shape or bent shape for water.

 

                   O
                 ∕    \
               H    H

 

LABORATORY PREPARATION OF WATER 

When dry hydrogen gas is lighted in air. It burns with a faint blue flame to give steam, which condenses when it comes in contact with any cold surface to form water.

 

PHYSICAL PROPERTIES OF WATER

1.  Water has a boiling point of  100^oC and freezes at 0^oC

2.  It has a maximum density of 1gcm^-3 at 4^oC

3.  It is neutral to litmus.

 4. It is a liquid at room temperature

 

CHEMICAL PROPERTIES

1.     Water reacts with electropositive metals like K, Na and Ca to form alkali and liberate hydrogen gas. E.g                   Na_(s)+ H₂O_(aq) →NaOH_(aq) +  H_2(g)

  Mg & Zn react with steam to form an alkaline solution while Cu, Hg, Ag, Au, do not react with water

2.       Non-metal especially the halogens chlorine reacts with water to form acid solution

   i.   H₂O_(aq)  +  Cl_2(g) →HCl_(aq)  +  HOCl_(aq)

 ii.   H₂O_(aq)  +Br_2(g) →HBr_(aq)  +  HOBr_(aq)

    

TEST FOR WATER

When few drops of water are added to

1.    White anhydrous copper (II) tetraoxosulphate (VI), it turns blue.

2.     Blue cobalt (II) chloride, it turns pink.

NOTE: These two tests are not specific for water. They only indicate the presence of water. Any aqueous solution or substance containing water will give a positive test for water

 HARDNESS OF WATER

Hard water is any water that does not form lather (foam) readily with soap. 

There are two types of hardness / hard water 

I. Temporary hardness or temporary hard water

II. Permanent hardness/ permanent hard water.

I. Temporary hardness is caused by the presence of Ca(HCO3) or Mg(HCO3) in any water sample and these can be removed by boiling the water.

II. permanent hardness is caused by the presence of CaSO4 or MgSO4 or CaCl2 or MgCl2 it can not be removed by boiling

 

REMOVAL OF TEMPORARY HARDNESS

1.  Physical method: By boiling

                             _heat
    Ca(HCO₃)_2(aq) →     CaCO_3(s)   +   H₂O_(l)   +  CO_2(g)

2.      Chemical method: By using  calculated amount of slaked lime (calcium hydroxide solution)

              Ca(HCO₃)_2(aq) + Ca(OH)_2(aq)  →2CaCO_3(s)+ 2H₂O_(l)

3         Addition of washing soda :-

               Ca(HCO₃)_2(aq) + Na₂CO_3(aq) → CaCO_3(s)+ NaHCO_{3 (aq)}

EFFECTS OF TEMPORARY HARDNESS:  

Hard water causes

1.         Furring of kettles and boilers.

2.        Stalagmite and stalactites in caves.

 

Removal of permanent hardness: 

1. by physical method : Distillation 

2.By chemical method only

i.           Addition of washing soda

    Na₂CO_3(aq) + CaSO_4(aq) →CaCO_3(s) + Na₂SO_4(aq)

ii.   Addition of caustic soda

  2NaOH_(aq) +CaSO_4(aq) → Ca(OH)_2(s)+ Na₂SO_4(aq)

iii.    Ion exchange resin

       CaSO_4(aq) + Sodium zeolite →Calcium zeolite + NaSO_4(aq)
                                                                          (insoluble)

ADVANTAGES OF HARD WATER

i.  Hard water taste better than soft water because of the presence of ions

ii.  Calcium salts in it helps to build strong teeth and bones.

iii.  It provides CaCO₃, that crab and snail use to build their shells.

iv. It does not dissolve Lead, hence it can be supplied in lead pipes.

 

DISADVANTAGES OF HARD WATER

1.             It causes furring of kettles and boilers.

2.             It wastes soap.

3.             It cannot be used in dying and tanning.

4.             Effects is seen in stalactites and stalagmite

 

TREATMENT OF WATER FOR MUNICIPAL/ TOWN SUPPLY

The following are the processes involved in the treatment of river water for town supply

1.    Coagulation: Chemicals like potash alum, KAl(SO₄)₂, or sodium aluminate III, NaAlO₂ is added to water in a large settling tank.

 

2.    Sedimentation: The coagulated solid particles or flocs are allowed to settle in the settling tank to form sediments at the bottom of the tank.

 

3.     Filtration: The water above the sediment still contains some suspended particles.  The water is passed through a filter bed to remove the remaining fine dirt particles.

 

4.     Chlorination (Disinfection): Chemicals like chlorine is then added to the water to kill germs. 

5. Calculated amount of iodine to prevent goiter and fluorine to prevent tooth decay are added as food supplements to prevent goiter and tooth decay respectively. 

 The treated water is then stored in a reservoir and distributed to the town.

