pH at a glance

 

DEFINITION OF pH

pH is defined as the negative logarithm to the base 10 of the hydrogen ion [H⁺] concentration.

It is also defined as the degree of acidity or alkalinity of a solution

i.e. pH = -log₁₀ [H⁺].

Thus: If [H⁺] = 0.00001 or 10^-5.

log [H⁺] = log10^-5 = -5

pH= -log [H⁺] = - (-5) = 5.

If [H⁺] =10^-x

Therefore, pH= -log10^-x = - (-x) = x

If [H⁺] = 10^-2, pH = 2

 

DEFINITION OF pOH

pOH is defined as the negative logarithms of the hydroxide ion [OH^-] concentration to the base of 10.

i.e. pOH= -log [OH^-].

A solution with pH 7 is neutral.

A solution with pH less than 7 i.e. pH 6,5,4, 3, 2, 1 or 0 indicates increasing acidity as the numbers decreases.

A solution with pH greater than 7, i.e. pH 8,9,10, 11, 12, 13, or 14 indicate increasing alkalinity as the numbers increase.

 1< 2 <3< 4< 5< 6        7          8< 9<10< 11<12<13<14

Increasing acidity    Neutral     Increasing alkalinity

 

A solution with pH 1 is very acidic [with high concentration of H⁺]. A solution with pH 13 is very alkaline [with low concentration of H+, but high concentration of OH-].

Note that: If pH is 1, it has concentration of H+ 10 times greater than pH 2 and 100 times greater than pH 3 e.t.c.

pH 1 > pH 2 > pH 3.

Concentration of H⁺    10^-1      10^-2      10^-3.

                                      0.1       0.01     0.001.

 

Relationship between pH and pOH.

                 H₂O →   H⁺    +   OH^-

From conductivity measurement [H⁺] =10^-7moldm^-3, [OH^-]=10^-7moldm^-3.

[H⁺] [OH^-] = Kw=10^-7 x 10^-7=10^-14mol²dm^-6.

Taking logarithm of both sides

log ([H⁺] [OH^-]) = logKw

log [H⁺] + log[OH^-] =logKw

Subtracting both sides

-(log[H⁺]  +  [OH^-]) = -logKw

-log [H+] – log[OH^-] = -logKw

-log [H⁺] + (-log [OH^-]) = -logKw

pH + pOH = PKw

pKw = -log10^-14 = -(-14) = 14

Therefore, pH + pOH = 14.

 

Worked examples

1. Find the hydrogen and hydroxide ion concentrations in

(a) 0.01moldm^-3 tetraoxosulphate (vi) acid solution.

(b) 0.001moldm^-3 potassium hydroxide solution.

 

Solution

(a).  H₂SO_4(aq)→ 2H⁺_(aq)+SO₄^2-_(aq)

 

From the equation, 1 moldm^-3 H₂SO₄ ionizes to give 2moldm^-3 H⁺

Therefore, 0.01moldm^-3 H₂SO₄ would ionize to give (2x0.01) moldm^-3 H⁺

[H⁺] = 2x10^-2moldm^-3

[H⁺] [OH^-] = 10^-14

(2x10^-2) [OH^-] = 10^-14

[OH^-] =    10^-14

                2 x 10^-2

[OH-] = 0.5x (10^{-14- -2})

[OH-] =0.5 x10^-14+2

[OH-] =0.5x10^-12moldm^-3.

 

(b).     KOH_(aq)→ K⁺_(aq)+  OH^-_(aq)

From the equation,

1moldm^-3 of KOH ionizes to give 1moldm^-3 of OH-

10^-3moldm^-3 of KOH would ionize to give 10^-3moldm^-3 of OH

[OH^-]=10^-3moldm^-3.

[H⁺] [OH^-]=10^-14

[H⁺] (10^-3) = 10^-14.

[H⁺] = 10^-14

            10^-3

[H+] = 10^-14+3

[H+] = 10^-11moldm^-3

 

2.  A glass cup of orange juice is found to have a POH of 11.40. Calculate the concentration of the hydrogen ions in the juice.

Solution

pH + pOH = 14.

pH = 14 – 11.4.

pH = 2.6.

pH = -log [H⁺]

2.6 =-log [H⁺].

[H+] = Antilog (-2.6)

[H+] = 0.0025moldm^-3

[H+] = 2.5x10^-3moldm^-3.

 

Measuring pH of a solution.

We use the pH meter or a universal indicator to measure the pH of a solution. 

A universal indicator shows different colours at specific pH or hydrogen ion concentrations.

CHEMICAL BONDING at a glance

Chemical bonding deals with the various types of forces of attraction that binds atoms and molecules together.  There are two major types: -

Electrovalent bonding/Ionic bonding 

2.Covalent bond bonding.

I. Ordinary covalent

II. Dative covalent

Electrovalent bonding or IONIC BOND: - This is a type of bonding that involves the total transfer of electrons from one atom (usually a metal) to another atom (usually a non-metal).

It is also called ionic bonding because the compound splits into ions when dissolved in water or when molten

This type of bond is usually identified by looking at the two atoms involved in the bonding. It is usually between a metal and a non-metal. 

