F4 PAPER 1 ANSWER ( OCT 2012)

1 B 26 A 2 C 27 B 3 A 28 C 4 D 29 C 5 C 30 D 6 A 31 D 7 C 32 A 8 D 33 B 9 D 34 B 10 B 35 A 11 C 36 C 12 D 37 D 13 B 38 A 14 D 39 D 15 A 40 B 16 A 41 B 17 B 42 B 18 C 43 C 19 C 44 C 20 A 45 C 21 D 46 B 22 D 47 C 23 B 48 C 24 B 49 B 25 C 50 B

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+	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+
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.
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.

b) 100 cm³

preparation standard solution

Standard solution

•  A standard solution is a solution containing a precisely known concentration.

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                       

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                        :   

Titration set	Estimation	1	2	3
Final burette reading/ cm3
Initial burette reading/ cm3
Volume of hydrochloric acid needed/ cm3

 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                :

PEKA FORM 4

PEKA 1

ELECTROPLATING


PREPARE THE FORMAT OF REPORT AS BELOW:

Aim                           : To study the electroplating of an object with copper.

Problem statement  

Hypothesis               

Variables                

                                  

Materials                  

Apparatus               

Procedure

Observation           

Set	Electrode		Observation
	Anode	Cathode	Anode	Cathode
I	Iron spoon	Copper
II	Copper	Iron spoon

                                  

                 

Discussion

     

Conclusion    

EXAMPLE OF PEKA

Aim                           :To investigate the effect of concentration of ions on selective discharge of ions at the electrodes

Problem statement   :How does the concentration of ions in hydrochloric acid, HCl affect the discharge of ions at the anode?

Hypothesis               :When the concentration of chloride ion is higher, then the chloride ion will be selectively discharged at anode.

                                    

Variables                  :Manipulated variable  :  Concentration of chloride ion

                                   Responding variable   :    Ion discharged

                                   Constant variable        :    Type of electrolyte, type of electrode, duration of electrolysis

                                  

Materials                   1.0 mol dm^-3 hydrochloric acid, 0.001 mol dm^-3 hydrochloric acid

Apparatus                batteries, carbon electrodes, connecting wires with crocodile clips, ammeter, electrolytic cell, test tubes, litmus paper and splinter

Procedure                 1. A electrolytic cell is filled with 1.0 mol dm^-3 hydrochloric acid until it is half full.

                                   2. The switch is turned on.

                                   3. Gas produced at anode is collected and tested with moist litmus paper and glowing splinter.

                                   4. Observations are recorded.

                                   5. Steps 1 to 4 are repeated by using 0.001 mol dm^-3 hydrochloric acid to replace 1.0 mol dm^-3 hydrochloric acid.

 Observation

Electrolyte	Observation at anode
1.0  mol dm-3 hydrochloric acid	A greenish-yellow gas with pungent smell is released. The gas turns the blue litmus paper to red then to white.
0.001 dm-3 hydrochloric acid	Gas bubbles are released. A colourless gas relight a glowing splinter.

                  
                                  

                                   

Discussion                  :

1.      the aqueous solution of hydrochloric acid consists of hydrogen ions, H+, chloride ions, Cl^- and hydroxide ions, OH^- tat move freely.

2.      During the electrolysis, the Cl^- ions and OH^- ions move to the anode.

a)      Electrolysis of 0.001 dm^-3 hydrochloric acid

The OH^- ions are selectively discharged at anode to form oxygen and water. This is because OH^-ion is lower than Cl^- ion in the electrochemical series.

4OH^-     →  O₂  +  2H₂O  +  4e

b)      Electrolysis of 1.0 mol dm^-3 hydrochloric acid

The Cl^- ions are selectively discharged at anode to form chlorine gas. This is because the concentration of Cl^- ions are higher than OH^- ions.

2Cl^-    → Cl₂  +  2e

Conclusion     : During electrolysis of 1.0 mol dm^-3 hydrochloric acid, the Cl^- ions are selectively discharged at anode instead of OH-.to form chlorine gas due tohigher concetration of Cl^-.The hypothesis is accepted.

ANSWER FOR PAPER 1 FORM 4 CHEMISTRY (AUG 2012)

 

1	B		26	C /D
2	D		27	B
3	B		28	C
4	C		29	C
5	D		30	A
6	A		31	C
7	B		32	C
8	B		33	B
9	B		34	C
10	D		35	A
11	D		36	B
12	C		37	D
13	C/D		38	A
14	A		39	B
15	C		40	A
16	B		41	C
17	C		42	C
18	B		43	D
19	D		44	D
20	C		45	C
21	A		46	B
22	D		47	D
23	B		48	C
24	C		49	B
25	D		50	B

Electrolysis of aqueous solution

Electrolysis of aqueous solution

1.     Electrolysis of potassium nitrate solution, KNO₃ using carbon electrodes.

	Anode (+)	Cathode (-)
Ion presents	NO-3, OH-	K+, H+
Half equation	4OH- → 2H2O + O2 +4e	2H+ + 2e → H2
Observation	Colourless gas bubbles are released	Colourless gas bubbles are released
Product	Oxygen gas	Hydrogen gas

2.     Electrolysis of silver nitrate solution, AgNO₃ using carbon electrodes.

	Anode (+)	Cathode (-)
Ion presents	NO-3, OH-	Ag+, H+
Half equation	4OH- → 2H2O + O2 +4e	Ag+ + e → Ag
Observation	Colourless gas bubbles are released	A grey solid is formed
Product	Oxygen gas	Silver metal

3.     Electrolysis of copper(II) sulphate solution, CuSO₄ using carbon electrodes.

	Anode (+)	Cathode (-)
Ion presents	SO2-4, OH-	Cu2+, H+
Half equation	4OH- → 2H2O + O2 +4e	Cu2+ + 2e → Cu
Observation	Colourless gas bubbles are released	Brown solid is formed
Product	Oxygen gas	Copper metal

4.     Electrolysis of concentrated potassium chloride solution, KCl using carbon electrodes.

	Anode (+)	Cathode (-)
Ion presents	Cl- , OH-	K+, H+
Half equation	Cl- → Cl2 + 2e	2H+ + 2e → H2
Observation	Greenish-yellow gas bubbles are released	Colourless gas bubbles are released
Product	Chlorine gas	Hydrogen gas

5.     Electrolysis of silver sulphate solution, Ag₂SO₄ using silver electrode as anode and carbon electrode as cathode.

	Anode (+)	Cathode (-)
Ion presents	-
Half equation	Ag→ Ag+ +e	Ag+ + e → Ag
Observation	Silver electrode corrodes and become thinner	A grey solid is deposited and electrode becomes thicker.
Product	Silver ion	Silver metal

6.     Electrolysis of copper(II) nitrate solution, Cu(NO₃)₂ using copper electrode as anode and carbon electrode as cathode.

	Anode (+)	Cathode (-)
Ion presents	-	-
Half equation	Cu → Cu2+ +2e	Cu2+ + 2e → Cu
Observation	Copper electrode corrodes and become thinner	Brown solid is deposited and electrode becomes thicker.
Product	Copper(II) ion	Copper metal