Chem prac IV.docx

SIR MALAIKA MASTER OF ALL ACTUAL PRACTICALS TABORA
SIR MALAIKA .K. KABATE PHONE NO 0678426804 @ 24 HRS ALI HASSAN MWINYI ISLMC. SEC
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ALI HASSAN MWINYI HIGH SCHOOL
MALAIKA FOUNDATION LABORATORY
chemistry ACTUAL PRACTICAL
volumetric & rate of reaction
AND
qualitative analysis
SIR. MALAIKA KASSIM KABATE
TEL NO; 0678 42 68 04 / 0672849212/ 0687 200 911
“The purpose of education is to replace an empty mind with the new one”
Acknowledgemant

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I feel extremely happy to come up with this chemistry pamphlet and i would like to extend my
thanks and happiness to all individuals who supported me in all forms.
I wish to thanks all students in the classes at ALI Hassan Mwinyi High School. I have ever
through for their contributions in classes. This enabled me a better understanding of the
subject matter and how to deliver this content to each specific or individual students groups.
I thank @ll@h (Sw) or GOD, almighty @ll@h of Heavens and Earth for keeping me a live, giving
me courage and wisdom to make this compilation what it is. @ll@h , I also request that you
keep an open the minds of whoever reads this pamphlet to take the best out of it.
Finally, my appreciations go to my family members;
 Mr. RICHARD KABATE MAGESA (Manager at ALI HASSAN MWINYI ISLAMIC SEC
SCHOOL and MUHUMBU ISLAMIC SEC SCHOOL) for ensuring me that I get the basic
education in life.
 Mrs. NASMA RICHARD MAGESA
 GODFREY MAKENE (BROTHER)
 KABATE SADIKI MAGESA (YOUNG BROTHER)
 MALAIKA SADIKI KABATE
 ABUBAKARI SADIKI KABATE
 ABDUL-AZIZ SADIKI MAGESA
 AHMEDI RICHARD KABATE
 RAHMA SADIKI KABATE
Malaikakassim99@gmail.com 0678426804 / 0672849212
1. VOLUMETRIC ANALYSIS

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NECTA 2012
1. You are provided with the following solutions;
TZ: Containing 3.5g of impure sulphuric acid in 500 cm3 of solution;
LO: Contain 4g of sodium hydroxide in 1000 cm3 of solution
Phenolphthalein and methyl orange indicators.
Questions
a) i/ What is the suitable indicator for the titration of the given solution? Give a
reason for your answer
ii/ write a balanced chemical equation for the reaction between TZ and LO.
iii/ Why it is important to swirl or shake the contents of the flask during the
addition of acid?
b) Titrate the acid (in a burette) against the base (in a conical flask) using two drops
of your indicator and obtained three titre values.
c) i/ Cm3 of acid required Cm3 of base for complete reaction.
ii/ Showing your procedure clearly, calculate the percentage purity of TZ.
SOLUTION
a) i/ The suitable indicator fo the titration of the given solution is either of the two
indicators, it can be either methyl orange (mo) or phenolphthalein (p. o. p) indicators
because the titration involves strong acid against strong base.
ii/ H2SO4 (aq) + 2NaOH (aq) Na2SO4 (aq) + H2O (L)
iii/ It is important to swirl or shake the contents of the flask during the addition of the
acid to allow reaction to take place properly or completely as well as allowing easy
recognition of sharp end point during titration.
b) Table of results
Titration number Pilot 1 2 3
Final burette reading (cm3)
25.90 25.10 25.40 25.00
Initial burette reading (cm3)
0.00 0.00 0.00 0.00
Volume used or titre value
(cm3)
25.90 25.10 25.40 25.00
Average titre value = V1 + V2 + V3
3

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Va = 25.10 + 25.40 + 25.00
3
Va = 25.16 cm3
c) i/ 25.16 cm3 of acid solution required 25.00 cm3 of base for complete reaction.
ii/ Calculate for the percentage purity of TZ
Given that
Concentration (g/dm3) = Conc in g/dm3
So, its’ Molarity (mol/dm3) = Conc in g/dm3
Molar mass g/mol
= 4g/dm3
40g/mol
Molarity of solution LO (base) = 0.1mol/dm3
Molarity of solution TZ (mol/dm3) can be obtained from the relation
Ma Va = na
Mb Vb nb
Where:
Ma = Molarity of Acid (TZ)
Va =Volume of Acid (TZ)
Mb =Molarity of Base (LO)
Vb=Volume of Base (LO)
Na =Number of Acid (TZ)
Nb =Number of Base (LO)
We have
Ma = ?
Va =25.16
Mb = 0.1
Vb=25
Nb =2
Na =1

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H2SO4 (aq) + 2NaOH (aq) Na2SO4 (aq) + H2O (L)
Na= 1 Nb= 2
Ma Va = na
Mb Vb nb
Ma = Mb X Vb X na
Va X nb
Ma = 0.1 X 25 X 1
25.16 X 2
Ma = 0.05M
Molarity of solution TZ (acid) = 0.05M.
Then,
Calculate the percentage purity of TZ
Percentage purity = Concentration of pure X 100%
Concentration of impure
Therefore
Find the concentration of pure in solution TZ
Concentration of pure = Molarity X molar mass
Conc of pure = 0.05 X 98
Concentration of pure = 4.9g/dm3
Find the concentration of impure in solution TZ

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Concentration of impure = Mass
Volume
Conc of impure = 3.5g (Volume = 500 = 0.5 dm3)
0.5 1000
Concentration of impure = 7g/dm3
Percentage purity = 4.9g/dm3 X 100%
7g/dm3
Percentage purity = 70%
 Percentage purity of solution TZ = 70%
Solution AA Prepared by diluting 100cm3
of 1M Hydrochloric Acid
Solution BB is sodium hydroxide
Phenolphthalein indicator (POP)
Procedure:

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Pipette 20 cm3
or 25cm3
of solution BB into a titration flask. Add two drops of POP
indicator. Titrate solution BB against solution AA from the burette until a colour change is
observed. Note the burette reading. Repeat the procedure to obtain three more readings.
Record your results as shown bellow.
a) Table of results
i) Burette readings
Titration Pilot 1 2 3
Final reading (cm3
)
Intial reading (cm3
)
Volume used (cm3
)
ii) The volume of piprtte used was Cm3
iii) The colour change at the end point was from to
iv) Cm3 of solution AA was required to neutralize Cm3of solution BB
b) Write a balanced chemical equation for the neutralization of sodium hydroxide by
hydrochloric acid
c) Calculate the:
(i) Molarity of solution AA
(ii) Concetration in mole/dm3 of solution BB
(iii) Concentration in g/dm3 of solution BB
SOLUTION
a) Table of results
i) Burette readings
Final reading (cm3
) 24.00 48.10 23.70 47.00
Intial reading (cm3
) 0.00 24.00 0.00 23.70
Volume used (cm3
) 24.00 24.10 23.70 23.30
Average volume of acid (AA) = V1 + V2 + V3
3

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Va = 24.10 + 23.70 + 23.30
3
Va = 23.70 cm3
ii) The volume of pipette used was 25.00 Cm3
v) The colour change at the end point was from Pink to Colourless
vi) 25.00 Cm3 of solution AA was required to neutralize 23.70 Cm3of solution BB
b) Balanced chemical equation
HCl (aq) + NaOH (aq) NaCl (aq) + H2O (L)
c)
(i) Molarity of solution AA
From the formula
McVc = MdVd
Where
Mc = Molarity of conc. Acid (AA)
Vc = Volume of conc. Acid (AA)
Md = Molarity of diluted Acid (AA)
Vd = Volume of diluted or final volume of Acid AA
We have
Mc =1M
Vc =100cm3
Vd =1000cm3
Md= ?
McVc = MdVd
1Mx100cm3 = Md x 1000cm3
Md = 1M x 100cm3
1000cm3
= 0.1M

