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BASIC PRINCIPLES OF CHEMISTRY PRACTICALS

Introduction/Rationale

Chemistry is a science.

Chemistry practical all over the world is emphasized to all candidates sitting for a Chemistry paper.

There are about seven main basic universal emphasis for all chemistry candidates sitting for a chemistry paper;

(i)Titration /volumetric analysis

(ii)Thermochemistry(energy changes)

(iii)Chemical kinetic(rates of reaction)

(iv)Qualitative analysis(organic/inorganic)

(v)Solubility and solubility curves

(vi)Flame test

(vii)Physical / general chemistry

  1. Titration/volumetric analysis

Titration is determining the end point of the burette contents that react with fixed (usually 25.0cm3 from a pipette) conical flask contents.

As evidence of a titration actually done examining body require the candidate to record their burette readings before and after the titration.

For KCSE candidates burette readings must be recorded in a titration table in the format
provided by the Kenya National Examination Council.

As evidence of all titration actually done Kenya National Examination Council require the candidate to record their burette readings before and after the titration to complete the titration table in the format
provided.

ecolebooks.com

Sample Titration table format

Final burette reading (cm3)

24.0

24.0

24.0

Initial burette reading (cm3

0.0

0.0

0.0

Volume of solution used(cm3)

24.0

24.0

24.0

 

Calculate the average volume of solution used

24.0 + 24.0 + 24.0 = 24.0 cm3

  3

As evidence of understanding the degree of accuracy of burettes ,all readings must be recorded to a decimal point.

As evidence of accuracy in carrying the out the titration ,candidates value should be within 0.2 of the school
value .

The school value is the teachers readings presented to the examining body/council based on the concentrations of the solutions s/he presented to her/his candidates.

Bonus mark is awarded for averaged reading within 0.1 school value as Final answer.

Calculations involved after the titration require candidates thorough practice mastery on the:

(i)relationship among the mole, molar mass, mole ratios, concentration, molarity.

(ii) mathematical application of 1st principles.

Very useful information which candidates forget appear usually in the beginning of the paper as:

“You are provided with…”

All calculation must be to the 4th decimal point unless they divide fully to a lesser decimal point.

Never round off answers.

b)Thermochemistry/energy changes

Energy is the capacity to do work which is measured in Joules(J) or(kJ) .

Chemical/physical changes take place with absorption (Endothermic ) or evolution/ production (Exothermic)of heat.

Practically:

(i)endothermic changes show absorption of heat by a fall / drop in temperature and has a +∆H

(ii)exothermic changes show evolution/ production of heat by a rise in temperature and has a -∆H

(iii)temperature is measure using a thermometer.

(iv)a school thermometer is either coloured (alcohol) or colourless(mercury)

(v) For accuracy ,candidates in the same practical session should use the same type of thermometer.

(vi) fall / drop (+∆H) in temperature is movement of thermometer level downward.

(vii) rise (-∆H) in temperature is movement of thermometer level upwards.

Physical changes changes mainly involve melting/freezing/fussion and boiling /vapourization.

Chemical changes changes mainly involve displacement ,dissolving , neutralization

a).Energy changes in physical processes

Melting/freezing/fusion/solidification and boiling/vaporization/evaporation are the two physical processes.

Melting /freezing point of pure substances is fixed /constant.

The boiling point of pure substance depends on external atmospheric pressure.

Melting/fusion is the physical change of a solid to liquid. Freezing/fusion is the physical change of a liquid to solid.

Melting/freezing/fusion/solidification are therefore two opposite but same reversible physical processes. i.e

A (s) ========A(l)

Boiling/vaporization/evaporation is the physical change of a liquid to gas/vapour. Condensation/liquidification is the physical change of gas/vapour to liquid. Boiling/vaporization/evaporation and condensation/liquidification are therefore two opposite but same reversible physical processes. i.e

B (l) ========B(g)

Practically

  1. Melting/liquidification/fusion involves heating a solid to weaken the strong bonds holding the solid particles together.

Solids are made up of very strong bonds holding the particles very close to each other (Kinetic Theory of matter

On heating these particles gain energy/heat from the surrounding heat source to form a liquid with weaker bonds holding the particles close together but with some degree of freedom.

Melting/fusion is an endothermic (+∆H)process that require/absorb energy from the surrounding.

(ii)Freezing/fusion/solidification involves cooling a a liquid to reform /rejoin the very strong bonds to hold the particles very close to each other as solid and thus lose their degree of freedom (Kinetic Theory of matter).

Freezing /fusion / solidification is an exothermic (∆H)process that require particles holding the liquid together to lose energy to the surrounding.

(iii)Boiling/vaporization/evaporation involves heating a liquid to completely break/free the bonds holding the liquid particles together.

Gaseous particles have high degree of freedom (Kinetic Theory of matter). Boiling /vaporization / evaporation is an endothermic (+∆H) process that require/absorb energy from the surrounding.

(iv)Condensation/liquidification is reverse process of boiling /vaporization / evaporation.

It involves gaseous particles losing energy to the surrounding to form a liquid.It is an exothermic(+∆H) process.

The quantity of energy required to change one mole of a solid to liquid or to form one mole of a solid from liquid at constant temperature is called molar enthalpy/latent heat of fusion. e.g.

H2O(s) -> H2O(l)   ∆H = +6.0kJ mole-1 (endothermic process)

H2O(l) -> H2O(s)   ∆H = -6.0kJ mole-1 (exothermic process)

The quantity of energy required to change one mole of a liquid to gas/vapour or to form one mole of a liquid from gas/vapour at constant temperature is called molar enthalpy/latent heat of vapourization. e.g.

H2O(l) -> H2O(g)   ∆H = +44.0kJ mole-1 (endothermic process)

H2O(g) -> H2O(l)   ∆H = -44.0kJ mole-1 (exothermic process)

  • To determine the boiling point of water

Procedure:

Measure 20cm3 of tap water into a 50cm3 glass beaker. Determine and record its temperature.Heat the water on a strong Bunsen burner flame and record its temperature after every thirty seconds for four minute

Sample results

Time

(seconds)

0

30

60

90

120

150

180

210

240

Temperature(oC)

25.0

45.0

85.0

95.0

96.0

96.0

96.0

97.0

98.0

Questions

1.Plot a graph of temperature against time(y-axis)

Sketch graph of temperature against time

 

 

 

 

 

 

 

Image From EcoleBooks.com

 

 

 

 

 

 

 

 

Image From EcoleBooks.com

 

2.From the graph show and determine the boiling point of water

Note:

Water boils at 100oC at sea level/one atmosphere pressure/101300Pa but boils at below 100oC at higher altitudes.

