ANSWERS for ENRICHMENT QUESTIONS Sem 1

8 A sample of 0.1000 g of NaOH is dissolved in 25 mL of water and titrated stepwise
until 30 mL of 0.100 M HCl is added. Using these data; describe the titration process
from beginning till the end. Sketch a graph showing the titration curve for this
process.

If HCl is replaced by a weak monoprotic acid, HY, sketch the expected titration curve
on the same graph as above.

ANSWER

NaOH is a strong base and acts as analyte while HCl is strong acid and acts as
titrant.
The pH start out higher because the analyte is strong base. The initial pH of
NaOH is 13.

Initial pH

NaOH(aq) + H2O(l) → Na+(aq) + H3O+(aq)
2.5×10-3
ni (mol) 2.5×10-3 2.5×10-3

nNaOH = 2.5×10-3mol

[NaOH] = 2.5103 mol = 0.1 M
25 103 L

pOH = - log [OH-] = - log (0.1 M) = 1

pH = 13

The pH is gradually decreases as base is being neutralised by the added acid

until it reached the equivalence point.

At equivalence point,
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

Number of mol of OH- are stoichiometry equivalence to number of mol of H3O+

NaCl(aq) → Na+(aq) + Cl-(aq)

Na+ is weak conjugate acid while Cl- is weak conjugate base. Both cannot be
hydrolysed. Hence pH of salt depends on ionisation of water.

H2O  H2O +H3O+ OH-

[OH-] = [H3O+] = 1 x 10-7 M
pH = 7

NaCl is neutral salt.

Volume of HCl at equivalence point,
n of OH- = n of H3O+ = 2.5×10-3mol

50

V of HCl added = 2.5103 mol = 0.025 L = 25 mL
0.1M

The pH jumps for this titration is 3-11.
The type of titration is strong base and strong acid.

Beyond this sharp portion, the pH decreases slowly again as more acid is added.
When continue added the acid until the volume of HCl become 30 ml, just
strong acid and salt are left. On that time, the pH of solution depends on pH
HCl.

ni (mol) NaOH(aq) + HCl(aq) → NaCl(aq) -
nΔ + H2O(l) -
3×10-3 0
(mol) 2.5×10-3 -2.5×10-3 +2.5×10-3 -
-2.5×10-3
nf (mol) 0.5×10-3 2.5×10-3
0

0.5 x 10-3
[HCl] =

0.055
 9.09 × 10-3 molL-1
pH = ( - log 9.09× 10-3 )
2

The final point approaching pH≈ 1.

[HCl] = 0.1 M
pH = ( - log 0.1)
1

pH

13

equivalence point

7 Strong base vs Weak acid

2 Strong base vs Strong acid
1

25 mL 30 mL V of HCl added (mL)

51

SUGGESTED ANSWER PSPM 1
SESI 2015/2016

1. Compound Q contains carbon, hydrogen and nitrogen. Combustion of 0.250 g of Q produces
0.344 g of water, H2O, and 0.558 g of carbon dioxide, CO2. Determine
a) the empirical formula of Q.

Given:
Mass of Q = 0.250 g
H2O = 0.344 g
CO2 = 0.558 g

Empirical formula

Mass of C = 12g mol1  0.558g = 0.152 g
44g mol1

Mass of H = 2g mol1  0.344 g = 0.0382 g
18g mol1

Mass of N = 0.250 g – (0.152 + 0.0382)g = 0.0598 g

Element C H N
Mass(g) 0.152 0.0382
No. of mole 0.0127 0.0382 0.0598
Mole ratio 2.9 4.27×10─3
8.9
3 9 1

1

∴ Empirical formula = C3H9N

b) the molar mass of Q if the molecular formula is the same as the empirical formula
and

Molar mass of Q

Molecular formula = empirical formula = C3H9N
Molar mass = 3(12 g mol─1) + 9 (1 g mol─1) + 1 (14 g mol─1)

= 59 g mol─1

52

c) the number of hydrogen atoms present in the above sample Q.