OBJECTIVE QUESTIONS

1.  Treated town water undergoes the following steps except

 (A). coagulation    

 (B). precipitation

 (C). sedimentation

 (D). chlorination

2.   Water is temporarily hard because it contains

 A. CaSO₄              

B MgSO₄    

 C. Chlorine

 D. Ca(HCO₃)₂

3.    Temporary hardness of water is removed by the use of one of the following

A. boiling

 B. use of use of Ca(OH)₂

C. use of Na₂CO₃

D. use of alum

4. A substance that turns white anhydrous CuSO₄ blue is

A. water

B. liquid ammonia

C. hydrochloric acid

D. molten Sulphur

5.   Distilled water is different from deionized water because

 A. distilled water is a product of condensed steam while deionized water is filtered laboratory water

B. distilled water is always pure and sold in packs while deionized is not packaged for consumption

C. distilled water is condensed steam but deionized water is produced using ion-exchange resins which absorbs undesired ions.

D. distilled water is man-made while deionized water is both natural and artificial

6.

 

 

THEORY QUESTIONS

1.a i    Mention two compounds that causes permanent hardness in water

  ii.    State two ways of removing permanent hardness in water

  iii.     List two advantages of hard water

  b.(ai )    State the steps involved in the treatment of river water for town supply.

       ii. Write two equations to show the removal of permanent hardness of water.

      iii. Name two cations that causes hardness of water?

c.i. Give two methods of removal removal removal 

 

 

SALTS at a glace

SALTS

A salt is a compound formed when all or part of the ionizable or replaceable hydrogen ion in an acid is replaced by a metallic or ammonium ion e.g.

i. HCl_(aq) + NaOH_(aq) →      NaCl_(aq) + H₂O_(l)

ii.  H₂SO_4(aq) + KOH_(aq) → KHSO_4(aq) + H₂O_(l)

TYPES OF SALTS

There are five main types of salts namely:

1. Normal salt.

2. Acid salts

3. Basic salts

4. Double salts.

5. Complex salts.

1. Normal salts: are the salts formed when all the replaceable hydrogen ion in the acid has been completely replaced by a metal or an ammonium ion e.g. NaCl, K₂SO₄, Na₃PO₄, NaNO₃ etc. 

Normal salts are neutral to litmus and does not contain any replaceable hydrogen ion (H+) 

i. HCl_(aq)+NaOH_(aq) → NaCl_(aq) + H₂O_(l)

ii. H₂SO_4(aq) + KOH_(aq) → K₂SO_4(aq) +   H₂O_(aq)

2. Acid salts: Acid salts are salts formed when the replaceable hydrogen atoms of an acid are only partially replaced by a metal or an ammonium ion. e.g. NaHSO₄, Na₂HPO₄, NaH₂PO₄, NaHCO₃. 

They are usually formed from acids which contain more than one replaceable hydrogen ion. Acids with two replaceable hydrogen ions can form only one acid salt while acids with three replaceable hydrogen ions can form two different acid salts.

H₂SO_4(aq) + NaOH_(aq) → NaHSO_4(aq)+ H₂O_(l)
                                         sodiumhydrogentetraoxosulphate (VI)

2H₃PO_4(aq) + NaOH_(aq) → NaH₂PO_4(aq) 
                                            monosodiumhydrogentetraoxophosphate (V)

NaH₂PO_4(aq)  + 2NaOH_(aq) → Na₂HPO_4(aq) +H₂O_(l)

disodiumhydrogentetraoxophosphate (V)

Na₂HPO_4(aq + NaOH_(aq) → Na₃PO_4(aq) +H₂O_(l) 
                                       sodiumhydrogentetraoxophosphate (VI)                                        

Properties of Acid salts 

i. Acid salts turn blue litmus red. 

ii. Acid salts react with bases to form salts 

 KHSO_4(aq) + KOH_(aq) → K₂SO_4(aq) + H₂O_(l)

3. Basic salts: Basic salts that still contain replaceable hydroxide ions.

Thay are are formed when only part of the hydroxide ions of a base are replaced by the negative ions from an acid.

e.g Zn(OH)Cl,   Mg(OH)Cl, Mg(OH)NO₃, Bi(OH)₂NO₃ e.t.c.

i. Zn(OH)_2(aq) + HCl_(aq)  → Zn(OH)Cl_(aq) +  H₂O_(l)

ii. Ca(OH)(aq) + HNO3 →    Ca(OH)NO3(aq) + H2O(l)

Properties of basic salts 

i. basic salts turn red litmus blue. 

ii. basic salts react with more acid to form a normal salt and water only.

     Mg(OH)NO_3(aq) +HNO_3(aq) → Mg(NO₃)_2(aq) + H₂O_(l)

4. Double salts: Double salts are salt which ionize to produce three different types of ions in solution. Usually, two of these are positively charged (metallic or NH4+ ion) while the other is negatively charged e.g. (NH₄)2Fe(SO₄)₂.6H₂O, KAl(SO₄)₂.12H₂O, KCr(SO₄)₂.12H₂O.

(NH₄)2Fe(SO₄)₂.6H₂O:  Ammonium iron (II) tetraoxosulphate (VI) hexahydrate.

KAl (SO₄)₂.12H₂O:  Aluminium Potassium tetraoxosulphate (V) dodecahydrate (Potash alum).