The sample results above are from Kiriari Girls High School-Embu County on the slopes of Mt Kenya in Kenya. Water here boils at 96oC.

3.Calculate the molar heat of vaporization of water.(H= 1.0,O= 16.O)

Working:

Mass of water = density x volume => (20 x 1) /1000 = 0.02kg

Quantity of heat produced

= mass of water x specific heat capacity of water x temperature change

=>0.02kg x 4.2 x ( 96 – 25 ) = -5.964kJ  

 

Heat of vaporization of one mole H2O

  = Quantity of heat

Molar mass of H2O

 

  => -5.964kJ = -0.3313 kJ mole -1

18

To determine the melting point of candle wax

Procedure

Weigh exactly 5.0 g of candle wax into a boiling tube. Heat it on a strongly Bunsen burner flame until it completely melts.

Insert a thermometer and remove the boiling tube from the flame. Stir continuously. Determine and record the temperature after every 30seconds for four minutes.

Sample results

Time

(seconds)

0

30

60

90

120

150

180

210

240

240

Temperature

(oC)

93.0

85.0

78.0

70.0

69.0

69.0

69.0

67.0

65.0

65.0

 

Questions

1.Plot a graph of temperature against time(y-axis)

Image From EcoleBooks.com

b)Energy changes in chemical processes

(i)Standard enthalpy/heat of displacement ∆Hᶿd

(ii)Standard enthalpy/heat of neutralization ∆Hᶿn

(iii)Standard enthalpy/heat of solution/dissolution ∆Hᶿs

  1. Standard enthalpy/heat of displacement ∆Hᶿd

The molar standard enthalpy/heat of displacement may be defined as the energy/heat change when one mole of substance is displaced /removed from its solution at standard conditions

Some displacement reactions

(i)Zn(s) + CuSO4(aq) -> Cu(s) + ZnSO4(aq)

Ionically: Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+ (aq)

(ii)Fe(s) + CuSO4(aq) -> Cu(s) + FeSO4(aq)

Ionically: Fe(s) + Cu2+(aq) -> Cu(s) + Fe2+ (aq)

(iii)Pb(s) + CuSO4(aq) -> Cu(s) + PbSO4(s)

This reaction stops after some time as insoluble PbSO4(s) coat/cover unreacted lead.

(iv)Cl2(g) + 2NaBr(aq) -> Br2(aq) + 2NaCl(aq)

Ionically:

Cl2(g)+ 2Br(aq) -> Br2(aq) + 2Cl(aq)

To determine the molar standard enthalpy/heat of displacement(∆Hᶿd) of copper

Procedure

Place 20cm3 of 0.2M copper(II)sulphate(VI)solution into a 50cm3 plastic beaker/calorimeter.

Determine and record the temperature of the solution T1.

Put all the Zinc powder provided into the plastic beaker. Stir the mixture using the thermometer.

Determine and record the highest temperature change to the nearest 0.5oC- T2 .

Repeat the experiment to complete table 1 below

Sample results Table 1

Experiment

I

II

Final temperature of solution(T2)

30.0oC

31.0oC

initial temperature of solution(T1)

25.0oC

24.0oC

Change in temperature(∆T)

5.0

6.0

Questions


1.(a) Calculate:

 (i)average ∆T

Average ∆T = change in temperature in experiment I and II

=>5.0 + 6.0 = 5.5oC

  2

(ii)the number of moles of solution used


Moles used = molarity x volume of solution = 0.2 x 20 = 0.004 moles

1000 1000


(iii) the enthalpy change ∆H for the reaction

Heat produced ∆H = mass of solution(m) x specific heat capacity (c)x ∆T

=> 20 x 4.2 x 5.5 = 462 Joules = –0.462 kJ

 1000 1000


(iv)State two assumptions made in the above calculations.

Density of solution = density of water = 1gcm-3

Specific heat capacity of solution=Specific heat capacity of water =4.2 kJ-1kg-1K

This is because the solution is assumed to be infinite dilute.

2. Calculate the enthalpy change for one mole of displacement of Cu2+ (aq) ions.

Molar heat of displacement ∆Hd = Heat produced ∆H

Number of moles of fuel

=> 0.462 kJ = –115.5 kJmole-1

0.004

3.Write an ionic equation for the reaction taking place.

 Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+(aq)

4.State the observation made during the reaction.

Blue colour of copper(II)sulphate(VI) fades/becomes less blue/colourless.

Brown solid deposits are formed at the bottom of reaction vessel/ beaker.

5.Illustrate the above reaction using an energy level diagram.

Image From EcoleBooks.com

8. The enthalpy of displacement ∆Hd of copper(II)sulphate (VI) solution is 12k6kJmole-1.Calculate the molarity of the solution given that 40cm3 of this solution produces 2.204kJ of energy during a displacement reaction with excess iron filings.

Number of moles = Heat produced ∆H

Molar heat of displacement ∆Hd

 

=> 2.204 kJ =  0.0206moles

    126 moles

Molarity of the solution = moles x 1000

 Volume of solution used  

 
 

= 0.0206moles x 1000 = 0.5167 M

40

Graphical determination of the molar enthalpy of displacement of copper

Procedure:

Place 20cm3 of 0.2M copper(II)sulphate (VI) solution into a calorimeter/50cm3 of plastic beaker wrapped in cotton wool/tissue paper.

Record its temperature at time T= 0.Stir the solution with the thermometer carefully and continue recording the temperature after every 30 seconds .

Place all the (1.5g) Zinc powder provided after 1 ½ minutes.

Stir the solution with the thermometer carefully and continue recording the temperature after every 30 seconds for five minutes.

Determine the highest temperature change to the nearest 0.5oC.

Image From EcoleBooks.comQuestions

1.Show and determine the change in temperature ∆T

From a well constructed graph ∆T= T2 –T1 at 150 second by extrapolation

∆T = 36.5 – 25.0 = 11.5oC

2.Calculate the number of moles of copper(II) sulphate(VI)used given the molar heat of displacement of Cu2+ (aq)ions is 125kJmole-1

Heat produced ∆H = mass of solution(m) x specific heat capacity (c)x ∆T

=> 20 x 4.2 x 11.5 = 966 Joules = –0.966 kJ

1000

Number of moles = Heat produced ∆H Molar heat of displacement ∆Hd

=> 0.966 kJ =  –0.007728moles

    125 moles –7.728 x 10-3moles

3. What was the concentration of copper(II)sulphate(VI) in moles per litre.

Molarity = moles x 1000

Volume used

  =>7.728 x 10-3moles x 1000 = 0.3864M

20


4.The actual concentration of copper

(II) Sulphate (VI) solution was 0.4M. Explain the differences between the two.

