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Introduction to Chemistry Practicals /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

Chemistry Practicals: 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.

Sample Titration table format

Final burette reading (cm3)24.024.024.0
Initial burette reading (cm30.00.00.0
Volume of solution used(cm3)24.024.024.0

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.

Practicals in 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

  • 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)0306090120150180210240
Temperature(oC)25.045.085.095.096.096.096.097.098.0

Questions

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

Sketch graph of temperature against time

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)0306090120150180210240240
Temperature (oC)93.085.078.070.069.069.069.067.0 65.065.0

Questions

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

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

  • 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

ExperimentIII
Final temperature of solution(T2)30.0oC31.0oC
initial temperature of solution(T1)25.0oC24.0oC
Change in temperature(∆T)5.06.0

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.

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.

Questions

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 OH–  ions).

(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 OH–  ions) 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)22.522.5
Temperature of base T2 (oC)22.023.0
Final temperature of solution T4(oC)35.536.0
Initial temperature of solution T3(oC)22.2522.75
Temperature change( T5)13.2513.75

  (a)Calculate T6 the average temperature change                                   T =           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 ofsodium hydroxide = molarity x volume

                                              1000

                 =>  0.5 M x 20cm3  = 0.01 moles

                            1000                              

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

                                      Molessodium 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 volume – volume 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)25.025.025.025.025.025.025.025.025.025.025.0
Volume of alkali(cm3)05.010.015.020.025.030.035.040.035.040.0
Final temperature(oC)22.024.026.028.028.027.026.025.024.025.024.0
Initial temperature(oC)22.022.022.022.022.022.022.022.022.022.022.0
Change in temperature0.02.04.06.06.05.04.03.02.03.02.0

 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

(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.

(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

Practical: Chemical 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

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.

Chem Practicals: 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  

  •       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 7to 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 saltsorange
Sodium/ sodium saltsyellow
Potassium/potassium saltsPurple/lilac
Lithium/Lithium saltsDeep red/crimson
Calcium/ calcium saltsred
Copper/copper saltsBlue/ 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)