KCr(SO₄)₂.12H₂O:   Chromium (III) Potassium tetraoxosulphate (VI) dodecahydrate (Chrome alum).

5. Complex salts: Complex salts contain complex ion i.e ion consisting of a charged group of atoms e.g. Na₂Zn(OH)₄, K₄Fe(CN)₆, NaAl(OH)₄.

Na₂Zn(OH)₄:     Sodium tetrahydroxozincate (II)

K₄Fe (CN)₆:      Potassium hexacyanoferrate (II)

NaAl(OH)₄:       Sodium tetrahydroxoaluminate (III)

Na₂Zn(OH)₄  → 2Na⁺  + [Zn(OH)₄]^2-

K₄Fe(CN)₆  → 4K⁺ + [Fe(CN)₆]^4-.

Properties of complex salts

i. they are soluble in water

HYDROLYSIS OF SALT

Some salts when dissolved in water, undergoes hydrolysis to give an acidic or alkaline solution.

 e.g. Na₂CO₃, NaHCO₃, AlCl₃, Na₂S, NH₄Cl, CH₃COONa e.t.c. It is like the reverse of neutralization. A salt dissolves in water to give the initial acid and alkali or hydroxide from which it was formed. for example 

1. Na₂CO_3(s) +H₂O_(l) → Na⁺ + CO3^2-
 From water   OH^-     

    H⁺
                       strong.    weak

                        base.      acid                                              pH < 7

2. AlCl_3(s) + H₂O_(l) → Al³⁺ + Cl^-
 From water             3(OH)^-    H⁺
                             weak.    strong

                               base      acid                                           pH < 7

3. (NH₄)₂CO_3(s) +H₂O_(l) ⇌ NH₄⁺ + CO₃^2-

     from water       OH^-        H⁺
                            Weak      weak

                             base.       acid                                         pH = 7

Hydrolysis of salt occurs when a salt reacts with water e.g, salt of strong acid and weak base to give an acidic solution. The change in pH of solution is due to hydrolysis.

USES OF SALTS

	SALT	USES
1.	NH4Cl	is used as an electrolyte in dry cell (Leclanché cell)
2.	CaCO3	is used as medicine to neutralize acidity in the stomach.
3.	CaCl2	i. is used as antifreeze while fused CaCl2 is used as a drying agent and also in desiccators. ii.  is used in dyeing and calico printing.
4.	CaSO4	is used for making plaster of Paris.
5.	MgSO4	is used as a laxative.
6.	KNO3	is used for making gunpowder, matches and soil fertilizer.
7.	NaCl	is used for preserving food and in glazing pottery.
8.	ZnCl2	is used in petrol

METHODS OF PREPARATION OF SALTS

The method of preparing  a salts in general depends on its:

i. Solubility in water

ii. Stability to heat.

It is important for us to know the simple rules of solubility indicated above. If we know  the solubility of a salt, it will enables us to determine which method will be used for its preparation

SOLUBLE SALT

Soluble salts can be prepared by any one of the following method:

1. Neutralization of an acid by an alkali

2. Action of dilute acid on a metal.

3. Action of dilute acid on an insoluble base.

4. Action of dilute acid on trioxocarbonate (IV).

OBTAINING SOLUBLE SALTS FROM SOLUTION

This can be done by:

1. Heating to dryness (Evaporation): This is used to recover soluble salts which do decomposed or destroyed by heat e.g. most chlorides such as NaCl, ZnCl2, FeCl2 and FeCl3 are recovered by heating.

2. Crystallization: This method is used to prepare salts which are easily decomposed or destroyed by heating to dryness. All trioxonitrate (V) salts and tetraoxosulphate (VI) are recovered by crystallization.

INSOLUBLE SALTS

Insoluble salts can be prepared by the following method:

1. Double decomposition or precipitation.

i. Pb (NO₃)_2(aq) + 2NaCl_(aq) → 2NaNO_3(aq) + PbCl_2(s)

ii. AgNO_3(aq) + NH₄Cl_(aq) →NH₄NO_3(aq) + AgCl_(s)

2. Direct combination of 2 elements.

i. Fe_(s) + S_(s) → FeS_(s)

ii. 2Fe_(s) + 3Cl_2(g) →2FeCl_3(s)

ANHYDROUS AND HYDRATED SALT

Anhydrous salts: These are salts which do not contain water of crystallization. They cannot be crystallized out from aqueous solution. 

Hydrated salts are salts which contain water of crystallization, when heated, such salt loses their water of crystallization.

Water Of Crystallization: This is a specific amount of water molecules that is embedded in crystals of salts as they form during crystallization.

Cu(NO₃)₂.3H₂O:  Copper (II) trioxonitrate (V) trihydrate.

MgSO₄.7H₂O:      Magnesium tetraoxosulphate (VI) heptahydrate.

FeSO₄.7H₂O:       Iron (ii) tetraoxosulphate (VI) heptahydrate.

Calculations of water of crystallization

1.  14g of hydrated H₂C₂O₄.xH₂O was heated to give an anhydrous salt weighing 9.99g.