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 Molarity of solution AA = 0.1M
(ii) Concentration in mole/dm3 of solution BB
From the relation
Ma Va = na
Mb Vb nb
Where:
Ma = Molarity of Acid (AA)
Va =Volume of Acid (AA)
Mb =Molarity of Base (BB)
Vb=Volume of Base (BB)
Na =Number of Acid (AA)
Nb =Number of Base (BB)
HCl (aq) + NaOH (aq) NaCl (aq) + H2O (L)
Na= 1 Nb= 1
We have
Ma = 0.1
Va =23.70
Mb = ?
Vb=25
Nb =1
Na =1
Mb = Ma X VaX nb
Vb X na
Mb = 0.1 X 23.70 X 1
25X 1
Mb = 0.0948 M = 0.1M

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 Concentration of BB in mol/dm3 = 0.1M.
(iii) Concentration in g/dm3 of solution BB
From the relation
Molarity (mol/dm3) = Conc in g/dm3
Molar mass g/mol
0.1 = Conc in g/dm3
40
Concentration g/dm3 = Molarity X molar mass
Conc = 0.1 x 40
= 4 g/dm3
 Concentration in g/dm3 of solution BB = 4 g/dm3
 Solution E: Containing 3.65g of Hydrochloric Acid per dm3
of solution.
 Solution K: Containing 3.575g of pure hydrated sodium carbonate, Na2CO3.xH2O per
0.25dm3
of solution.
 Methyl orange indicator solution

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Procedure:
Put the acid solution into the burette. Pipette 20 cm3
or 25cm3
of solution K into a titration
flask. Add few drops of methyl orange indicator. Titrate this base solution against the acid
solution until an end point is reached. Record your results as shown bellow and Repeat the
titration.
a) Table of results
Final reading (cm3
)
Intial reading (cm3
)
Volume used (cm3
)
The volume of pipette used was Cm3
i/ The colour change at the end point was from to
a) i/ The volume of acid solution E needed for complete neutralization was Cm3
ii/ Write down the balanced chemical equation for the reaction
b) Calculate the:
i/ Molarity of Acid solution E
ii/ Molarity of base solution K
iii/ Find the value of water crystallization in Na2CO3. XH2O
SOLUTION
The volume of pipette used was 20.00 Cm3
a) Table of results

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Final reading (cm3
) 20.00 39.80 19.60 40.00
Intial reading (cm3
) 0.00 20.00 0.00 20.00
Volume used (cm3
) 20.00 19.80 19.60 20.00
Average volume of acid (E) = V1 + V2 + V3
3
Va = 19.80 + 19.60 + 20.00
3
Va = 19.93 cm3
b) i/ The colour change at the end point was from Yellow to pink
ii/ Balanced equation for the reaction
2HCl (aq) + Na2CO3(aq) 2NaCl(aq) + H2O (L) + CO2(g)
C) Calculate the;
(i) Molarity of Acid solution E
From the relation
Molar mass g/mol
Then
Find the concentration of solution E
Concentration of solution E = Mass
Volume
Conc of E = 3.65g
1dm3

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Conc of E= 3.65g/dm3
Therefore;
Molar mass g/mol
Molarity = 3.65g/dm3
36.5g/mol
Molarity = 0.1 mol/dm3
 Molarity of Acid of solution E = 0.1 mol/dm3
(ii) Molarity of base solution K
From the relation
Ma Va = na
Mb Vb nb
Where:
Ma = Molarity of Acid (E)
Va =Volume of Acid (E)
Mb =Molarity of Base (K)
Vb=Volume of Base (K)
Na =Number of Acid (E)
Nb =Number of Base (K)
2HCl (aq) + Na2CO3(aq) 2NaCl (aq) + H2O (L) + CO2(g)
Na= 2 Nb= 1
We have
Ma = 0.1M
Va =19.93 cm3
Mb = ?
Vb=20.00 cm3

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Nb =1
Na =2
Mb = Ma X VaX nb
Vb X na
Mb = 0.1 X 19.93 X 1
20X 2
Mb = 0.0498
Mb = 0.05 M
 Molarity of base (Mb) = 0.05M.
(iii) The value of water crystallization in Na2CO3. XH2O
Find the Concentration in g/dm3 of Na2CO3
From the formular
Molar mass g/mol
Then
Conc. Of Na2CO3 = Molarity X Molar mass
Conc. Of Na2CO3 = 0.05M X 106g/mol
Concentration of solution K = 5.3g/dm3
Then
Find the molar mass of Na2CO3. XH2O
From the formula
Molar mass g/mol
Then

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Molar mass (g/mol) = Conc in g/dm3
Molarity (mol/dm3)
But
Concentration of solution K = Mass
Volume
Conc of k = 3.575g
0.25dm3
Conc of K= 14.3g/dm3
Also
Molarity (mol/dm3)
Molar mass = 14.3g/dm3
0.05mol/dm3
= 286g/mol
Molar mass of Na2CO3.xH2O is 286g/mol
Therefore,
Na2CO3.xH2O = 286
(23 X 2 +12+16X3 +(2+16) = 286
46 + 12 + 48 + 18x
106 + 18x = 286
18x = 286- 106
18x = 180
18x =180
18 18
X = 10
 The value of x in Na2CO3.xH2O is 10
TECNIC QUESTION
Solution EE: Containing 0.2 M of sodium hydroxide
Solution FF: Containing 4.50g of acid H2Q which were dissolved in 250 cm3
of solution
Phenolphtalein indicator and Methyl orange indicator

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Procedure;
(i) Measure exactly 75 cm3 of acid by using measuring cylinder and then transfer it into
an empty clean beaker.
(ii) Measure exactly 75 cm3 of distilled water and then put it into the solution of the acid
in part (i) above.
(iii) Stir well the solution mixture in part (ii) above by means of a clean and dry
glass rod
Questions
a. (i) Why is it important to clean a conical flask at the end of each titration
before staring the subsequent titration?
(ii)State the role of white tile or white paper on which conical flask and its
contents is put during titration.
b. Titrate solution EE against solution FF three times
c. Cm3 of sodium hydroxide required cm3 of acid H2Q for complete
neutralization.
d. Identify H2Q acid by providing its name and formula.
e. Classify acid H2Q according to its strength, basicity and origin respectively
and sate why in each case.
f. i/ Was methyl orange also a suitable indicator for this titration? Give
reasons.
ii/ What is the significance of the indicator in this experiment?
iii/ Why there a colour change when enough acid has been added into the
base
SOLUTION
a. (i) It is important to clean a conical flask at the end of each titration before
starting the subsequent titration in order to avoid the effect of impurities in
the conical flask which will turn the effect the titre value the acid used. Or to
avoid effect of diluting base solution in the next experiment due to waste

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(ii)The role of white tile or white paper on which conical flask and its contents
is put during titration was to increase clear visibility end point of colour.
b. Table of results of burette reading
Final reading (cm3
) 20.00 39.80 19.70 40.00
Intial reading (cm3
) 0.00 20.00 0.00 20.00
Volume used (cm3
) 20.00 19.80 19.00 20.00
3
Va = 19.80 + 19.70 + 20.00
3
Va = 19.97 cm3
c. (i) 20.00cm3 sodium hydroxide required 19.97cm3 of acid H2Q for complete
neutralization
(ii)Colour change from Pink to colourless point of equivalent
d. Identification of acid H2Q and name it’s the formula
Step 1: calclat molarity of diluted acid H2Q.
From the relation
Ma Va = na
Mb Vb nb
Where:
Ma = Molarity of Acid (FF)
Va =Volume of Acid (FF)
Mb =Molarity of Base (EE)
Vb=Volume of Base (EE)
Na =Number of Acid (FF)
Nb =Number of Base (EE)

