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CHEMISTRY FORM FOUR NOTES

 

ACIDS, BASES AND SALTS

 

A.ACIDS AND BASES

At a school laboratory:

(i)An acid may be defined as a substance that turn litmus red.

(ii)A base may be defined as a substance that turn litmus blue.

Litmus is a lichen found mainly in West Africa. It changes its colour depending on whether the solution it is in, is basic/alkaline or acidic.It is thus able to identify/show whether

1. An acid is a substance that dissolves in water to form H+/H3O+ as the only positive ion/cation. This is called the Arrhenius definition of an acid. From this definition, an acid dissociate/ionize in water releasing H+ thus:

 HCl(aq) ->   H+ (aq)   +   Cl(aq)

 HNO3(aq) ->   H+ (aq)   +   NO3(aq)

 CH3COOH(aq)  ->   H+ (aq)   +   CH3COO(aq)

 H2SO4(aq) ->   2H+ (aq)   +   SO42-(aq)

 H2CO3(aq) ->   2H+ (aq)   +   CO32-(aq)

 H3PO4(aq) ->   3H+ (aq)   +   PO43-(aq)

2.A base is a substance which dissolves in water to form OH as the only negatively charged ion/anion.

This is called Arrhenius definition of a base.

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From this definition, a base dissociate/ionize in water releasing OH thus:

 KOH(aq) ->  K+(aq)  +  OH(aq)

 NaOH(aq) ->  Na+(aq)  +  OH(aq)

 NH4OH(aq) ->  NH4+(aq)  +  OH(aq)

 Ca(OH)2(aq)  ->  Ca2+(aq)  +  2OH(aq)

 Mg(OH)2(aq)  ->  Mg2+(aq)  +  2OH(aq)

3. An acid is a proton donor.

A base is a proton acceptor.

This is called Bronsted-Lowry definition of acids and bases.

From this definition, an acid donates H+ .

H+ has no electrons and neutrons .It contains only a proton.

Examples

I. From the equation:

HCl(aq)  +   H2O(l)  ===  H3O+(aq)  +  Cl(aq)

(a)(i)For the forward reaction from left to right, H2O gains a proton to form H3O+ and thus H2O is a proton acceptor .It is a Bronsted-Lowry base

(ii) For the backward reaction from right to left, H3O+ donates a proton to form H2O and thus H3O+ is an ‘opposite’ proton donor. It is a Bronsted-Lowry conjugate acid

(b)(i)For the forward reaction from left to right, HCl donates a proton to form Cl and thus HCl is a proton donor .

It is a Bronsted-Lowry acid

(ii) For the backward reaction from right to left, Cl gains a proton to form HCl and thus Cl is an ‘opposite’ proton acceptor.

It is a Bronsted-Lowry conjugate base.

Every base /acid from Bronsted-Lowry definition thus must have a conjugate product/reactant.

II. From the equation:

HCl(aq)  +   NH3(aq)  ===  NH4+(aq)  +  Cl(aq)

(a)(i)For the forward reaction from left to right, NH3 gains a proton to form NH4+ and thus NH3 is a proton acceptor .

It is a Bronsted-Lowry base

(ii) For the backward reaction from right to left, NH4+ donates a proton to form NH3 and thus NH4+ is an ‘opposite’ proton donor.

It is a Bronsted-Lowry conjugate acid

 

(b)(i)For the forward reaction from left to right, HCl donates a proton to form Cl and thus HCl is a proton donor .

It is a Bronsted-Lowry acid

(ii) For the backward reaction from right to left, Cl gains a proton to form HCl and thus Cl is an ‘opposite’ proton acceptor.

It is a Bronsted-Lowry conjugate base.

4. Acids and bases show acidic and alkaline properties/characteristics only in water but not in other solvents e.g.

(a)Hydrogen chloride gas dissolves in water to form hydrochloric acid Hydrochloric acid dissociates/ionizes in water to free H+(aq)/H3O+(aq) ions. The free H3O+(aq) / H+(aq) ions are responsible for:

 (i)turning blue litmus paper/solution red.

 (ii)show pH value 1/2/3/4/5/6

 (iii)are good electrolytes/conductors of electricity/undergo electrolysis.

 (iv)react with metals to produce /evolve hydrogen gas and a salt. i.e.

 Ionically:

 -For a monovalent metal: 2M(s)  + 2H+(aq)   -> 2M+(aq) + H2(g)

 -For a divalent metal: M(s)  + 2H+(aq)   -> M2+(aq) + H2(g)

 -For a trivalent metal: 2M(s)  + 6H+(aq)  -> 2M3+(aq) + 3H2(g)

 Examples:

 -For a monovalent metal: 2Na(s) + 2H+(aq) -> 2Na+(aq) + H2(g)

 -For a divalent metal: Ca(s)   + 2H+(aq) -> Ca2+(aq) + H2(g)

 -For a trivalent metal: 2Al(s)   + 6H+(aq) -> 2Al3+(aq) + 3H2(g)

 (v)react with metal carbonates and hhydrogen carbonates to produce /evolve carbon(IV)oxide gas ,water and a salt. i.e.

 Ionically:

 -For a monovalent metal: M2CO3(s)+ 2H+(aq) -> 2M+(aq) + H2O (l)+ CO2(g)

MHCO3(s)+ H+(aq) -> M+(aq) + H2O (l)+ CO2(g)

 -For a divalent metal: MCO3(s)+ 2H+(aq) -> M2+(aq) + H2O (l)+ CO2(g)

  M(HCO3) 2(aq)+2H+(aq) ->M2+(aq)+2H2O(l)+2CO2(g)

 Examples:

 -For a monovalent metal: K2CO3(s)+ 2H+(aq) -> 2K+(aq) + H2O (l)+ CO2(g)

  NH4HCO3(s)+ H+(aq) -> NH4+(aq) + H2O (l)+ CO2(g)

 -For a divalent metal: ZnCO3(s)+ 2H+(aq) -> Zn2+(aq) + H2O (l)+ CO2(g)

 Mg(HCO3) 2(aq)+2H+(aq) ->Mg2+(aq)+2H2O(l)+2CO2(g)

(vi)neutralize metal oxides/hydroxides to salt and water only. i.e.

 Ionically:

 -For a monovalent metal: M2O(s) + 2H+(aq) -> 2M+(aq) + H2O (l)

MOH(aq) + H+(aq) -> M+(aq) + H2O (l)

 -For a divalent metal: MO(s) + 2H+(aq) -> M2+(aq) + H2O (l)

  M(OH) 2(s) + 2H+(aq) -> M2+(aq) + 2H2O(l)

-For a trivalent metal: M2O3(s) + 6H+(aq) -> 2M3+(aq) + 3H2O (l)

  M(OH) 3(s) + 3H+(aq) -> M3+(aq) + 3H2O(l)

 Examples:

 -For a monovalent metal: K2O(s) + 2H+(aq) -> 2K+(aq) + H2O (l)

   NH4OH(aq) + H+(aq) -> NH4+(aq) + H2O (l)

 -For a divalent metal: ZnO (s) + 2H+(aq) -> Zn2+(aq) + H2O (l)

  Pb(OH) 2(s) + 2H+(aq) -> Pb2+(aq) + 2H2O(l)

(b)Hydrogen chloride gas dissolves in methylbenzene /benzene but does not dissociate /ionize into free ions.

It exists in molecular state showing none of the above properties.

(c)Ammonia gas dissolves in water to form aqueous ammonia which dissociate/ionize to free NH4+ (aq) and OH(aq) ions.

This dissociation/ionization makes aqueous ammonia to:

 (i)turn litmus paper/solution blue.

 (ii)have pH 8/9/10/11

 (iii)be a good electrical conductor

 (iv)react with acids to form ammonium salt and water only.

NH4OH(aq) + HCl(aq) -> NH4Cl(aq) + H2O(l)

(d)Ammonia gas dissolves in methylbenzene/benzene /kerosene but does not dissociate into free ions therefore existing as molecules

6. Solvents are either polar or non-polar.

A polar solvent is one which dissolves ionic compounds and other polar solvents.

Water is polar solvent that dissolves ionic and polar substance by surrounding the free ions as below:

  H ð+ H ð+ O ð-

H ð+ H ð+

H ð+ O ð- H ð+ H ð+

  H ð+

 O ð-  H+  O ð- O ð- Cl
  O ð-

  H ð+   Hð+

H ð+ O ð- H ð+   H+  H ð+

 

H ð+ H ð+ O ð-

 

 

 

 

 

 

 

 

Beaker

Cl

Cl H+ water

H+

Cl H+ Free ions

Note:Water is polar .It is made up of :

Oxygen atom is partially negative and two hydrogen atoms which are partially positive.

They surround the free H+ and Clions.

A non polar solvent is one which dissolved non-polar substances and covalent compounds.

If a polar ionic compound is dissolved in non-polar solvent ,it does not ionize/dissociate into free ions as below:

  H-Cl

H-Cl methyl benzene H-Cl H-Cl Covalent bond

7. Some acids and bases are strong while others are weak.

(a)A strong acid/base is one which is fully/wholly/completely dissociated / ionized into many free H+ /OH ions i.e.  

I. Strong acids exists more as free H+ ions than molecules. e.g.

HCl(aq) H+(aq)  +  Cl (aq)   (molecules)   (cation)   (anion)

HNO3(aq) H+(aq)  +  NO3(aq)   (molecules)   (cation)   (anion)

H2SO4(aq) 2H+(aq)  +  SO42-(aq)   (molecules)   (cation)   (anion)

II. Strong bases/alkalis exists more as free OH ions than molecules. e.g.

KOH(aq) K+(aq)  +  OH (aq)   (molecules)   (cation)   (anion)

NaOH(aq) Na+(aq)  +  OH(aq)   (molecules)   (cation)   (anion)

 

(b) A weak base/acid is one which is partially /partly dissociated /ionized in water into free OH (aq) and H+(aq) ions.

I. Weak acids exists more as molecules than as free H+ ions. e.g.

 CH3COOH(aq) H+(aq)  +  CH3COO (aq) (molecules)   (cation)   (anion)

H3PO4(aq) 3H+(aq)  +  PO43-(aq)   (molecules)   (cation)   (anion)

H2CO3(aq) 2H+(aq)  +  CO32-(aq)   (molecules)   (cation)   (anion)

II. Weak bases/alkalis exists more as molecules than free OH ions. e.g.

NH4OH(aq) NH4+(aq)  +  OH (aq)   (molecules)   (cation)   (anion)  Ca(OH)2(aq) Ca2+(aq)  +  2OH(aq)   (molecules)   (cation)   (anion)

Mg(OH)2(aq) Mg2+(aq)  +  2OH(aq)   (molecules)   (cation)   (anion)

8. The concentration of an acid/base/alkali is based on the number of moles of acid/bases dissolved in a decimeter(litre)of the solution.