(a). Calculate the value of x.

(b). Give the formula of the hydrated salt.

(c). Calculate the % of water of crystallization.

Solution

(a).  Mass of hydrated salt      = Molar mass of hydrated salt
       Mass of water molecule       Molar mass of water molecule
               

                mass of water lost = (14-9.99) = 4.01
                     14 = (90+18x)
                     4.01        18x

        14(18x) = 4.01 (90 + 18x)

        252x = 360.9 + 72.18x

        252x – 72.18x = 360.9

        179.82x = 360.9

        x =    360.9
                 179.82
                   x = 2.007

        x = 2 to the nearest whole number.

(b). Formula of hydrated salt = H₂C₂O₄.2H₂O.

(c) To calculate the % of water of crystallization:

% of water of crystallization = Mass of water x 100%
                                                         Total mass
                       = 36 x 100
                          (90 + 36)
            
                =   36    x   100   = 28.57%
                          126

   

EFFLORESCENCE, DELIQUESCENCE AND HYGROSCOPIC

When certain compounds are exposed to the air, they either lose some or all of their water of crystallization or they absorb moisture from their surroundings to become either moist or form solutions. The term efflorescent, deliquescent and hygroscopic are used to describe such compound/ or phenomenon.

EFFLORESCENCE: This is a phenomenon whereby some salts/ compounds when exposed to the atmosphere loss all or part of their water of crystallization.

EFFLORESCENT SAALTS: are substances which on exposure to air, lose some or all of their water of crystallization. The phenomenon or process is efflorescence. There is loss of weight or mass of the substances.

e.g Na₂CO₃.10H₂O →   Na₂CO₃.H₂O + 9H₂O

Other examples are Na₂SO₄.10H₂O, MgSO₄.7H₂O and CuSO₄.5H₂O e.t.c

DELIQUESCENCE: This a phenomenon whereby some salts when exposed to air absorbs so much water from the air that they form a solution.

DELIQUESCENTS SALTS: are substances that absorb so much water from air and form a solution e.g. NaOH, CaCl₂, FeCl₃, MgCl₂, KOH and P₄O₁₀. There is a gain in weight.

HYGROSCOPIC SUBSTANCES: are substances which absorb moisture on exposure to the atmosphere without forming a solution but only become sticky or wet. If they are solids, no solution will be formed but if a liquid, they absorb water and become diluted  e.g Conc. H₂SO₄, NaNO₃, CuO, CaO and anhydrous Na₂CO₃.

DRYING AGENTS

These are substances which have high affinity for water or moisture. They are either deliquescent or hygroscopic substances. They remove water molecules attached to wet substances to effect physical change. Drying agents are different from dehydrating agents which removes elements of water i.e hydrogen and oxygen atoms or intra-molecular water.

Drying agents which react with gases are not used to dry the gas e.g conc. H₂SO₄ is not used to dry NH₃ and H₂S gas.

NH_3(g) + H₂SO_4(aq) → (NH₄)₂SO_4(aq)

H₂S_(g) + H₂SO_4(aq) → 2H₂O_(l) + SO_2(g) + S_(s)

Drying agent For Gases

Concentrated H₂SO₄ is used to dry All gases except NH₃ & H₂S

Fused CaCl₂ is used to dry All gases except NH₃

CaO (quicklime) is used to dry Ammonia 

P₂O₅ All gases except Ammonia

Silica gel All gases

Salts are usually placed inside desiccators to dry

             a desicator

OBJECTIVE QUESTIONS

1. A substance is said to be hygroscopic if it absorbs 

a. water from the atmosphere to form a solution

b. heat from the surrounding 

c. carbon (iv) oxide from the atmosphere

d. moisture from the atmosphere 

2. The gas given off when NH4Cl is heated with an alkali is 

a. H₂

b. Cl₂

c. N₂ 

d. NH₃

3. A major factor considered in selecting a suitable method for preparing a simple salt is its

a. crystalline form 

b. melting point 

c. reactivity with dilute acids 

d. solubility in water

4. Which of the following salts solutions will have a pH greater than 7

a. NaCl_(aq)

b. Na₂CO_3(aq)

c. Na₂SO_4(aq)

d.NaHSO_4(aq)

5.  Which of the following compound will leave a metal residue when heated 

a. Cu(NO₃)₂

b. AgNO₃

c. K₂CO₃

d.CaCO₃

6. 

THEORY QUESTIONS

1. Give one example of the following salts 

i. Hydrated salt  

ii. Acidic salt

iii. Basic salt 

2.(a)(i) State three methods of preparing salts, giving one example in each case of a salt so prepared.

    (ii). What type of salt is each of the following?   

 i. NaH₂PO₄; 

ii. (CH₃COO)₂Pb; 

iii. KAl(SO₄).12H₂O

b. Write an equation for the reaction between dilute HCl and a solution of AgNO3 

3.(a) Rock salt is an impure form of sodium chloride.

  (i). Outline a suitable procedure for preparing a pure sample of sodium chloride from rock salt.