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H2Q (aq) +2 NaOH(aq) Na2Q (aq) + 2H2O (L) + CO(g)
Na= 1 Nb= 2
We have
Ma = ?
Va =19.97 cm3
Mb = 0.2 M
Vb=20.00 cm3
Nb =2
Na =1
Ma= Mb X VbX na
Va X nb
Ma = 0.2 X 20.00 X 1
19.97X 2
Ma = 0.1 M
Ma= 0.1 M
 Molarity of diluted acid = 0.1M.
But from
Find the molarity before diluting acid
From the formular
McVc = MdVd
Where
Mc = Molarity of conc. Acid (FF)
Vc = Volume of conc. Acid (FF)
Md = Molarity of diluted Acid (FF)
Vd = Volume of diluted or final volume of Acid FF
We have

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Mc =?
Vc =75cm3
Vd =150cm3
Md=0.1
McVc = MdVd
MC x75cm3 = 0.1 x 150cm3
Mc = 0.1M x 150cm3
75cm3
= 0.2M
Molarity of conc of solution FF = 0.2M
Given that
4.50g = 250cm3
? = 1000
Concentration = 18g/dm3
Step 2:
Find the molar mass of H2Q
Molarity (mol/dm3)
Molar mass = 18g/dm3
0.2mol/dm3
Molar mass = 90g/mol
So,
H2Q =90
2 + Q=90
Q=90-2
Q= 88

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Q= 88 (This is molecular mass of Q)
 Acid H2Q is most probably oxalic acid with the formula H2C2O4
e. Acid H2Q which is H2C2O4 is weak acid, because its degree of dissociation is
low. For example dissociate partially compared to strong acids like
hydrochloric acid which dissociate fully.
 Basically of H2Q is two (2) because it contain two (2) number of
hydrogen atoms
 Original of oxalic acid or ethanedioic is organic acid or organic
compound with structure
C O
OH
C O
OH
f. i/ Methyl orange indicators are not suitable indicator because the titration
involves weak acid against strong base and if applied the colour will change
before equivalent or end point.
ii/ The significance of the indicator in this experiment was to show the end
point of the reaction. The indicator changes colour when the end point
reached. The colour changes from Yellow to pink
iii/ When enough acid has been added into the base the reaction will complete
or will neutralize base and reaction take place properly and the colour change to
show end point of titration.
NECTA 2015
G: Containing 0.1 mole hydrochloric acid per dm3 of solution;
B: Contain 2.65g M2CO3 per 0.5dm3 of solution
Methyl orange indicator.

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Questions;
(a) Titrate G (in burette) against B (in conical flask) using two drops of your indicator
obtain three titre values. Record your data in a tabular form.
(b) (i) cm3 of B required cm3 of G for complete reaction.
(ii) Write a balanced chemical equation for the reaction between B and G and the
corresponding ionic equation with state symbols.
(iii)Showing your procedures clearly; calculate the molar mass of M2CO3 and hence
identify element M
SOLUTION
(a) Table of result
Final reading (cm3
) 26.00 25.00 25.10 25.00
Intial reading (cm3
) 0.00 0.00 0.00 0.00
Volume used (cm3
) 26.00 25.00 25.10 25.00
3
Va = 25.00 + 25.10 + 25.00
3
Va = 25.03 cm3
(b)(i) 25.00cm3 of B required 25.03cm3 of G for complete reaction
(ii) 2Hcl (aq) + M2CO3 (aq) 2MCl (aq) + CO2 (g) + H2O (L)
2H+(aq) + 2Cl- (aq) + 2M+ (aq) + CO3+ (aq) 2M+ (aq) + 2Cl- (aq) + CO2 (g) + H2O (L)

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Ionic equation: 2H+ (aq) + CO3+ (aq) CO2 (g) + H2O (L)
(iii) Calculate the molar mass of M2CO3 and hence identify element M
Apply for the formula,
Ma Va = na
Mb Vb nb
Where:
Ma = Molarity of Acid
Va =Volume of Acid
Mb =Molarity of Base
Vb=Volume of Base
Na =Number of Acid
Nb =Number of Base
2Hcl (aq) + M2CO3 (aq) 2MCl (aq) + H2O (L) + CO2(g)
Na= 2 Nb= 1
We have
Ma = 0.1M
Va =25.03 cm3
Mb = ?
Vb=25.00 cm3
Nb =1
Na =2
Mb = Ma X VaX nb
Vb X na
Mb = 0.1 X 25.03 X 1
25.00X 2
Mb = 0.05 M
Mb = 0.05 M
Molarity of base (Mb) = 0.05M.

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Then
Concentration of solution M2CO3 = Mass
Volume
Conc of M2CO3 = 2.65g
0.5dm3
Conc of M2CO3 = 5.3g/dm3
But
Molar mass M2CO3 = Conc in g/dm3
Molarity (mol/dm3)
Molar mass M2CO3 = 5.3g/dm3
0.05mol/dm3
 Molar mass M2CO3 = 106g/mol
Therefore
M2CO3 = 106
2M + 12 + (16 X 3) = 106
2M + 12 + 48 = 106
2M + 60 = 106
2M= 106 – 60
2M = 46
2M=46
2 2
Atomic mass of M = 23
 Molar mass of M2CO3 = 106g/mol and element M is Sodium (Na)
NECTA 2016
Q: Containing 36.5g of hydrochloric acid in 1 litre of solution;
P: Contain 4.0g of impure ammonium hydroxide per 0.25dm3 of solution
Methyl orange indicators.

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Procedure;
(i)Measure exactly 10 cm3 of Q by using 10 cm3 measuring cylinder and pour
into100cm3 measuring cylinder. Carefully add distilled water to 100 cm3 mark then
stir. Fill the resulting into a burette.
(ii)Titrate Q against P using two drops of the indicators obtain three accurate values.
Record your results in a tabular form.
Questions;
(a) What if phenolphthalein indicator was used in place of methyl orange indicator for
the titration of the given solutions? Give reasons for your answer
(b) (i) cm3 of P required cm3 of Q for complete reaction.
(ii) Write a balanced chemical equation for the reaction between Q and P
(c) Showing your procedures clearly; calculate the percentage by weight of the impurity
in the ammonium hydroxide.
SOLUTION
Table of results
Final reading (cm3
) 21.80 40.00 20.10 40.80
Intial reading (cm3
) 0.00 20.00 0.00 20.00
Volume used (cm3
) 20.00 20.00 20.10 20.80
Average volume of acid (Q) = V1 + V2 + V3
3
Va = 20.00 + 20.10 + 20.80
3
Va = 20.30 cm3
(a) Phenolphthalein indicator is not suitable indicator for this titration the reaction
involves strong acid against weak base, so if phenolphthalein indicator was

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used in place of methyl orange indicator we could have not get end point and
get wrong volume of acid.
(b) (i) 20 cm3 of P required 20.30 cm3 of Q for complete reaction
(ii) HCl (aq) + NH4 OH(aq) NH4Cl (aq) + H2O (L)
(c) Calculate for the percentage by weight of the impurity in the ammonium
hydroxide.
Step 1:
Calculate molarity of acid Q (molarity of concentrated Q (Mc).
Molarity of conc. (MC) = Conc in g/dm3
Molar mass g/mol
Molarity of conc. (Mc) = 36.5g/dm3
36.5g/mol
Molarityof conc. (MC) = 1 mol/dm3
Molarity of conc. Acid (MC) of solution Q = 1 mol/dm3
Then
Calculate Molarity of diluted acid Q can be calculated
From the formula
McVc = MdVd
Where
Mc = Molarity of conc. Acid (Q)
Vc = Volume of conc. Acid (Q)
Md = Molarity of diluted Acid (Q)
Vd = Volume of diluted or final volume of Acid Q
We have
Mc =1M
Vc =10cm3
Vd =100cm3
Md= ?
McVc = MdVd