An acid/base/alkali with more acid/base/alkali in a decimeter(litre) of solution is said to be concentrated while that with less is said to be dilute.

9. (a) (i)strong acids have pH 1/2/3 while weak acids have high pH 4/5/6.

(ii)a neutral solution have pH 7

(iii)strong alkalis/bases have pH 12/13/14 while weak bases/alkalis have pH 11/10 /9 / 8.

(b) pH is a measure of H+(aq) concentration in a solution.

The higher the H+(aq)ions concentration

-the higher the acidity

-the lower the pH

-the lower the concentration of OH(aq)

-the lower the alkalinity

At pH 7 , a solution has equal concentration of H+(aq) and OH(aq).

Beyond pH 7,the concentration of the OH(aq) increases as the H+(aq) ions decreases.

10.(a) When acids /bases dissolve in water, the ions present in the solution conduct electricity.

The more the dissociation the higher the yield of ions and the greater the electrical conductivity of the solution.

A compound that conducts electricity in an electrolyte and thus a compound showing high electrical conductivity is a strong electrolyte while a compound showing low electrical conductivity is a weak electrolyte.

 

(b) Practically, a bright light on a bulb ,a high voltage reading from a voltmeter high ammeter reading from an ammeter, a big deflection on a galvanometer is an indicator of strong electrolyte(acid/base) and the opposite for weak electrolytes(acids/base)

11. Some compounds exhibit/show both properties of acids and bases/alkalis.

A substance that reacts with both acids and bases is said to be amphotellic.

The examples below show the amphotellic properties of:

 (a) Zinc (II)oxide(ZnO) and Zinc hydroxide(Zn(OH)2)

(i)When ½ spatula full of Zinc(II)oxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e.

 (i) when reacting with nitric(V)acid, the oxide shows basic properties by reacting with an acid to form a simple salt and water only.

Basic oxide + Acid   -> salt + water

Examples:

Chemical equation

ZnO(s) + 2HNO3(aq) -> Zn(NO3) 2 (aq) + H2O(l)

ZnO(s) + 2HCl(aq) -> ZnCl2 (aq) + H2O(l)

ZnO(s) + H2SO4(aq) -> ZnSO4 (aq) + H2O(l)

Ionic equation

ZnO(s) + 2H+ (aq) -> Zn 2+ (aq) + H2O(l)

(ii) when reacting with sodium hydroxide, the oxide shows acidic properties by reacting with a base to form a complex salt.

Basic oxide + Base/alkali   + Water -> Complex salt

Examples:

Chemical equation

1.When Zinc oxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxozincate(II) complex salt.

ZnO(s) + 2NaOH(aq) + H2O(l) -> Na2Zn(OH) 4(aq)

2.When Zinc oxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxozincate(II) complex salt.

 ZnO(s) + 2KOH(aq) + H2O(l) -> K2Zn(OH) 4(aq)

Ionic equation

 ZnO(s) + 2OH(aq) + H2O(l) -> 2[Zn(OH) 4]2- (aq)

(ii)When Zinc(II)hydroxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e.

 (i) when reacting with nitric(V)acid, the hydroxide shows basic properties. It reacts with an acid to form a simple salt and water only.

Basic hydroxide + Acid   -> salt + water

Examples:

Chemical equation

Zn(OH) 2 (s) + 2HNO3(aq) -> Zn(NO3) 2 (aq) + 2H2O(l)

Zn(OH) 2 (s) + 2HCl(aq) -> ZnCl2 (aq) + 2H2O(l)

Zn(OH) 2 (s) + H2SO4(aq) -> ZnSO4 (aq) + 2H2O(l)

Ionic equation

Zn(OH) 2 (s) + 2H+ (aq) -> Zn 2+ (aq) + 2H2O(l)

(ii) when reacting with sodium hydroxide, the hydroxide shows acidic properties by reacting with a base to form a complex salt.

Basic hydroxide + Base/alkali -> Complex salt

Examples:

Chemical equation

1.When Zinc hydroxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxozincate(II) complex salt.

Zn(OH) 2 (s) + 2NaOH(aq) -> Na2Zn(OH) 4(aq)

2.When Zinc hydroxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxozincate(II) complex salt.

  Zn(OH) 2 (s) + 2KOH(aq) -> K2Zn(OH) 4(aq)

Ionic equation

  Zn(OH) 2 (s) + 2OH(aq) -> 2[Zn(OH) 4]2- (aq)

(b)
Lead (II)oxide(PbO) and Lead(II) hydroxide (Pb(OH)2)

(i)When ½ spatula full of Lead(II)oxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e.

 (i) when reacting with nitric(V)acid, the oxide shows basic properties by reacting with an acid to form a simple salt and water only. All other Lead salts are insoluble.

Chemical equation

PbO(s) + 2HNO3(aq) -> Pb(NO3) 2 (aq) + H2O(l)

Ionic equation

PbO(s) + 2H+ (aq) -> Pb 2+ (aq) + H2O(l)

(ii) when reacting with sodium hydroxide, the oxide shows acidic properties by reacting with a base to form a complex salt.

Chemical equation

1.When Lead(II) oxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoplumbate(II) complex salt.

PbO(s) + 2NaOH(aq) + H2O(l) -> Na2Pb(OH) 4(aq)

2.When Lead(II) oxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoplumbate(II) complex salt.

 PbO(s) + 2KOH(aq) + H2O(l) -> K2Pb(OH) 4(aq)

Ionic equation

 PbO(s) + 2OH(aq) + H2O(l) -> 2[Pb(OH) 4]2- (aq)

 

(ii)When Lead(II)hydroxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e.

 (i) when reacting with nitric(V)acid, the hydroxide shows basic properties. It reacts with the acid to form a simple salt and water only.

Chemical equation

Pb(OH) 2 (s) + 2HNO3(aq) -> Pb(NO3) 2 (aq) + 2H2O(l)

Ionic equation

Pb(OH) 2 (s) + 2H+ (aq) -> Pb 2+ (aq) + 2H2O(l)

(ii) when reacting with sodium hydroxide, the hydroxide shows acidic properties. It reacts with a base to form a complex salt.

Chemical equation

1.When Lead(II) hydroxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoplumbate(II) complex salt.

Pb(OH) 2 (s) + 2NaOH(aq) -> Na2Pb(OH) 4(aq)

2.When Lead(II) hydroxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoplumbate(II) complex salt.

  Pb(OH) 2 (s) + 2KOH(aq) -> K2Pb(OH) 4(aq)

Ionic equation

  Pb(OH) 2 (s) + 2OH(aq) -> 2[Pb(OH) 4]2- (aq)


(c)Aluminium(III)oxide(Al2O3) and Aluminium(III)hydroxide(Al(OH)3)

(i)When ½ spatula full of Aluminium(III)oxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e.

 (i) when reacting with nitric(V)acid, the oxide shows basic properties by reacting with an acid to form a simple salt and water only.

Chemical equation

Al2O3 (s) + 6HNO3(aq) -> Al(NO3)3 (aq) + 3H2O(l)

Al2O3 (s) + 6HCl(aq) -> AlCl3 (aq)  + 3H2O(l)

Al2O3 (s) + 3H2SO4(aq) -> Al2(SO4)3 (aq) + 3H2O(l)

Ionic equation

Al2O3 (s) + 3H+ (aq) -> Al 3+ (aq) + 3H2O(l)

 

(ii) when reacting with sodium hydroxide, the oxide shows acidic properties by reacting with a base to form a complex salt.

Chemical equation

1.When Aluminium(III) oxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoaluminate(III) complex salt.

 Al2O3 (s) + 2NaOH(aq) + 3H2O(l) -> 2NaAl(OH) 4(aq)

 

2.When Aluminium(III) oxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoaluminate(II) complex salt.  

 Al2O3 (s) + 2KOH(aq) + 3H2O(l) -> 2NaAl(OH) 4(aq)

Ionic equation

 Al2O3 (s) + 2OH(aq) + 3H2O(l) -> 2[Al(OH) 4] (aq)

(ii)When Aluminium(III)hydroxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e.

 (i) when reacting with nitric(V)acid, the hydroxide shows basic properties. It reacts with the acid to form a simple salt and water only.

Chemical equation

Al(OH) 3 (s) + 3HNO3(aq) -> Al(NO3)3 (aq) + 3H2O(l)

Al(OH)3 (s) + 3HCl(aq) -> AlCl3 (aq)   + 3H2O(l)

2Al(OH)3 (s) + 3H2SO4(aq) -> Al2(SO4)3 (aq) + 3H2O(l)

Ionic equation

Al(OH)3 (s) + 3H+ (aq) -> Al 3+ (aq) + 3H2O(l)

(ii) when reacting with sodium hydroxide, the hydroxide shows acidic properties. It reacts with a base to form a complex salt.

Chemical equation

1.When aluminium(III) hydroxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoaluminate(III) complex salt.

Al(OH) 3 (s) + NaOH(aq) -> NaAl(OH) 4(aq)

2.When aluminium(III) hydroxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoaluminate(III) complex salt.

  Al(OH) 3 (s) + KOH(aq) -> KAl(OH) 4(aq)

Ionic equation

  Al(OH) 3 (s) + OH(aq) -> [Al(OH) 4] (aq)

Summary of amphotellic oxides/hydroxides  

Oxide

Hydroxide

Formula of simple salt from nitric (V)acid

Formula of complex salt

from sodium hydroxide

ZnO

Zn(OH)2

Zn(NO3)2

Na2Zn(OH)4

[Zn(OH)4]2-(aq)

Sodium tetrahydroxozincate(II)

PbO

Pb(OH)2

Pb(NO3)2

Na2Pb(OH)4

[Pb(OH)4]2-(aq)

Sodium tetrahydroxoplumbate(II)

Al2O3

Al(OH) 3

Al(NO3)3

NaAl(OH)4

[Al(OH)4](aq)

Sodium tetrahydroxoaluminate(II)

12.(a) A salt is an ionic compound formed when the cation from a base combine with the anion derived from an acid.

