Value Added Series I ( Preparation Of Salts )

Most pupil are puzzled regarding the use of various salt-making methods.

(1) why can't titration be used for all the salts?

(2) Since the first reaction for the topic 'Acids-Bases' introduced 'acids react with REACTIVE metals to give salt and hydrogen gas, can we use this method instead?

Discussion:

There are many methods to make a salt, HOWEVER, we need to see if the method is feasible ( workable).

For the case of 'reactive metals and acids'

Of course, you can use reactive metals and acids to make a salt.

BUT, it will be too dangerous if you are using a REACTIVE metal.

Eg: To make Potassium chloride (KCl), it will be DANGEROUS if you are going to react Potassium with Hydrochloric acid ! Potassium explodes with cold water. [Can You Imagine the reaction of potassium with acids?]

Hence, this method ( acids with reactive metals) can only be used for metals that are moderately reactive! example: magnesium, zinc, etc.

If you want to form Copper(II) sulphate, you cannot use this method because copper is unreactive to acids!
Hence reacting copper(II) oxide/carbonate with acids is the ONLY method!

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For the case of 'insoluble oxides/insoluble carbonates with acids'

Ever wonder why it has to be insoluble reagents in the first place?

Back to basics: 'what are acids made of' ?

'An acid is a substance that produces H+ ions when dissolve in water.'

In short, there is water in an acid.

If the oxides and carbonates are soluble in water, then how do you ensure that your acids will be FULLY reacted?

remember in our procedure for heating insoluble oxides/carbonate with acids, step #2

(1) Warm the acid.
(2) add EXCESS insoluble oxides and carbonate and stir to dissolve.
(3) FILTER the mixture ........

If your acids did not REACT with the insoluble oxides/carbonate FULLY, then the acid will contaminate the salt that is formed in the process!

In short, for the usage of heating acids with reactive metals/ insoluble oxide/ insoluble carbonates, the REASONS behind it include:

(1) the oxides /carbonates will be insoluble in water, hence will definitely be able to react with the acids FULLY. Upon complete reaction with acids, all excess metal/insoluble oxides/carbonates will remained and this is an indication that all acids had fully reacted.

(2) the metals used are moderatly reactive and is suitable to react with warm acids.

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What about 'titration'?

There are a few reasons for the usuage of titration.

The first reason is because the metal is too reactive and using their hydroxides is more appropriate.

Eg: to make sodium nitrate. We cannot use sodium metal as it is far too reactive with acids.
So, we use sodium HYDROXIDE and nitric acid.

In this case, we cannot use heating method as both are soluble and we cannot tell when neutralization takes place.

Hence, for the second reason:

Second reason: If you use soluble oxides and carbonates for the method of heating, one problem you foresee would be unable to know when neutralization occurs. Reason: all the reagents are soluble, you can't tell. ( recall for the case of heating insoluble oxides/carbonates, the reagent is insoluble so you can see when the reagent no longer dissolves in the acid).

But titration is able to tell because of the use of INDICATOR.

Hence, we will be able to know the EXACT amount of alkali needed to neutralize the acid to form the salt when the indicator CHANGES COLOUR.

Please do not state this reason as your answer for 'why titration is used': 'because if we use insoluble oxides/carbonates, it will not be able to pass through the burette' !!!!

I mentioned this point to the class because i wanted you to remember insoluble hydroxides and oxides cannot be used for titration in the LABORATORY. How can an insoluble reagent flow through a burette?

The answer why 'titration is used instead of heating method' because: ** both the reagents are soluble and there is NO INDICATION when neutralization takes place. For titration, indicator can be used to identify the point of neutralization.

Now, is it much clearer?!

2007 EYE PURE CHEM PAPER SOLUTION

SECTION A (MCQ)

1. (B)

2. (C)

3. Redox not covered in sec 3, anyway, answer is (A)

4. (A)

5. given no. of moles of Al2O3 = 1.0 mole
mole ratio of Al2O3:O2 is 2:3
no. of moles of oxygen is 3/2 x 1 = 1.5 moles
mass = mole x Mr of (O2)
= 1.5 x 2(16)
= 48 g (B)

6. chlorine = 2,8,7
when chlorine gains 1 electron to form chloride ion, it becomes 2,8,8 and it has 3 SHELLS
fluorine = 2,7
when fluorine gains 1 electron to form fluoride ion, it becomes 2,8 and it has 2 SHELLS
Magnesium = 2,8,2
when magnesium lose 2 electrons to form magnesium ion, it becomes 2,8 and it has 2 SHELLS
Oxygen = 2,6
when oxygen gains 2 electrons to form oxide ion, it becomes 2,8 and it has 2 SHELLS

so form the largest radius.... which means greater 'circumference' = chlorine
(A)