Number of hydrogen atom in Q

Mol of Q = 0.25g = 4.24×10─3 mol
59g mol1

1 mol Q ≡ 9 mol H
4.24×10─3 mol Q ≡ 0.0381 mol H

∴number of H atom = (0.0381 mol)(6.02×1023 atom/mol)
= 2.296×1022 atoms

2. An atom X has 5 valence electrons, X reacts with fluorine gas to form XF3 and XF5
compounds.
a) For each compound,
i. draw the Lewis structure,

XF3
Total valence electrons = 5 + 3(7) = 26

FXF

F

XF5
Total valence electrons = 5 + 5(7) = 40

FF

FXF

F

ii. predict the electron pair geometry and the molecular geometry and

XF3
 Number of electron pair around central atom = 4
 Electron pair arrangement = tetrahedral
 Based on VSEPR theory, electron pair will be located as far as

possible in order to minimize the repulsion between them.
 The strength of lone pair-bonding pair repulsion > bonding pair –

bonding pair repulsion.
 Molecular geometry = trigonal pyramidal

53

XF5
 Number of electron pair around central atom = 5
 Electron pair arrangement = trigonal bipyramidal
 Based on VSEPR theory, electron pair will be located as far as

possible in order to minimize the repulsion between them.
 The strength bonding pair – bonding pair repulsions are equal.
 Molecular geometry = trigonal bipyramidal

iii. draw the molecular geometry and state the bond angle(s).

XF3

X
FF

F

Bond angle = <109.5o

XF5

FF

FXF

F

Bond angle = 120o and 90o

b) Predict the change in hybridization (if any) of the X atom in the following reaction:

XF3  F2  XF5
XF3 = sp3 to XF5 = sp3d

3. a) A gas mixture contains 82% (w/w) methane, CH4, and 18% (w/w) ethane, C2H6. If a
15.50 g sample of the gas mixture is placed in a 15-L container at 20.0oC, calculate
Given:

% w/w of methane = 82%, % w/w of ethane = 18%, Mass of gas mixture = 15.50 g
V = 15 L, T = 20.0oC = 293.15 K

i. the total moles of gases,

Mass of methane = 82 15.5g = 12.71 g

100

Mole of methane = 12.71g = 0.794 mol
16g mol1

Mass of ethane = 18 15.5g = 2.79 g

100

Mole of ethane = 2.79 g = 0.093 mol
30g mol1

54

∴ nT = 0.794 mol + 0.093 mol = 0.887 mol
ii. the total pressure (atm) and

PT  nTRT
V

 0.887 mol 0.08206 atm L mol1 K1 293.15K

 15L

= 1.423 atm

iii. the partial pressure of each gas in the container.

PCH4   0.794 mol  1.423atm = 1.274 atm
0.887 mol

PC2H6  0.093mol 1.423atm = 0.149 atm
0.887 mol

b) Explain each of the following observations:
i. Methane, CH4 (16 g mol−1) has lower boiling point than propane, C3H8
(44 g mol−1) whereas water, H2O (18 g mol−1) has higher boiling point than
hydrogen sulphide, H2S (34 g mol−1).
Boiling point of methane < propane.
Reason:
 Molecular mass of methane < propane.
 Strength of Van der Waals forces of methane < propane.

Boiling point of H2S< H2O.
Reason:
 H2O can form hydrogen bond between molecules while H2S can

only form weak Van der Waals forces between molecules.
 Strength of Van der Waals forces < hydrogen bond.

ii. Carbon dioxide, CO2, molecule and diamond are both covalently bonded.
However, solid CO2, is softer and exhibits lower melting point as compared to
diamond.

CO2 solid is softer and has lower melting point than diamond.
Reason:
 CO2 molecules are held by weak Van der Waals forces.
 Each C atoms in diamond are held by strong covalent bond to

form tetrahedral arrangement in gigantic covalent network.

55

4. a) FIGURE 1 shows a titration curve for 20.0 mL of an unknown concentration of NaOH
solution titrated against a standard solution of 1.00 M HCl.

pH

Z

22.5 25.0 Volume of HCl (mL)

FIGURE 1
i. Define standard solution, equivalent point and end point.

Standard solution
A solution of known concentration for use in volumetric analysis.

Oxford Dictionary

A solution of accurately known concentration. Raymond Chang

A solution of known concentration. Brown Le. May

Equivalent Point
The point in a titration at which reaction is complete. Oxford Dictionary

The point at which the acid has completely reacted with or been

neutralized by the base. Raymond

Chang

The point in a titration when the number of moles of the added species is

stoichiometrically equivalent to the other species. Silberberg

The point in a titration at which the added solute reacts completely with

the solute present in the solution. Brown Le. May

End Point

The point in a titration at which reaction is complete as shown by the

indicator. Oxford Dictionary

The pH at which the indicator changes colour. Raymond Chang

The point in a titration at which the indicator changes colour.
Silberberg

56

ii. What is the pH at equivalent point, Z for this titration?