Practical value is lower than theoretical

. Heat/energy loss to the surrounding and that absorbed by the reaction vessel decreases ∆T hence lowering the practical number of moles and molarity against the theoretical value

(c)Standard enthalpy/heat of neutralization ∆Hᶿn

The molar standard enthalpy/heat of neutralization ∆Hᶿn is defined as the energy/heat change when one mole of a H+ (H3O+)ions react completely with one mole of OH ions to form one mole of H2O/water.

Neutralization is thus a reaction of an acid /H+ (H3O+)ions with a base/alkali/ OH ions to form salt and water only.

Strong acids/bases/alkalis are completely/fully/wholly dissociated to many free ions(H+ /H3O+ and OHions).

(ii) for strong acid/base/alkali neutralization, no energy is used to dissociate /ionize since molecule is wholly/fully dissociated/ionized into free H+ H3O+ and OH ions.

The overall energy evolved is comparatively higher / more than weak acid-base/ alkali neutralizations.

For strong acid-base/alkali neutralization, the enthalpy of neutralization is constant at about 57.3kJmole-1 irrespective of the acid-base used.

This is because ionically:


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

for all wholly/fully /completely dissociated acid/base/alkali

Weak acids/bases/alkalis are partially dissociated to few free ions(H+ (H3O+ and OHions) and exist more as molecules.

Neutralization is an exothermic(-∆H) process.

The energy produced during neutralization depend on the amount of
free ions (H+ H3O+ and OH)ions existing in the acid/base/alkali reactant:

 (i)for weak acid-base/alkali neutralization,some of the energy is used to dissociate /ionize the molecule into free H+ H3O+ and OH ions therefore the overall energy evolved is comparatively lower/lesser/smaller than strong acid / base/ alkali neutralizations.

Practically ∆Hᶿn can be determined as in the examples below:

To determine the molar enthalpy of neutralization ∆Hn of Hydrochloric acid

Procedure

Place 50cm3 of 2M hydrochloric acid into a calorimeter/200cm3 plastic beaker wrapped in cotton wool/tissue paper.

Record its temperature T1.

Using a clean measuring cylinder, measure another 50cm3 of 2M sodium hydroxide.

Rinse the bulb of the thermometer in distilled water.

Determine the temperature of the sodium hydroxide T2.

Average T2 andT1 to get the initial temperature of the mixture T3.

Carefully add all the alkali into the calorimeter/200cm3 plastic beaker wrapped in cotton wool/tissue paper containing the acid.

Stir vigorously the mixture with the thermometer.

Determine the highest temperature change to the nearest 0.5oC T4 as the final temperature of the mixture.

Repeat the experiment to complete table 1.

(ii)enthalpy change ∆H of neutralization.

∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T(T6) => (50 +50) x 4.2 x 13.5 = 5670Joules = 5.67kJ

 (iii) the molar heat of neutralization the acid.

∆Hn = Enthalpy change ∆H => 5.67kJ  = 56.7kJ mole-1

  Number of moles 0.1moles

(c)Write the ionic equation for the reaction that takes place

 OH(aq)+ H+(aq) -> H2O(l)

(d)The theoretical enthalpy change is 57.4kJ. Explain the difference with the results above.

The theoretical value is higher

Heat/energy loss to the surrounding/environment lowers ∆T/T6 and thus ∆Hn

Heat/energy is absorbed by the reaction vessel/calorimeter/plastic cup lowers ∆T and hence ∆Hn

Sample results

Experiment

I

II

Temperature of acid T1 (oC)

22.5

22.5

Temperature of base T2 (oC)

22.0

23.0

Final temperature of solution T4(oC)

35.5

36.0

Initial temperature of solution T3(oC)

22.25

22.75

Temperature change( T5)

13.25

13.75


(a)Calculate T6 the average temperature change   T6 =  13.25 +13.75 = 13.5 oC   2


(b)Why should the apparatus be very clean?

Impurities present in the apparatus reacts with acid /base lowering the overall temperature change and hence ∆Hᶿn.  

(c)Calculate the:

(i)number of moles of the acid used

 number of moles = molarity x volume => 2 x 50 = 0.1moles 1000 1000

(e)Compare the ∆Hn of the experiment above with similar experiment repeated with neutralization of a solution of:

(i) potassium hydroxide with nitric(V) acid

The results would be the same/similar.

Both are neutralization reactions of strong acids and bases/alkalis that are fully /wholly dissociated into many free H+ / H3O+ and OH ions.

(ii) ammonia with ethanoic acid

The results would be lower/∆Hn would be less.

Both are neutralization reactions of weak acids and bases/alkalis that are partially /partly dissociated into few free H+ / H3O+ and OH ions. Some energy is used to ionize the molecule.

(f)Draw an energy level diagram to illustrate the energy changes

Theoretical examples

1.The molar enthalpy of neutralization was experimentary shown to be 51.5kJ per mole of 0.5M hydrochloric acid and 0.5M sodium hydroxide. If the volume of sodium hydroxide was 20cm3, what was the volume of hydrochloric acid used if the reaction produced a 5.0oC rise in temperature?

Working:

Moles of
sodium hydroxide = molarity x volume

  1000

  => 0.5 M x 20cm3 = 0.01 moles

1000  

Enthalpy change
∆H = ∆Hn =>
51.5 = 0.515kJ

Moles
sodium hydroxide 0.01 moles

Mass of base + acid = Enthalpy change
∆H in Joules

Specific heat capacity x ∆T

=> 0.515kJ x 1000 = 24.5238g

  4.2 x 5

Mass/volume of HCl = Total volumevolume of NaOH

=>24.5238 – 20.0 = 4.5238 cm3

Graphically ∆Hn can be determined as in the example below:

Procedure

Place 8 test tubes in a test tube rack .

Put 5cm3 of 2M sodium hydroxide solution into each test tube. Measure 25cm3 of 1M hydrochloric acid into 100cm3 plastic beaker.

Record its initial temperature at volume of base =0.

Put one portion of the base into the beaker containing the acid.

Stir carefully with the thermometer and record the highest temperature change to the nearest 0.5oC.