Examples include the bonding between sodium and chlorine.

Using Lewis electron dot symbol:  Lewis dot electron symbol uses the symbol of the element surrounded by the valence electrons of the 

Characteristics of Electrovalent bonds

1.   They have high melting and high boiling points.

II.   They are soluble in polar solvents (e.g water)

III.   They conduct electricity when molten or in solution

IV.   They are crystalline in nature

V.    They are solids at room temperature.

Covalent bonding 

Covalent bonding can be of two types

 i. ordinary covalent bonding and

 ii. dative or coordinate covalent bonding.

Ordinary COVALENT BONDING: - This is a type of bonding that involves the sharing of electrons between two atoms (usually non-metals).

Ordinary covalent bond can be identified by looking at the two atoms involved in the bonding. If the two atoms involved in the bonding are both non-metals then the bond is possibly covalent bond.

Ordinary covalent bonding can be grouped into three types 

i. Single covalent bond: - In this type of covalent bond each atom donates one electron each to the shared pair. Examples include the bonding between two hydrogen atoms to form a hydrogen molecule or two chlorine atoms to form a chlorine molecule 

i.

ii.

                     

Here you can see only one dash (representing the b0ond between the hydrogen atoms and the chlorine atoms).

ii. Double Covalent Bond: - In this type of covalent bonding each atom donates two electrons to the shared electron pair. Examples include the bonding between two Oxygen atoms to form an Oxygen molecule, and also the bonding between carbon and oxygen to form Carbon (IV) oxide 

i.

      O      +      O ---------->      O=O      ( O2)

Covalent bonding does not exist only between atoms of the same element, it can occur between atoms of different elements. For example, the combination of carbon and oxygen to form carbon (IV) oxide 

using Lewis dot electron symbol

      C + 2 O   ---------->   O==C==O

Here there are two lines/dash (representing double covalent bonds) between the the two atoms 

iii. Triple Covalent bond: - In this type of bonding, each atom donates three electrons each to the shared electron pair. an example includes the bonding between nitrogen atoms to form the molecule (N2)

                                        

1.   N + N → N2

    

Note: In all these examples of covalent combinations, each atom is contributing the same number of electrons it needs to acquire an octet or a duplet (in the case of hydrogen) structure.

Characteristics of Covalent bond

I. Solids have low melting and low boiling points.

2. They are insoluble in polar solvents but soluble in organic solvents.

3. They are mainly volatile liquid or gases at room temperature.

4. They form molecules not ions.

5. They are non-electrolytes.

DATIVE COVALENT BOND: - This is a type of covalent bond that involves the sharing of electrons between two atoms, but unlike ordinary covalent bonding the shared electron pair is donated by only one of the two atoms.

Example of dative covalent bond is the bond in ammonium ion NH₄⁺ and hydroxonium ion H₃O⁺.

So, the basic difference between ordinary covalent and dative covalent bond is that the shared electron pair is donated by only one of the two atoms involved in the bonding.

METALLIC BONDING: - This is the attraction between the positive nuclei of metals and the electron cloud or sea of electrons enveloping the nucleus

.

HYDROGEN BONDING: - Hydrogen bonding is a type of force (bond) that arises when hydrogen is bonded covalently to a strongly electronegative element.

Hydrogen bond can only exist in compounds where hydrogen is bonded to another nonmetal which is highly electronagative. Example of electronegative elements include Flourine, Oxygen, Chlorine, Nitrogen.

Electronegativity is the tendency of an element to attract electrons to itself in a molecule

H-F.......H-F,,,,,,H-F   the dotted lines represent hydrogen bonding. hydrogen bonding gives added properties to compounds they occur in such as higher boiling points; and such covalent compounds usually exist in the liquid state. example is water.

 OBJECTIVE QUESTIONS 

1. What is responsible for metallic bonding?                        

 (a). Shearing of electrons between the metal atoms.          

 (b). Attraction between the atomic nuclei and cloud of electrons.        

(c). Transfer of electrons from one atom to another                   

 (d).  Attraction between negative and positive ions.

2. Electrostatic forces of attraction between sodium ion and halide are greatest in                                

 (a) NaCl.            

 (b) NaBr.                  

 (c) NaF .                      

 (d) NaI

3. Which of the following compounds is covalent?      

 (a). CaCl2     

(b).  MgO 

 (c).  NaH      

 (d).  CH4.

4. The type of chemical bond that exist between potassium and oxygen in potassium oxide is     

 (a) Ionic

 (b) metallic 

(c) covalent 

(d) dative

5. Which type of bond is represented by the dotted lines in the following structure?        

H-F''''''''H-F''''''''H-F'''''''H-F

(a) Covalent bond

(b) Dative bond 

(c) Electrovalent bond 

(d) Hydrogen bond.