pH at point Z = 7

iii. Calculate the molarity of the NaOH solution used in the titration.
molarity of NaOH used

mol of HCl = (0.0225 L)(1.00 M) = 0.0225 mol

1 mol HCl ≡ 1 mol H+
0.0225 mol HCl ≡ 0.0225 mol H+

at equivalent point,
mol of H+ ion ≡ mol of OH─ ion
0.0225 mol of H+ ion ≡ 0.0225 mol of OH─ ion

1 mol NaOH ≡ 1 mol OH─
0.0225 mol NaOH ≡ 0.0225 mol OH─

Molarity of NaOH =  0.0225 mol  = 1.125 M
 0.02 L 

iv. Calculate the pH of the solution after the addition of 25.0 mL of HCl.
pH of solution after addition 25.0 mL of HCl.

Mol of HCl = (0.025 L)(1.00 M) = 0.025 mol

NaOH(aq)  HCl(aq)  NaCl(aq)  H2O(l)

ni 0.0225 0.025 - -
∆n -0.0225 -0.0225 +0.0225 -
2.5×10─3 0.0225 -
nf 0

[HCl] = 2.5103 mol = 0.056 M
0.045 L

1 mol HCl ≡ 1 mol H+
0.056 M HCl ≡ 0.056 M H+

pH = -log (0.056 M) = 1.26

57

b) Pyridine, C5H5N, is a weak base. This base ionises in water to produce C5H5NH+ and
OH−. Calculate the pH of a 1.00 M pyridine solution if the dissociation constant, Kb, of
pyridine is 1.50×10−9.

C5H5N(aq) + H2O(l) → C5H5NH+(aq) + OH─(aq)

[ ]i 1.00 - - -

[ ]c -x - +x +x

[ ]e 1-x -x x

Kb = C5H5NH  OH 

C5H5N

1.50×10─9 = xx
1 x

Since α <<<< 0.5, therefore, 1-x≈1.0

1.50×10─9 = xx
1

x = [OH─] = 3.87×10─5 M

pOH = - log (3.87×10─5 M) = 4.41
pH = 9.59

5. a) Calculate the first three wavelengths of the possible transitions of an electron in the
Paschen series for a hydrogen atom. Show these wavelengths in a sketch of the line
spectrum for this emission series. Explain how these transitions occurred.

First transition, ni = 4 to nf = 3

1  RH  1  1   n2
  n12 , n1
 n 2 
2

  1 1 
1.097 107 m1  32  42 

  1.88×10─6 m = 1880 nm

Second transition, ni = 5 to nf = 3

1  RH  1  1   n2
  n12 , n1
 n 2 
2

  1 1 
1.097 107 m1  32  52 

  1.282×10─6 m = 1282 nm

58

Third transition, ni = 6 to nf = 3

1  RH  1  1   n2
  n12 , n1
 n 2 
2

  1 1 
1.097 107 m1  32  62 

  1.094×10─6 m = 1094 nm

Line spectrum

st 2nd line 3rd line

1 line

1.88 1.28 1.09

wavelength,  increase (x10 -6 m)

Explanation:
 When gas in discharge tube is heated, the electrons of the gaseous atom

will absorbed the heat supply and excited from lower energy level to
higher energy level.
 The electrons in excited state is unstable and will fall back to lower energy
level, n=3 by releasing specific amount of energy with specific wavelength
that produce line in the Paschen series.

b) The following data are given for atomic and ionic radii of halogens and halides

respectively.

TABLE 1

Atom Ion

Species F Cl Br I F─ Cl─ Br─ I─

Radii 0.72 0.99 1.14 1.35 1.36 1.81 1.95 2.16
(Å)

Discuss the trend of radii in terms of nuclear charge and valence electron of the
species.
Explain the ionization energy and electron affinity trends of these species going
down the group in the periodic table.

Size of I > Br > Cl > F
 All atoms are in the group 17.
 Down group 17 from F to I, the number of shell is increased.
 Thus, shielding effect in F < Cl < Br < I.
 Strength of nucleus attraction towards valence electrons in F > Cl > Br >

I.

59

Size of atom < its respective ion
 When electron is added into the particular atom, the mutual repulsion

of the atom will increased.
 Expand the electron cloud of atom and enlarge the size.

Ionization energy of I < Br < Cl < F
 Size of atom I > Br > Cl > F.
 Outermost electron is located further from the nucleus.
 Strength of nucleus attraction towards valence electron in atom

I < Br < Cl < F.
 Energy needed to remove the outermost electron in I < Br < Cl < F.