Repeat the procedure above with other portions of the base to complete table 1 below

Volume of acid(cm3)

25.0

25.0

25.0

25.0

25.0

25.0

25.0

25.0

25.0

25.0

25.0

Volume of alkali(cm3)

0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

35.0

40.0

Final temperature(oC)

22.0

24.0

26.0

28.0

28.0

27.0

26.0

25.0

24.0

25.0

24.0

Initial temperature(oC)

22.0

22.0

22.0

22.0

22.0

22.0

22.0

22.0

22.0

22.0

22.0

Change in temperature

0.0

2.0

4.0

6.0

6.0

5.0

4.0

3.0

2.0

3.0

2.0


Complete the table to determine the change in temperature.

Plot a graph of volume of sodium hydroxide against temperature change.

From the graph show and determine :

(i)the highest temperature change ∆T

∆T =T2-T1 : highest temperature-T2 (from extrapolating a correctly plotted graph) less lowest temperature at volume of base=0-T1

=> 28.7 – 22.0 = 6.7 0 oC

Image From EcoleBooks.com

(ii) the volume of sodium hydroxide used for complete neutralization

From correctly plotted graph = 16.75 cm3

(iii) Calculate the number of moles of the alkali used

Moles NaOH = molarity x volume ()Vn=

1000

=> 2 x 16.75 = 0.0335 moles

1000

(iv)Calculate ∆H for the reaction.

  ∆H = mass of solution mixture x c x ∆T

  => (25.0 + 16.75) x 4.2 x 6.7

= 1174.845 J = 1.174845 kJ

1000

(iii) Calculate the molar enthalpy of the alkali:

 ∆Hn = Heat change =>
1.174845 kJ

number of moles 0.0335 moles

  = 35.0699kJ mole-1  

(i) Standard enthalpy/heat of solution/dissolution ∆Hᶿs

The standard enthalpy of solution ∆Hᶿs is defined as the energy change when one mole of a substance is dissolved in excess distilled water to form an infinite dilute solution.

An infinite dilute solution is one which is too dilute to be diluted further.

Practically the heat of solution is determined by dissolving a known mass /volume of a solute in known mass/volume of water/solvent and determining the temperature change.

To determine the heat of dissolution of ammonium nitrate(V)

Place 100cm3 of distilled water into a plastic cup/beaker/calorimeter

Put all the 5.0g of ammonium nitrate(v)/potassium nitrate(V)/ ammonium chloride into the water.

Stir the mixture using the thermometer and record the temperature change after every ½ minute to complete table1.

Continue stirring throughout the experiment.

 

Image From EcoleBooks.com

(a)From the graph show and determine:

(i)the highest temperature change ∆T

∆T =T2-T1 : highest temperature-T2 (from extrapolating a correctly plotted graph) less lowest temperature at volume of base=0-T1

=> 18.7 – 22.0 = 3.3 oC ( not -3.3 oC)

(b) Calculate the total energy change
∆H during the reaction

∆H
= mass of water x c x ∆T

=>∆H=100 x4.2 x 3.3 oC = + 1386 J = + 1.386 kJ

  1000

(c) Calculate the number of moles of ammonium nitrate (v) used

Moles = mass => 5.0 = 0.0625 moles


molar mass 80

(d)What is the molar heat of dissolution of ammonium nitrate(V)

∆H = Heat change = + 1.386 kJ = + 22.176 kJmole-1

Number of mole 0.0625 moles

(e)What would happen if the distilled water is heated before experiment was performed .

The ammonium nitrate(V) would take less time to dissolve.

Increase in temperature reduces lattice energy causing endothermic dissollution to be faster.

(e)Illustrate the above process on an energy level diagram

 

Image From EcoleBooks.com


c)Chemical Kinetic/Rate of reaction

The rate of a chemical reaction can be defined as the time taken for a known amount of reactants to form known amount of products.

Some reactions are too slow to be determined e.g weathering others are instantaneous

The SI unit of time is seconds. Minutes and hours are also common .

Time is determined using a stop watch/clock

Candidates using stop watch/clock should learn to:

(i)Press start button concurrently with starting off determination of a reaction using one hand each.

(ii)Press stop button when the reaction is over.

(iii)Record all times in seconds unless specified.

(iv)Press reset button to begin another timing

(v)Ignore time beyond seconds for stop clock/watch beyond this accuracy

(vi)Avoid accidental pressing of any button before recording

It can be very frustrating repeating a whole procedure

The following factors theoretically and practically alter/influence/affect/determine the rate of a chemical reaction:

(a)Concentration

(b)Temperature

(a)Concentration

An increase in concentration increases the rate the rate of reaction by reducing the time taken to completion.

Theoretically, increase in concentration is a decrease in distance between reacting particles which increases their collision frequency.

Practically decreasing concentration is diluting/adding water

To demonstrate the effect of concentration on reaction rate

You are provided with

(i) sodium thiosulphate containing 40gdm3 solution labeled A

(ii) 2M hydrochloric acid labeled solution B

You are required to determine the rate of reaction between solution A and B

Procedure

Measure 40cm3 of solution A into 100 cm3 glass beaker. Place it on top of a pen-mark “X”. Measure another 40cm3 of solution B. Simultaneously put solution B into solution A and start off a stop watch/clock. Determine the time taken for the pen-mark “X” to be invisible/obscurred from above. Repeat the procedure by measuring 35cm3 of solution B and adding 5cm3 of water. Complete the table 1 below by using other values os solution B and water

Image From EcoleBooks.com

 

Image From EcoleBooks.com

Sample questions

(i)Explain the shape of the graph

(Straight line graph from the origin)

Decrease in concentration decreases the rate of reaction. The higher the concentration of solution B the less time taken for mark x to be obscurred/invisible due to increased collision frequency between the reacting particles.

(ii)From the graph determine the time taken for the mark to be invisible at 37cm3

At 37cm3 then 1/t => 1/ 37 = 0.027

From a well plotted graph:

 1/t = 0.027 => 16.2602 seconds

(ii)From the graph determine the volume of solution B at 100 seconds

100 seconds => 1/t = 1 / 1000 = 0.01

From a well plotted graph:

 At 1/t = 0.01 => the volume of B = 17.0cm3

(iii) State another factor that would alter the rate of the above reaction.

Temperature

(iii) State another factor that would not alter the rate of the above reaction.

 Surface area

 Pressure

 Catalyst

(b) Temperature

An increase in temperature increases the rate of reaction.

An increase of 10 oC/10K practically doubles the rate of a chemical reaction/reduces time of completion by 1/2.

An increase in temperature increase the kinetic energy of reacting particles increasing their collision frequency

Practically ,increase in temperature involves heating the reactants

The results and presentation should be as in the effect of concentration.