A salt is therefore formed when the hydrogen ions in an acid are replaced wholly/fully or partially/partly ,directly or indirectly by a metal or ammonium radical. (b) The number of ionizable/replaceable hydrogen in an acid is called basicity of an acid.

Some acids are therefore:

(i)monobasic acids generally denoted HX e.g.

HCl, HNO3,HCOOH,CH3COOH.

(ii)dibasic acids generally denoted H2X e.g.

H2SO4, H2SO3, H2CO3,HOOCOOH.

(iii)tribasic acids generally denoted H3X e.g.

H3PO4.

(c) Some salts are normal salts while other are acid salts.

(i)A normal salt is formed when all the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical.

  (ii)An acid salt is formed when part/portion the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical.

Table showing normal and acid salts derived from common acids

Acid name

Chemical formula

Basicity

Normal salt

Acid salt

Hydrochloric acid

HCl

Monobasic

Chloride(Cl)

None

Nitric(V)acid

HNO3

Monobasic

Nitrate(V)(NO3)

None

Nitric(III)acid

HNO2

Monobasic

Nitrate(III)(NO2)

None

Sulphuric(VI)acid

H2SO4

Dibasic

Sulphate(VI) (SO42-)

Hydrogen sulphate(VI)

(HSO4)

Sulphuric(IV)acid

H2SO3

Dibasic

Sulphate(IV) (SO32-)

Hydrogen sulphate(IV)

(HSO3)

Carbonic(IV)acid

H2CO3

Dibasic

Carbonate(IV)(CO32-)

Hydrogen carbonate(IV)

(HCO3)

Phosphoric(V)

acid

H3PO4

Tribasic

Phosphate(V)(PO43-)

Dihydrogen phosphate(V)

(H2PO42-)

 

Hydrogen diphosphate(V)

(HP2O42-)

The table below show shows some examples of salts.

Base/alkali

Cation

Acid

Anion

Salt

Chemical name of salts

NaOH

Na+

HCl

Cl

NaCl

Sodium(I)chloride

Mg(OH)2

Mg2+

H2SO4

SO42-

MgSO4

Mg(HSO4)2

Magnesium sulphate(VI)

Magnesium hydrogen sulphate(VI)

Pb(OH)2

Pb2+

HNO3

NO3

Pb(NO3)2

Lead(II)nitrate(V)

Ba(OH)2

Ba2+

HNO3

NO3

Ba(NO3)2

Barium(II)nitrate(V)

Ca(OH)2

Ba2+

H2SO4

SO42-

MgSO4

Calcium sulphate(VI)

NH4OH

NH4+

H3PO4

PO43-

(NH4 )3PO4

(NH4 )2HPO4

NH4 H2PO4

Ammonium phosphate(V)

Diammonium phosphate(V)

Ammonium diphosphate(V)

KOH

K+

H3PO4

PO43-

K3PO4

Potassium phosphate(V)

Al(OH)3

Al3+

H2SO4

SO42-

Al2(SO4)2

Aluminium(III)sulphate(VI)

Fe(OH)2

Fe2+

H2SO4

SO42-

FeSO4

Iron(II)sulphate(VI)

Fe(OH)3

Fe3+

H2SO4

SO42-

Fe2(SO4)2

Iron(III)sulphate(VI)

(d) Some salts undergo hygroscopy, deliquescence and efflorescence.

(i) Hygroscopic salts /compounds are those that absorb water from the atmosphere but do not form a solution.

Some salts which are hygroscopic include anhydrous copper(II)sulphate(VI), anhydrous cobalt(II)chloride, potassium nitrate(V) common table salt.

(ii)Deliquescent salts /compounds are those that absorb water from the atmosphere and form a solution.

Some salts which are deliquescent include: Sodium nitrate(V),Calcium chloride, Sodium hydroxide, Iron(II)chloride, Magnesium chloride.

(iii)Efflorescent salts/compounds are those that lose their water of crystallization to the atmosphere.

Some salts which effloresces include: sodium carbonate decahydrate, Iron(II)sulphate(VI)heptahydrate, sodium sulphate (VI)decahydrate.

(e)Some salts contain water of crystallization.They are hydrated.Others do not contain water of crystallization. They are anhydrous.

Table showing some hydrated salts.

Name of hydrated salt

Chemical formula

Copper(II)sulphate(VI)pentahydrate

CuSO4.5H2O

Aluminium(III)sulphate(VI)hexahydrate

Al2 (SO4) 3.6H2O

Zinc(II)sulphate(VI)heptahydrate

ZnSO4.7H2O

Iron(II)sulphate(VI)heptahydrate

FeSO4.7H2O

Calcium(II)sulphate(VI)heptahydrate

CaSO4.7H2O

Magnesium(II)sulphate(VI)heptahydrate

MgSO4.7H2O

Sodium sulphate(VI)decahydrate

Na2SO4.10H2O

Sodium carbonate(IV)decahydrate

Na2CO3.10H2O

Potassium carbonate(IV)decahydrate

K2CO3.10H2O

Potassium sulphate(VI)decahydrate

K2SO4.10H2O

(f)Some salts exist as a simple salt while some as complex salts. Below are some complex salts.

Table of some complex salts

Name of complex salt

Chemical formula

Colour of the complex salt

Tetraamminecopper(II)sulphate(VI)

Cu(NH3) 4 SO4 H2O

Royal/deep blue solution

Tetraamminezinc(II)nitrate(V)

Zn(NH3) 4 (NO3 )2

Colourless solution

Tetraamminecopper(II) nitrate(V)

Cu(NH3) 4 (NO3 )2

Royal/deep blue solution

Tetraamminezinc(II)sulphate(VI)

Zn(NH3) 4 SO4

Colourless solution

(g)Some salts exist as two salts in one. They are called double salts.

 

Table of some double salts

Name of double salts

Chemical formula

Trona(sodium sesquicarbonate)

Na2CO3 NaHCO3.2H2O

Ammonium iron(II)sulphate(VI)

FeSO4(NH4) 2SO4.2H2O

Ammonium aluminium(III)sulphate(VI)

Al2(SO4) 3(NH4) 2SO4.H2O

(h)Some salts dissolve in water to form a solution. They are said to be soluble. Others do not dissolve in water. They form a suspension/precipitate in water.

Table of solubility of salts

Soluble salts

Insoluble salts

All nitrate(V)salts

 

All sulphate(VI)/SO42- salts except

Barium(II) sulphate(VI)/BaSO4

Calcium(II) sulphate(VI)/CaSO4

Lead(II) sulphate(VI)/PbSO4

All sulphate(IV)/SO32- salts except

Barium(II) sulphate(IV)/BaSO3

Calcium(II) sulphate(IV)/CaSO3

Lead(II) sulphate(IV)/PbSO3

All chlorides/Clexcept

Silver chloride/AgCl

Lead(II)chloride/PbCl2(dissolves in hot water)

All phosphate(V)/PO43-

 

All sodium,potassium and ammonium salts

 

All hydrogen carbonates/HCO3

 

All hydrogen sulphate(VI)/ HSO4

 

Sodium carbonate/Na2CO3,

potassium carbonate/ K2CO3,

ammonium carbonate (NH4) 2CO3

except All carbonates

All alkalis(KOH,NaOH, NH4OH)

except All bases

13 Salts can be prepared in a school laboratory by a method that uses its solubility in water.

  1. Soluble salts may be prepared by using any of the following methods:

(i)Direct displacement/reaction of a metal with an acid.

By reacting a metal higher in the reactivity series than hydrogen with a dilute acid,a salt is formed and hydrogen gas is evolved.

Excess of the metal must be used to ensure all the acid has reacted.

When effervescence/bubbling /fizzing has stopped ,excess metal is filtered.

The filtrate is heated to concentrate then allowed to crystallize.

Washing with distilled water then drying between filter papers produces a sample crystal of the salt. i.e.

M(s) + H2X  -> MX(aq) + H2(g)

Examples

Mg(s) + H2SO4(aq)  -> MgSO4 (aq)  + H2(g)

Zn(s) + H2SO4(aq)  -> ZnSO4 (aq)  + H2(g)

Pb(s) + 2HNO3(aq)  -> Pb(NO3) 2(aq)  + H2(g)

Ca(s) + 2HNO3(aq)  -> Ca(NO3) 2(aq)  + H2(g)

Mg(s) + 2HNO3(aq)  -> Mg(NO3) 2(aq)  + H2(g)

Mg(s) + 2HCl(aq)  -> MgCl 2(aq)  + H2(g)

Zn(s) + 2HCl(aq)  -> ZnCl 2(aq)  + H2(g)

(ii)Reaction of an insoluble base with an acid

By adding an insoluble base (oxide/hydroxide )to a dilute acid until no more dissolves, in the acid,a salt and water are formed. Excess of the base is filtered off. The filtrate is heated to concentrate ,allowed to crystallize then washed with distilled water before drying between filter papers e.g.

PbO(s) + 2HNO3(aq)  -> Pb(NO3) 2(aq)  + H2O (l)

Pb(OH)2(s) + 2HNO3(aq)  -> Pb(NO3) 2(aq)  + 2H2O (l)

CaO (s) + 2HNO3(aq)  -> Ca(NO3) 2(aq)  + H2O (l)

MgO (s) + 2HNO3(aq)  -> Mg(NO3) 2(aq)  + H2O (l)

MgO (s) + 2HCl(aq)  -> MgCl 2(aq)  + H2O (l)

ZnO (s) + 2HCl(aq)  -> ZnCl 2(aq)  + H2O (l)

Zn(OH)2(s) + 2HNO3(aq)  -> Zn(NO3) 2(aq)  + 2H2O (l)

CuO (s) + 2HCl(aq)  -> CuCl 2(aq)  + H2O (l)

CuO (s) + H2SO4(aq)  -> CuSO4(aq)  + H2O (l)

Ag2O(s) + 2HNO3(aq)  -> 2AgNO3(aq)  + H2O (l)

Na2O(s) + 2HNO3(aq)  -> 2NaNO3(aq)  + H2O (l)

(iii)reaction of insoluble /soluble carbonate /hydrogen carbonate with an acid.

By adding an excess of a soluble /insoluble carbonate or hydrogen carbonate to adilute acid, effervescence /fizzing/bubbling out of carbon(IV)oxide gas shows the reaction is taking place. When effervescence /fizzing/bubbling out of the gas is over, excess of the insoluble carbonate is filtered off. The filtrate is heated to concentrate ,allowed to crystallize then washed with distilled water before drying between filter paper papers e.g.