Electron affinity of I < Br < Cl < F
 Size of atom I > Br > Cl > F.
 Outermost electron is located further from the nucleus.
 Strength of nucleus attraction towards valence electron in atom

I < Br < Cl < F.

6. Name the bonding theory that explains orbital hybridization.

Valence bond theory

Using orbital diagram, describe the hybridization of the central atom in sulphur
hexafluoride, SF6 and formaldehyde, H2CO, molecules.

SF6

Total valence electron = 6 + 6(7) = 48

0

F F0

0

F S0 F

0

0F F

0

 Number of electron pairs around central atom : 6

 Type of hybrid of S : sp3d2

Valence orbital diagram,
F ground state:

2s 2p Sexcited state: Shybrid:
Sground state:

3s 3p 3s 3p 3d sp 3d2

60

H2CO

Total valence electron = 2 + 4 + 6 = 12

0
0O

H C0 H

0

 Number of electron pairs around central atom : 3

 Type of hybrid of C : sp2

 Type of hybrid of O : sp2

Valence orbital diagram,
H ground state:

1s

Oground state: Oexcited state: Ohybrid:

2s 2p 2s 2p sp 2 2p
Cground state: Cexcited state: Chybrid:

2s 2p 2s 2p sp2 2p

Draw the orbitals overlaps of each molecule and state the expected bond angles.

2p

F 2p sH
F 

 s p2
2p
 
F
F  sp3d2 2p  sp2 C 
s p2
s p3d2 s p3d2 s
H
S s p2 s p2

 sp3d2 s p3d2 O

s p3d2 s p2

F 
2p F

2p Bond angle: 120o

Bond angle: 90o

61

7. a) Crystalline solid is classified based on the assembly of their respective basic particles.
State four types of crystalline solid and discuss the nature of the particle assembly in
each crystalline solid. Give one example for each type of the crystalline solid.

4 types of crystalline solids:
Ionic crystal

 Composed of positively charged particle (cation) and negatively charged
particle (anion).

 This particles are hold by electrostatic attraction/ionic bond.
 Example: NaCl.

Covalent crystal
 Atoms are held together by covalent bond to form giant covalent network.
 Example: diamond.

Simple covalent crystal
 Consist of molecules that are held together by van der Waals force or
hydrogen bond.
 Atoms in molecule are held by covalent bond.
 Example: solid I2.

Metallic crystal
 Metal atoms in a crystal can be imagined as an array of positive ions
immersed in a sea of delocalised electrons.
 Both positive ions and delocalised electrons are attracted by metallic
bond.
 Example: solid Na.

b) Consider the following reaction system;

2NO(g)  O2 (g)  2NO2(g)

This system is made up from 2.0×10─3 mol of each gaseous component in a 2-L closed
vessel. If the equilibrium constant, Kc, for the system is 7.7×107 at 600 K, prove that

the system is not in equilibrium at this temperature.

 [NO] = [O2] = [NO2] =
2.0103 mol = 1×10−3 M

2L

Qc =  NO2 2
NO2 O2 

 1103 M 2
   = 1103 M 2 1103 M

= 1000

∴Qc < Kc, thus the system is not in equilibrium.

62

When the system is cooled to 450 K and left to reach equilibrium, the final amount of
O2 detected in the vessel is 9.6×10─4 M. Calculate the equilibrium constant, Kc, of the

system at 450 K.

At 450 K, the equilibrium position will move to right because amount of O2 was
decreased after reach equilibrium at this temperature.

2NO(g) + O2(g) → 2NO2(g)
[ ]I 1×10−3 1×10−3 1×10−3

[ ]c -2x -x +2x
[ ]e 1×10−3 – 2x 1×10−3-x 1×10−3+2x

9.6×10−4 M = 1×10−3 M – x
x = 4×10−5M

[NO] = 9.2×10−4 M [O2] = 9.6×10−4 M [NO2] = 1.08×10−3 M

Kc at 450 K =  NO2 2
NO2 O2 

 1.08103 M 2
   = 9.2104 M 2 9.6104 M

= 1435.49

8. a) Milk of magnesia consists of saturated solution Mg(OH)2 and gelatinous Mg(OH)2.
Determine the molarity of Mg2+ in a saturated solution of Mg(OH)2 and in the
saturated solution of Mg(OH)2 containing 0.01 M NaOH.
[Given: Ksp Mg(OH)2 = 8.9×10─12]