6. What type of bond will be formed between elements P and Q if their electronegativity values are 0.8 and 4.0 respectively

(a) Covalent bonding 

(b) Co-ordinate covalent bonding 

(c) Electrovalent bonding 

(d)Hydrogen bonding

7. A solid substance with high melting and boiling points is likely to be a/an   

(a) covalent compound 

(b) dative covalent compound 

(c) electrovalent compound 

(d) non-metal

8. Which of the following molecules has a triple bond  

(a) CH4   

(b) NH3   

(c) N2   

(d) O2

9. Which of the following molecules has a linear shape? 

(a) CH4 

(b) CO2 

(c) H2S 

(d) NH3.

10. Hydrogen bonds are formed between molecules containing a hydrogen atom bonded to a 

(a) strongly electronegative atom 

(b) non-polar species 

(c) diatomic element 

(d) complex ion.

11. Element X with electronic configuration 2,8,2 and an element Y with electronic configuration 2,8,7 are likely to combine by

(a) metallic bonding 

(b) covalent bonding

(c) electrovalent bonding 

(d) dative bonding

12. Which of the following statements is correct about sodium chloride in the solid state?

(a) It exists as aggregates of ions 

(b) it conducts electricity

(c) its melting point is below 100⁰C

(d) it exists as discrete molecules.

13. If an element with high electron affinity combines with another element with low ionization energy, the bond formed will be mainly 

(a) covalent 

(b) ionic 

(c) dative 

(d) metallic

14. The bond formed when two electrons that are shared between two atoms are donated by only one of the atoms is

 (a) covalent dative 

(b) dative 

(c) ionic 

(c) metallic.

15. When element 20A combines with element 8Y.

 (a) a covalent compound AY is formed 

(b) an ionic compound AY is formed 

(c) an ionic compound A2Y is formed 

(d) a covalent compound AY2 is formed.

16. In metallic solids the forces of attractions are between the mobile valence electrons and 

(a) Atoms

 (b) Neutrons

 (c) The negative electrons 

(d) positively charged nuclei.

17. which of the following compounds has hydrogen bonds between its molecules

(a).  HF      

(b).  HBr        

(c).  HCl        

(d).  HI

18. The bonds in crystalline ammonium chloride are

 (a) covalent and dative 

(b) ionic and covalent 

(c) ionic, covalent and dative 

(d) ionic, covalent and hydrogen bond.

19. Which of the following properties of covalent compounds is not correct? They

 (a) are non-electrolytes 

(b) are mostly gaseous and volatile liquids.  

(c) Have low melting points 

(d) have high boiling points.

20. Which of the following molecules is not linear in shape? 

(a) CO2 

(b) O2 

(c) NH3  

(d)HCl.

21. In bonded atoms, increase in electronegativity difference, 

(a) increases polarity 

(b) decreases polarity 

 (c) has no effect on polarity 

(d) brings the polarity to zero.

22. If the difference between the electronegativities of two element is large, the type of bond that can be formed between them is 

(a) covalent    

(b) dative

(c) ionic

 (d) metallic.

23. Which of the following species does not contain a co-ordinate bond? 

 (a) Al2Cl3

(b) CCl4

(c) H3O⁺ 

(d) NH4⁺

24. Which of the following compounds has hydrogen bonds between its molecules? 

(a) HF 

(b) HBr

(c) HCl

(d) HI

25. What type of chemical bonding is involved in the formation of NH4+ from a molecule of  ammonia and a proton? 

(a). Covalent bonding 

(b). Co-ordinate covalent bonding 

(c). Electrovalent bonding 

(d). Hydrogen bonding 

26.

THEORY QUESTIONS

1a) Define i. electrovalent bond

ii. Give two properties of electrovalent bond

iii. With the aid of a diagram show the bonding between sodium and chlorine to form sodium chloride.

1(b)i define covalent bond

ii. Give three properties of covalent bond

iii. With the aid of a diagram show how ammonia molecule is formed

1(c). State the type of bonds that exist in

i. ammonia

ii. ammonium ion.

2a) Define the following terms 

 i. electrovalent bonding  

 ii. ordinary covalent bonding

iii. give two properties of electrovalent bond

iv. with the aid of a diagram show the bonding between magnesium and chlorine to form magnesium chloride.

bi) differentiate between ordinary covalent and dative covalent bond

ii. Define hydrogen bonding

3. List three types of chemical bonds

ACIDS at a glance

 

ACIDS

Definition: 

By Arrhenius: -An acid is a substance which in aqueous solution produces hydrogen ion (H⁺)or  hydroxonium ion (H₃O⁺) as the only positive ion. 

By Bronsted-Lowry: -An acid is a proton donor.

By  Lewis:- an acid is a substance that donates a pair of electron    

CLASSES OF ACIDS

There are two classes of acids:

(1). Organic acids: - this are naturally occurring acids found in plants and animal material.

	ORGANIC ACIDS	SOURCES
1.	Ethanoic Acid	vinegar
2.	Citric acid	Lime, lemon
3.	Amino acids	proteins
4.	Fatty acids	Fats and Oils
5.	Ascorbic acids (vitamin C)	oranges

(2). Inorganic acid: Inorganic acid can be prepared from mineral elements or inorganic matter.