PbCO3 (s) + 2HNO3(aq)  -> Pb(NO3) 2(aq)  + H2O (l)+ CO2(g)

ZnCO3 (s) + 2HNO3(aq)  -> Zn(NO3) 2(aq)  + H2O (l)+ CO2(g)

CaCO3 (s) + 2HNO3(aq)  -> Ca(NO3) 2(aq)  + H2O (l)+ CO2(g)

MgCO3 (s) + H2SO4(aq)  -> MgSO4(aq)  + H2O (l)+ CO2(g)

Cu CO3 (s) + H2SO4(aq)  -> CuSO4(aq)  + H2O (l) + CO2(g)

Ag2CO3 (s) + 2HNO3(aq)  -> 2AgNO3(aq)  + H2O (l) + CO2(g)

 Na2CO3 (s) + 2HNO3(aq)  -> 2NaNO3(aq)  + H2O (l) + CO2(g)

 K2CO3 (s) + 2HCl(aq)   -> 2KCl(aq)  + H2O (l) + CO2(g)

NaHCO3 (s) + HNO3(aq)  -> NaNO3(aq)  + H2O (l) + CO2(g)

 KHCO3 (s) + HCl(aq)   -> KCl(aq) + H2O (l) + CO2(g)

(iv)neutralization/reaction of soluble base/alkali with dilute acid

By adding an acid to a burette into a known volume of an alkali with 2-3 drops of an indicator, the colour of the indicator changes when the acid has completely reacted with an alkali at the end point. The procedure is then repeated without the indicator .The solution mixture is then heated to concentrate , allowed to crystallize ,washed with distilled water before drying with filter papers. e.g.

NaOH (aq) + HNO3(aq)   -> NaNO3(aq)  + H2O (l)

KOH (aq) + HNO3(aq)   -> KNO3(aq)  + H2O (l)

 KOH (aq) + HCl(aq)   -> KCl(aq)  + H2O (l)

 2KOH (aq) + H2SO4(aq)   -> K2SO4(aq)  + 2H2O (l)

 2 NH4OH (aq) + H2SO4(aq)   -> (NH4)2SO4(aq) + 2H2O (l)

 NH4OH (aq) + HNO3(aq)   -> NH4NO3(aq) + H2O (l)

(iv)direct synthesis/combination.

When a metal burn in a gas jar containing a non metal , the two directly combine to form a salt. e.g.

2Na(s)  +  Cl2(g)  ->  2NaCl(s)

2K(s) +  Cl2(g)  ->  2KCl(s)

Mg(s) +  Cl2(g)  ->  Mg Cl2 (s)

Ca(s) +  Cl2(g)  ->  Ca Cl2 (s)

Some salts once formed undergo sublimation and hydrolysis. Care should be taken to avoid water/moisture into the reaction flask during their preparation.Such salts include aluminium(III)chloride(AlCl3) and iron (III)chloride(FeCl3)

1. Heated aluminium foil reacts with chlorine to form aluminium(III)chloride that sublimes away from the source of heating then deposited as solid again

 2Al(s) +  3Cl2(g)  ->  2AlCl3 (s/g)

Once formed aluminium(III)chloride hydrolyses/reacts with water vapour / moisture present to form aluminium hydroxide solution and highly acidic fumes of hydrogen chloride gas.

AlCl3(s)+  3H2 O(g)  ->  Al(OH)3 (aq) + 3HCl(g)

2. Heated iron filings reacts with chlorine to form iron(III)chloride that sublimes away from the source of heating then deposited as solid again

 2Fe(s) +  3Cl2(g)  ->  2FeCl3 (s/g)

Once formed , aluminium(III)chloride hydrolyses/reacts with water vapour / moisture present to form aluminium hydroxide solution and highly acidic fumes of hydrogen chloride gas.

FeCl3(s)+  3H2 O(g)  ->  Fe(OH)3 (aq) + 3HCl(g)

(b)Insoluble salts can be prepared by reacting two suitable soluble salts to form one soluble and one insoluble. This is called double decomposition or precipitation. The mixture is filtered and the residue is washed with distilled water then dried.

CuSO4(aq) + Na2CO3 (aq)  ->  CuCO3 (s) + Na2 SO4(aq)

BaCl2(aq)   + K2SO4 (aq) ->  BaSO4 (s) + 2KCl (aq)

 Pb(NO3)2(aq) + K2SO4 (aq) ->  PbSO4 (s) + 2KNO3 (aq)

 2AgNO3(aq)   + MgCl2 (aq) ->  2AgCl(s) + Mg(NO3)2 (aq)

 Pb(NO3)2(aq) + (NH4) 2SO4 (aq)  ->  PbSO4 (s) + 2NH4NO 3(aq)

 BaCl2(aq)   + K2SO3 (aq) ->  BaSO3 (s) + 2KCl (aq)

14. Salts may lose their water of crystallization , decompose ,melt or sublime on heating on a Bunsen burner flame.

The following shows the behavior of some salts on heating gently /or strongly in a laboratory school burner:

 (a)effect of heat on chlorides

All chlorides have very high melting and boiling points and therefore are not affected by laboratory heating except ammonium chloride. Ammonium chloride sublimes on gentle heating. It dissociate into the constituent ammonia and hydrogen chloride gases on strong heating.

NH4Cl(s) NH4Cl(g) NH3(g) + HCl(g)

  (sublimation) (dissociation)

 (b)effect of heat on nitrate(V)

(i) Potassium nitrate(V)/KNO3 and sodium nitrate(V)/NaNO3 decompose on heating to form Potassium nitrate(III)/KNO2 and sodium nitrate(III)/NaNO2 and producing Oxygen gas in each case.

2KNO3 (s)   -> 2KNO2(s) + O2(g)

2NaNO3 (s)   -> 2NaNO2(s) + O2(g)

(ii)Heavy metal nitrates(V) salts decompose on heating to form the oxide and a mixture of brown acidic nitrogen(IV)oxide and oxygen gases. e.g.

2Ca(NO3)2 (s)   -> 2CaO(s) + 4NO2(g) + O2(g)

2Mg(NO3)2(s)   -> 2MgO(s) + 4NO2(g) + O2(g)

 2Zn(NO3)2(s)   -> 2ZnO(s) + 4NO2(g) + O2(g)

2Pb(NO3)2(s)   -> 2PbO(s) + 4NO2(g) + O2(g)

2Cu(NO3)2(s)   -> 2CuO(s) + 4NO2(g) + O2(g)

2Fe(NO3)2(s)   -> 2FeO(s) + 4NO2(g) + O2(g)

(iii)Silver(I)nitrate(V) and mercury(II) nitrate(V) are lowest in the reactivity series. They decompose on heating to form the metal(silver and mercury)and the Nitrogen(IV)oxide and oxygen gas. i.e.

 2AgNO3(s)   -> 2Ag (s) + 2NO2(g) + O2(g)

 2Hg(NO3)2 (s)   -> 2Hg (s) + 4NO2(g) + O2(g)

(iv)Ammonium nitrate(V) and Ammonium nitrate(III) decompose on heating to Nitrogen(I)oxide(relights/rekindles glowing splint) and nitrogen gas respectively.Water is also formed.i.e.

NH4NO3(s)   -> N2O (g) + H2O(l)

NH4NO2(s)   -> N2 (g) + H2O(l)

 

(c) effect of heat on nitrate(V)

Only Iron(II)sulphate(VI), Iron(III)sulphate(VI) and copper(II)sulphate(VI) decompose on heating. They form the oxide, and produce highly acidic fumes of acidic sulphur(IV)oxide gas.

  2FeSO4 (s)   -> Fe2O3(s) + SO3(g) + SO2(g)

  Fe2(SO4) 3(s) -> Fe2O3(s) + SO3(g)

  CuSO4 (s)   -> CuO(s) + SO3(g)

 (d) effect of heat on carbonates(IV) and hydrogen carbonate(IV).

(i)Sodium carbonate(IV)and potassium carbonate(IV)do not decompose on heating.

(ii)Heavy metal nitrate(IV)salts decompose on heating to form the oxide and produce carbon(IV)oxide gas.
Carbon (IV)oxide gas forms a white precipitate when bubbled in lime water. The white precipitate dissolves if the gas is in excess. e.g. CuCO3 (s)   -> CuO(s) + CO2(g)

CaCO3 (s)   -> CaO(s) + CO2(g)

PbCO3 (s)   -> PbO(s) + CO2(g)

FeCO3 (s)   -> FeO(s) + CO2(g)

ZnCO3 (s)   -> ZnO(s) + CO2(g)

(iii)Sodium hydrogen carbonate(IV) and Potassium hydrogen carbonate(IV)decompose on heating to give the corresponding carbonate (IV) and form water and carbon(IV)oxide gas. i.e.

2NaHCO 3(s) -> Na2CO3(s) + CO2(g) + H2O(l)

2KHCO 3(s) -> K2CO3(s) + CO2(g) + H2O(l)

(iii) Calcium hydrogen carbonate (IV) and Magnesium hydrogen carbonate(IV) decompose on heating to give the corresponding carbonate (IV) and form water and carbon(IV)oxide gas. i. e.

Ca(HCO3) 2(aq) -> CaCO3(s) + CO2(g) + H2O(l)

 Mg(HCO3) 2(aq) -> MgCO3(s) + CO2(g) + H2O(l)

15. Salts contain cation(positively charged ion) and anions(negatively charged ion).When dissolved in polar solvents/water.

The cation and anion in a salt is determined/known usually by precipitation of the salt using a precipitating reagent.

The colour of the precipitate is a basis of qualitative analysis of a compound.

16.Qualitative analysis is the process of identifying an unknown compound /salt by identifying the unique qualities of the salt/compound.

It involves some of the following processes.

 (a)Reaction of cation with sodium/potassium hydroxide solution.

Both sodium/potassium hydroxide solutions are precipitating reagents.

The alkalis produce unique colour of a precipitate/suspension when a few/three drops is added and then excess alkali is added to unknown salt/compound solution.

NB: Potassium hydroxide is not commonly used because it is more expensive than sodium hydroxide.