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1Mx100cm3 = Md x 1000cm3
Md = 1M x 10cm3
100cm3
= 0.1M
Molarity of diluted acid Q = 0.1M
Step 2:
Find the molarity of base of solution P (Mb)
From the relation
Ma Va = na
Mb Vb nb
Where:
Ma = Molarity of Acid (Q)
Va =Volume of Acid (Q)
Mb =Molarity of Base (P)
Vb=Volume of Base (P)
Na =Number of Acid (Q)
Nb =Number of Base (P)
HCl (aq) + NH4 OH (aq) NH4Cl (aq) + H2O (L)
Na= 1 Nb= 1
We have
Ma = 0.1M
Va =20.30 cm3
Mb = ?
Vb=20.00 cm3
Nb =1
Na =1
Mb = Ma X VaX nb

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Vb X na
Mb = 0.1 X 20.30 X 1
20X 1
Mb = 0.1
Mb = 0.1 M
Molarity of base (Mb) of solution P = 0.1M
Step 3:
Calculation for percentage purity of base of solution P
Concentration in g/dm3 of pure P = molarity X molar mass of (NH4OH)
Concentration in g/dm3 of pure P= 0.1 X 35
Concentration in g/dm3 of pure P = 3.5g/dm3
But
Concentration in g/dm3 of impure solution P
Concentration of solution P = Mass
Volume
Conc of P = 4.0g
0.25dm3
Concentration of impure of solution P = 16g/dm3
From
Percentage purity = 3.5g/dm3 X 100%
16g/dm3

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Percentage purity = 21.875%
Percentage purity of solution TZ = 21.875%
So,
Percentage impure % = 100% % purity
Percentage impure % = 100% 21.875%
= 78.125%
 Percentage by weight of the impurity in the ammonium hydroxide = 78.13%
NECTA 2018
X: A 0.5 M sulphuric acid solution;
Y: A monovalent metal hydroxide made by dissolving 4.00g of MOH in distilled
water to make up to 1000cm3 of solution;
MO: Methylorange indicator.
Procedure
(i) Measure exactly 20 cm3 of X by using 100 cm3 measuring cylinder. Carefully
add distilled water to 100 cm3 mark then stir. Call this solution X1. Fill X1 into
the burrette.
(ii) Titrate X1 against 20.00 cm3 or 25.00 cm3 of Y using two drops of the
indicator; obtain three accurate values. Record your results in a tabular form.
Questions
(a) (i) The colour change at the end point was from to
(iii) Cm3 of Y required cm3 of X1 for complete reaction.
(b) Write a balanced chemical equation for the reaction between X1 and Y.
(c) Calculate the following:
i. Molarity of X1 and Y;
ii. Molar mass of Y;
iii. Atomic mass of M.

Page 29
(d) Identify metal M and write its electronic configuration.
(e) State two properties of each of X and Y.
SOLUTION
(a) (i) The colour change at the end point was from Yellow to Pink
(ii) 25.00 Cm3 of Y required 12.50 cm3 of X1 for complete reaction.
(b) Balanced chemical reaction
H2SO4 (aq) + 2MOH (aq) M2SO4 (aq) + 2H2O (L)
(c)
i/ Molarity of X1 and Y;
Step 1;
Find the molarity of X1 after dilution
From the formula of dilution
McVc = MdVd
Where
Mc = Molarity of conc. Acid (X)
Vc = Volume of conc. Acid (X)
Md = Molarity of diluted Acid (X1)
Vd = Volume of diluted or final volume of Acid (X1)
We have
Mc =0.5M
Vc =20cm3
Vd =100cm3
Md= ?
McVc = MdVd
0.5Mx20cm3 = Md x 100cm3
Md = 0.5M x 20cm3
100cm3
= 0.1M

Page 30
Molarity of diluted acid X1 = 0.1M
Step 2;
Find the molarity of Y
From the relation
Ma Va = na
Mb Vb nb
Where:
Ma = Molarity of Acid (X1)
Va =Volume of Acid (X1)
Mb =Molarity of Base (Y)
Vb=Volume of Base (Y)
Na =Number of Acid (X1)
Nb =Number of Base (Y)
H2SO4 (aq) + 2MOH(aq) M2SO4(aq) + 2H2O (L)
Na= 1 Nb= 2
We have
Ma = 0.1M
Va =12.50 cm3
Mb = ?
Vb=25.00 cm3
Nb =2
Na =1
Mb = Ma X VaX nb
Vb X na
Mb = 0.1 X 12.50 X 2
25X 1
Mb = 0.1

Page 31
Mb = 0.1 M
Molarity of base (Mb) of solution Y = 0.1M
 Molarity of X1 is 0.1M and of Y is 0.1M
ii/ Molar mass of Y
From the relation
Molar mass of Y = Conc in g/dm3
Molarity (mol/dm3)
Molar mass of Y = 4.00g/dm3
0.1mol/dm3
 Molar mass M2CO3 = 40g/mol
iii/ Atomic mass of Y
Therefore
MOH = 40
M + 16 + 1 = 40
M + 17= 40
M= 40 – 17
M = 23
 Atomic mass of M = 23
(d)The metal M is Sodium (Na) and the electronic configuration is 2: 8: 1
(e) Properties of conc. Sulphuric acid
i/ Sulphuric acid is a thick, colourless, oily fluid.
ii/ Concentrated Sulphuric acid is a very strong dehydrating agent.
iii/ It is good oxidizing agent, it can oxides both non-metals as well as
metals.
Properties of Sodium hydroxide
i/ it is soluble in polar solvents such as water, ethanol and methanol and
insoluble in organic solvents.

Page 32
ii/ it is odorless, white crystalline solid with a density of 2.13 g/mL and
melting point of 318 0C.
ALL QUESTIONS OF CONCETRATION AND TEMPERATURE IN ACTUAL PRACTICAL
1. What is the aim of the whole experiment
2. Complete table
3. Give reasons (s), identify the experiment in which the reaction was
i/ Fast

Page 33
ii/ Slow
4. With state symbols write balanced chemical equation for the reaction
5. List four factors which can affect rate of chemical reaction
6. Write the electronic configuration of the production which causes the solution to
cloud letter X
7. What can you conclude from the data obtain
8. Write a balanced ionic equation for the reaction
9. What does 1/time represent?
10. From obtained data what can you conclude the effect increasing temperature on the
rate of chemical reaction?
11. Cause the smell or irritant chocking smell
12. Is the graph is a straight line or curve
13. How was the factor of concentration varied in this experiment
14. Use the graph to explain how variation of concentration affects the rate of chemical
reaction
15. State the sources of errors in this experiment
16. From obtained data what conclusion can you make about the effect of increasing
temperature on the rate of the reaction
17. Why did the cross disappear?
18. Write two uses of the product which obscured the cross
19. Record the room temperature
20. State the product which causes the solution to cloud letter X
21. Plot the graph
22. Determine the slope
23. How was the factor of temperature varied in this experiment
NECTA 2012
2. You are provided with the following:
ZO: A solution of 0.13M of Na2S2O3 (Sodium thiosulphate)
UU: A solution of 2M HCl;
Thermometer
Heat source/ burner
Stop watch

Page 34
Procedure
(i) Place a 500cm3 beaker, which is half- filled with water, on the heat source as a
water bath.
(ii) Measure 10cm3 of ZO and 10 cm3 of UU into two separate test tubes
(iii) Put the two test tubes containing ZO and UU solution into a water bath.
(iv) When the solutions attain a temperature of 600C, remove the test tubes from the
water bath and pour both solutions into 100 cm3 empty beaker and immediately
start the stop watch.
(v) Place the beaker with the contents on top of a piece of paper marked X
(vi) Note the time taken for themark X to disappear
(vii) Repeat procedures (i) to (vi) at temperature 700C, 800C and 900C.
(viii) Record your results as in table 1.
Table 1
Experiment Temperature Time (s)
1 600C
2 700C
3 800C
4 900C
QUESTIONS
a. Write a balance chemical equation for reaction between UU and ZO
b. What is the product which causes the solution to cloud the letter X?
c. Plot a graph of temperature against time
d. What conclusion can you draw from your graph?
SOLUTION