In saturated solution of Mg(OH)2

Mg(OH)2 (s)  Mg 2 (aq)  2OH

xM xM 2x M

Ksp = [Mg2+][OH-]2
8.9×10−12 = (x)(2x)2

x = 1.305×10-4 M

∴ [Mg2+] = 1.305×10−4 M

In saturated solution containing 0.01 M NaOH

63

NaOH (aq)  Na (aq)  OH  (aq)
0.01 M 0.01 M 0.01 M

Mg(OH)2 (s)  Mg 2 (aq)  2OH

y M y M 0.01 + 2y M

Ksp = [Mg2+][OH-]2
8.9×10−12 = (y)(0.01+2y)2
Since k is too small, assume 2y + 0.01 M ≈ 0.01 M

8.9×10−12 = (y)(0.01)2
x = 8.9×10−8 M

∴ [Mg2+] = 8.9×10−8 M

b) A buffer solution is prepared by dissolving 0.125 mol of sodium nitrite, NaNO2, in 500
mL of 0.25 M nitrous acid, HNO2. What is the pH of the solution?
The above buffer solution can resist the change in pH when a small amount of strong

acid or base is added into it. Explain how this buffer solution maintain its pH. If an

amount of 0.001 mol of NaOH is added to the solution, determine the change in the

pH value of the buffer solution.
[Given: Ka HNO2 = 5.0×10─4]

HNO 3 (aq)  H2O(l)  H3O (aq)  NO3 (aq)

0.125mol 
[NaNO2] = 0.5L

= 0.25 M

pH = -log ka + log NaNO2 
HNO2 

= - log (5.1×10−4) + log 0.25
0.25

= 3.29

When small amount of strong acid is added into the solution. H+ ion of strong
acid will be neutralized by conjugate base, NO 2 - ion in the buffer solution and

produce HNO2.

NO2 (aq)  H (aq)  HNO 2 (aq)

When small amount of strong base added in the solution. OH- of strong base will
be neutralized by weak acid, HNO2 in the buffer solution and produce NO2- and

water.

64

HNO 2 (aq)  OH (aq)  NO2 (aq)  H2O (l)

If 0.001 mol NaOH added.

HNO 2 (aq)  OH (aq)  NO2 (aq)  H2O (l)

ni 0.125 0.001 0.125 -

nc -0.001 -0.001 +0.001 -

nf 0.124 0 0.126 -

pH = -log ka + log NaNO2 
HNO2 

 = - log (5.1×10−4) + log
0.126 45

 45
0.248

= 3.30

∴ change in the pH value of the buffer solution = 0.01

65

SUGGESTED ANSWER PSPM 1
SESI 2016/2017

1. (a) Sodium carbonate, Na2CO3 dissolves in ethanol to give 2.5 M solution.
Calculate the molarity of the solution if the density of the solution is 1.430 g mL-1.

Molarity = Mol of solute(mol)__ = 2.5 M @ mol L-1
Volume of solution(L)

Assume that Volume of solution = 1L = 1000 mL
Mol of solute (Na2CO3) = 2.5 mol
Mass of solute (Na2CO3) = 2.5 mol x 106 gmol-1
= 265 g

Density = Mass of solution (g)__ =1.430 g mL-1
Volume of solution(mL)

Mass of solution(Na2CO3)= 1430g

Mass of solution = mass of solute (Na2CO3) + mass of solvent(Ethanol)
mass of solvent(Ethanol) = 1430 g – 265 g
= 1165 g
= 1.165 g

Molality = Mol of solute (Na2CO3) (mol)
Mass of solvent (Ethanol)(Kg)

= 2.5 mol
1.165 Kg
= 2.146 m @ molal

(b) The following redox reaction occurs in acidic condition.

Zn + NO3-  Zn2+ + NH4+

i. Balance the redox equation using ion electron method.

4Zn → 4 Zn2+ + 8e

10H+ + NO3- + 8e → NH4+ + 3H2O______

4Zn +10H+ + NO3- → 4 Zn2+ + NH4+ + 3H2O

66

ii. A 4.0 g impure sample of zinc reacted completely with 25.0 mL of 0.5 M
HNO3 solution. Calculate the percentage purity of the Zinc sample.

Mol of NO3- = 25.0mL x 0.5M
1000

= 0.0125 mol

1 mol of NO3- ≡ 4 mol of Zn
0.0125 mol of NO3- ≡ 0.0125 x 4

1

= 0.05 mol of Zn

Mass of Zn = 0.05 mol x 65.4 gmol-1
= 3.27 g

% purity of the Zinc sample = 3.27 g x 100%
4.0 g

= 81.75%

2. (a) An atom X combines with oxygen atoms to form two covalent molecules, XO2 and
XO3. XO2 is identified as a polar molecule while XO3 is a nonpolar molecule.
i. Draw a possible molecular geometry for each of the molecule.