The table below shows the observations, inferences / deductions and explanations from the following test tube experiments:

Procedure

Put about 2cm3 of MgCl2, CaCl2, AlCl3, NaCl, KCl, FeSO4, Fe2(SO4) 3, CuSO4, ZnSO4NH4NO3, Pb(NO3) 2, Ba(NO3) 2 each into separate test tubes. Add three drops of 2M sodium hydroxide solution then excess (2/3 the length of a standard test tube).

Observation

Inference

Explanation

No white precipitate

Na+ and K+

Both Na+ and K+ ions react with OHfrom 2M sodium hydroxide solution to form soluble colourless solutions

 

Na+(aq)
+ OH(aq) ->
NaOH(aq)

K+(aq)
+ OH(aq) ->
KOH(aq)

No white precipitate then pungent smell of ammonia /urine

NH4+ ions

NH4+ ions react with 2M sodium hydroxide solution to produce pungent smelling ammonia gas

 

NH4+ (aq)
+ OH(aq) ->
NH3 (g) + H2O(l)

White precipitate insoluble in excess

Ba2+ ,Ca2+, Mg2+ ions

Ba2+ ,Ca2+ and Mg2+ ions react with OHfrom 2M sodium hydroxide solution to form insoluble white precipitate of their hydroxides.

 

Ba2+(aq)
+ 2OH(aq) ->
Ba(OH) 2(s)

Ca2+(aq)
+ 2OH(aq) ->
Ca(OH) 2(s)

Mg2+(aq)
+ 2OH(aq) ->
Mg(OH) 2(s)

White precipitate soluble in excess

Zn2+ ,Pb2+, Al3+ ions

Pb2+ ,Zn2+ and Al3+ ions react with OHfrom 2M sodium hydroxide solution to form insoluble white precipitate of their hydroxides.

 

Zn2+(aq)
+ 2OH(aq) ->
Zn(OH) 2(s)

Pb2+(aq)
+ 2OH(aq) ->
Pb(OH) 2(s)

Al3+(aq)
+ 3OH(aq) ->
Al(OH) 3(s)

 

The hydroxides formed react with more OHions to form complex salts/ions.

 

Zn(OH) 2(s) + 2OH(aq) -> [ Zn(OH) 4]2-(aq)

Pb(OH) 2(s) + 2OH(aq) -> [ Pb(OH) 4]2-(aq)

Al(OH) 3(s) + OH(aq) -> [ Al(OH) 4](aq)

 


 

  

Blue precipitate insoluble in excess

Cu2+

Cu2+ ions react with OHfrom 2M sodium hydroxide solution to form insoluble blue precipitate of copper(II) hydroxide.

 

Cu2+(aq)
+ 2OH(aq) ->
Cu(OH) 2(s)

Green precipitate insoluble in excess

 

On adding 3cm3 of hydrogen peroxide, brown/yellow solution formed

Fe2+

 

 

 

 

Fe2+ oxidized to Fe3+

Fe2+ ions react with OHfrom 2M sodium hydroxide solution to form insoluble green precipitate of Iron(II) hydroxide.

 

Fe2+(aq)
+ 2OH(aq) ->
Fe(OH) 2(s)

Hydrogen peroxide is an oxidizing agent that oxidizes green Fe2+ oxidized to brown Fe3+

 

Fe(OH) 2(s) + 2H+ ->
Fe(OH) 3(aq)

Brown precipitate insoluble in excess

Fe3+

Fe3+ ions react with OHfrom 2M sodium hydroxide solution to form insoluble brown precipitate of Iron(II) hydroxide.

Fe3+(aq)
+ 3OH(aq) ->
Fe(OH) 3(s)

(b)Reaction of cation with aqueous ammonia

Aqueous ammonia precipitating reagent that can be used to identify the cations present in a salt.

Like NaOH/KOH the OH ion in NH4OH react with the cation to form a characteristic hydroxide .

Below are the observations ,inferences and explanations of the reactions of aqueous ammonia with salts from the following test tube reactions.

Procedure


Put about 2cm3 of MgCl2, CaCl2, AlCl3, NaCl, KCl, FeSO4, Fe2(SO4) 3, CuSO4, ZnSO4NH4NO3, Pb(NO3) 2, Ba(NO3) 2 each into separate test tubes.

Add three drops of 2M aqueous ammonia then excess (2/3 the length of a standard test tube).

 

Observation

Inference

Explanation

 


 

No white precipitate

Na+ and K+

NH4+,Na+ and K+ ions react with OHfrom 2M aqueous ammonia to form soluble colourless solutions

NH4+ (aq)
+ OH(aq) ->
NH4+OH(aq)

Na+(aq)
+ OH(aq) ->
NaOH(aq)

K+(aq)
+ OH(aq) ->
KOH(aq)

White precipitate insoluble in excess

Ba2+ ,Ca2+, Mg2+ ,Pb2+, Al3+, ions

Ba2+ ,Ca2+,Mg2+ ,Pb2+ and Al3+, ions react with OHfrom 2M aqueous ammonia to form insoluble white precipitate of their hydroxides.

 

Pb2+ (aq)
+ 2OH(aq) ->
Pb(OH) 2(s)

Al3+ (aq)
+ 3OH(aq) ->
Al(OH) 3(s)

Ba2+ (aq)
+ 2OH(aq) ->
Ba(OH) 2(s)

Ca2+ (aq)
+ 2OH(aq) ->
Ca(OH) 2(s)

Mg2+ (aq)
+ 2OH(aq) ->
Mg(OH) 2(s)

White precipitate soluble in excess

Zn2+ ions

Zn2+ ions react with OHfrom 2M aqueous ammonia to form insoluble white precipitate of Zinc hydroxide.

 

Zn2+(aq)
+ 2OH(aq) ->
Zn(OH) 2(s)

The Zinc hydroxides formed react NH3(aq) to form a complex salts/ions.

Zn(OH) 2(s) + 4NH3(aq)

->[ Zn(NH3) 4]2+(aq)+ 2OH(aq)

Blue precipitate that dissolves in excess ammonia solution to form a deep/royal blue solution

Cu2+

Cu2+ ions react with OHfrom 2M aqueous ammonia to form blue precipitate of copper(II) hydroxide.

 

Cu2+(aq)
+ 2OH(aq) ->
Cu(OH) 2(s)

The copper(II) hydroxide formed react NH3(aq) to form a complex salts/ions.

 

Cu(OH) 2 (s) + 4NH3(aq)

->[ Cu(NH3) 4]2+(aq)+ 2OH(aq)

Green precipitate insoluble in excess.

 

On adding 3cm3 of hydrogen peroxide, brown/yellow solution formed

Fe2+

 

 

 

 

Fe2+ oxidized to Fe3+

Fe2+ ions react with OHfrom 2M aqueous ammonia to form insoluble green precipitate of Iron(II) hydroxide.

Fe2+(aq)
+ 2OH(aq) ->
Fe(OH) 2(s)

 

Hydrogen peroxide is an oxidizing agent that oxidizes green Fe2+ oxidized to brown Fe3+

Fe(OH) 2(s) + 2H+ ->
Fe(OH) 3(aq)

Brown precipitate insoluble in excess

Fe3+

Fe3+ ions react with OHfrom 2M aqueous ammonia to form insoluble brown precipitate of Iron(II) hydroxide.

Fe3+(aq)
+ 3OH(aq) ->
Fe(OH) 3(s)

Note

(i) Only Zn2+ ions/salts form a white
precipitate that dissolve in excess of both 2M sodium hydroxide and 2M aqueous ammonia.

(ii) Pb2+ and Al3+ ions/salts form a white
precipitate that dissolve in excess of 2M sodium hydroxide but not in 2M aqueous ammonia.

(iii) Cu2+ ions/salts form a blue
precipitate that dissolve to form a deep/royal blue solution in excess of 2M aqueous ammonia but only blue insoluble precipitate in 2M sodium hydroxide

(c)Reaction of cation with Chloride (Cl)ions

All chlorides are soluble in water except Silver chloride and Lead (II)chloride (That dissolve in hot water).When a soluble chloride like NaCl, KCl, NH4Cl is added to about 2cm3 of a salt containing Ag+ or Pb2+ions a white precipitate of AgCl or PbCl2 is formed. The following test tube reactions illustrate the above.

Experiment

Put about 2cm3 of silver nitrate(V) andLead(II)nitrate(V)solution into separate test tubes. Add five drops of NaCl /KCl / NH4Cl/HCl. Heat to boil.

Observation

Inference

Explanation

(i)White precipitate does not dissolve on heating

Ag+ ions

Ag+ ions
reacts with Clions from a soluble chloride salt to form a white precipitate of AgCl

(ii)White precipitate dissolve on heating

Pb2+ ions

Pb2+ ions
reacts with Clions from a soluble chloride salt to form a white precipitate of PbCl2. PbCl2 dissolves on heating.

Note

Both Pb2+ and Al3+ ions forms an insoluble white precipitate in excess aqueous ammonia. A white precipitate on adding Clions/salts shows Pb2+.

No white precipitate on adding Clions/salts shows Al3+.

Adding a chloride/ Clions/salts can thus be used to separate the identity of Al3+ and Pb2+.

(d)Reaction of cation with sulphate(VI)/SO42- and sulphate(IV)/SO32- ions


All sulphate(VI) and sulphate(IV)/SO32- ions/salts are soluble/dissolve in water except Calcium sulphate(VI)/CaSO4, Calcium sulphate(IV)/CaSO3, Barium sulphate(VI)/BaSO4, Barium sulphate(IV)/BaSO3, Lead(II) sulphate(VI)/PbSO4
and Lead(II) sulphate(IV)/PbSO3.When a soluble sulphate(VI)/SO42- salt like Na2SO4, H2SO4, (NH4)2SO4 or Na2SO3 is added to a salt containing Ca2+, Pb2+, Ba2+ ions, a white precipitate is formed.

The following test tube experiments illustrate the above.

Procedure

Place about 2cm3 of Ca(NO3)2, Ba(NO3)2, BaCl2 and Pb(NO3)2, in separate boiling tubes. Add six drops of sulphuric(VI)acid /sodium sulphate(VI)/ammonium sulphate(VI)solution. Repeat with six drops of sodium sulphate(IV).

Observation

Inference

Explanation

White precipitate

Ca2+, Ba2+, Pb2+ ions

CaSO3 and CaSO4 do not form a thick precipitate as they are sparingly soluble.