Page 35
Table 1
Experiment Temperature Time (s)
1 600C 11.72
2 700C 7.62
3 800C 6.00
4 900C 3.22
a. 2HCl (aq) + Na2S2O3(aq) 2NaCl(aq) + H2O(L) + SO2(g) + S(s)
b. The product which causes the solution to cloud the letter X was Sulphur (S).
C. The graph of temperature against time
d. conclusion
As temperature of the reactant increases the time taken for letter X to
disappear decreases or temperature is inversely proportional to the time but the
graph was straight line or increase.
NECTA 2015
U: A solution containing 79g of Na2S2O3.5H2O
V: A solution contain 0.1mol/dm-1 of dilute HCL
T: Distilled water

Page 36
Stop watch/clock
Procedure
(i) Place a 100cm3 beaker on top of letter X on a plain paper provided
(ii) Measure 8.0cm3 of U and 2cm3 of T and put them in the beaker in (i)
(iii) Measure 10cm3 of V and put it into a beaker containing U and T; immediately
start the stop watch and observe the changes from above
(iv) Record the time taken for the disappearance of letter X
(v) Repeat steps (i) to (iv) using the data shown in table 1.
Table 1
Number of
experiment
Volume of
U(cm3)
Volume of
V(cm3)
Volume of
T(cm3)
Time (t) in
second
1/t(sec-1)
1 8 10 2
2 6 10 4
3 4 10 6
4 2 10 8
5 1 10 9
Questions:
a) What is the aim of the whole experiment
b) Complete table 1
c) Give reasons (s), identify the experiment in which the reaction was
i. Fast
ii. Slow
d) With state symbols write balanced chemical equation for the reaction between U
and V
e) List four factors which can affect rate of chemical reaction
f) Write the electronic configuration of the production which causes the solution to
cloud letter X

Page 37
g) What can you conclude from the data obtain
SOLUTION
a) The aim of the whole experiment is to show the effect of concentration on the rate of
chemical reaction.
b) Table of results
Table 1
Number of
experiment
Volume of
U(cm3)
Volume of
V(cm3)
Volume of
T(cm3)
Time (t)
in second
1/t(sec-1)
1 8 10 2 21.12 0.047
2 6 10 4 28.62 0.035
3 4 10 6 44.78 0.022
4 2 10 8 102.31 0.010
5 1 10 9 295.65 0.003
C)
i. The experiment in which the reaction was fast was experiment number 1
because it has higher concentration and the time taken for disappearance of
letter X was the shortest compared to others.
ii. The experiment in which the reaction was slow was experiment number 5
because it has lower concentration and the time taken for disappearance of
letter X was the longest compared to others.
d)
2HCl(aq) + Na2S2O3(aq) 2NaCl(aq) + H2O(L) + SO2(g) + S(s)
e)
i/ Concentration
ii/ Temperature
iii/ Pressure
iv/ Catalyst
f) The electronic configuration of the product which cause the solution to cloud letter
X , the product was Sulphur (S) and its electronic configuration = 2:8:6

Page 38
g) Conclusion
As concentration decreases, the times taken for the disappearance of letter X
increases. Or concentration is inversely proportional to the time but the
concentration is directly proportional to the rate of reaction.
NECTA 2016
BB: A solution containing 79g of Na2S2O3.5H2O
DD: A solution contain 0.1mol/dm-1 of dilute HCL
A stop watch/ clock; a white plain paper with a cross and thermometer
Procedure
(i) Place a 100cm3 beaker on top of a cross on a plain paper provided such that the
cross is visible through the solution when viewed from above
(ii) Prepare a water bath using a 250 cm3 or 300 cm3 beaker
(iii) Measure exactly 10cm3 of BB and 10 cm3 of DD pour into separate boiling
test tubes
(iv) Put the two boiling test tubes into the water bath in (ii) above and warm the
contents to 400C.
(v) Immediately pour the hot solutions BB and DD into the 100 cm3 beaker in (i)
above and simultaneous start the stop watch/clock. Record the time taken in
seconds, for the cross to disappear completely.
(vi) Repeat procedures (iii) to (v) at different temperature, 500C, 600C and 700C.
Record your reading in a tabular form as shown in table 1.
Table 1
Temperature Time (sec) 1/time (sec)
40
50
60

Page 39
70
Questions
a. (i) Record the room temperature
(ii) Complete the table by filling the blank columns
b. Plot a graph of temperature against time from your results
c. Write a balanced ionic equation for the reaction between the dilute
acid and sodium thiosulphate
d. What does 1/time represernt?
e. From obtained data what can you conclude the effect increasing
temperature on the rate of chemical reaction?
SOLUTION
a. (i) Room temperature = 210C
(ii) Table of results
40 17.22 0.06
50 10.03 0.10
60 6.47 0.15
70 3.62 0.28
b. The graph of temperature against time
C. Na2S2O3 (aq) + 2HCl (aq) 2NaCl(aq) + H2O (L) + SO2 (g) + S (s)
In ionic equation
2Na+ + S2O32- + 2H+ + 2Cl- 2Na+ + 2Cl- + H2O(L) + SO2(g) + S(s)
S2O32- + 2H+ H2O (L) + SO2 (g) + S (s)

Page 40
d. 1/time represent rate of chemical reaction
e. As temperature of the reactants increases, the rate of chemical reaction is increases
Or temperature is directly proportional to the rate of chemical reaction.
NECTA 2017
E: A solution containing 79g of Na2S2O3.5H2O
F: A solution contain 0.1mol/dm-1 of dilute HCL
G: Distilled water

Page 41
A stop watch/ clock; a white plain paper and thermometer.
Procedure
(i) Put a beaker (100cm3) on top of a cross on the given sheet of paper.
(ii) Measure 25 cm3 of E using measuring cylinder and pour into a beaker in (i).
(iii) Using another measuring cylinder measure 5 cm3 of F and pour it into a
beaker containing E and instantly, start a stop watch.
(iv) Stir the mixture with a glass rod while you keep on observing the cross from
above; record the time taken for the cross to disappear.
(v) Repeat the procedures for different concentrations of E by taking 20 cm3, 15
cm3, 10 cm3 and 5 cm3 of the original E and making the total volume up to 25
cm3 by adding G.
Repeat the results as shown in the table 1.
Table 1
Volume of E (cm3) Volume of G (cm3) Time taken for the cross to
disappear (sec)
25
20
15
10
5
Questions
a) Complete filling the Table 1
b) (i) Using the data in the table, plot a volume time graph (volume on the y-axis
and time in second on the x-axis)
(ii) What does the shape of the graph indicate?
c) Write down the ionic equation of the reaction between E and F.
d) Why did the cross disappear?

Page 42
e) Write two uses of the product which obscured the cross.
SOLUTION
a)
Volume of E (cm3) Volume of G (cm3) Time taken for the cross to
disappear (sec)
25 0 27.27
20 5 39.94
15 10 52.10
10 15 65.40
5 20 140.00
b) (i) The shape of the graph is a curve
c)
. Na2S2O3 (aq) + 2HNO3 (aq) 2NaNO3(aq) + H2O (L) + SO2 (g) + S (s)
In ionic equation
2Na+ (aq) + S2O32- (aq) + 2H+ (aq) + 2NO3- (aq) 2Na+ + 2NO3- + H2O (L) + SO2 (g) + S(s)
S2O32- (aq) + 2H+(aq) H2O (L) + SO2 (g) + S (s)
d) Cross disappear because of Sulphur (S) or Precipitation of sulphur or sulphur
deposite
e)
1. Sulphur is used in the manufacture of carbon disulphate, sodium
thiosulphate, gun powder, matches and in fireworks.
2. Sulphur is used on vulcanization of rubber. Natural rubber is soft and sticky.
Heating it with sulphur makes it hard non-sticky and more elastic. This
process of heating of natural rubber with sulphur is known as vulcanization.