	Inorganic Acids	Formula	constituents
1.	Hydrochloric acid	HCl	Hydrogen and chlorine
2.	Tetraoxosulphate  acid	H2SO4	Hydrogen, oxygen and Sulphur
3	Trioxonitrate (V) acid me	HNO3	Hydrogen, Nitrogen and Oxygen
4.	Trixocarbonate (IV) acid	H2CO3	Carbon, Hydrogen and Oxygen
5.	Tetraoxophosphate (V) acids	H3PO4	Hydrogen, Oxygen and Phosphorus

An acid is also defined as a substance which produces hydroxonium ion as the only positive ion when dissolved in water.

  H₂O_(l) →  H3O⁺_(aq)

Acids can be dilute or concentrated depending on the amount of water it contains.

A dilute acid is acid is one that contains a large amount of water added to a small amount of the acid.

 A concentrated acid is acid one that contains only a little amount of water added to a relatively large amount of the acid.

STRENGTH OF AN ACID

The strength of an acid can either be weak or strong.

1.  Strong acids: are acids which ionize completely in aqueous solution. 

Such acid solutions have a high concentration of H+. Examples are HCl, H2SO4 and HNO3.

H₂SO_4(aq) → 2H⁺_(aq) + SO₄^2-_(aq)

HNO_3(aq) → H⁺_(aq)+ NO₃^-_(aq)

HCl_(aq) → H⁺_(aq) + Cl^-_(aq)

2.  Weak acids: are acids which ionize or dissociate slightly or partially in aqueous solution. 

Such acid solutions have a low concentration of hydrogen ions.

        Examples of weak acids are ethanoic acid CH₃COOH, H₂CO₃, H₃PO₄, H₂SO₃.

H₂CO_3(aq) → 2H⁺_(aq) + CO₃^2-_(aq)

H₃PO_4(aq) → 3H⁺_(aq) + PO₄^3-_(aq)

CH₃COOH_(aq) → H⁺_(aq) + CH₃COO^-_(aq)

H₂SO_3(aq) → 2H⁺_(aq) +   SO₃^2-_(aq)

BASICITY OF AN ACID

The basicity of an acid is the number of replaceable hydrogen ions, H+, in one molecule of the acid

Examples: -

1. HCl_(aq) →    H⁺_(aq) + Cl^-_(aq)

2. HNO_3(aq) → H⁺_(aq)+ NO₃^-_(aq)

3. H₂SO_4(aq) → 2H⁺_(aq) + SO₄^2-_(aq)

4. H₂CO_3(aq)→ 2H⁺_(aq) + CO₃² _(aq)

5. H₃PO_4(aq)→ 3H⁺_(aq) + PO₄^2-_(aq)

_{That is the number of hydrogen ions that dissociates or separates out or ionizes.}

The Table below shows some acids and thier Basicity 

	ACID	BASICITY
1.	Hydrochloric acid	1 or Monobasic
2.	Trioxonitrate V acid	1 or Monobasic
3	Tetraoxosulphate VI acid	2 or Dibasic
4	Trioxocarbonate IV acid	2 or Dibasic
5	Trioxophosphate V acid	3 or tribasic

PHYSICAL PROPERTIES OF ACID

1. They have a sour taste.

2. They are corrosive in nature especially the strong acid.

3. In aqueous solution, they conduct electricity.

4. Most acids are colourless

CHEMICAL PROPERTIES OF ACID

1.      Reaction with metals: They react with metals to liberate hydrogen gas and salt of metal i.e

QAcid    +    Metal → Salt + Hydrogen gas.

 E.g.  2HCl_(aq) + Zn_(s)→ZnCl_2(aq) + H_2(g)

 H₂SO_4(aq) + Mg_(s)→MgSO_4(aq) + H_2(g)

2. They react with soluble bases to form salt and water only. This reaction is known as neutralization.

      Acid + Base → salt + water

  E.g  

H₂SO_4(aq) + 2KOH_(aq) → K₂SO_4(aq) + 2H₂O_(l)

   2HCl_(aq) + CaO_(s) →CaCl_2(aq) + H₂O_(l)

3.  They react with  trioxocarbonates (iv) salts to liberate carbon (iv) oxide, salt and water i.e.

    Acid +  trioxocarbonate (iv) → Salt + Water + CO_2(g)

  E.g  2HCl_(aq) + Na₂CO_3(aq) → 2NaCl_(aq) + H₂O_(l) + CO_2(g)

PREPARATION OF ACIDS

Acid can be prepared by using the following methods:

1.      Dissolving an acid anhydride in water: Acid anhydride is oxides of non-metal that dissolve in water to produce the corresponding acids e.g SO₂, CO₂, NO₂, SO₃.

SO_2(g) + H₂O_(l)   →     H₂SO_3(aq)

CO2 +  H₂O_(l) →    H₂CO_3(aq)

SO_3(g) + H₂O_(l) →    H₂SO_4(aq

2. Combination of constituent elements.

(a). Burning hydrogen in chlorine, in the presence of activated charcoal as the catalyst, yields HCl gas which dissolves readily in water to give HCl acid.