XO2 XO3

X O
OO
X
OO

ii. Comment on the polarity of X-O bond. Give a possible reason.

X-O bond is polar.
O and X both are from group 16 but O is located at higher position than
O. Since electronegativity decreases going down the group, then O is more
electronegative than X.

iii. Explain why XO2 is polar while XO3 is nonpolar.

XO2 XO3

X O
OO
X
OO

67

XO2 XO3
 Molecular geometry is bent.  Molecular geometry is symmetrical.
 The polarity of bond is identical and
 Vectors cannot cancel out.
in opposite direction, therefore can
 The combined vectors is not cancel one another out.
zero, µ≠0  The sum vectors is zero, µ = 0.

(b) Phosporyl chloride, POCl3, can be prepared by the reaction of phosphorus trichloride,
PCl3 with oxygen at 20-50oC.
Draw the most stable Lewis structure of PCl3 and POCl3, with P as the central atom.

PCl3 POCl3

Cl Cl
Cl P Cl Cl P Cl

O

i. What is the hybrid orbital of P atoms in both compounds?
sp3

ii. Show the overlapping of orbitals in PCl3.

P ground state 3p Cl: 3p
3s 3p 3s

P excited state
3s

P hybrid state

sp3

68

iii. Compare the Cl-P-Cl bond angles in PCl3 and POCl3. Explain your answer.

PCl3 POCl3
 Has 4 pairs of e- around the  Has 4 pairs of e- around

central atom, e- arrangement is the central atom, e-
tetrahedral. arrangement is
tetrahedral.
 Has 3 bonding pair and 1 lone  All are bonding pair e-.
pair e-.
 According to VSEPR, the
 According to VSEPR, repulsion repulsion is equal.
between lone pair-bonding pair
is greater than bonding pair-  Bond angle is 109.5o
bonding pair.

 The bonding pair e- is pushed
closer towards each other.

 Bond angle is <109.5o

3. (a) A sample of potassium chlorate, KClO3, was decomposed upon heating, producing

potassium chloride, KCl, and oxygen gas, O2. The volume of gas collected by
displacement of water was 0.250 L at 26oC and a pressure of 765 mmHg. Given
water vapour pressure at 26oC is 25.2 mmHg, calculate

i. The partial pressure of O2.

PT = PO2 + PH2O
PH2O = 765 – 25.2 mmHg

= 739.8 mmHg

ii. The moles of O2 collected.

PV = nRT
n = (738/760 atm) (0.250 L)___
(0.8206 L mol-1 K-1) (26+273.15 K)
= 9.91 x 10-3 mol

iii. The amount of KClO3 (in g) decomposed.

2 KClO3 (s)  2 KCl (s) + 3 O2 (g)

3 mol O2 ≡ 2 mol KClO3
nKClO3 = ⅔ x 9.91 x 10-3

= 6.61 x 10-3 mol

mass KClO3 = 6.61 x 10-3 x [ 39.1 + 35.5 + (3x16)]
= 0.810 g

69

(b) Nitrogen gas is a real gas that deviates from ideality. It is the most abundant gas
found in the atmosphere.
i. Why does nitrogen behave as a real gas at room temperature?

Nitrogen gas behave as a real gas at room temperature because nitrogen

gas has mass and volume. Its occupy space. So the intermolecular forces
can’t be ignored and it has volume.

At low temperature, nitrogen gas has low kinetic energy and moves at low
speed. Therefore the intermolecular forces can’t be ignored.

ii. State the conditions under which nitrogen gas can be liquefied.

At very high pressure and very low temperature.

iv. How can nitrogen gas be made to behave as an ideal gas?

At very low pressure, the nitrogen molecules a very far apart. Hence the
intermolecular force can be ignored. And at very low pressure the
volume of container is huge. Hence the volume of gas is not significant
and consider volume of gas equal to volume of container.

At very high temperature, the nitrogen gas molecules has very high
kinetic energy and move at very high speed. Therefore, the intermolecular
forces can be ignored.