Page 43
3. Sulphur is used for medicine.
It is used for its cleansing properties in the treatment of
infections.
4. Sulphur is used for skin care or curing skin diseases.
5. Sulphur is used for give specific shapes to the hair.
6. Sulphur is used for agriculture
Sulphur is necessary for the growth and development of plant life used
in fertilizers
NECTA 2018
H1: A solution of 0.25 mol/dm-3 Na2S2O3;
H2: A solution of 2M HCL;
Distilled water; stop watch; Thermometer and a plain paper marked X.

Page 44
Procedure
Place 150cm3 of water in the 250cm3 beaker and use this as your water bath; heat
the water 800C. Measure 10cm3 of H1 and 25cm3 of water and pour the cotents into the
100cm3 beaker. Put the beaker with the contents ito a hot water bath. When the contents
attain a temperature of 700C, place the beaker on top of the marker X on the paper
provided. Add 10cm3 of H2 and immediately start the stop watch. Swirl the beaker twice
and look vertically on the top of the beaker so as to see X through the bottomof the beaker.
Stop thr clock when X is invisible. Record the time taken for X to disapper completely.
Repeat the experiment at different temperatures as shown in the table 1.
Table 1
Exp. No. Temperature of H1
(0C)
Time (sec)
Rate 1/ t (S-1)
1 70
2 60
3 50
4 40
5 Room temperature
QUESTIONS
a) Fill Table 1.
b) Write a balance equation for reaction between H1 and H2
c) Which product causes the solution to cloud the letter X?
d) Plot a graph of rate (1/t) against temperature
e) What conclusion can you draw from this experiment?
SOLUTION
a)
Table 1
Exp. No. Temperature of H1
(0C)
Time (sec)
Rate 1/ t (S-1)

Page 45
1 70 2.56 0.39
2 60 4.05 0.25
3 50 9.23 0.11
4 40 14.08 0.07
5 30 29.15 0.03
b) 2HCl(aq) + Na2S2O3(aq) 2NaCl(aq) + H2O(L) + SO2(g) + S(s)
c) The product which causes the solution to cloud the letter X was Sulphur (S)
d) The graph of rate (1/t) against temperature
e) conclusion
As temperature of the reactant decreases the time taken for letter X to
disappear increases or temperature is inversely proportional to the time but
the graph was increase.
TECHNIC QUESTION
Z: A solution containing 32.24g/dm3 of Na2S2O3.5H2O
U: A solution contain 2 M HCL
Distilled water
Stop watch/clock

Page 46
Procedure
(i) Mark letter A on a piece of paper with black ink
(ii) Measure 50 cm3 of Z and pour it into a conical flask
(iii) Measure 10cm3 of U by using different measuring cylinder immediately start the
stop watch and observe the changes from above
(iv) Stand the conical flask on the paper above the cross. Look down vertically
through the mouth of the conical flask and make sure you can see the cros at the
bottom of the flask.
(v) Swirl the flask while keep on observing until the letter A disappears
(vi) Record the time taken for the disappearance of letter A
(vii) Repeat the whole procedure using 40 cm3, 30 cm3, 20 cm3 and finally 10 cm3.
Table 1
Number of
experiment
Volume of Z
(cm3)
Volume of
distilled
H2O (cm3)
Volume of Na2S2O3
after addind water
(mol/dm3)
Time (t)
in
second
Rate of
reaction(S-1)
1 50 0
2 40 10
3 30 20
4 20 30
5 10 40
Questions
a) Complete filling the table of results showing your working for the fifth column
(conc. In mol/dm3)
b) Write the balance chemical equation between Z and U
c) Plot the graph of time (y-axis) against Concentration of Na2S2O3 (x- axis)
d) What substance was produced during the reaction which obscured the mark
A

Page 47
e) What conclusion can you draw from the result of this experiments
f) Briefly explain the smell and color observed as the reaction take place
g) Explain how concentration changes affect the rate of chemical reaction
SOLUTION
a) Table of results
Number of
experiment
Volume of Z
(cm3)
Volume of
distilled
H2O (cm3)
Volume of Na2S2O3
after adding water
(mol/dm3)
Time (t)
in
second
Rate of
reaction(S-1)
1 50 50 0.13 10.22 0.10
2 40 50 0.104 13.63 0.07
3 30 50 0.078 17.22 0.06
4 20 50 0.052 25.34 0.04
5 10 50 0.026 55.50 0.02
Molarity = Concentration g/dm3
Molar mass
Molarity = 32.24g/dm3
248. 17g/mol
Molarity = 0.13mol/dm3
Experiment 1
0.13 x 50 = 6.5
6.5 ÷ 50 = 0.13
Experiment 2
0.13 x 40 = 5.2
5.2 ÷ 50 = 0.104
Experiment 3
0.13 x 30 = 3.9

Page 48
3.9 ÷ 50 = 0.078
Experiment 4
0.13 x 20 = 2.6
2.6 ÷ 50 = 0.052
Experiment 5
0.13 x 10 = 1.3
1.3 ÷ 50 = 0.026
b)
Na2S2O3(aq + 2HCl(aq) 2NaCl(aq) + H2O(L) + SO2(g) + S(s)
c) The graph of time (y-axis) against Concentration of Na2S2O3 (x- axis)
d) Substance that was produced during the reaction which obscured the mark A
was Sulphur (S) (precipitation of sulphur)
e) Conclusion
As concentration of sodium thiosulphate decreases the time taken for
disappearance of letter A increases.
f) The smell or irritating chocking smell produced the production of Sulphur
dioxide (SO2)
g) Increase in concentration of sodium thiosulphate increases the rate of
chemical reaction or concentration is directly proportional to the rate of
chemical reaction.
3. QUALITATIVE ANALYSIS
Qualitative analysis it contain two questions only but (NECTA) it gives one Question
only

Page 49
GUIDED UN GUIDED
Every student must be a Qualitative sheet guide – CSEE (Certificate of secondary
education examination).
REAGENTS MUST BE
 Conc. HCl
 Dilute HCl
 Dilute HNO3
 Conc. H2SO4 IZI REAGENTS NI LAZIMA MWALI
 NaOH (solution) MU WAKO ATAZIWEKA KWENYE MEZA
 NH3 (solution) NI KWA AJILI YA (PRELIMINARY TESTS)
 Distilled Water (H2O)
ZIFUATAZO NI REAGENTS AMBAZO UKIZIONA MWALIMU WAKO
KAZIONGEZA ZIKAUNGANAA NA REAGENTS (7) ZA HAPO JUU JUMLA
ZIKAWA (10) BASI ZITAKUPA JIBU SAHIHII KUWA UMELETEWA
CHUMVI YA AINA GANII (SAMPLE SALTS).
Cation
 Ammonium oxalate (Ca2+ Confirmed) or Perform flame test
 Potassium iodide (KI) (Pb2+ confirmed) or Potassium dichromate(K2CrO4)
Anion
 Iron (ii) sulphate (FeSO4) (NO3- Confirmed)
 Magnesium sulphate (MgSO4) (CO32- Confirmed)
HIYO NI MIFANO ILA UKIHITAJI UELEWE ZAIDI NENDA KWENYE
(CONFIRMATORY TEST Kwa kutumia Qualitative shhets za BALAZA
ujuwe reagents mbali mbalii za KUWEZA KUJUA NI AINA GANII YA
SAMPLE SALT UMELETEWAA KABLA HUJA FANYA MAJARIBIO
(TEST).
FOREXAMPLE OF UN GUIDED QUESTIONS
3. Sample F is a simple salt contain one cation and one anion. Carefully carry out the
experiments described below and record all your observations, inferences and hence
identify the cation and anion present sample salt F.