                 _{Activated charcoal}
  H2+ Cl_2(g)      →       2HCl_(g)

                            _Platinum
     H_2(g) + Br_2(g)      →      2HBr_(g)

(3). By displacement of a weak or more volatile acid from its salt by a stronger or less volatile acid. For Example

Displacement of the more volatile hydrogen chloride from metallic chloride by the less volatile concentrated tetraoxosulphate (VI) acid.

NaCl_(s) + H₂SO_4(aq) → NaHSO_4(aq) + HCl_(aq)

Uses of acid

(1) Acids are useful chemicals, used in many industries to make other consumer chemicals such as fertilizers, detergent and drugs.

(2) They are used in industrial process as drying agents, oxidizing agents and catalysts.

USES OF ORGANIC AND INORGANIC ACID

	Acids	uses
1.	HCl	i.   Needed by industries to make chemicals used to remove rust. ii.  it is used in the production of batteries
2.	H2SO4	Used as an electrolyte in lead-acid accumulators
3	HNO3	Needed by industries for making fertilizers, explosives etc.
4	Boric acid	Used as mild antiseptic or germicide.  Used as mild antiseptic or germicide.
5	Tartaric acid	Used in making baking soda, soft drinks and health salts
6	Acetic acid (ethanoic acid)	Used in preserving food.
7	Citric acid	Used in making fruits juice.
8	Fatty acid	Used in the manufacture of soap. This process is known as saponification.
9.	Phosphoric acids	Used in most soft drinks as the main ingredients
10.

OBJECTIVE QUESTIONS

1. The colour of phenolphthalein indicator in dilute HNO3 is

a.  Colourless

b. Orange

c. Pink

d. Purple

2. When concentrated H2SO4 is added to NaCl(s), the gas evolved 

a. Bleaches damp blue litmus paper

b. Forms a white precipitate with AgNO3(aq)

c. Forms a white precipitate with BaCl2(aq)

d. Turns moist blue litmus paper blue

3. Which of the following processes involves neutralization?

a. Hardening of oils 

b. Souring of milk

c. Charring of sugar

d. Liming of soils

4. Bees inject an acid substance when they sting. Which of of the following chemical compound could be used to treat bee stings?

a. CH3COOH

b. NaHCO3

c. C2H5OH

d. H2O2

5. Which of the following acid is dibasic

a. Hydrochloric acid

b. Trioxisulphate (IV) acid

c. Dioxonitrate V acid

d. Ethanoic acid

6.

THEORY QUESTIONS

1(a)i. What is an acid ?

 ii. What is the name of the process used for the industrial preparation of tetraoxosulphate VI acid?

(iii). State the catalyst used in the process mentioned above 

 iv. List three chemical properties of tetraoxosulphate (VI) acid.

2(a) What is the acid anhydride of each of the following acids? 

(i). H2SO4  (ii). HNO3 

 

WATER

TOPIC: WATER

CONTENT

·        Sources, Types, Uses of water

·        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 regarded as a universal solvent, because it can dissolve almost all other substances.

SOURCES OF WATER

The following are the sources of water:

1.             Natural water: - Rainwater, Well water, Spring water and Sea water

2.             Treated water: - Distilled water, Pipe-borne water and chlorinated water.

STRUCTURE OF WATER

In a molecule of water, H₂O, the central atom is Oxygen. Oxygen has the following electronic configuration: 1s² 2s² 2p⁴.

There are two lone pairs electrons in the valence shell of oxygen (2s²2p²) and two unpaired electrons (2p_y¹2p_z¹). Each unpaired electron forms a covalent bond with one electron from a hydrogen atom. The water molecule has two lone pairs and two bonded pairs of electrons in the valence shell of its central atom, thereby satisfying the octet rule for stability.

Ideally, the four electron pairs should be directed towards corners of a tetrahedron. However, when lone pairs of electrons are located near another lone pair, the repulsion between them is so great that they tend to push the two bond pairs of electrons closer together. As a result, the bond angle in water is compressed to approximately 105^o, such that the structure of the water molecule is V-shaped or angular shape.

                                                                            O

                                                                   H              H

LABORATORY PREPARATION OF WATER 

To prepare water in laboratory, dry hydrogen gas is ignited in air. It burns with a faint blue flame to give steam, which will condense on contact with any cold surface to form water.

PHYSICAL PROPERTIES OF WATER

1.   Water boils at 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.

CHEMICAL PROPERTIES

1.  Water reacts with electropositive metals to form alkalis and liberate hydrogen gas. E.g

                              Na_(s) + H₂O_(aq)→ NaOH_(aq)+ H_2(g)

  Mg & Zn react with steam to also liberate hydrogen gas

 Mg_(s) + H₂O_(g) → Mg(OH)_2(s) + H_2(g)

       Cu, Au, Ag, Hg do not react with water to form alkaline solution

2.   Non-metal like chlorine reacts with water to form acid solution.

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

TEST FOR WATER

Both white anhydrous copper II tetraoxosulphate (VI) and blue colbalt II chloride paper are used to test for the presence of water. When few drops of water are added to

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

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

 

TREATMENT OF WATER FOR MUNICIPAL SUPPLY

The following are the processes of treating river water for town supply

1. Coagulation (or Floculation): Chemicals like potash alum, KAl(SO₄)₂, or sodium aluminate III, NaAlO₂ is added to water in a large settling tank. This causes the large dirt to coagulate or clog together

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 of layers of sand to remove the remaining fine dirt particles.