 (b). Classify each of the following as normal salt/ acid salt/basic salt/double salt

(i). Sodium hydrogen trioxocarbonate (IV) 

(ii). Iron (III) chloride 

(iii). Sodium ethanoate 

4. When a sample of a crystalline salt X was exposed to air, there was a loss in mass.

i. What phenomenon was exhibited by X ?

ii. Suggest two substances which X could be. 

iii. On heating 5.00g of a fresh sample of X to constant mass, 1.80g was lost in the form of water vapour.  Calculate the number of moles of water of crystallization in one molecule of X [ H=1.00, O=16.00; anhydrous form of X=160g/mol]

BALANCING IONIC EQUATIONS at a glance

 

Ionic equations

Ionic equations are equations that show only the oxidized and the reduced species. For example, given the reaction below 

1.       Zn_(s) + CuSO_4(aq) → ZnSO4_(aq) + Cu_(s)
                  blue                              colourlesss

          the oxidation number of Zn changes from 0 to +2 (R.A) while the O.N of Cu changes from +2 to 0 (oxidizing agent); but the O.N’s of the other elements (S and O) remains unchanged and so will cancel out.

                 Zn_(s) + CuSO_4(aq) → ZnSO4_(aq) + Cu_(s)
                       ^{blue                colourlesss}

 Ionic equations show only the oxidized and the reduced specie. So, the ionic equation for the reaction above can be written as

               Zn_(s) + Cu²⁺_(aq) →   Zn²⁺_(aq) + Cu_(s)

2        The reaction between potassium iodide and iron (III) tetraoxosulphate (VI) is another example. It results in the formation of brown a mixture due to the formation of iodine. Here, iodide ions are oxidized to iodine while iron (III) ions are reduced to iron (II) ions.

          Fe₂(SO₄)_3(s) + KI_(aq) → FeSO_4(aq) + K₂SO_4(aq) + I_2(l)

          The oxidation number of iodine changes from (–1) to 0 due to loss of electrons while the Fe³⁺ gains electron to become Fe²⁺. There is no change in the O.N of K

          Ionically, the above equation is written as

          Fe³⁺_(aq)  +  I^-_(aq) → Fe²⁺_(aq) + I_2(l)

 

Half equations

Half-equations are equations shows only the oxidation half or the reduction half equation. Using both reactions discussed above that is,

         Zn_(s) + Cu²⁺_(aq) →Zn²⁺_(aq)  + Cu_(s)    

             R. A      O.A

                   Zn_(s) → Zn²⁺             Oxidation half

                   Cu²⁺_(aq) → Cu_(s)         Reduction half

 

Similarly

            Fe³⁺_(aq)  +  I^-_(aq) → Fe²⁺_(aq) + I_2(l)

 

R.A        O.A  

We have

1     I^-_(aq)   →   I_2(l)              Oxidation half

2     Fe³⁺_(aq) → Fe²⁺_(aq)         Reduction half

 

Balancing redox equations

 It is important to make sure that the number of electrons lost by the reducing agent equals the number of electrons gained by the oxidizing agent in redox reactions.

There are two main methods that can be used to balance redox equations, they are

1.   The half-equation method and

 2.  The oxidation number method 

1      The half-equation method  

          b)      Balancing redox equation using half-equation methods

                   Rules for balancing redox equation using half-equation method

                   Solution

     Step I : Write down the oxidation number of every atom present

    Step II: Divide the equation into two half-equations (the oxidation half and the reduction half)

   Step III:       Balance each half-equation thus

                   a)       Balance other atoms apart from O and H, then balance O and finally balance the H

                   b)      Balance the charges by adding electrons to the left side for reduction half-equation                                   (gain) and to the right side for oxidation half-equation (loss).

  Step IV:     Cross-multiply each half-equation by the electron co-efficient of the other half-equation                            (to balance electron gain and loss)

  Step V:  Add the two half-equations to get the net-balanced equation; then include the state of matter.

                   Example: Balance the following ionic equation using half-equation methods

                   i)      Zn_(s) + Cu²⁺_(aq) →Zn²⁺_(aq)  + Cu_(s) 

                       ii)      Br^-_(aq)  +  Cl_2(g) → Cl^–_(aq) + Br_2(l)

 

          Solution

          I        Step I

                   a)      Write out the O.N of every atom present

                    0       +2              +2               0
                   ↑         ↑                 ↑               ↑
                   Zn_(s) + Cu²⁺_(aq) →Zn²⁺_(aq)  + Cu_(s)

                  

                   b)      Divide the equation into 2 half-equations (oxidation-half and the reduction half)

                          Zn_(s) →   Zn²⁺             Oxidation half

                         Cu²⁺_(aq) → Cu_(s)           Reduction half

 

                   c)      Balance the atoms and then the charges using appropriate coefficients

                     Zn → Zn²⁺ + 2e^–   [Add electrons to the product side for oxidation]

                     Cu²⁺ + 2e^– → Cu   [ Add electrons to the reactant side for reduction.]