7. chemical analysis not tested in sec 3, anyway, answer is (B)

8. for graphite, each carbon atom is COVALENT bonded to each other. However, the carbon atoms form a HEXAGON RING. Each ring is connected to another ring by weak VAN DER WAALS ( weak VDW) forces of attraction. Graphite is used as lubricant because its layers can SLIDE over each other. (D)

9. (D)
All these elements are from group 2. All have 2 electrons in the outermost shell.

10. (B)

11. REDOX not teated in sec 3, anyway, answer is (D)

12. option (A) look like group 1 metals , option (B) look like macromolecules. copper is a metal, so it will have HIGH MP, HIGH DENSITY, GOOD ELECTRICAL CONDUCTIVITY (D)

13. (B)

14. empirical formula is the SIMPLEST ratio but MOLECULAR FORMULA is the formula multiply by a factor. In order to calculate the Molecular Formula, we need the relative molecular masses. Eg: the Mr of methane, CH4 is 16 while the Mr of (CH4 )2 is 32. (B)

15. given conc of acid = 1.0 M and mass of CaCO3 = 1.0 g.

no. of moles of CaCO3 = 1/100 = 0.01 mol

mole ratio of HCl: CaCO3 is 2:1

no. of moles of HCl = 0.01 x 2 = 0.02

vol = 0.02 / 1.0 = 0.02 dm3 = 20 cm3 (B)

16. H2SO4 + MgO ---> MgSO4 + H2O ;

H2SO4 + MgCO3 ---> MgSO4 + H2O + CO2

so, in both reactions, water is formed. (C)

17. CHEMICAL ANALYSIS not tested in sec 3, anyway, the answer is (A)

18. (A)

19. option B is very interesting. Covalent compounds are made up by atoms.... but the BETTER word for ' Nature of Particles' should be MOLECULES. Anyway, the formula for option B is wrong too. answer(C)

20. you need knowledge from Chemical Analysis, not tested in sec 3. Anyway, option (D)

21. (C)

22. if X is noble gas, then X,Y,Z is consecutive. So, Y should be in group 1, Z should be group 2.(D)

23. IMPT

Cl2 (g) + 2KBr ( aq) ---> 2KCl (aq) + Br2 (aq) ; colour of KBr and KCl = colourless; but aqueous Br2 = reddish brown

Cl2 (g) + 2KI (aq) ---> 2KCl (aq) + I2 (aq) ; colour of KI and KCl = colourless; but aqueous iodine is dark purple or blue (D)

24. although both potassium (2,8,1) and calcium (2,8, 2) have more electrons than aluminium ( 2,8,3 ). But aluminium donates MORE electrons into the sea : Al (s) ---> Al3+ (aq) + 3e- so answer (A)

25. total no. of electrons share = no. of BONDS in the molecule: 8 bonds x 2 e- = 16 e- answer (C)

26. (C) reason: the formula should be XO

27. when ammnium chloride is heated, it forms ammonia gas and hydrogen chloride gas: NH4Cl (s) ---> NH3 (g) + HCl (g) option (B)

28. notice the no. of protons and elecytrons for both X and Y is not equal! by right, in an neutral atom, X should have 4 electrons and 4 protons while Y should have 20 protons and 20 electrons. So this shows that both X and Y lose electrons. Only Metals lose electrons. Hence option (D)

29. R has 4 electrons involved in bonding. Hence R is in group 4. option (B)

30. given X contains acid ---> from universal indicator, X should have a pH less than 7. As for the reaction between KI and Pb(NO3)2, it will give a yellow ppt of PbI2 :

2 KI + Pb(NO3)2 ---> 2 KNO3 + PbI2 ( PbI2 is yellow ) option (A)

SECTION B (STRUCTURED )

1a) ammonia b) Chlorine
c) Lead (II) oxide
d) Magnesium chloride
e) Brass

2(a) chemical analysis Not tested in sec 3

a) zinc hydroxide / white / soluble in excess giving colourless solution. [1.5]
iron(II) hydroxide / dark green / insoluble in excess [1.5]

b) (i) Magnesium, sodium, caesium and beryllium [½]
(ii) Sodium, aluminium, magnesium and chlorine [½]
(iii) Hydrogen, sodium and caesium [½]
(iv) Carbon and tin [½]