Increased temperature reverses the table I time results

i.e less time as temperature increases.

d)Qualitative analysis

Process of identifying unknown compounds

Compounds may be:

(i)Inorganic

(ii)organic

Inorganic analysis:

This involve mainly identification of ionic compounds containing cations and anions.

Cations present in an ionic compounds are identified by adding a precipitating reagent that forms a precipitate unique to the cation/s in the compound.

The main precipitating reagents used are:

 2M NaOH and/or 2M NH3(aq)

When using 2M sodium hydroxide:

(i)No white precipitate is formed if K + and Na + ions are present

(ii) No white precipitate is formed if NH4+ ions are present but a clourless gas with pungent smell of urine is produced which may not be recognized in a school laboratory examination setting.

(iii)White precipitate that dissolves / soluble in excess if Zn2+ Pb2+ Al3+ ions are present.

(iv)White precipitate that do not dissolves/insoluble in excess if Ba2+ Mg2+ Ca2+
ions are present.

(v)Blue precipitate that do not dissolves /insoluble in excess if Cu2+ ions are present.

(vi)Green precipitate that do not dissolves/insoluble in excess if Fe2+ ions are present.

(vii)Brown precipitate that do not dissolves/insoluble in excess if Fe3+ ions are present.

When using 2M aqueous ammonia

(i)No white precipitate is formed if K + ,NH4+ Na + ions are present

(ii)White precipitate that dissolves / soluble in excess if Zn2+
ions are present.

(iii)White precipitate that do not dissolves/insoluble in excess if Ba2+ Mg2+ Ca2+ Pb2+ Al3+
ions are present.

(iv)Blue precipitate that dissolves /soluble in excess to form a deep/royal blue solution in excess if Cu2+ ions are present.

(v)Green precipitate that do not dissolves/insoluble in excess if Fe2+ ions are present.

(vi)Brown precipitate that do not dissolves/insoluble in excess if Fe3+ ions are present.

Anions present in an ionic compounds are identified by adding a specific precipitating reagent that forms a precipitate unique to the specific anion/s in the compound.

(i)Lead(II)nitrate(V) solution

Lead forms insoluble PbSO4 ,PbSO3 ,PbCO3, PbS, PbI2,PbCl2

PbS is a black precipitate,

PbI2 is a yellow precipitate.

All the others are white precipitates.

(a)If a Lead(II)nitrate(V) solution is added to a substance/ solution/ compound :

(i)A yellow ppt shows presence of I ions

(ii)A black ppt shows presence of S2- ions

(iii) A white ppt shows presence of SO42- ,SO32- ,CO32- Cl

(b)If the white precipitate is added dilute nitric(V) acid:

(i)It dissolves to show presence of SO32- ,CO32-

(ii)It persist/remains to show presence of SO42-, Cl

(c)If the white precipitate in b(i) is added acidified potassium manganate(VII)/ dichromate(VI)

(i) acidified potassium manganate(VII) is decolorized /orange colour of acidified potassium dichromate(VI) turns to green to show presence of SO32-

(ii) acidified potassium manganate(VII) is not decolorized /orange colour of acidified potassium dichromate(VI) does not turn to green/remains orange to show absence of SO32- /presence of CO32-

(c)If the white precipitate in b(ii) is boiled:

(i)It dissolves to show presence of Cl

(ii)It persist/remains to show presence of SO42-

(ii)Barium(II)nitrate(V)/Barium chloride solution

Barium(II)nitrate(V)/Barium chloride solution precipitates BaSO4 ,BaSO3 , BaCO3, from SO42- ,SO32- ,CO32- ions.

Inorganic qualitative analysis require continous practice discussion

Sample presentation of results

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

1(a) Appearance

Observations inference (1mark)

 

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

 

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

Observations inference (1mark)

 

Colourless droplets formed on the cooler Hydrated compound/compound

part of the test tube   containing water of crystallization

Solid remains a white residue  

 

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

Observation Inference  (1mark)

Solid dissolves to form  Polar soluble compound

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

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

 Observation Inference  (1mark)

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

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

 Observation Inference  (1mark)

White ppt, insoluble in excess   Pb2+ , Al3+

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

 Observation Inference  (1mark)

No white ppt   Al3+

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

 Observation Inference  (1mark)

White ppt   CO32-, SO42-, SO32-, Cl,

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

 Observation Inference  (1mark)

White ppt persist/remains   SO42-, Cl,

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

 Observation Inference  (1mark)

White ppt persist/remains   SO42-,

Organic analysis:

This involve mainly identification of the functional group:


(i) – C = C- / = C = C= / C – C

 

  1. R OH

     

(iii) R COOH / H+

 

These functional groups can be identified by:

(i)burning-a substance which “catches fire” must reduce in amount.

Candidates should not confuse burning with flame coloration/test

(ii)Decolorization of bromine water/chlorine water/acidified KMnO 4 / to show presence of

 

  • C = C – / – C = C – and R – OH

 

(iii)Turning orange acidified K 2 Cr 2 O 7
to green to show presence as in above.

(iii)pH 1/2/3 for strongly acidic solutions. pH 4/5/6 for weakly acidic solutions

(iv)Turning blue litmus paper red. red litmus paper remaining red show presence of H+ ions

d)Flame test

The colour change on a clear colourless Bunsen flame is useful in identifying some cations / metals.

A very clean metallic spatula is recommended since dirt obscures /changes the correct coloration distinct flame coloration of some compounds

Barium/barium salts

orange

Sodium/ sodium salts

yellow

Potassium/potassium salts

Purple/lilac

Lithium/Lithium salts

Deep red/crimson

Calcium/ calcium salts

red

Copper/copper salts

Blue/ green

(e)Physical chemistry

Chemistry is a science subject that incorporate many scientific techniques.

Examining body/council, require tabulated results/data from the candidate.

This tabulated results is usually then put in a graph.

The general philosophy of methods of presentation of chemistry practical data is therefore availability of evidence showing:

(i)Practical done(complete table)

(ii)Accuracy of apparatus used(decimal point)

(iii)Accuracy/care in doing experiment to get collect trend(against teachers results)

(iv)Graphical work(use of mathematical science)

(v)Calculations (Scientific mathematical integration)

 

 

(f)Sample practicals

Name ………………..………………….Class…………..Index No……………..

Candidate’s signature………………………………..

Date done………………Date marked………… .……Date revised…………….…..….

233/3

CHEMISTRY Paper 3

PRACTICAL.

Pre-KCSE Practice 1: 2013

MARKS SCHEME

 

Instruction to Candidate

Write your name and index number in the spaces provided above.

