the point at which the acid has been neutralised. This process is done in three stages. The acid is poured into the burette. A known volume of the alkali is transferred using a pipette into four separate conical flasks. Two or three drops of suitable indicator are added to each of the conical flasks. The acid is run into the flask until the indicator just changes colour. The difference between the initial and final burette readings gives the volume of the acid used commonly known as the titre. The first titration usually gives and approximate end point and is treated as the trial. The experiment is repeated for each of the other conical flask, to try to obtain the end point accurately. This is usually done by running out the acid to a point one unit away from the trial, the going drop-by-drop until an accurate end point is obtained. The volume of the acid required for each change is read off and recorded in a table similar to the one below. PROCEDURE DURING TITRATION
Titrations require continuous shaking of the conical flask and its contents. Check the quantity of the pipette as indicated on the bulb of the pipette and remember to record it in the appropriate place. As you release the liquid from a pipette into a conical flask, one should not blow out the last drop remaining in the jet. Before you take any reading from the burette after filling it, first allow the solution to run out to fill the tap and jet of the burette, and then you begin taking your readings. Always write the reading immediately you take them. When completing the table of results, you will be expected to complete all columns as accurately as the as the limits of the apparatus can allow e.g. burette used is usually read to the nearest read to the nearest 0.05 cm 3, pipette is accurate to 0.05 cm 3 (1 drop). Use minimum amount of indicator possible (2 – 3 drops) and recognize the end point has been reached e.g. when the colour just changes. Show the values that can be averaged to obtain an acceptable value for use in calculations (only those values within 0.2 cm 3 should be averaged). Attempt to work out the questions from the first principles and not use the formula method, which has its own limitations.


when the colour just changes. Show the values that can be averaged to obtain an acceptable value for use in calculations (only those values within 0.2 cm 3 should be averaged). Attempt to work out the questions from the first principles and not use the formula method, which has its own limitations. GENERAL NOTES FOR QUALITATIVE AND QUANTITIVE ANALYSIS
Indicators
An indicator is a substance used in titrations, which has one colour in the presence of an excess of one reagent and a different colour in the presence of an excess of the other. Examples of indicators include;
Methyl orange
Phenolphthalein
Methyl red
Bromothymol blue
Indicator colour changes
Candidates may be asked to carry out exercises involving:
Simple quantitative experiments involving the measurement of volumes:
Speeds of reaction. Measurement of temperature based on a thermometer with 10C graduations. Problems of an investigatory nature, possibly including suitable organic compounds. Filtration. Identification of ions and gasses as specified in the curriculum. QUALITATIVE ANALYSIS
Qualitative analysis is mainly about identification of substances. No emphasis is laid on amount or quantity. In qualitative analysis exercises, candidates should use approximately 1cm depth of a solution (1-2cm3) for each test and add reagents slowly, ensuring good mixing, until no further change is seen. Candidates should indicate at what stage a change occurs, writing any deductions alongside the observations on which they are based. Answers should include details of colour changes and precipitates formed and the names and chemical tests for any gases evolved. Marks for deductions or conclusions can only be gained if the appropriate observations are recorded. QUALITATIVE ANALYSIS NOTES
TABLE 1:TESTING FOR GASES
TABLE 2: TEST FOR ANIONS
TABLE 3: TESTS FOR CATIONS
TABLE 4: ORGANIC TESTS
TABLE 5: MISCELLANEOUS TESTS
TABLE 6. FLAME TESTS
A tungsten wire loop is first dipped into some concentrated hydrochloric acid to dissolve any oxides and hence clean the wire. It is then dipped into salt powder and introduced into a colourless Bunsen burner flame. The colour is then observed and inferences made.