4 (a) Methylamine, CH3NH2, is a weak base. This base has the degree of ionization of 4.7%

at 0.16 M concentration. Calculate
i. the concentrations of OH-, CH3NH3+ and H3O+,

() () () ()

[ ]i/ M 0.16 -- -
/M
-x +x +x
[ ]f/ M (0.16 – x) xx

M

Therefore, [ ] = [ ]=

[ ] [H3O+] = 1 x 10-14 M2
 [H3O+] = 1.33 x 10-12 M

ii. the pH and

pH = -log [H3O+]
= -log (1.33 x 10-12)
= 11.90

70

iii. the base dissociation constant, Kb.
[]

() )
(
)
( )
(

(b) Would lead(II) iodide, PbI2, precipitate (Ksp PbI2 = 8.5 x 10-9) be produced when a
10.0 mL solution of 1.0 M lead(II) nitrate, Pb(NO3)2 and 40.0 mL solution of 2.0 x 10-3
M sodium iodide, NaI, is mixed?

Pb(NO3)2 (aq) + 2NaI (aq)  PbI2 (aq) + 2NaNO3 (aq)

[Pb(NO3)2] = 0.2 M
[NaI] = 1.6 x 10-3M

PbI2(s) Pb2+(aq) + 2I-(aq)

0.2 M 1.6 x 10-3M

Qsp= [Pb2+][I-]2
= 5.12 x 10-7

Qsp > Ksp
- Solution is supersaturated.
- Precipitate will form.

5 (a) State the difference between the ground state and excited state of an electron in an

atom. An electron of a hydrogen atom is excited to n = 6 and falls to a lower energy
level forming the Paschen series. Calculate, in kJ mol -1, the energy of electron at the

excited level and the energy emitted as the result of the transition. Determine the

shortest wavelength in nanometres (nm) in the Paschen series for the atom

hydrogen atom.

A ground state electron of an element is an outer orbital electron that is in the
lowest energy state possible for that electron while an excited state electron has
absorbed additional energy and exists at a higher energy level than a ground
state electron.

Energy of electron at the excited level

En = -RH/n2
= -2.18 x 10-18/62
=-6.055 x 10-20 J/1000
= -6.055 x 10-23kJ x (6.02x 1023)
=-36.45 kJ/mol

71

Energy emitted

E = RH(1/ni2-1/nf2)
= 2.18 x 10-18(1/62 -1/32) x (6.02 x 1023)/1000
= -109.6 kJ/mol

Shortest wavelength

1/ = RH (1/n12 -1/n22) n1 < n2
1/ = 1.097 x 107 (1/32-1/2)
 = 8.2 x 10-7 x 109

 = 820.45 nm

(b) Define the first ionization energy.
Sketch a graph to show the energies involved in the removal of the first four
electrons of an aluminium atom. Explain your answer.

Ionization energy is minimum energy required to remove 1 mol electron from 1
mol of gaseous atom to form 1 mol of positively charge gaseous ion.

Energy

12 3 4 No. of electron

removed

Based on the graph above, the sharp increase occur between lE3 to lE4. 3
electrons in outermost shell and fourth electron in inner shell.

6 (a) CH3F and CF4 are tetrahedral molecules. Using the Lewis structure and dipole
moment of these molecules, deduce their polarity.

For CH3F molecule:

Element No. of valence electrons, e-
C 4e- x 1 = 4e-
H
F 1e- x 3 = 3e-
7e- x 1 = 7e-

TOTAL = 14e-

72

Lewis structure:

Number of electron groups = 4
Electron pair arrangement = tetrahedral

Molecular shape = tetrahedral

Each C-H bond and C-F bond are polar. However, C-F bond has greater dipole
moment than C-H bond due to F atom is more electronegative than H atom.

The four dipole moments do not cancel each other.
The resultant dipole moment, µ is not equal to zero.
Therefore, the molecule of CH3F is a polar molecule.
For CF4 molecule:

Element No. of valence electrons, e-
C 4e- x 1 = 4e-
F 7e- x 4 = 28e-

TOTAL = 32e-

Lewis structure:

73

Number of electron groups = 4
Electron pair arrangement = tetrahedral

Molecular shape = tetrahedral

Each C-F bond is polar and has the same dipole moment.
Since the molecule is symmetrical, each C-F bond dipole moment cancel each
other.

The resultant dipole moment, µ is equal to zero.
Therefore, the molecule of CF4 is a non-polar molecule.

(b) Iodine pentafluoride, IF5, is a colourless liquid with a pungent odour. By using VSEPR
theory, predict the geometry and bond angles of this compound.