4.  Chlorination (Disinfection): Chemicals like chlorine is then added to the water to kill germs. Iodine and fluorine are also 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.

 

REMOVAL OF TEMPORARY HARDNESS

1.  Physical method: By boiling

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

2.  Chemical method: By using 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)  +  2NaHCO_{3 (aq)}

EFFECTS OF TEMPORARY HARDNESS:  It causes

1. Furring of kettles and boilers.

2. Stalagmites and stalactites in caves.

PERMANENT HARDNESS

Permanent hardness in water is caused by the presence of Calcium and Magnesium ions in the form of soluble tetraoxosulphate (VI) and chlorides (i.e. CaSO₄, MgSO₄, MgCl₂, CaCl₂)

Removal of permanent hardness: 

1. Physical method: by distillation.

By chemical method only

1. Addition of washing soda          Na₂CO_3(aq) + CaSO_4(aq) → CaCO_3(s) + Na₂SO_4(aq)

2.  Addition of caustic soda

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

3. Ion exchange resin

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

           (insoluble)

ADVANTAGES OF HARD WATER

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

ii. Provides calcium for the development of bones and teeth of animals

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

iv. Reduce heart disease

v.  It does not dissolve lead, hence, it can be supplied in lead pipes. (it is less likely to cause corrosion in pipes.

DISADVANTAGES OF HARD WATER

i. It causes furring of kettles and boilers.

ii.  It wastes soap.

iii. It cannot be used in dying and tanning.

iv. Effects is seen in stalactites and stalagmite

v. It cause dry itchy skin

   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. When a sample of water was boiled, it lathered more readily with soap. it can be concluded that the sample most likely contained 

(a). magnesium and tetraoxosulphate (VI) 

(b). suspended solids 

(c). organic impurities 

(d). calcium hydrogentetraoxocarbonate (IV)

7.  What are the gases associated with the formation of acid rain? 

(a). CO₂ and HCl 

(a)  CO₂ and NO

(c). SO₂ and NO 

(d). which of the following processes will pollute water? 

(a) Exposure of a body of water to ultraviolet rays

(b) Discharge of industrial effluents into waterways

(c). Passage of river water through a sand bed 

(d). Addition of measured quantity of chlorine                                                    

SECTION B

1.     State the steps involved in the treatment of river water for town supply ?

2.     Give two ions that causes hardness of water?

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

4.      Mention two compounds that causes permanent hardness in water ?

5.      State two ways of removing permanent hardness in water?

6.       List two advantages of hard water and two disadvantages of hard water?

RELATIVE MOLECULAR MASS, MOLAR MASS AND PERCENTAGE COMPOSITION

RELATIVE MOLECULAR MASS: -

The relative atomic mass or the relative molecular mass of an element is the number of times one mole of the element is as heavy as one-twelfth (1/12) the mass of one atom of carbon -12. It has no unit.

The relative molecular mass of a compound is the sum of the masses of all the atoms present in one molecule of the compound. e.g.

For NaCl, the relative molecular mass= (23 +35.5) = 58.5

For ethanol = C₂H₅OH (carbon=12, H=1, O =16)

The relative molecular mass of ethanol

= C₂H₅OH

  (12×2) + (1×5) + (16) + (1)

  24 + 5 + 16 +1 = 46

activities 

1. Calculate the relative molecular mass of the following compounds 

a. NaOH     b. CaCO₃   c. Al₂(SO₄)₂ 

[ Na = 23, O =16, S = 32, Ca = 40, Al = 27, C = 12, H = 1]

 Solution 

a.  NaOH 

(Na x 1) + (O x 1) + (H x 1)

(23 x 1) + (16 x 1) + (1 x 1)

 23 + 16 + 1 =

  40

b.  CaCO₃

(Ca x 1) + (C x 1) + (O x 3)

(40 x 1) + (12 x 1) + (16 x3)

 40 + 12 + 48 = 

    100

c. Al₂(SO₄)₃

   (Al x 2) + (S x 3) + (O x 12)

    (27 x 2) + (32 x 3) + (16 x 12)

      54 + 96 + 192 =

             342

THE MOLAR MASS

This is the relative molecular mass expressed in grams. That is, expressing the molecular mass of a compound in grammes.

E.g. the molar mass of ethanol is 46gmol^-1

In 12g of carbon-12, there are 6 × 10²³ atoms of carbon. This is one mole of carbon -12.

THE MOLE CONCEPT

The mole is the amount of a substance that contains as many elementary particles as there are atoms in 12 g of C-12.

Carbon -12 isotope has been taken as an arbitrary standard, and it contains 6×10²³ particles of C-12 atoms and it weighs 12grammes and so the amount of any substance that contains 6× 10²³ is equal to 1 mole of the substance, and the mass will be equivalent to the molar mass of the substance. 