FLAME TESTS
A tungsten wire loop is first dipped into some concentrated hydrochloric acid to dissolve any oxides and hence clean the wire. It is then dipped into salt powder and introduced into a colourless Bunsen burner flame. The colour is then observed and inferences made. TABLE 7: HEAT ANALYSIS
TABLE 8: SOLUBLE AND INSOLUBLE SALTS
TABLE 9: GENERAL PRELIMINARY TESTS
TABLE 10 : IGNITION ANALYSIS
Candidates may be asked to heat an unknown alone in an ignition tube. The colour changes or the identity of any gases evolved may provide evidence as to the identity of the unknown. TABLE 10: RESIDUE WHEN HOT AND COLD
Residue remaining after ignition may have a different colour when hot and cold
TABLE 11: HEATING A SOLID
Gases or vapours may be evolved during heating of the solid
TABLE 12: SOLUBILITY
The patterns of solubility for various types of salts
KEY
Ppt = precipitate
Sol = solution
Insol = insoluble
Xs = excess
Dil = dilute
1.HYDROGEN CHLORIDE
Physical Properties
Chemical Properties
Footnotes:
Dilute hydrochloric acid is one of the three common dilute acids used in the laboratory. Concentrated hydrochloric acid is used in the manufacture of many chemicals. 2.CARBON DIOXIDE
Physical Properties
Chemical Properties
Footnotes:
Makes up 0.04% of the gases in the air. Rainwater is slightly acidic because carbon dioxide dissolves in it. Needed by green plants during photosynthesis. Product of respiration in living things. Product of combustion
Is one of the "greenhouse" gases present in the atmosphere. Used in fire extinguishers since it is heavier than air and forms a "blanket" around the fire. This prevents combustion since carbon dioxide does not support it and the presence of carbon dioxide stops oxygen reaching the combustible material. 3.CHLORINE GAS
Physical Properties
Chemical Properties
Footnotes:
One of the "family" of halogen gases (iodine and bromine are in the same family)
Used in water purification. Used to make bleaching powder, disinfectants and antiseptics
Also used to make some explosives, poison gases and pesticides.


This prevents combustion since carbon dioxide does not support it and the presence of carbon dioxide stops oxygen reaching the combustible material.

3.CHLORINE GAS
Physical Properties
Chemical Properties
Footnotes:
One of the "family" of halogen gases (iodine and bromine are in the same family)
Used in water purification.

Used to make bleaching powder, disinfectants and antiseptics
Also used to make some explosives, poison gases and pesticides.

4.HYDROGEN CHLORIDE GAS
Physical Properties
Chemical Properties
Footnotes:
Dilute hydrochloric acid is one of the three common dilute acids used in the laboratory.

Concentrated hydrochloric acid is used in the manufacture of many chemicals.

5.HYDROGEN GAS
Physical Properties
Chemical Properties
Footnotes:
The lightest gas known.

Once used in airships but replaced by helium which is not explosive.

Used to make ammonia which is needed in the manufacture of fertilizers and explosives.

6.NITROGEN
Physical Properties
Chemical Properties
Footnotes:
Makes up around 79% of the gases in the air.

Used to make ammonia gas, which in turn is used to make explosives and fertilizers.

Used in light bulbs and thermometers because it is not reactive.

Also used as the atmosphere in rooms where explosives are stored.

7.OXYGEN
Physical Properties
Chemical Properties
Footnotes:
Makes up 20% of the gases in the air.

Needed by the majority of living organisms for respiration
Is produced by green plants as a by-product of photosynthesis.

8.SULPHUR DIOXIDE GAS
Physical Properties
Chemical Properties
Footnotes:
It is used as a bleaching agent.

It is one of the gases responsible for air pollution.


18.0.0 ACIDS, BASES AND SALTS (25 LESSONS) 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. 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 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 BronstedLowry 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.


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.


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.


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 Cl- ions. 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.


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.


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.


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.


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.


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.


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.


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.


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.


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.


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.


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.


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.


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) H3PO4 Tribasic Phosphate(V)(PO43-) Dihydrogen
acid phosphate(V) (H2PO42-) Hydrogen diphosphate(V) (HP2O42-) The table below show shows some examples of 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) H3PO4 Tribasic Phosphate(V)(PO43-) Dihydrogen
acid 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.


(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) H3PO4 Tribasic Phosphate(V)(PO43-) Dihydrogen
acid 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.


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) H3PO4 Tribasic Phosphate(V)(PO43-) Dihydrogen
acid 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.


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.


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 OH- from 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 OH- from 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 OH- from 2M sodium hydroxide solution to form insoluble white precipitate of their hydroxides.


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 OH- from 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 OH- from 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 OH- from 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 OH- ions to form complex salts/ions.