Ca2+(aq)+ SO32-(aq) -> CaSO3(s)

Ca2+(aq)+ SO42-(aq) -> CaSO4(s)

 

Ba2+(aq)+ SO32-(aq) -> BaSO3(s)

Ba2+(aq)+ SO42-(aq) -> BaSO4(s)

 

Pb2+(aq)+ SO32-(aq) -> PbSO3(s)

Pb2+(aq)+ SO42-(aq) -> PbSO4(s)

(e)Reaction of cation with carbonate(IV)/CO32- ions

All carbonate salts are insoluble except sodium/potassium carbonate(IV) and ammonium carbonate(IV).

They dissociate /ionize to release
CO32- ions. CO32- ions produce a white precipitate when the soluble carbonate salts is added to any metallic cation.

Procedure

Place about 2cm3 of Ca(NO3)2, Ba(NO3)2, MgCl2 ,Pb(NO3)2 andZnSO4 in separate boiling tubes.

Add six drops of Potassium /sodium carbonate(IV)/ ammonium carbonate (IV)solution.

Observation

Inference

Explanation

Green precipitate

Cu2+ ,Fe2+,ions

 

CO32-(aq)

Copper(II)carbonate(IV) and Iron(II) carbonate (IV) are precipitated as insoluble green precipitates.

 

Cu2+(aq)+ CO32-(aq) -> CuCO3(s)

Fe2+(aq)+ CO32-(aq) -> FeCO3(s)

 

When sodium carbonate(IV)is added to CuCO3(s) the CO32-(aq) ions are first hydrolysed to produce CO2(g) and OH(aq)ions.

 

CO32-(aq) + H2O (l) -> CO2 (g) + 2OH (aq)

The OH(aq) ions further react to form basic copper(II) carbonate(IV). Basic copper(II) carbonate(IV) is the only green salt of copper.

Cu2+(aq)+ CO32-(aq)+2OH (aq)

->CuCO3.Cu(OH)2 (s)


 

White precipitate

CO32-

White ppt of the carbonate(IV)salt is precipitated

Ca2+(aq) + CO32- (aq)
-> CaCO3(s)

Mg2+(aq) + CO32- (aq)
-> MgCO3(s)

Pb2+(aq) + CO32- (aq)
-> PbCO3(s)

Zn2+(aq) + CO32- (aq)
-> ZnCO3(s)

Note

(i)Iron(III)carbonate(IV) does not exist.

(ii)Copper(II)Carbonate(IV) exist only as the basic CuCO3.Cu(OH) 2

(iii)Both BaCO3 and BaSO3 are insoluble white precipitate. If hydrochloric acid is added to the white precipitate;

I. BaCO3 produces CO2 gas. When bubbled/directed into lime water solution,a white precipitate is formed.

II. I. BaSO3 produces SO2 gas. When bubbled/directed into orange acidified potassium dichromate(VI) solution, it turns to green/decolorizes acidified potassium manganate(VII).

(f) Reaction of cation with sulphide / S2- ions

All sulphides are insoluble black solids/precipitates except sodium sulphide/ Na2S/ potassium sulphide/K2S.When a few/3drops of the soluble sulphide is added to a metal cation/salt, a black precipitate is formed.

Procedure

Place about 2cm3 of Cu(NO3)2, FeSO4, MgCl2,Pb(NO3)2 and ZnSO4 in separate boiling tubes.

Add six drops of Potassium /sodium sulphide solution.

Observation

Inference

Explanation

Black ppt

S2- ions

CuS, FeS,MgS,PbS, ZnS are black insoluble precipitates

Cu2+(aq) + S2-(aq) -> CuS(s)

Pb2+(aq) + S2-(aq) -> PbS(s)

Fe2+(aq) + S2-(aq) -> FeS(s)

Zn2+(aq) + S2-(aq) -> ZnS(s)

Sample qualitative analysis guide

You are provided with solid Y(aluminium (III)sulphate(VI)hexahydrate).Carry out the following tests and record your observations and inferences in the space provided.

1(a) Appearance

Observations inference (1mark)

 

White crystalline solid  Coloured ions Cu2+ , Fe2+ ,Fe3+ absent

 

(b)Place about a half spatula full of the solid into a clean dry boiling tube. Heat gently then strongly.

Observations inference (1mark)

 

Colourless droplets formed on the cooler Hydrated compound/compound

part of the test tube   containing water of crystallization

Solid remains a white residue  

(c)Place all the remaining portion of the solid in a test tube .Add about 10cm3 of distilled water. Shake thoroughly. Divide the mixture into five portions.

Observation Inference  (1mark)

Solid dissolves to form  Polar soluble compound

a colourless solution  Cu2+ , Fe2+ ,Fe3+ absent

(i)To the first portion, add three drops of sodium hydroxide then add excess of the alkali.

 Observation Inference  (1mark)

White ppt, soluble in excess  Zn2+ , Pb2+ , Al3+

 

(ii)To the second portion, add three drops of aqueous ammonia then add excess of the alkali.

 Observation Inference  (1mark)

White ppt, insoluble in excess   Pb2+ , Al3+

(iii)To the third portion, add three drops of sodium sulphate(VI)solution.

 Observation Inference  (1mark)

No white ppt   Al3+

(iv)I.To the fourth portion, add three drops of Lead(II)nitrate(IV)solution. Preserve

 Observation Inference  (1mark)

White ppt   CO32-, SO42-, SO32-, Cl,

II.To the portion in (iv) I above , add five drops of dilute hydrochloric acid.

 Observation Inference  (1mark)

White ppt persist/remains   SO42-, Cl,

III.To the portion in (iv) II above, heat to boil.

 Observation Inference  (1mark)

White ppt persist/remains   SO42-,

Note that:

(i)From test above, it can be deduced that solid Y is hydrated aluminium(III)sulphate(VI) solid

(ii)Any ion inferred from an observation below must be derived from previous correct observation and inferences above. e.g.

 Al3+ in c(iii) must be correctly inferred in either/or in c(ii) or c(i)above

SO42- in c(iv)III must be correctly inferred in either/or in c(iv)II or c(iv)I above

(iii)Contradiction in observations and inferences should be avoided.e.g.

“White ppt soluble in excess” to infer presence of Al3+ ,Ba2+ ,Pb3+

(iv)Symbols of elements/ions should be correctly capitalized. e.g.

 “SO4-2” is wrong, “sO42-” is wrong, “cu2+” is wrong.

Sample solutions of salt were labeled as I,II, III and IV. The actual solutions, not in that order are lead nitrate, zinc sulphate potassium chloride and calcium chloride.

a)When aqueous sodium carbonate was added to each sample separately, a white precipitate was formed in I, III and IV only. Identify solution II.

b)When excess sodium hydroxide was added to each sample separately, a white precipitate was formed in solutions III and I only.

Identify solution I

17.When solids/salts /solutes are added to a solvent ,some dissolve to form a solution.

 Solute  +  Solvent  ->  Solvent

If a solution has a lot of solute dissolved in a solvent ,it is said to be concentrated.

If a solution has little solute dissolved in a solvent ,it is said to be dilute.

 

There is a limit to how much solute can dissolve in a given /specified amount of solvent/water at a given /specified temperature.

The maximum mass of salt/solid/solute that dissolve in 100g of solvent/water at a specified temperature is called solubility of a salt.

When no more solute can dissolve in a given amount of solvent at a specified temperature, a saturated solution is formed.

For some salts, on heating, more of the salt/solid/solute dissolve in the saturated solution to form a super saturated solution.

The solubility of a salt is thus calculated from the formula

Solubility = Mass of solute/salt/solid x 100

   Mass/volume of water/solvent

Practice examples

(a)Calculate the solubility of potassium nitrate(V) if 5.0 g of the salt is dissolved in 50.0cm3 of water.

Solubility = Mass of solute/salt/solid x 100 =>( 5.0 x 100 ) = 10.0 g /100g H2O

   Mass/volume of water/solvent   50.0

(b)Calculate the solubility of potassium chlorate(V) if 50.0 g of the salt is dissolved in 250.0cm3 of water.

Solubility = Mass of solute/salt/solid x 100 =>( 50.0 x 100 ) = 20.0 g /100g H2O

   Mass/volume of water/solvent   250.0

(c)If the solubility of potassium chlorate(V) is 5g/100g H2O at 80oC,how much can dissolve in 5cm3 of water at 80oC .

Mass of solute/salt/solid = Solubility x Mass/volume of water/solvent

100

=> 5 x 5 = 0.25g of KClO3 dissolve

  100

(d)If the solubility of potassium chlorate(V) is 72g/100g H2O at 20oC,how much can saturate 25g of water at 20oC .

Mass of solute/salt/solid = Solubility x Mass/volume of water/solvent

100

=> 72 x 25 = 18.0g of KClO3 dissolve/saturate

  100

(e) 22g of potassium nitrate(V) was dissolved in 40.0g of water at 10oC. Calculate the solubility of potassium nitrate(V) at 10oC.

Solubility = Mass of solute/salt/solid x 100 =>( 22 x 100 ) = 55.0 g /100g H2O

   Mass/volume of water/solvent   40.0.

(f)What volume of water should be added to 22.0g of water at 10oC if the solubility of KNO3 at 10oC is 5.0g/100g H2O?


Solubility is mass/100g H2O => 22.0g + x = 100cm3/100g H2O

X= 100 – 22 = 78 cm3 of H2O

 

18. A graph of solubility against temperature is called solubility curve.

It shows the influence of temperature on solubility of different substances/solids/salts.

Some substances dissolve more with increase in temperature while for others dissolve less with increase in temperature

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Note:

(i)solubility of KNO3 and KClO3 increase with increase in temperature.

(ii)solubility of KNO3 is always higher than that of KClO3 at any specified temperature.

(iii)solubility of NaCl decrease with increase in temperature.

(iv)NaCl has the highest solubility at low temperature while KClO3 has the lowest solubility at low temperature.

(v)At point A both NaCl and KNO3 are equally soluble.

(vi)At point B both NaCl and KClO3 are equally soluble.

(vii) An area above the solubility curve of the salt shows a saturated /supersaturated solution.

(viii) An area below the solubility curve of the salt shows an unsaturated solution.

19.(a) For salts whose solubility increases with increase in temperature, crystals form when the salt solution at higher temperatures is cooled to a lower temperature.

  1. For salts whose solubility decreases with increase in temperature, crystals form when the salt solution at lower temperatures is heated to a higher temperature.

The examples below shows determination of the mass of crystals deposited with changes in temperature.