Page 50
S/N Experiment Observation Inference
(a) i/ The cation present in sample F is………….
ii/ The anion present in sample F is………….
(b) Write the chemical formula of sample F……………
(c) Write the name of sample F ……………..
SOLUTION

Page 51
1 Appearance of sample F White powder salt NH4+, Na+, Ca2+,
Zn2+, Pb2+ May be
present.
Or
Transition metals
Fe2+,Fe3+,Cu2+ may
be absent.
2 Action of heat Colourless gas
evolves, which
turn lime water
milk and wet
litmus paper from
blue to red
CO32+,HCO3- May
be present
3 Action of dilute HCl acid on a
solid sample
Effervescence of
colourless gas
evolves, which
turns lime water
milky and wet
litmus paper from
blue to red
CO32-,HCO3- May
be present
4 Action of concentrated H2SO4 on
a solid sample
Effervescence of
colourless gas
evolves. The gas
turns lime water
milky and wet
litmus paper from
blue to red
CO32-,HCO3- May
be presen
5 Flame test Golden yellow
flame
Na+ May be
present
6 Solubility of solid samples Soluble forming
colourless
solution
CO32-, HC03-, of
Na+, NH4+ May be
present
7 Action of NaOH solution on a
sample solution
No precipitate is
formed, even on
warming
Na+ May be
present
8 Action of NH3 solution on a
sample solution
No precipitate is
formed
Ca2+, Na+, NH4+
May be present
9 Confirmatory test for CO32- White precipitate
is formed before
warming the
contents
CO32- Confirmed
(a) i/ The cation present in sample F is Na+

Page 52
ii/ The anion present in sample F is CO32-
(b) The chemical formula of sample F is Na2CO3
(c) The name of sample F is Sodium carbonate
NECTA 2015
3. Sample M contains one cation and one anion. Using systematic qualitative analysis
procedures carry out the experiment and record carefully your observations,
inferences and finally identify the cation and anion present sample M. Record your
work in a tabular form as shown in Table 2.
Table 2
Conclusion
(i) The cation sample M is
(ii) The anion sample M is
(iii)The chemical formula of sample M is
SOLUTION

Page 53
1 Appearance of sample M White crystalline
salt
NH4+, Na+, Ca2+,
Zn2+, Pb2+ May be
present.
Or
Transition metals
Fe2+,Fe3+,Cu2+ may
be absent.
2 Action of heat Cracking sound
with brown gas
NO3- of Pb2+ May
be present
3 Action of dilute HCl acid on a
solid sample
Effervescence of
colourless gas
evolves, which
turns lime water
milky
CO32-,HCO3- May
be present
4 Action of concentrated H2SO4 on
a solid sample
Evolution of
brown fumes,
which turn wet
litmus paper from
blue to red
NO3- May be
presen
5 Flame test Blue-white (pale-
blue) flame
Pb2+ May be
present
6 Solubility of solid samples Insoluble in cold
water but soluble
in hot water
Cl- of Pb2+ May be
present
7 Action of NaOH solution on a
sample solution
White precipitate
is formed, soluble
in excess
Zn2+, Pb2+ May be
present
8 Action of NH3 solution on a
sample solution
White precipitate
is formed,
insoluble in
excess
Pb2+ May be
present
9 Confirmatory test for NO3- Brown rng is
formed at the
junction of the
liquids
NO3- Confirmed
10 Confirmatory test for Pb2+ Yellow precipitate
which disappears
on warming but
re-appears on
cooling
Pb2+ Confirmed
Conclusion
(i) The cation sample M is Pb2+

Page 54
(ii)The anion sample M is NO3-
(iii)The chemical formula of sample M is Pb(NO3)2
FOREXAMPLE OF GUIDED QUESTIONS
NECTA 2016
3. You are provided sample R containing one cation and one anion. Carryvout the
guided systematic procedure in the table 2 to identify the cation and anion present in
the sample R.
Table 2
S/n Experiments Observation Inference
(a) Observe sample R
(b) Heat sample R in a dry test tube
(c) Prepare a stock solution of sample
R. Devide the resulting solution
into six portions then add:
(i) dilute HCl solution in small
quantities then in excess to the first
portion
(ii) Small amount of concentration
H2SO4 to the second portion, then
warm
(iii) NaOH solution to the third
portion, drop wise till excess
(iv) dilute NH4OH in small amount
then in excess to the fourth portion
(v) FeSO4 solution followed by
Conc.H2SO4 to the fifth portion
(vi) KI solution to the sixth portion,
warm then cool the mixture.
Conclusion
(i) The cation sample R is .
(ii) The anion sample R is .
(iii)The chemical formula of sample R is .
(iv) Write the equestions for the reactions that took place at experiments (b) and
(c) (ii).
SOLUTION

Page 55
S/n Observation Inference
(a)
White crystalline salt
NH4+, Na+, Ca2+, Zn2+, Pb2+
May be present.
Or
Transition metals
Fe2+,Fe3+,Cu2+ May be
absent.
(b) Cracking sound with brown gas NO3- of Pb2+ May be
present
(c)
(i)
Insoluble in cold water but soluble in hot
water
Cl- of Pb2+ May be present
Effervescence of colourless gas evolves,
which turns lime water milky
CO32-,HCO3- May be
present
(ii) Evolution of brown fumes, which turn wet
litmus paper from blue to red
NO3- May be presen
(iii) White precipitate is formed, soluble in
excess
Zn2+, Pb2+ May be present
(iv) White precipitate is formed, insoluble in
excess
Pb2+ May be present
(v) Brown rng is formed at the junction of the
liquids
NO3- Confirmed
(vi)
Yellow precipitate which disappears on
warming but re-appears on cooling Pb2+ Confirmed
Conclusion
(i) The cation of sample R is Pb2+
(ii)The anion of sample R is NO3-
(iii)The chemical formula of sample R is Pb(NO3)2
(iv) PbNO3 (aq) + H2SO4 (aq) PbSO4 (aq) + NO2 (g) + H2O (L)
NECTA 2017

Page 56
3. Sample Q is a simple salt containing one cation and one anion. Carefully carry out
all the experiments described in the Table 2. Record all your observation and
appropriate inferences to identify the ions present in sample Q.
Table 2
(a) Observe sample Q
(b) Put a spatulaful of sample Q in a
test tube and add distilled water.
(c) Transfer 0.5g of sample Q in a test
tube, add dilute HCl
(d) Transfer 0.5g of sample Q in a test
tube, add concentrated sulphuric
acid.
(e) Dissolve sample Q then divide the
resulting solution into three
portions.
(i) To the first portion add sodium
hydroxide solution.
(ii) To the second portion add
ammonia solution
(iii) To the third portion add
potassium ferricyanide solution
[K3Fe(CN)6].
(iv) To the fourth portion add lead
acetate solution followed by acetic
acid solution.
Conclusion
(a) (i) The cation sample Q is
(ii) The anion sample Q is
(iii)The Compound Q is
(b) Write the reaction equation that took place at experiments (c)
(c) State three chemical properties of metal in Q
(d) State two uses of Q
SOLUTION

Page 57
S/n Observation Inference
(a) Pale or light green Fe2+ May be present.
(b)
Insoluble
CO32- of Ca2+, Pb2+ , Zn2+,
Fe2+, Fe3+ , Cu2+ May be
present
SO42-, of Ca2+, Pb2+ May
be present
(c) Effervescence of colourless gas evolves, which
turns lime water milky
CO32-,HCO3- May be
present
(d) No gas evolves SO42- May be present
(e)
(i)
Green precipitate is formed, insoluble in
excess, which turns brown on standing
Fe2+ May be presen
(ii) Green precipitate is formed, insoluble in
excess
Fe2+ May be present
(iii) Deep blue precipitate Fe2+ Confirmed
(iv) White precipitate insoluble SO42- Confirmed
Conclusion
(a) (i) The cation sample Q is Fe2+
(ii) The anion sample Q is SO42-
(iii)The Compound Q is FeSO4
(b) Write the reaction equation that took place at experiments (c)
H+ + Cl- HCl (aq)
(c) Three chemical properties of iron (Fe)
i/ Iron combines easily with oxygen to produce rust.
ii/ Iron both malleable it can be shaped into thin wires or sheets
iii/ Iron has a high tensile strength so it can be stretched without breaking
iv/ Iron dissolves in dilute acids
(d)Two uses of iron (Fe)
i/ Making steel
ii/Agriculture
iii/ Food and medicine
NECTA 2018