Sign and write the date of examination in the spaces provided above

Answer all questions in the spaces provided.

Mathematical tables and electronic calculators may be used.

All working must be clearly shown where necessary.

This paper consist of 8 printed pages.

Candidates should check the question paper to ascertain that all the pages are printed and indicated and that no questions are missin

For examiners use only

Question

Maximum score

Candidates core

 

1

 

20

 

20

 

2

 

10

 

10

 

3

 

10

 

10


 

 

Total score

 

40

 

40


 

1.You are provided with:

 (i)solution L containing 5.0g per litre of a dibasic organic acid H2X.2H2O.

 (ii)solution M which is acidified potassium manganate(VII)

 (iii)solution N a mixture of sodium ethanedioate and ethanedioic acid

 (iv)0.1M sodium hydroxide solution P

 (v)1.0M sulphuric(VI)

You are required to:

 (i)standardize solution M using solution L

(ii)use standardized solution M and solution P to determine the % of sodium ethanedioate in the mixture.

Procedure 1

Image From EcoleBooks.comFill the burette with solution M. Pipette 25.0cm3 of solution L into a conical flask. Heat this solution to about 70oC(but not to boil).Titrate the hot solution L with solution M until a permanent pink colour just appears .Shake thoroughly during the titration. Repeat this procedure to complete table 1.

Table 1

1

2

3

Final burette reading (cm3)

20.0

20.0

20.0

Initial burette reading (cm3)

0.0

0.0

0.0

Volume of N used (cm3)

20.0

20.0

20.0

(2marks)


(a)Calculate the average volume of solution L used (1mk)

20.0 + 20.0+ 20.0   = 20.0cm3

 3


(b)Given that the concentration of the dibasic acid is 0.05molesdm-3.determine the value of x in the formula H2X.2H2O (H=1.0,O=16.0) (1mark)

Molar mass H2X.2H2O = mass / litre =>

  moles / litre

  5.0g/litre   = 100 g

  0.05molesdm-3

H2X.2H2O =100

X = 100 – ((2 x1) + 2 x (2 x1) + (2 x 16) => 100 – 34 = 62


(c) Calculate the number of moles of the dibasic acid H2X.2H2O.  (1mark)

Moles = molarity x pipette volume =>

1000

    0.05 x 25 = 0.00125 / 1.25 x10 -3 moles

  1000


(d)Given the mole ratio manganate(VII)(MnO4): acid H2X is 2:5, calculate the number of moles of manganate(VII) (MnO4) in the average titre.  (1mark)

Moles H2X = 2/5 moles of MnO4

=> 2/5 x 0.0125/1.25 x10 -2 moles = 0.0005 / 5.0 x 10 -4 moles

(e)Calculate the concentration of the manganate(VII)(MnO4) in moles per litre.

(1mark)

Moles per litre/molarity =   moles x 1000

  average burette volume

=> 0.0005/5.0 x 10 -4moles x 1000  = 0.02083 moles l-1 / M

 24.0

Procedure 2

Image From EcoleBooks.comWith solution M still in the burette ,pipette 25.0cm3 of solution N into a conical flask. Heat the conical flask containing solution N to about 70oC.Titrate while hot with solution M. Repeat the experiment to complete table 2.

Table 2 (2marks)

1

2

3

Final burette reading (cm3)

12.5

12.5

12.5

Initial burette reading (cm3)

0.0

0.0

0.0

Volume of N used (cm3)

12.5

12.5

12.5

(a)Calculate the average volume of solution L used (1mk)

12.5 + 12.5 + 12.5 = 12.5 cm3

  3

(b)Calculations:

(i)How many moles of manganate(VII)ions are contained in the average volume of solution M used? (1mark)

Moles = molarity of solution M x average burette volume

1000

=>   0.02083 molesl-1/ M x 12.5 = 0.00026 / 2.6 x 10-4 moles

1000

(ii)The reaction between manganate(VII)ions and ethanedioate ions that reacted with is as in the equation:

2MnO4(aq) + 5C2O42- (aq) + 16H+ (aq) -> 2Mn2+(aq) + 10CO2(g) + 8H2O(l)

Calculate the number of moles of ethanedioate ions that reacted with manganate (VII) ions in the average volume of solution M.  (1mark)

From the stoichiometric/ionic equation:

mole ratio MnO4(aq): C2O42- (aq) = 2:5

=> moles C2O42- = 5/2 moles MnO4

=> 5/2 x 0.00026 / 2.5 x 10-3 moles

= 0.00065 / 6.5 x10-4 moles

(iii)Calculate the number of moles of ethanedioate ions contained in 250cm3 solution N. (1mark)

25cm3 pipette volume -> 0.00065 /6.5 x10-4 moles

250cm3 ->
0.0065 /6.5 x10-3 moles x 250 = 0.0065 / 6.5 x10-3 moles

25

Procedure 3

Image From EcoleBooks.comRemove solution M from the burette and rinse it with distilled water. Fill the burette with sodium hydroxide solution P. Pipette 25cm3 of solution N into a conical flask and add 2-3 drops of phenolphthalein indicator. Titrate this solution N with solution P from the burette. Repeat the procedure to complete table 3.

Table 3

1

2

3

Final burette reading (cm3)

12.5

12.5

12.5

Initial burette reading (cm3)

0.0

0.0

0.0

Volume of N used (cm3)

12.5

12.5

12.5

(2 mark)

(a)Calculate the average volume of solution L used    (1mk)

12.5 + 12.5 + 12.5 = 12.5 cm3

3


(b)Calculations:

(i)How many moles of sodium hydroxide solution P were contained in the average volume? (1mark)

Moles = molarity of solution P x average burette volume

  1000

=> 0.1 molesl-1 x 24.9 = 0.00249 / 2.49 x 10-3 moles

  1000

(ii)Given that NaOH solution P reacted with the ethanedioate ions from the acid only and the equation for the reaction is:

2NaOH
(aq) + H2C2O4 (aq) -> Na2C2O4(g) + 2H2O(l)

Calculate the number of moles of ethanedioic acid that were used in the reaction.(1 mk)

 

From the stoichiometric equation,mole ratio

 NaOH(aq): H2C2O4 (aq) = 2:1

=> moles H2C2O4 = 1/2 moles NaOH

=> 1/2 x 0.00249 / 2.49 x 10-3 moles

=0.001245/1.245 x10-3 moles.