 

                   d)      Cross-multiply the electron coefficient to balance the electron gain or loss.

                            Zn  →  Zn²⁺  + 2e^–

                            Cu²⁺  + 2e^–  →  Cu

 

                   e)      Add the two half-equations to get the net balanced equation.

                            Zn_(s) + Cu²⁺_(aq) →Zn²⁺_(aq)  + Cu_(s) 

 

          II       Example 2

                   Write down the O.N of each atom or ion present

                   a)      Br^– + Cl₂  →  Cl^–  +  Br₂

         

                   b)      Divide the equation into two half-equations (oxidation half and the reduction half)

                            Br^– →  Br₂       Oxidation

                            Cl₂  +  2e-  →  Cl^–   Reduction

 

                         c)      Balance each half equation by suing appropriate number of atoms and number of electrons.

                            Br^– → Br₂ + 2e

                            Cl₂ + 2e → 2Cl^–

 

                   d)      Cross-multiply the electron coefficient to balance the electron gain or loss.

                            2Br^–  →  Br₂  +  2e^–

                              Cl₂  +  2e^–   →  2Cl^–

                  

                   e)      Add the 2–half-equation   to get the net balanced equation.

                            2Br^-_(aq)  +  Cl_2(g)  →  2Cl^–_(aq) +  Br_2(l)

 

The oxidation number method

          This method involves or follows five steps to balance redox equation. These steps are:

          a)       Assign O.N to all the elements in the reaction

          b)      Identify the oxidized species and the reduced specie from the O.N assigned.

          c)       Deduce the number of electrons lost by the R.A in the oxidation and gained by the O.A in the reduction (use lines to show the changes)

          d)      Multiply these numbers by appropriate factors (numbers) to make the electron lost equal the electron gained using these factors as balancing coefficients.

          e)       Inspect the equation to complete the balancing assigning states be sure it is complete hen add the states of matter.

 

                   b)      ZnS_(s)  +  O_2(g)  →  ZnO  +  SO_2(g)

 

                   Solution

                   i)       Step I

                            Assign O.N for all the atoms present            

             +2   -2            0             +2   -2      +4 -2
              ↑   ↑          ↑            ↑   ↑       ↑ ↑
              ZnS_(s)  +  O_2(g)  →  ZnO  +  SO_2(g)

                   ii)      Step II

                            Identify the oxidized specie and the reduced specie

                                 Loses 6 electrons

                            ZnS  +  O₂  →  ZnO  +  SO₂

                                       Gains 6e^–

                   ZnS is oxidized; the O.N of S increased from –2 in ZnS to +4 in SO₂ (thereby losing 6 electrons) and O.N of O decreased from 0(zero) in O₂ to –2 in ZnO and SO₂.

 

                    iii)     Step III

                            Deduce the electron gain and electron loss and indicate with connecting lines.

 

                    iv)     Step IV

                            Multiply the electron gain and loss with co-efficient to make them equal. The sulphur atom loses 6 electrons and each O atom in O₂ gains 2 electrons, as well as O in ZnO for a total of 6-electrons. If we put the coefficient in front of O₂ it will give 3-atoms of O on the right-hand side of the equation; balancing the total number of electron gain and loss.

                            2ZnS + 3O₂ → 2ZnO + SO₂

 

                   v)      Step V

                            inspect the balancing to see if it is complete. The atoms are balanced but our coefficients must be whole numbers and so we multiply through by 2.

                            2ZnS_(s) + 3O_{2(g) →} 2ZnO_(s) + 2SO_2(g)

                       Check (Zn=2, S = 2, O = 6)         (Zn = 2, S = 2, O = 6)

                                                                 

 

 Example I

          Balance the following redox equations by oxidation number method

          i)       Cu_(s) + HNO₃ →Cu(NO₃)₂ + 2H₂O + O₂

          ii)      ZnS_(s) + O_2(g) → ZnO_(s) + SO_2(g)

 

          Solution

          a)      Step I

                   Assign oxidation number to all elements in the reaction.

      0             +1   +5 –2               +2     +5 –2           +1   +2           +4 –2
      |             |   |   |           |     |  |         |    |        |   |  
     Cu  +  2 H N O₃  → Cu(NO₃)₂  +  H₂O   +  NO₂

 

          b)      Step II

                   Identify the oxidized and the reduced species from the change in the oxidation numbers. That is O.N of Cu increased from 0 to +2, so Cu was oxidized and the O.N of N decreased from +5 (in HNO₃) to +4 (in NO₂); HNO₃ was reduced. 

 3        Step III

          Deduce electron lost and electron gained using connecting lines between the atoms. 2 electrons were lost by Cu in the oxidation and 7 electron was gained by N in the reduction.

                     loss 2e^–

          Cu + HNO₃ →Cu(NO₃)₂ + H₂O + NO₂

                                        gain 1e^– 

4        Step IV

          Multiply each equation by electron coefficient to balance electron gain or loss. So we will multiply the electron gain by the N by 2 and then put the coefficient of 2 in front of NO₂ and HNO₃.

