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Wednesday, 17 October 2012
F4 PAPER 1 ANSWER ( OCT 2012)
Tuesday, 18 September 2012
MODULE ANSWER
MODUL ANSWER
Preparation of a
solution by dilution (pg 128)
1(a) Standard solution is a solution in which its
concentration is accurately known.
(b) -molar mass of NaOH = 23+16+1 = 40 g mol-1
-
mol NaOH = 500 ×
1.0 /1000 = 0.5 mol
-
Mass of NaOH = 0.5 mol ×
40 g mol-1= 20.0 g
Preparation of 500 cm3
1.0 mol dm-3 sodium hydroxide
-
20.0 g weighing
bottle
-
20.0 g distilled
water
-
Volumetric flask
-
Rinse distilled
water volumetric flask
-
Distilled water volumetric
flask calibration mark
-
Volumetric flask stopper inverted
Calculate the volume
of 1 mol dm-3 sodium hydroxide used:
-
M2
× V2 =
0.1 × 250 = 25 cm3
M1 1
Preparation of 250 cm3 1.0 mol dm-3 sodium
hydroxide
-
25 cm3
-
25 cm3 volumetric flask
-
Distilled water volumetric
flask calibration mark
-
Volumetric flask stopper inverted
1(a) (i)
weak acid : An acid that
partially ionizes in water to produce low concentration of hydrogen ion, H+.
Strong acid : An acid that
completely ionizes in water to produce high concentration of hydrogen ion, H+.
(ii)
-
higher
-
stong acid completely higher
-
weak acid partially
-
CH3COO- + H+
(iii)
-
higher lower
-
lower higher
b)
- molecules ionise molecules molecules neutral
Hydrogen
-
ionises partially ethanoate hydrogen
2a(i) solution P
(ii) solution U
b(i) Q
(ii) R
(iii) T
(iv) P
(v) U
(vi) S
c(i) P/Q/R and T/U
(ii) P/Q
3.
80 g dm-3
4.
0.5 mol dm-3
5.
2 mol dm-3
6.
10 g
7.
0.4 mol dm-3
8.
50 cm3
9.
a)
Number of mol of sulphuric acid
= 50 × 1
1000
= 0.05 mol
H2SO4 → 2H+ + SO42-
From the equation,
1 mol of H2SO4: 2 mol of H+
0.05 mol of H2SO4 : 0.1 mol of H+
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Number of mol of hydrocloric acid
= 50 × 1
1000
= 0.05 mol
HCl →
H+ + Cl-
From the equation,
1 mol of HCl: 1 mol of H+
0.05
mol of HCl : 0.05 mol of H+
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The
number of H+ in 50 cm3
of 1 mol dm-3 of sulphuric acid is twice of the number of in 50 cm3
of 1 mol dm-3 of hydrochloric acid.
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Sulphuric
acid is diprotic acid whereas hydrochloric acid is monoprotic acid.
1 mol of sulphuric acid ionises to 2 mol of H+ whereas 2 mol of hydrochloric acid ionizes to
1 mol of H+ . The number of H+ in the same volume and same concentration is
doubled in sulphuric acid compared to hydrochloric acid.
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b) 100
cm3
Friday, 7 September 2012
Methods of Preparing Salts
Methods of Preparing Salts
1. The method used to prepare a salt will
depend on its solubility in water, that
is, whether it is soluble or insoluble in water.
Q1. Suggest suitable chemicals required to prepare the following salts and write down the chemical equations involved.
a. Potassium nitrate
b. Copper(II) sulphate
c. Aluminium nitrate
d. Iron(II) chloride
e. Ammonium nitrate
e. Ammonium nitrate
PEKA 2 (TITRATION)
Topic : Acids and Bases
Aim :To
find the end point in the titration of hydrochloric acid and sodium hydroxide
solution using an acid-base indicator
Materials :0.1 mol dm-3
hydrochloric acid, 0.1 mol dm-3 sodium hydroxide solution,
phenolphthalein
Apparatus :Burette, 25 cm3
pipette, pipette filler, retort stand and clamp, white tile, 250 cm3
conical flask
Procedure : (Must draw the diagram and list out the procedure)
Result :
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Titration set
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Estimation
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1
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2
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3
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Final burette reading/ cm3
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Initial burette reading/ cm3
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Volume of hydrochloric acid needed/ cm3
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Discussion :
Able to write
the chemical equations for the reactions correctly
Able to show the calculation
Able to state a observation and inference correctly
Able
to state precaution
Conclusion :
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