1.The solubility of KClO3

at 100oC is 60g/100g water .What mass of KClO3 will be deposited at:

(i)75 oC if the solubility is now 39g/100g water.

At 100oC   = 60.0g

Less at 75oC = – 39.0g

Mass of crystallized out 21.0g

(i)35 oC if the solubility is now 28 g/100g water.

At 100oC   = 60.0g

Less at 35oC = – 28.0.0g

Mass of crystallized out 32.0g

2. KNO3 has a solubility of 42 g/100g water at 20oC.The salt was heated and added 38g more of the solute which dissolved at100oC. Calculate the solubility of KNO3 at 100oC.

Solubility of KNO3 at 100oC = solubility at 20oC + mass of KNO3 added

  => 42g + 38g = 80g KNO3 /100g H2O

3. A salt solution has a mass of 65g containing 5g of solute. The solubility of this salt is 25g per 100g water at 20oC. 60g of the salt are added to the solution at 20oC.Calculate the mass of the solute that remain undissolved.

Mass of solvent at 20oC = mass of solution – mass of solute

=>  65 – 5 = 60g

Solubility before adding salt = mass of solute x 100

Volume of solvent

=> 5 x 100 = 8.3333g/100g water

60

Mass of solute to equalize with solubility = 25 – 8.3333g = 16.6667g

Mass of solute undissolved = 60.0 – 16.6667g = 43.3333 g

4. Study the table below

Salt


 

Solubility in gram at

50oC

20oC

KNO3

90

30

KClO3

20

6

(i)What happens when the two salts are dissolved in water then cooled from 50oC to 20oC.

(90 – 30) = 60.0 g of KNO3 crystals precipitate

(20 – 6) = 14.0 g of KClO3 crystals precipitate

(ii)State the assumption made in (i) above.

Solubility of one salt has no effect on the solubility of the other.


5. 10.0 g of hydrated potassium carbonate (IV) K2CO3.xH2O on heating leave 7.93 of the hydrate.

(a)Calculate the mass of anhydrous salt obtained.

 Hydrated on heating leave anhydrous = 7.93 g

(b)Calculate the mass of water of crystallization in the hydrated salt

 Mass of water of crystallization = hydrated – anhydrous

=> 10.0 – 7.93 = 2.07 g

(c)How many moles of anhydrous salt are there in 10of hydrate? (K= 39.0,C=12.0.O= 16.0)

Molar mass K2CO3= 138

Moles K2CO3 = mass of K2CO3 => 7.93 = 0.0515 moles

Molar mass K2CO3   138

(d)How many moles of water are present in the hydrate for every one mole of K2CO3 ? (H=1.0.O= 16.0)

Molar mass H2O
= 18

Moles H2O
= mass of H2O
=> 2.07 = 0.115 moles

Molar mass H2O   18

Mole ratio H2O : K2CO3 = 0.115 moles 2 = 2

0.0515 moles   1

(e)What is the formula of the hydrated salt?

K2CO3 .2 H2O

6. The table below shows the solubility of Potassium nitrate(V) at different temperatures.

Temperature(oC)

5.0

10.0

15.0

30.0

40.0

50.0

60.0

mass KNO3/ 100g water

15.0

20.0

25.0

50.0

65.0

90.0

120.0

(a)Plot a graph of mass of in 100g water(y-axis) against temperature in oC

 

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(b)From the graph show and determine

 (i)the mass of KNO3 dissolved at:

I. 20oC

From a correctly plotted graph = 32g

II. 35oC

From a correctly plotted graph = 57g

III. 55oC

From a correctly plotted graph = 104g

 (ii)the temperature at which the following mass of KNO3 dissolved:

I. 22g

From a correctly plotted graph =13.0oC

II. 30g

From a correctly plotted graph =17.5oC

III.100g

From a correctly plotted graph =54.5oC

(c)Explain the shape of your graph.

Solubility of KNO3
increase with increase in temperature/More KNO3 dissolve as temperature rises.

(d)Show on the graph the supersaturated and unsaturated solutions.

 Above the solubility curve write; “supersaturated”

 Below the solubility curve write; “unsaturated”

(e)From your graph, calculate the amount of crystals obtained when a saturated solution of KNO3 containing 180g of the salt is cooled from 80oC to:

 I. 20oC

Solubility before heating = 180 g

 Less Solubility after heating(from the graph) = 32 g

Mass of KNO3crystals   = 148 g

 II. 35oC

Solubility before heating = 180 g

 Less Solubility after heating(from the graph) = 58 g

Mass of KNO3crystals   = 122 g

 III. 55oC

Solubility before heating = 180 g

 Less Solubility after heating(from the graph) = 102 g

Mass of KNO3crystals   = 78 g

7. The table below shows the solubility of salts A and B at various temperatures.

 

Temperature(oC)

0.0

10.0

20.0

30.0

40.0

50.0

60.0

70.0

80.0

Solubility of A

28.0

31.0

34.0

37.0

40.0

43.0

45.0

48.0

51.0

Solubility of B

13.0

21.0

32.0

46.0

64.0

85.0

110.0

138.0

169.0

(a)On the same axis plot a graph of solubility (y-axis) against temperature for each salt.

(b)At what temperature are the two salts equally soluble.

The point of intersection of the two curves = 24oC

(c)What happens when a mixture of 100g of salt B with 100g if water is heated to 80oC

From the graph, the solubility of B at 80oC is 169g /100g water. All the 100g crystals of B dissolve.


(d)What happens when the mixture in (c) above is then cooled from 50oC to 20oC.

Method I.

Total mass before cooling at 50oC = 100.0 g

 (From graph) Solubility/mass after cooling at 20oC  = 32.0 g

 Mass of crystals deposited 68.0 g

Method II.

Mass of soluble salt crystals at 50oC added   = 100 g

(From graph)Solubility/mass before cooling at 50oC = 85.0 g

Mass of crystals that cannot dissolve at 50oC 15.0 g

(From graph) Solubility/mass before cooling at 50oC   = 85.0 g

(From graph) Solubility/mass after cooling at 20oC   = 32.0 g

Mass of crystals deposited after cooling 53.0 g

Total mass of crystals deposited = 15.0 + 53.0  = 68.0 g

(e)A mixture of 40g of A and 60g of B is added to 10g of water and heated to 70oC.The solution is then allowed to cool to 10oC.Describe clearly what happens.

I.For salt A

Solubility of A before heating = mass of A x 100

Volume of water added

  => 40 x 100 = 400g/100g Water

  10

(Theoretical)Solubility of A before heating   = 400 g

Less (From graph ) Solubility of A after heating at 70oC  = 48g

Mass of crystals that can not dissolve at70oC = 352 g

(From graph ) Solubility of A after heating at 70oC  = 48g

Less (From graph ) Solubility of A after cooling to 10oC  = 31g

Mass of crystals that crystallize out on cooling to10oC = 17 g

Mass of crystals that can not dissolve at70oC = 352 g

Add Mass of crystals that crystallize out on cooling to10oC = 17 g

Total mass of A that does not dissolve/crystallize/precipitate = 369 g

 

I.For salt B

Solubility of B before heating = mass of B x 100

Volume of water added

  => 60 x 100 = 600g/100g Water

  10

(Theoretical)Solubility of B before heating   = 600 g

Less (From graph ) Solubility of B after heating at 70oC  = 138g

Mass of crystals that cannot dissolve at70oC = 462 g

(From graph ) Solubility of B after heating at 70oC  = 138g

Less (From graph ) Solubility of B after cooling to 10oC  = 21g

Mass of crystals that crystallize out on cooling to10oC = 117 g

Mass of crystals that cannot dissolve at70oC = 462 g

Add Mass of crystals that crystallize out on cooling to10oC = 117 g

Total mass of A that does not dissolve/crystallize/precipitate = 579 g

(f)State the assumption made in (e)above

Solubility of one salt has no effect on the solubility of the other

8. When 5.0 g of potassium chlorate (V) was put in 10cm3 of water and heated, the solid dissolves. When the solution was cooled , the temperature at which crystals reappear was noted. Another 10cm3 of water was added and the mixture heated to dissolve then cooled for the crystals to reappear .The table below shows the the results obtained

Total volume of water added(cm3)

10.0

20.0

30.0

40.0

50.0

Mass of KClO3

5.0

5.0

5.0

5.0

5.0

Temperature at which crystals appear

80.0

65.0

55.0

45.0

30.0

Solubility of KclO3

50.0

25.0

16.6667

12.5

10.0

(a)Complete the table to show the solubility of KclO3 at different temperatures.

(b)Plot a graph of mass of KClO3 per 100g water against temperature at which crystals form.

(c)From the graph, show and determine

(i)the solubility of KClO3 at

 I. 50oC

From a well plotted graph = 14.5 g KClO3/100g water

 II. 35oC

From a well plotted graph = 9.0 g KclO3/100g water

 (ii)the temperature at which the solubility is:

I.10g/100g water

From a well plotted graph = 38.0 oC

II.45g/100g water

From a well plotted graph = 77.5 oC

 

(d)Explain the shape of the graph.

Solubility of KClO3 increase with increase in temperature/more KclO3dissolve as temperature rises.

(e)What happens when 100g per 100g water is cooled to 35.0 oC

Solubility before heating = 100.0

(From the graph)  Solubility after cooling = 9.0

Mass of salt precipitated/crystallization = 91.0 g

9. 25.0cm3 of water dissolved various masses of ammonium chloride crystals at different temperatures as shown in the table below.

Mass of ammonium chloride(grams)

4.0

4.5

5.5

6.5

9.0

Temperature at which solid dissolved(oC)

30.0

50.0

70.0

90.0

120.0

Solubility of NH4Cl

16.0

18.0

22.0

26.0

36.0


(a)Complete the table

(b)Plot a solubility curve

(c)What happens when a saturated solution of ammonium chloride is cooled from 80oC to 40oC.

(From the graph )Solubility at 80oC = 24.0 g

Less (From the graph )Solubility at 40oC = 16.8 g

Mass of crystallized/precipitated = 7.2 g

20. Solubility and solubility curves are therefore used


(i) to know the effect of temperature on the solubility of a salt

(ii)to fractional crystallize two soluble salts by applying their differences in solubility at different temperatures.

(iii)determine the mass of crystal that is obtained from crystallization.