          Cu + 2HNO₃ →Cu(NO₃)₂ + H₂O + 2NO₂

 

5        Step V

          Complete the balancing (process) by inspection:  The N has become 4 on the right-hand side of the equation and so a 4 should be placed in front of HNO₃.

          Cu + 4HNO₃ → Cu(NO₃)_2(q) + H₂O_(l) + 2NO_2(g)

          We also place a 2 in front of H₂O on the right-hand side of the equation to balance the H; and then add the states of matter.

          Cu_(s) + 4HNO₃ → Cu(NO₃)_2(aq) + 2H₂O_(l) + 2NO_2(g)
       Check        
               Reactants      →                          Product

    [Cu = 1,  N = 4, O = 12, H = 4]          [Cu = 1, N = 4, O = 12, H = 4]

 

Balancing complex ionic equations

To balance complex ionic equations; we take into consideration the reaction medium; where the reaction is occurring in an acidic medium or in an alkaline medium, and the half-equation method works best

1        To balance complex ionic equation occurring in acidic medium:  The rules are the same as stated earlier for balancing redox reactions but for some few additions as shown below

          2 Rules for balancing complex ionic equation occurring in an acidic medium

                  

     Step I : Write down the oxidation number of every atom present

    Step II: Divide the equation into two half-equations (the oxidation half and the reduction half)

   Step III:       Balance each half-equation thus

                   a)      To the reduction half-equation, add hydrogen ions (H+) to the LHS and water H2O to the RHS of the equation. Now balance the atoms using appropriate coefficients  

                   b)      Balance the charges by adding electrons to the left side for reduction half-equation                                   (gain) and to the right side for oxidation half-equation (loss).

  Step IV:     Cross-multiply each half-equation by the electron co-efficient of the other half-equation                            (to balance electron gain and loss)

  Step V:  Add the two half-equations to get the net-balanced equation; then include the state of matter.

Example 

1. Balance the following ionic equations occurring in and acidic medium

ALKANOIC ACID at a glance

 

ALKANOIC ACIDS

The alkanoic acids are a homologous series of organic compounds containing a carbonyl (-CO) group attached to the hydroxyl group (-OH).   They have the general molecular formula of C_nH_2n+1COOH and a functional group of R– COOH.

They are named by replacing the ending ‘e’ of the corresponding parent alkane with – oic acid. 

In the IUPAC method we take into consideration the functional group

    O
    ||
(–C–OH) and the positions of other substituents on the carbon chain.  The lowest number is given to the C- atom carrying the functional group.  

NAMING OF ALKANOIC ACIDS

The formular of the first 10 members of the series shown in the table below. Members. By applying the general molecular formular (C_nH_2n+1COOH) we have

When n=	General Molecular Formulae  CnH2n+1COOH	Name
0.	C0H2x0+1COOH =     HCOOH	Methanoic acid
1.	C1H2x1+1COOH   = CH3COOH	Ethanoic acid
2.	C2H2x2+1COOH    = C2H5COOH	Propanoic acid
3.	C3H2x3+1COOH    = C3H7COOH	Butanoic acid
4.	C4H2x4+1COOH     = C4H9COOH	Pentanoic acid
5.	C5H2x5+1COOH     = C5H11COOH	Hexanoic acid
6.	C6H2x6+1COOH     = C6H13COOH	Heptanoic acid
7.	C7H2x7+1COOH     = C7H15COOH	Octanoic acid
8.	C8H2x8+1COOH     =   C8H17COOH	Nonanoic acid
9.	C9H2x9+1COOH     =   C9H19COOH	Decanoic acid
10.	C10H2x10+1COOH  =  C10H21COOH	Undacanoic acid

 MOLECULAR STRUCTURES OF ALKANOIC ACIDS

N	ALKANOIC ACID	STRUCTURAL FORMULAR	MOLECULAR FORMULAR
1.	HCOOH Methanoic acid	O           ||          C -OH           |           H	HCOOH
2.	CH3COOH Ethanol	H   O         |   ||    H-C-C-OH         |           H	CH3COOH
3.	C2H5COOH Propanoic acid	H H  O       |   |   ||  H-C-C-C-OH       |   |          H H	CH3CH2COOH
4.	C3H7COOH Butanoic acid	H H H O         |   |   |   ||   H-C-C-C-C-OH         |   |   |           H H H	CH3(CH2)2COOH
5.	C5H11COOH Pentanoic acid	H H H H  O        |   |    |   |   ||   H-C-C-C-C-C-OH         |   |   |   |            H H H H	CH3(CH2)3COOH
6.	C6H13COOH Hexanoic acid	H H H H H  O        |   |   |    |   |   ||   H-C-C-C-C-C-C-OH        |   |   |    |   |            H H H H H	CH3(CH2)4COOH
7.	C7H15COOH Heptanoic acid	H H H H H H  O       |   |    |   |   |   |   ||  H-C-C-C-C-C-C-C-OH        |   |   |   |   |   |           H H H H H H	CH3(CH2)5COOH
8.	C8H17COOH Octanoic acid	H H H H H H H  O        |   |   |    |   |   |   |   ||   H-C-C-C-C-C-C-C-C-OH         |   |   |   |   |   |   |            H H H H H H H	CH3(CH2)6COOH
9.	C9H19COOH Nonanoic acid	H H H H H H H H  O       |    |   |   |   |   |    |   |   ||  H-C-C-C-C-C-C-C-C-C-OH       |   |   |    |    |   |   |   |          H H H H H H H H	CH3(CH2)7COOH
10.	C10H21COOH Decanoic acid	H H H H H H H H H  O      |   |    |  |    |   |   |   |   |   || H-C-C-C-C-C-C-C-C-C-C-OH       |   |   |   |   |   |   |   |    |   |       H H H H H H H H H H	CH3(CH2)8COOH