Page 58
3. Sample N is a simple salt containing one cation and one anion. Carefully carry out
all the experiments described in the Table 2. Record all your observations and
appropriate inferences to identify the cation and anion present in sample N.
Table 2
(a) Appearance of the solid sample N
(b) Put a spatulaful of sample N in a
test tube and heat.
(c) Add dilute hydrochloric acid to the
solid sample N
(d) Perform flame test on the sample.
(e) Dissolve the sample in water.
Divide the resulting solution into
two portions:
(i) Add sodium hydroxide solution
to the first portion.
(ii) Add few drops of magnesium
sulphate solution to the second
portion, then warm.
Conclusion
(i) The cation is and the anion is .
(ii) The sample N is .
(iii) Write chemical equations (with state symbols) to show reactions which took
place in the experiments (b) and (c).
(iv) Describe three uses of sample N
SOLUTION
S/N Observation Inference

Page 59
(a)
White powder salt
NH4+, Na+, Ca2+, Zn2+, Pb2+
May be present.
Or
Transition metals
Fe2+,Fe3+,Cu2+ may be
absent.
(b) Colourless gas evolves, which turn lime water
milk and wet litmus paper from blue to red
CO32+,HCO3- May be
present
(c) Effervescence of colourless gas evolves, which
turns lime water milky and wet litmus paper
from blue to red
CO32-,HCO3- May be
present
(d) Golden yellow flame Na+ May be present
(e) Soluble forming colourless solution CO32-, HC03-, of Na+, NH4+
May be present
(i) No precipitate is formed, even on warming Na+ May be present
(ii) White precipitate is formed before warming
the contents CO32- Confirmed
Conclusion
(i) The cation is Na+ and the anion is CO32-
(ii) The sample N is Sodium carbonate (Na2CO3)
(iii) Na2CO3 (aq) + 2HCl (aq) 2NaCl (aq) + H2O (L) + CO2 (g)
(iv) Three uses of Sodium Carbonate
i/ Manufacturing of glass and production of chemicals: Sodium
carbonate reduces the amount of energy required for the production of
glass and reduces the cost of manufacture.
ii/ It can remove alcohol and grease stains from clothing, coffee pots.
iii/ Used in medicine principally for its acid neutralizing.
* THE END* FOR SOLVING*
MWANAFUNZI; NAKUKARIBISHA MDA WOWOTE ULE KAMA UTAONA HUELEWI
ACTUAL PRACTICAL KWA SOMO LOLOTE LILEE.
*** MALAIKA FOUNDATION LABORATORY%%%
PONGEZI ZANGU ZA DHATI KWA BINT ***LATIFA ALLY MZEE%%%
QUIZ (VOLUME NIMEKUPATIA) FANYA JARIBIO

Page 60
NECTA 2011
2. You are provided with the following;
AA: Containing 0.2 mole HN03 per dm3 of solution;
BB: Contain 4.2g NaxCO3 per 0.5dm3 of solution
MO is Methyl orange indicator.
Procedure:
Put solution AA into the burette. Pipette 20 cm3
or 25cm3
of solution BB into a titration flask.
Add two drops of methyl orange indicator into the titration flask. Titrate solution BB against
solution AA from the burette until a colour change is observed. Note the burette reading.
Repeat the procedure to obtain three more readings and Record your results as in a tabular
form.
Questions:
(a) (i) Calculate the average titre volume.
(ii) Summary 25.00 cm3
of solution BB required 12.53 cm3
of solution AA for complete
reaction.
(b) If the mole ratio for the reaction is 1:1 find:
(i) Concentration of NaxCO3 in mol/dm3
and g/dm3
.
(ii) Molecular mass of NaxCO3.
(iii) Atomic mass of x and replace it in the formula NaxCO3.
(c) Write a balanced chemical equation for the reaction in this experiment.
(d) What is the significance of the indicator in this experiment?
(e) Why is there a colour change when enough acid has been added to the base?
NECTA 2014
H: Containing 6.3 g of hydrated oxalic acid, (COOH)2.XH2O in 1 dm3 of solution.
M: Contain 1.4 g of potassium hydroxide in 0.5dm3 of solution.
Phenolphthalein indicator.
Questions;
(a) Titrate the acid (in a burette) against the base (in conical flask) using two drops
of the indicator and obtain three titre values.
(b) (i) 25.00 cm3 of M required 12.50 cm3 of H for complete reaction.
(ii) The colour changed at the end point was from to
(iii) Is the use of methyl orange indicator in this experiment as suitable as the use
phenolphthalein? Give a reason for your answer

Page 61
(c) Showing your procedures clearly, determine the value of X in the form
(COOH)2.XH2O Give that the equation for the reaction
(COOH)2(aq) + 2KOH(aq) (COOK)2(aq) + 2H2O(L)
(d) State any four precautions you would observe to ensure accuracy in this
experiment.
NECTA 2017
PP: A solution of 0.1M hydrochloric acid;
RR: A solution of 1.39g of impure sodium carbonate anhydrous dissolved in 250 cm3
of solution;
MO: Methyl orange indicator.
Questions
a) (i) Titrate solution PP against 20 cm3 or 25 cm3 of RR until the colour change.
Record the burette readings. Repeat the procedure to obtain three accurate readings
and record your results in a tabular form.
(ii) Why did the colour of the solution changed.
(iii) Determine the average titre volume.
(iv) 25.00 cm3 of solution RR required 25.40 cm3 of solution PP for complete
reaction.
(v) Assume that sulphuric acid of the same molarity was used in the place of
hydrochloric acid, would it be difference in the titre volume used? Give reason.
b) (i) Name the apparatus you used for measuring volume of PP
(ii) Wht the apparatus in (b) (i) is the best recommended for its function?
c) Write a balanced chemical equation of the reaction between the solutions PP and
RR.
d) Calculate the following and write your answer in two decimal places
(i) Molarity of RR.
(ii) Percentage purity of RR.
e) State two applications of volumetric analysis.
P: A solution containing 32.24g/dm3 of Na2S2O3.5H2O
Q: A solution contain 2M HCL

Page 62
A stop watch/ clock; a white plain paper with a cross and thermometer
Procedure
(i) Measure 50cm3 of solution P into a 250 cm3 conical flask.
(ii) Place the flask on top of gauze on a tripod stand. Heat the flask with a gentle
Bunsen burner flame.
(iii) Draw a cross on to paper stir the solution in the conical flask until the
temperature reaches 300C
(iv) Stand the flask on to paper above the cross and immediately add 5 cm3 of
solution Q at the same time start the stop clock stir the mixture up the
thermometer.
(v) Observe the cross through the mouth of the flask as before. Stop the clock
immediately as soon as the cross is obscured. Record the time. Read the
temperature in a table.
(vi) Repeat procedure three more times temperature of 400C, 500C, 600C and
700C. Record your reading in a tabular form as shown in table 1.
Table 1
30
40
50
60
70
Questions
a) Plot the graph of temperature against rate of reaction
b) Determine the slope of the graph
c) The graph is straight line or curve?

Page 63
d) Dial the time until the cross disappeared, increase or decrease as the
temperature was raised?
e) How was the factor of concentration varied in this experiment
f) Use the graph to explain how variation of concentration affects the rate of
chemical reaction
SOLUTION
30 30.43 0.03
40 14.56 0.07
50 12.47 0.08
60 8.10 0.13
70 6.92 0.17

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