3 a) ionic / ionic / covalent / covalent / covalent / covalent [2]
b) giant / giant / simple molecular [2]

c) MgO is ionic and giant structure is strong and stable, hence it has higher
melting point. whereas the SO3 is covalent bonding and a simple
molecular structure. [2]

d) Na2O(s) + H2O(l) ---> 2NaOH(aq) [1]
SO3(g) + H2O (l) ---> H2SO4(aq) [1]

4 ( chemical analysis not tested in sec 3 )
a) A Copper(II) carbonate [1]
B Carbon dioxide [1]
C Copper(II) chloride [1]
D Silver chloride [1]
E Copper(II) nitrate [1]

b) CuCO3 (s) + 2HCl(aq) ---> CuCl2(aq) +CO2(g)+ H2O(l)
CuCO3(s) + 2H+ ---> Cu2+(aq) +CO2(g)+ H2O(l) [2]

5 (oxidation and reduxtion not tested in sec 3)
a) i) +6 ii) +3 [2]
b) Aluminium [1]
c) gain of oxygen or increase in oxidation state [1]
d) Cr2O3 is oxidizing agent. [1]

6 a) i) yes, they all have similar formulae
ii) yes, peroxides and super oxides are likely formed at the
elements below the group.
b) O- , O2-
c) i) H+(aq) + OH-(aq) ---> H2O(l)
ii) 2I-(aq) + Cl2(g) ---> 2Cl-(aq) + I2(s)
iii) Zn(s) + 2H+(aq) ---> Zn2+(aq) + H2(g)

7 a) i. P6O4 ii. P3O2 [2]
b) It is an non-metal oxide
it is covalent bonding
it has no free electron and no mobile ions when molten. [3]

SECTION C (FREE RESPONSE)

1 a) i) Mg(s) + S(s) ---> MgS(s) [1]

iii) Mg :24 + S: 32 à MgS : 56
3 4
12g 16g

Actual: 16g of Mg is used, Sulphur will be used up first in the reaction. [1]
Hence sulphur is the limiting agent. [1]

iv) S MgS
32 56
16g 28g 28g of MgS formed [1]
b) CaCO3 (s) + H2SO4(aq) à CO2(g) + CaSO4(s) + H2O (l) [1]
100 24000
y 1200 [1]
y = 100 x 1200/ 24000
= 5 g (CaCO3 is reacted) [1]

Percentage purity = 5/6 x 100%
= 83.3 % [1]

2 a) i) % of Na = 12.5/45 x 100%
= 27.8%

ii) Na H C O

27.8 12. 14.3 56.7

27.8 1.2 14.3 56.7
23 1 12 16

1.2 1.2 1.19 3.54

1 1 1 3

Empirical formula = NaHCO3
iii) Mr of empirical formula (NaHCO3 )= 23+1+12+ 16x3 = 84
Mr of molecular of NaHCO3 =84
n = 84/84 = 1, Hence, molecular formula = NaHCO3 [1]

b) i) NaHCO3(aq) + HCl à NaCl (aq) + CO2(g) [1]
ii) 84 24
45 x

Conc. Of HCl = 91.3/36.5
= 2.5 mol/dm3
mol of HCl used = 280/1000 x 2.5
= 0.7 mol
hence, vol of CO2 produced = 0.7 x 24dm3
= 16.8 dm3

c) [ chemical analysis, not in syllabus ]
Add equal volume of sodium hydroxide and small spatula of
aluminium powder into a sample solution. [1]
Warm till the aluminium dissolves. [1]
Effervescence and pungent smell gas produced turns red litmus
paper blue. [1]

3 a) methyl orange [1]
b) 22.8 cm3 21.5 cm3 28.7 cm3 [1]
√ √ [1]
c) 25.0 cm3 [1]
d) H2SO4 + 2NaOH à Na2SO4 + 2H2O [1]
e) average of sulphuric acid used = (22.8 + 21.5) / 2 = 22.2 cm3 [1]

mol of H2SO4 / mol of NaOH = 1 / 2
mol of H2SO4 = ½ x mol of NaOH
0.05 x 22.2/1000 = ½ x 25/1000 x mol of NaOH
mol of NaOH = 0.0444 mol/dm3

Disclaimer: only the MCQ answers are from miss pang, answers from section B and C are done by another teacher. Miss Pang is not responsible for the accuracy of the answers!

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