                                                                                                    

The alkanoic acids like the alkanols are classified into groups based on the number of caboxyl group present in the molecule. Thus we have  

1. Monocarboxylic acids: These are carboxylic acid which have only one -COOH per molecule. Examples include 

ii.   H   O

        |   ||
   H-C-C-OH
        |   
       H

2. Dicarboxylic acids: alkanoic acids with two -COOH groups. Examples include 

 (i)    C₁OOH   

         |              Ethanedioic acid              C₁OOH

    

(ii)  C₂OOH   

         |                              

        C₂H₂
                                   
                                                             C₃OOH
                                                            Propane-1,3-dioic acid 
3.  Tricarboxylic acids:-  These are carboxylic acids containig three carboxylic groups per molecule           

                  C₁OOH
|
                  C₂H – CH₃
|
          CH₃-C₃ – CH₃
|       
                  C₄OOH
            2,2,3- trimethyl butan-1,3- dioic acid

In this chapter we will be concentrating on monocarboxylic acids.

-They are colourless liquid at room temperatures

-lower members behave as typical acids, but as the number of carbon atom increases their solubility in water as well as their acidic nature decreases

-they have higher boiling points than normal because of the presence of hydrogen bonding

The first two members of the series are methanoic acid HCOOH and ethanoic acid with general formula CH₃COOH.

ETHANOIC ACID :- This is the second member of the series, it is a liquid at room temperature. it has a characteristic pungent smell.

LABORATORY PREPARATION OF ETHANOIC ACID

Ethanoic acid can be prepared in the laboratory in two ways or stages by oxidation of ethanol with potassium hexaoxodichromate (iv) (K₂Cr₂O₇) acidified with tetraoxosulphate (vi) (H₂SO₄)

STAGE 1:              K₂Cr₂O₇
               C₂H₅OH   →     CH₃CHO + H₂O
                Ethanol               Ethanal

                                     

STAGE 2:                      K₂Cr₂O₇ 
                       CH₃CHO   →     CH₃COOH
                        Ethanal                  ethanoic acid

PHYSICAL PROPERTIES OF ETHANOIC ACID

1. It is a colourless liquid

2. It has a pungent and characteristic  of vinegar odour

3. It has a boiling point of 118⁰C and freezes at temperature below 17⁰C (glass-like crystals known as glacier ethanoic acids

4. It is very soluble in water

5. It is weak electrolyte.

CHEMICAL PROPERTIES OF ETHANOIC ACID

1.      i).  It turns blue litmus paper red

2.      2) As an acid it reacts with alkalis and base to form salts called ethanoates (esters) and water    

 (i).  It reacts with sodium hydroxide (NaOH) to form sodium ethanoate (CH₃COONa) and water

          CH₃COOH + NaOH →CH₃COONa + H₂O

4.      ii) And with moderately reactive metals such as magnessium to liberate hydrogen gas.

→            2CH₃COOH + Mg   ^heat (CH₃COO)₂Mg+ H₂

5.      When heated with soda lime (NaOH) it forms methane gas (CH₄) and carbon (iv) oxide

          CH₃COOH + NaOH →   CH₄ + CO₂ 

3). ESTERIFICATION: -This is the process whereby alkanoic acids reacts with alkanol to form  alkanoate (ester) and water in the [presence of an acid as catalyst) 

RCOOH + ROH →   RCOOR + H2O        

 

 CH₃COOH + CH₃CH₂OH → CH₃COOCH₂CH₃ + H₂O
                                          Ethyl ethanoate

CH₃COOH + PCl5  → CH₃COCl(l) + HCl(l) +  PCl3(l)

USES OF ETHANOIC ACID

1. As vinegar for preserving food

2. For making cellulose ethanoate

3 . For making non-inflammable safety film

4. For making textile fibres such as rayon

5. For making vinylethanoate which is used in emulsion paints,

6. It is used in making adhesives for wood, glass and paper

7. It is used  to coagulate rubber latex.

TEST FOR ALKANOIC ACIDS

1. It has a characteristic Pungent and sharp smell and turns blue litmus paper red.

2. Put some of the unknown substance into a solution sodium hydrogentrioxocarbonate (IV) (NaHCO₃).  If there is effervescence and the of a gas is colourless, odourless and tasteless then the substance is ethanoic acid.