(iii)How many moles of ethanedioic acid were contained in 250cm3 of solution N? (1mark)

25cm3 pipette volume   -> 0.001245/1.245 x10-3

250cm3   -> 0.001245/1.245 x10-3 moles x 250     25

=   0.01245/1.245 x10-2 moles

(iii)Determine the % by mass of sodium ethanedioate in the mixture

(H= 1.0,O=16.0,C=12.0 and total mass of mixture =2.0 g in 250cm3 solution) (1mark)

Molar mass H2C2O4 = 90.0g

Mass of H2C2O4 in 250cm3 =

moles in 250cm3 x molar mass H2C2O4

=>0.01245/1.245 x10-2 moles x 90.0

  = 1.1205g  

% by mass of sodium ethanedioate

=(Mass of mixture – mass of H2C2O4) x 100%

 Mass of mixture

=> 2.0 – 1.1205 g = 43.975%

2.0

2. You are provided with 5.0 g solid B. You are to determine the molar mass of solid B.

Procedure

Image From EcoleBooks.com Place 100cm3 of liquid L into a plastic beaker. Determine its temperature and record it at time = 0 in Table 2 below. Stir continuously using the thermometer and record the highest temperature change to the nearest 0.5oC after every 30 seconds. After 120 seconds, add all solid B. Continue stirring and recording the temperature to complete table 2.

Table 2

Time (seconds)

0.0

30

60

90

120

150

180

210

240

270

300

Temperature(oC)

20

20

20

20

 

18

16

14

14

15

16

(2mark)

 

 

 

(a)Plot a graph of temperature against time(x-axis)(3marks)

Image From EcoleBooks.com

 

(b)From the graph show and determine (2 mark )

(i) the highest temperature change ∆T

  ∆T = T2 –T1 => 13.4 -20 = 6.6o C

Note ∆T is not – 6.6oC

(ii) the temperature of the mixture at 130 seconds

From extrapolation at 130 seconds = 19.2 oC

(iii)the time when all the solid first dissolved

From extrapolation of the lowest temperature = 220 Seconds

(d) Calculate the heat change for the reaction.(Assume density of liquid L is 1.0gcm-3 ) specific heat capacity is 4.2Jkg-1K-1(1mark)

 ∆H = mass of liquid L x c x ∆T =>100 x 4.2 x 6.6 = + 2772 J = + 2.772 kJ

1000

(e) Given the molar enthalpy of dissolution of Solid B in liquid L is + 22.176kJ mole-1,determine the number of moles of B used(1mark)

Moles of B = ∆H  => + 2.772 kJ = 0.125 moles

∆Hs   + 22.176kJ mole-1

(f)Calculate the molar mass of B (1mark)


Molar mass of B = Mass used =>   5.0 =>40 g

Moles used 0.125 moles

3(a)You are provided with solid Y. Carry out the following tests and record your observations and inferences in the space provided.

(i) Appearance


Observations inference   (1mark)

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

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

Observations inference (1mark)

Colourless droplets forms on the cooler parts of Hydrated compound/salt

test tube

Solid remain white

(ii)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 a colourless solution  Coloured Fe2+ , Fe3+ , Cu2+ ions absent

 

I. To the first portion add three drops of universal indicator. (1mark)

 Observation Inference

pH= 4 weakly acidic solution

 

 

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 Al3+ , Pb2+

 

III.To the third portion, add three drops of sodium sulpide solution.

 Observation Inference (1mark)

  No black ppt Al3+

 

IV.To the fourth portion, add three drops of acidified Lead(II)nitrate(IV)solution. Heat to boil

 Observation Inference (1mark)


White ppt , persist/remains on boiling SO42-

 

(b)You are provided with solid P. Carry out the following tests and record your observations and inferences in the space provided.

(i)Place a portion of solid P on a clean metallic spatula and introduce it on a Bunsen flame.

(1/2 mark)


Solid burns with a yellow sooty flame   C  C // C C bonds

(ii)Add all the remaining solid to about 10cm3 of water in a test tube and shake well. Divide the mixture into 4 portions. (1/2 mark)

Solid dissolves to form a clourless solution Polar organic compound

 

 

I. To the 1st portion, test with litmus papers (1/2 mark)

Red litmus paper remain red  H+ ions  Blue litmus paper turn blue

 

 

II. To the 2nd portion, add a little sodium hydrogen carbonate(1/2 mark)  

Effervescence/fizzing/bubbles H+ ions  

Colourless gas produced

 

III.To the 3rd portion, and three drops of solution M. Warm(1/2 mark)

Acidified KMnO4 is decolorized R OH, C   C // C C bonds

// solution M is decolorized IV.To the 4th portion, add three drops of bromine water (1/2marks) Bromine water is decolorized C   C // C C bonds

233/3 CHEMISTRY

Pre-KCSE 2013

Practice 1

Moi High School-Mbiruri

Requirements for each Candidates:

0.05 M Oxalic acid labeled Solution L

0.01M Potassium manganate (VII) labeled Solution M

0.03M oxalic acid labeled Solution N

0.1M Sodium hydroxide labeled Solution P

1.0M sulphuric(VI)acid.

15 0cm3 distilled water labeled Liquid L

50cm3 burette

25cm3 pipette

Two clean conical flasks

Pipette filler

-10 -1100C Thermometer

Stop watch/clock

200cm3 clean beaker

5.4g ammonium chloride/8.0g Ammonium nitrate(V) labeled Solid B weighed accurately

About 2.0 g of hydrated Aluminium sulphate labeled Solid Y

About 2.0g of Citric acid labeled Solid P

One boiling tube

Six clean dry test tubes

Pair of litmus papers(red and blue)

Clean metallic spatula

Access to bench reagents/apparatus

Means of heating

2M aqueous ammonia

0.1M acidified Lead(II)nitrate(V)

Bromine water

0.1M sodium sulphide

About 0.1g Sodium hydrogen carbonate

Universal indicator solution

pH chart

Name …………………….Index Number……………..

233/3  Candidates signature………………

CHEMISTRY Date………………………..

Paper 3

PRACTICAL

Practice 2012

21/4 hours

You are provided with :

  • Solution A containing an oxidizing agent A;
  • Solution B ,0.05M aqueous sodium thiosulphate;
  • Solution C containing a reducing agent C;
  • Aqueous Potassium iodide;
  • Solution D, starch solution.

You are required to determined the:

 Concentration of solution A

 Rate of reaction between the oxidizing agent A and the reducing agent C.

Procedure 1

  1. Using a pipette and pipette filter ,place 25.0cm3 of solution A into a 250ml conical flask.
  2. Meassure 10cm3 of aqueous potassium iodide and add it to solution A in the conical flask.