Observation Inference Explanation No white precipitate Na+ and K+ Both Na+ and K+ ions react with OH- from 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 OH- from 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 OH- from 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 OH- ions 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 OH- from 2M sodium hydroxide solution to form insoluble blue precipitate of copper(II) hydroxide.


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 OH- from 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 OH- ions 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 OH- from 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 OH- from 2M sodium hydroxide solution to form insoluble green precipitate of Iron(II) hydroxide.


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 OH- ions 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 OH- from 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 OH- from 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 OH- from 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 .


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 OH- from 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 OH- from 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 OH- from 2M aqueous ammonia to form insoluble white precipitate of their hydroxides.


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 OH- from 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 OH- from 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 OH- from 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.


Observation Inference Explanation No white precipitate Na+ and K+ NH4+,Na+ and K+ ions react with OH- from 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 OH- from 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 OH- from 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 Cu2+ Cu2+ ions react with OH- from 2M aqueous
that dissolves in excess ammonia solution to form a deep/royal blue solution 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.


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 Cu2+ Cu2+ ions react with OH- from 2M aqueous
that dissolves in excess ammonia solution to form a deep/royal blue solution 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 OH- from 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 OH- from 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.


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 OH- from 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 Cl- ions from a soluble chloride salt to form a white precipitate of AgCl (ii)White precipitate dissolve on heating Pb2+ ions Pb2+ ions reacts with Cl- ions 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.


Observation Inference Explanation (i)White precipitate does not dissolve on heating Ag+ ions Ag+ ions reacts with Cl- ions from a soluble chloride salt to form a white precipitate of AgCl (ii)White precipitate dissolve on heating Pb2+ ions Pb2+ ions reacts with Cl- ions 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 Cl- ions/salts shows Pb2+. No white precipitate on adding Cl- ions/salts shows Al3+. Adding a chloride/ Cl- ions/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 Ca2+, Ba2+, CaSO3 and CaSO4 do not form a thick precipitate as
precipitate Pb2+ ions they are sparingly soluble.


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 Ca2+, Ba2+, CaSO3 and CaSO4 do not form a thick precipitate as
precipitate Pb2+ ions 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) -> 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.


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.


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.


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.


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.


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 KNO3 KClO3 Saturated solution of KClO3 Solubility
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.


(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. (b) 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.


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?


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
(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.


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
(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.


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.


(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.


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.


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.


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.


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.


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.


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.


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 : (i) make glass (ii) for making soapless detergents (iii) softening hard water. (iv) 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 Sodium chloride and Trona dissolved in the sea Natural fractional crystallization Crystals of Trona (Day time) Crystals of sodium chloride(At night) Dredging /scooping/ digging Crushing Furnace (Heating) Carbon(IV) oxide Soda ash
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.


It is used as salt lick/feed for animals. Summary flow diagram showing the extraction of Soda ash from Trona Sodium chloride and Trona dissolved in the sea Natural fractional crystallization Crystals of Trona (Day time) Crystals of sodium chloride(At night) Dredging /scooping/ digging Crushing Furnace (Heating) Carbon(IV) oxide Soda ash
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.


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.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.


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.


(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.


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.


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:
(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
19.0.0 ENERGY CHANGES IN CHEMICAL AND PHYSICAL PROCESSES (25 LESSONS) 1.Introduction to Energy changes Energy is the capacity to do work. There are many/various forms of energy like heat, electric, mechanical, and/ or chemical energy.There are two types of energy: (i)Kinetic Energy(KE) ;the energy in motion. (ii)Potential Energy(PE); the stored/internal energy. Energy like matter , is neither created nor destroyed but can be transformed /changed from one form to the other/ is interconvertible. This is the principle of conservation of energy. e.g. Electrical energy into heat through a filament in bulb. Chemical and physical processes take place with absorption or evolution/production of energy mainly in form of heat The study of energy changes that accompany physical/chemical reaction/changes is called Thermochemistry. Physical/chemical reaction/changes that involve energy changes are called thermochemical reactions. The SI unit of energy is the Joule(J).Kilo Joules(kJ)and megaJoules(MJ) are also used. The Joule(J) is defined as the: (i) quantity of energy transferred when a force of one newton acts through a distance of one metre. 2 (ii) quantity of energy transferred when one coulomb of electric charge is passed through a potential difference of one volt.