A mole of any substance is the amount of that substance which contains 6× 10²³ particles of that substance e.g. One mole of ethanol has a mass of 46g and contains 6× 10²³ ethanol molecules.

The symbol of any element or the formular of any compound represents one mole of the element or compound. For example, 

     the symbol Na represents 1mole of sodium atom  

                        Na⁺ represents 1mole of sodium ions 

                       O2 represents 1mole od oxygen molecule 

                       H2O represents 1mole of water molecule

and all the examples above contain Avogadro's number of particles that is, 6× 10²³.

NOTE: The relative molecular mass has no unit, but the molar mass of any substance is expressed in grammes per mole (g/mol).

This number 6× 10²³ is known as Avogadro’s number or Avogadro’s constant

Calculations on mole concept 

1. How many moles are there in 20g of sodium atoms (Na)

Solution

1mole of Na =23 (since the symbol of an element represent 1mole)

     Number of moles = mass given 
                                      molar mass (relative atomic/molecular mass)

     i.e mass give = 20g, relative atomic mass of sodium = 23

number of moles = 20 = 0.87mols
                               23

the unit of the mole is mol

2. how many moles are there in 25g of CaCO3

   Solution 

1 mole of CaCO3= 40 + 12 + (16 x 3) = 100

 i.e, 1mole of CaCO3 =100g

        25g of CaCO3 will contain 

            number of moles = mass given                  25 = 0.25mols
                                          molecular mas             100

PERCENTAGE COMPOSITION OF A COMPOUND

The percentage composition of an element in a compound is calculated by dividing the total mass of the element in the compound multiply by 100.

Example

1.To calculate the percentage composition of the elements in ethanol whose molecular formula is C₂H₅OH, given the relative atomic masses of carbon, hydrogen and oxygen. [C= 12, H= 1, and O=16] 

First calculate the molar mass of C₂H₅OH  

Then determine the masses of C   H and O present.

Mass of carbon= 12×2=24g

Mass of hydrogen= 6×1 = 6g

Mass of oxygen = 16× 1 = 16g

(12×2) +(1×6) + (16) = 46gmol^-1

Molar mass of C₂H₅OH= 46g

Therefore, 

         % of C = 24 × 100 = 52.17%
                             46
       

           % of H = 6 × 100 = 13.04%
                               46
         

           % of O = 16 × 100 =    34.78%
                                46

 percentage composition of the elements in sodium hydroxide (Na2CO3)      [Na= 23, O=16, C=12]  

first you calculate the molecular mass of Na2CO3

(Na x 2) + (C x 1) + (O x 3)

(23 x 2) + (12 + 1) + (16 x 3)

  46 + 12 + 48 = 106

      %   of Na in Na2CO3 = 46 x100 = 43.40%
                                               106

     % of C in Na2CO3 = 12 x100 = 11.32%
                                           106

    % of O in Na2CO3 = 48 x100 = 45.28%
                                         106

       

OBJECTIVE QUESTIONS 

1. what is the percentage composition of Sulphur in Al2(SO4)3

a. 

EMPIRICAL AND MOLECULAR FORMULAE

 

 EMPIRICAL AND MOLECULAR FORMULAE

Empirical formula: - This is the simplest formula of a compound; it shows the elements present and the ratio to which they are combined together.

The empirical formula of a compound can be calculated from the percentage compositions and the relative atomic mass of each element of the compound.

CALCULATION OF THE EMPIRICAL  FORMULA FROM PERCENTAGE COMPOSITION BY MASS.h

Te simplest empirical formula of a compound can be calculated from the percentage compositions of the various elements that make up the compound. For example, the formula for anhydrous disodium trioxocarbonate (iv)  can be calculated if the percentage composition by mass of each element present is known.

That is,  the percentage composition of the compound was found to be Na=43.40%, C= 11.32% and O = 45.28%. This would mean that in every 100g of the compound, the masses of Na, C and O were 43.40g, 11.32g and 45.28g respectively.

The amount in moles of Na, C and O would be.

       Na                C              O

  43.40g           11.32        45.28
   23                    12             16

 1.88                  0.94        2.83

1.88, 0.94 and 2.83 are the number of moles of each element respectively.

Now we divide by the smallest to get the mole ratio

 1.89:             0.94:         2.83. 

 0.94.             0.94.        0.94

  2:                    1    :        3= 

             Na₂C O₃      

The simplest formula is therefore, Na₂CO3

Example 2. What is the empirical formula of an organic compound whose percentage composition is carbon = 52.2%, hydrogen= 13.1% and oxygen = 34.7% (C = 12, H = 1, O = 16). 

It is very important to note that the addition of all the percentage compositions must be equally to 100

SOLUTION: 

  Carbon    Hydrogen      Oxygen      

  52.2           13.1              34.7

divide by the atomic mass

 52.2          13.1        34.7 
  12             1              16 

4.35           13.1            2.17

Divide by the smallest

    4.35          13.1           2.17  
   2.17          2.17            2.17.