21.Natural fractional crystallization takes place in Kenya/East Africa at:

(i) Lake Magadi during extraction of soda ash(Sodium carbonate) from Trona(sodium sesquicarbonate)

(ii) Ngomeni near Malindi at the Indian Ocean Coastline during the extraction of common salt(sodium chloride).  

22.Extraction of soda ash from Lake Magadi in Kenya

Rain water drains underground in the great rift valley and percolate underground where it is heated geothermically.

The hot water dissolves underground soluble sodium compounds and comes out on the surface as alkaline springs which are found around the edges of Lake Magadi in Kenya.

Temperatures around the lake are very high (30-40oC) during the day.

The solubility of trona decrease with increase in temperature therefore solid crystals of trona grows on top of the lake (upto or more than 30metres thick)

 

A bucket dredger mines the trona which is then crushed ,mixed with lake liquor and pumped to washery plant where it is further refined to a green granular product called CRS.

The CRS is then heated to chemically decompose trona to soda ash(Sodium carbonate)

Chemical equation

2Na2CO3.NaHCO3.2H2O(s) -> 3Na2CO3 (s) + CO2(g) + 5H2O(l)

Soda ash(Sodium carbonate) is then stored .It is called Magadi Soda. Magadi Soda is used :

  1. make glass
  2. for making soapless detergents
  3. softening hard water.

Common salt is colledcted at night because its solubility decreases with decrease in temperature. It is used as salt lick/feed for animals.

Summary flow diagram showing the extraction of Soda ash from Trona

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23.Extraction of common salt from Indian Ocean at Ngomeni in Kenya

Oceans are salty.They contain a variety of dissolved salts (about 77% being sodium chloride).

During high tide ,water is collected into shallow pods and allowed to crystallize as evaporation takes place.The pods are constructed in series to increase the rate of evaporation.

At the final pod ,the crystals are scapped together,piled in a heap and washed with brine (concentrated sodium chloride).

It contains MgCl2 and CaCl2 . MgCl2 and CaCl2are hygroscopic. They absorb water from the atmosphere and form a solution.

This makes table salt damp/wet on exposure to the atmosphere.

24.Some water form lather easily with soap while others do not.

Water which form lather easily with soap is said to be “soft

Water which do not form lather easily with soap is said to be “hard

Hardness of water is caused by the presence of Ca2+ and Mg2+ ions.

Ca2+ and Mg2+ ions react with soap to form an insoluble grey /white suspension/precipitate called Scum/ curd. Ca2+ and Mg2+ ions in water come from the water sources passing through rocks containing soluble salts of Ca2+ and Mg2+ e.g. Limestone or gypsum

There are two types of water hardness:

 (a)temporary hardness of water

 (b)permanent hardness of water

(a)temporary hardness of water

Temporary hardness of water is caused by the presence of dissolved calcium hydrogen carbonate/Ca(HCO3)2 and magnesium hydrogen carbonate/Mg(HCO3)2

When rain water dissolve carbon(IV) oxide from the air it forms waek carbonic(IV) acid i.e.

CO2(g) + H2O(l) -> H2CO3(aq)

When carbonic(IV) acid passes through limestone/dolomite rocks it reacts to form soluble salts i.e.

In limestone areas; H2CO3(aq) + CaCO3(s) ->
Ca(HCO3)2 (aq)

In dolomite areas; H2CO3(aq) + MgCO3(s) ->
Mg(HCO3)2 (aq)

(b)permanent hardness of water

Permanent hardness of water is caused by the presence of dissolved calcium sulphate(VI)/CaSO4 and magnesium sulphate(VI)/Mg SO4 Permanent hardness of water is caused by water dissolving CaSO4 and MgSO4 from ground rocks.

Hardness of water can be removed by the following methods:

(a)Removing temporary hardness of water

(i)Boiling/heating.

Boiling decomposes insoluble calcium hydrogen carbonate/Ca(HCO3)2 and magnesium hydrogen carbonate/Mg(HCO3)2
to insoluble CaCO3
and MgCO3 that precipitate away. i.e

Chemical equation

Ca(HCO3)2(aq) -> CaCO3 (s) + CO2(g) + H2O(l)

Mg(HCO3)2(aq) -> MgCO3 (s) + CO2(g) + H2O(l)

(ii)Adding sodium carbonate (IV) /Washing soda.

Since boiling is expensive on a large scale ,a calculated amount of sodium carbonate decahydrate /Na2CO3.10H2O precipitates insoluble Ca2+(aq) and Mg2+(aq) ions as carbonates to remove both temporary and permanent hardness of water .This a double decomposition reaction where two soluble salts form an insoluble and soluble salt. i.e.

(i)with temporary hard water

Chemical equation

Na2CO3 (aq)
+ Ca(HCO3) 2 (aq) -> NaHCO3(aq) + CaCO3 (s)

Na2CO3 (aq)
+ Mg(HCO3) 2 (aq) -> NaHCO3(aq) + MgCO3 (s)

Ionic equation

  CO32-
(aq)
+ Ca2+

(aq) -> CaCO3 (s)

CO32-
(aq)
+ Mg2+

(aq) -> MgCO3 (s)

(ii)with permanent hard water

Chemical equation

Na2CO3 (aq)
+ MgSO4

(aq)  -> Na2SO4 (aq) + MgCO3 (s)

 Na2CO3 (aq)
+ CaSO4

(aq)  -> Na2SO4 (aq) + MgCO3 (s)

Ionic equation

  CO32-
(aq)
+ Ca2+

(aq) -> CaCO3 (s)

 CO32-
(aq)
+ Mg2+

(aq) -> MgCO3 (s)

(iii)Adding calcium (II)hydroxide/Lime water

Lime water/calcium hydroxide removes only temporary hardness of water from by precipitating insoluble calcium carbonate(IV).

Chemical equation

Ca(OH)2 (aq)
+ Ca(HCO3) 2 (aq) -> 2H2O(l) + 2CaCO3 (s)

Excess of Lime water/calcium hydroxide should not be used because it dissolves again to form soluble calcium hydrogen carbonate(IV) causing the hardness again.

 (iv)Adding aqueous ammonia

Aqueous ammonia removes temporary hardness of water by precipitating insoluble calcium carbonate(IV) and magnesium carbonate(IV)

Chemical equation

2NH3 (aq)
+ Ca(HCO3) 2 (aq) -> (NH4) 2CO3(aq) + CaCO3 (s)

 2NH3 (aq)
+ Mg(HCO3) 2 (aq) -> (NH4) 2CO3(aq) + MgCO3 (s)

(v)Use of ion-exchange permutit

This method involves packing a chamber with a resin made of insoluble complex of sodium salt called sodium permutit.

The sodium permutit releases sodium ions that are exchanged with Mg2+ and Ca2+ ions in hard water making the water to be soft. i.e.

 Na2X(aq) + Ca2+ (aq) -> Na+ (aq)  + CaX(s)

Na2X(aq) + Mg2+ (aq) -> Na+ (aq)  + MgX(s)

 

Hard water containing Mg2+ and Ca2+

 

 

 

 

 

 Ion exchange resin as

 Sodium permutit

 

 

 

 

 

 

 

 

 

 

——-   Na+ ions replace Mg2+


and Ca2+ to make the water soft.

When all the Na+ ions in the resin is fully exchanged with Ca2+ and Ng2+ ions in the permutit column ,it is said to be exhausted.

Brine /concentrated sodium chloride solution is passed through the permutit column to regenerated /recharge the column again.

Hard water containing Mg2+ and Ca2+

 

 

 

 

 

 Ion exchange resin as

 Sodium permutit

 

 

 

 

 

 

 

 

 

 

——-   Na+ ions replace Mg2+


and Ca2+ to make the water soft.

 

(vi)Deionization /demineralization

This is an advanced ion exchange method of producing deionized water .Deionized water is extremely pure water made only of hydrogen and oxygen only without any dissolved substances.

Deionization involve using the resins that remove all the cations by using:

(i)A cation exchanger which remove /absorb all the cations present in water and leave only H+ ions.

(ii)An anion exchanger which remove /absorb all the anions present in water and leave only OH ions.

The H+(aq) and OH
(aq) neutralize each other to form pure water.

Chemical equation

H+(aq) + OH
(aq) -> H2O(l)

When exhausted the cation exchanger is regenerated by adding H+(aq) from sulphuric(VI)acid/hydrochloric acid.

When exhausted the anion exchanger is regenerated by adding OH(aq) from sodium hydroxide.

Advantages of hard water

Hard water has the following advantages:

(i)Ca2+(aq) in hard water are useful in bone and teeth formation

  (ii) is good for brewing beer

  (iii)contains minerals that cause it to have better /sweet taste

  (iv)animals like snails and coral polyps use calcium to make their shells and coral reefs respectively.

  (v)processing mineral water

Disadvantages of hard water

Hardness of water:

 (i)waste a lot of soap during washing before lather is formed.

 (ii)causes stains/blemishes/marks on clothes/garments

(iii)causes fur on electric appliances like kettle ,boilers and pipes form decomposition of carbonates on heating .This reduces their efficiency hence more/higher cost of power/electricity.

 

 

 

 

Sample revision questions

In an experiment, soap solution was added to three separate samples of water. The table below shows the volumes of soap solution required to form lather with 1000cm3 of each sample of water before and after boiling.

 

Sample I

Sample II

Sample III

Volume of soap before water is boiled (cm3)

27.0

3.0

10.0

Volume of soap after water is boiled(cm3)

27.0

3.0

3.0

 a)  Which water sample is likely to be soft? Explain.  (2mks)

Sample II: Uses little sample of soap .

 c)  Name the change in the volume of soap solution used in sample III  (1mk)

On heating the sample water become soft bcause it is temporary hard.

2.Study the scheme below and use it to aanswer the questions that follow:

Image From EcoleBooks.com

(a)Write the formula of:

(i)Cation in solution K

 Al3+

(ii)white ppt L

 Al(OH)3

(iii) colourless solution M

 [Al(OH)4]

(iv) colourless solution N

AlCl3

(v)white ppt P

 Al(OH)3

 

(b)Write the ionic equation for the reaction for the formation of:

(i)white ppt L

Al3+(aq)
+ 3OH(aq) -> Al(OH)3(s)

(v)white ppt P

 Al3+(aq)
+ 3OH(aq) -> Al(OH)3(s)

(c)What property is illustrated in the formation of colourless solution M and N.

Amphotellic

 


 




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EcoleBooks | Chemistry Form 4 Notes : ACIDS, BASES AND SALTS

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