    Shake the mixture .Add 10cm3 of 2M sulphuric(VI)acid to the mixture and shake.

  3. Fill a burette with solution B and use it to titrate the mixture in the conical flask until it just turns orange yellow.Add 2cm3 of solution D to the mixture in the conical flask .Shake thoroughly. Continue titrating until the mixture just turns colourless. Record your results in table 1 below.
  4. Repeat the procedure and complete table 1.Retain the remainder of solution A and D for use in procedure II

    Table I

     

    I

    II

    III

    Final burette reading

    20.0

    20.0

    20.0

    Initial burette reading

    0.0

    0.0

    0.0

    Volume of solution B used (cm3)

    20.0

    20.0

    20.0

     

    (4mks)

(a)Calculate the:

 (i) average volume of solution B used(1mk)

20.0 + 20.0 +20.0 √ = 20.0 √cm3

3

(ii)number of moles of sodium thisulphate(1mk)

Moles = molarity x burette volume

1000

=> 0.05 x20.0 √ = 0.001 / 1.0 x10-3 √ moles

1000

(b)Given that one mole of A reacts with six moles of sodium thiosulphate ,calculate the:

(i)number of moles of A that were used (1mk)

Mole ratio A:B = 1:6 √

=> Moles A = 0.001 / 1.0 x10-3 moles = 0.00016/1.6 x10-4 √ moles

6

(ii)concentration of solution A in moles per litre(2mk)

Molarity of solution A = moles x 1000

Pipette volume

=> 0.00016/1.6 x10-4 moles x1000 √ = 0.008/8.0 x10-4 M√

 20

Procedure II

  1. Label six test tubes as 1,2,3,4 ,5 and 6 and place them on a test tube rack.
  2. Using a clean burette, measure the volumes of distilled water as shown in table 2 into the labeled test tubes.
  3. Using a burette ,measure the volumes of solution A shown in table 2 into each of the test tubes .
  4. Clean the burette and rinse it with about 5cm3 of solution C
  5. Using the burette ,measure 5cm3 of solution C and place it into a 100ml beaker.
  6. Using a 10ml measuring cylinder ,measure 5cm3 of solution D and add it to the beaker containing solution C .Shake the mixture.
  7. Pour the contents of test tube number 1 to the mixture in the beaker and immediately start off stop watch/clock. Swirl the contents of the beaker.Record the time taken for a blue colour to appear in table 2.
  8. Repeat steps 5 to7 using the contents of test-tube 2,3,4,5 and 6.
  9. Complete table 2 by computing Rate = 1 (S-1)

    Time

Table 2(Sample results)

Test-tube number

1

2

3

4

5

6

Volume of distilled

water(cm3)

0

2

3

5

6

7

Volume of solution A(cm3)

10

8

7

5

4

3

Time(seconds)

40.0

60.0

70.0

90.0

100.0

110.0

Rate = 1 (S-1)

time

0.025

2.5 x 10-2

0.0167

1.67 x 10-2

0.0143

1.43 x 10-2

0.0111

1.11 x 10-2

0.0.1

1.11 x 10-2

0.0083

8.3 x 10-3

 

Plot a graph of rate(y-axis )against volume of solution A(3mk)

Sketch graph of rate against time

Image From EcoleBooks.comImage From EcoleBooks.com

 

 

 

(b)What time would be taken for the blue colour to appear if the experiment was repeated using 4cm3 of distilled water and 6cm3 of solution A?(2mk)

From a correctly plotted graph

1/t at 6cm3 = 0.0125 √ => t = 1/0.0125 = 80seconds√

2. You are provided with solid E. Carry out the experiments below. Write your observations and inferences in the spaces provided

(a)Place all solid in a boiling tube .Add 20cm3 of distilled water and shakeuntil all the solid dissolves.Label the solution as solution E.Use solution E for experiment (i)and (ii)

(i)To 2cm3 of solution E in a test tube in each of experiment I,II,III and IV add:

I. two drops of aqueous sodium sulphate(VI)

Observation(1mk) Inferences(1mk)

 

White precipitate Pb2+ Ba2+ Ca2+

 

 

II.five drops of aqueous sodium chloride

Observations (1mk) Inferences(1mk)

 

White ppt  Ca2+   Ba2+

 

III.two drops of barium chloride

Observations(1mk) Inferences(1mk)  

No white ppt SO42- SO32- CO32-  
IV
.two drops of lead(II)nitrate(V)

Observations(1mk) Inferences(1mk)  

No white ppt Cl

(ii)
To 2cm3 of solution E in a test tube ,add 5 drops of aqueous sodium hydroxide .Add the piece of aluminium foil provided to the mixture and shake. Warm the mixture and test any gas produced with both blue and red litmus papers

Observations(1mk) Inferences(1mk)  

Blue litmus paper remain blue

Red limus paper turn blue  NO3

Effervescence /fizzing/ bubbles

 Note:

Solid E is Calcium nitrate(V) / Barium nitrate(V)  

 

3.You are provided with solid F. Carry out the following test.Write your observations and inferences in the spaces provided.

(a)Place all of solid F in a boiling tube. Add about 20cm3 of distilled water and shake until all the solid dissolves. Label the solution as solution F.

Add about half of the solid sodium hydrogen carbonate provided to 2cm3 of solution F

Observations(1mk) Inferences (1mk)

 

No effervescence/fizzing   H+ absent

 

 

 

(b)(i)Add about 10cm3 of dilute hydrochloric acid to the rest of solution F in the boiling tube. Filter the mixture. Wash the residue with about 2cm3 of distilled water.Dry the residue between filter papers. Place about one third of the dry residue on a metallic spatula and burn it in a Bunsen burner flame.

Observations(1mk) Inferences(1mk)

 

Solid burns with a yellow sooty flame  C C //  C C

 

 

(ii)Place all the remaining residue into a boiling tube. Add about 10cm3 of distilled water and shake thoroughly. Retain the mixture for the tests in (c)

Observations (1/2mk) Inferences(1/2mk)   Solid dissolves to a colourless solution Polar compound

(c)Divide the mixture into two portions:

 

(i)to the first portion ,add the rest of the solid sodium hydrogen carbonate

Observations (1mk) Inferences(1mk) Effervescence/fizzing H+

 

 

 

(ii)to the second portion ,add two drops of bromine water.

Observations (1mk) Inferences(1mk)   Bromine water decolorized C C // C C bonds

 

 


 




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EcoleBooks | Chemistry Form 4 Notes : BASIC PRINCIPLES OF CHEMISTRY PRACTICALS

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