 = 2           = 6               = 1

The empirical formular therefore = C₂H₆O.

Example 2. An organic compound has the following composition 55% of carbon, 9% hydrogen and 36% oxygen. Calculate the empirical formular for the  compound. (C= 12, H= 1, O= 16)

SOLUTION: 

  Carbon    Hydrogen Oxygen            

 55%            9%           36

Divide by atomic mass:

55             9                   36
12              1                   16

 4.58         9.00             2.25

Divide by the smallest: 

 4.58            9              2.25
 2.25           2.25           2.25

  2      :        4        :         1

The empirical formular = C₂H₄O.

Molecular formula of a compound is the actual formula of a compound, it shows the exact number of atoms present in one molecule of the compound.

Most molecular formulas are actually multiples of their empirical formulas, for example, a substance whose empirical formula is CH₂ have a molecular formula C₂H₄, C₄H₈ and so on. So the molecular formulas of a compound is calculated from the it's empirical formulary and it's molecular mass. 

I.e molecula formula = ( empirical formula)n

EXAMPLE 3: An organic compound contains carbon = 62.1%, hydrogen =10.3% and oxygen= 27.6% by mass.

(i) Find the empirical formula of the compound

(ii) If the molar mass of the compound is 58.0g, find its molecular formula. (C = 12, H = 1, O = 16).

SOLUTION: 

 Carbon    Hydrogen    Oxygen 

 62.1        10.3              27.6 
 12             1                  16

=5.1        = 10.3          = 1.8 

Divide by the smallest:

 4.35         13.1             1.8 
  1.8          1.8               1.8

  = 3          = 6                 = 1

Therefore, the empirical formular = C₃H₆O.

(ii) To calculate the molecular formula, relate the empirical formula to the molar mass   

(Empirical formula) n = Molar mass

     (C₃H₆O)n  = 58

(3x12 + 1x6 + 16x1)n   = 58

        58n   =   58
        58          58 

                 n= 1

The molecular formula = (C₃H₆O)n = C₃H₆O.

EXAMPLE 4: A hydrocarbon contains 20.80% of hydrogen and has a relative molar mass of 30, what is the

(i) empirical formula

(ii) molecular formula (C = 12, H = 1).

SOLUTION:

Hydrocarbons contain only two elements carbon and hydrogen. And the percentage composition of all elements in a compound must be equal to 100. Therefore, the percentage composition of carbon which is the second element contained by a hydrocarbon equals 79.20%. i.e 100 – percentage composition of hydrogen (20.80%). 

 Carbon              Hydrogen  

 79.20                  20.80 
  12                          1 

 =6.60                   = 20.80

  4.60                     20.80 
  6.60                      6.60 

    = 1                       = 3                      Therefore the empirical formular = CH₃.

(ii) To calculate the molecular formula, we relate the empirical formula to the molar mass.

(Empirical formula) n = Molar mass

       (CH₃)n     =     30

        (12 + 1×3)n = 30

          (15)n    =    30

               n    =    30 
                           15

                 n    =     2

The molecular formula = (CH₃)n = C₂H₆.

EXAMPLE 5: A carbohydrate contains 40% and hydrogen 6.72%, Calculate its empirical formula and the molecular formula, if the molar mass is 180 (C = 12, H = 1, O = 16).

SOLUTION:

Carbohydrate contains the elements carbon, hydrogen and oxygen, but from the question above oxygen is missing, hence the percentage composition of oxygen equals 100 – (percentage composition of carbon and hydrogen)

 = 100 – (40 + 6.72)

= 100 – 46.72 = 53.30% 

  Carbon     Hydrogen     Oxygen

         40          6.72             53.3 
         12            1                 16

   =3.33       = 6.72           = 3.33

Divide by the smallest:               

 3.33              6.72          3.33    
 3.33             3.33           3.33

  = 1             = 2             = 1

Therefore, the empirical formular = CH₂O.

 (ii) To calculate the molecular formula, we relate the empirical formula to the molar mass

.(Empirical formula)n = Molar mass

      (CH₂O)n    = 180

    (12 + 1x2 + 16)n   =  180

          (30)n  =    180

            n = ¹⁸⁰/₃₀

             n =  6

 The molecular formula = (CH₂O)n =(CH2O)6 = C₆H₁₂O₆.

The molecular formula = C₆H₁₂O_6.

OBJECTIVE QUESTIONS 

1. 

THEORY QUESTIONS 

1. A hydrocarbon contains 92.40% of carbon. If the vapour density of the hydrocarbon is 39. Find

(i) empirical formula

(ii) molecular formula (C = 12, H = 1).

2. Calculate the empirical formular of an organic compound containing 81.8% carbon and 18.2% hydrogen (C = 12, H = 1).

3. What is the empirical formular of an oxide of phosporius that contains 43.6% phosphorous and 56.4% oxygen (P = 31, O = 16)

4. Determine the empirical formula of an oxide of nitrogen containing 70% oxygen, if the relative molecular mass of the oxide is 92, deduce its molecular formular