Element No. of valence electrons, e-
I 7e- x 1 = 7e-
F 7e- x 5 = 35e-

TOTAL = 42e-

Lewis Structure :

Based on the Lewis structure, the electron pair arrangement is octahedral since
there are six electron pairs around the central atom I which is consist of five
bonding pairs and one lone pair.
According to the VSEPR theory, the electrons are located as far as possible to
minimise the repulsion between them. Hence the repulsion between lone pair

74

electrons and bonding pair electrons is greater than the repulsion between the
bonding pair electrons and bonding pair electrons.

In order to minimise the repulsions, the six electron pairs are arranged as
follows:

Therefore, the molecular geometry of IF5 molecule is square pyramidal with all
of the bond angles are similar which is less than 90º.

Show how the orbitals of the central atom undergo hybridisation. Draw a labelled
diagram showing the overlapping of orbitals in the molecule.

Valence orbital diagram for iodine, I:

Ground ↑↓ ↑↓ ↑↓ ↑
state: 5p

5s 5d

Excited ↑↓ ↑↑↑ ↑↑
state: 5p 5d
5s

Hybrid ↑↓ ↑↑↑↑ ↑
state: sp3d2orbitals

5d

The six sp3d2 hybrid orbitals are produced by the hybridisation of one 5s, three
5p and two 5d orbitals of the valence shell of the iodine atom.

Valence electron in fluorine, F:

Ground ↑↓ ↑↓ ↑↓ ↑
state : 2p

2s

Each I-F bond is formed by the overlapping of one sp3d2 orbital of the iodine
atom with one 2p orbital of the fluorine atom.

The results of these overlapping formed five σ bonds.
One of the sp3d2 orbitals of the iodine atom is occupied by lone pair electrons.

75

< 90°

Electron pair arrangement: Octahedral
Molecular geometry: Square pyramidal

7 (a) Ethane, C2H6, burns in the air according to the equation below:
2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g)

A mixture of C2H6 and O2 at 30 mmHg and 124 mmHg, respectively, was added into a
7.5L flask at a temperature of 27oC. The combustion reaction was carried out until
completion. Calculate the total pressure of the gas mixture obtained at the end of
the reaction, at the same temperature. Give your answer in mmHg.

PV = nRT
n of C2H6 =

=
= 0.012 mol (available)

PV = nRT
n of O2 =

=
= 0.0497 mol (available)

From balanced equation ;

2 mol C2H6 7 mol O2
0.012 mol C2H6 0.012 x 7/2

= 0.042 mol O2 (needed)

moles of O2 needed < moles of O2 available
O2 is excess reactant while Limiting reactant is C2H6

nO2 remained = 0.0497 mol – 0.042 mol
= 0.0077 mol

76

From balanced equation;

2 mol C2H6 4 mol CO2
0.012 mol C2H6 = 0.024 mol CO2

2 mol C2H6 6 mol H2O
0.012 mol C2H6 = 0.036 mol H2O

nT = nO2 + nCO2 + nH2O
= 0.0077 + 0.024 + 0.036
= 0.0677 mol

PT = nTRT
V

= 0.0677 mol x 0.08206 Latmmol-1K-1 x 300.15K
7.5 L

= 0.2223 atm x 760mmHg
1 atm

= 168.948 mmHg

(b) For the following reaction, the value of Kp is 6.4 x 10-6 at 227oC.
2Cl2(g) + 2H2O(g)  4HCl(g) + O2(g)

In a 2.0L container, the initial quantity of Cl2 is 0.50 mol, H2O is 0.40 mol, HCl is 0.50
mol and O2 is 0.015 mol. Determine whether the reaction is at equilibrium at this
temperature. If not, predict to which direction will the reaction proceed?

Kp = Kc (RT)n
Kc = ( )

= 6.4 x 10-6______
(0.08206 x 500.15)5-4

= 1.5594 x 10-7

[ Cl2 ] = 0.50 mol
2.0 L

= 0.25 M

[H2O] = 0.40 mol
2.0 L

= 0.20 M

[ HCl] = 0.50 mol
2.0 L

= 0.25 M

77

[O2] = 0.015 mol
2.0 L

= 0.0075M

Qc = [HCl]4[O2]
[Cl2]2[H2O]2

= (0.25)4(0.0075)
(0.25)2(0.2)2

= 0.0117

Qc > Kc
- The reaction is not at equilibrium.
- The equilibrium position will shift to left to reestablish the equilibrium.

8. Name the processes that occur when HNO2 solution is added separately into water, NaOH
solution and NaNO2 solution. Write the chemical equation for each process. In three
separate beakers, 20.0 mL of 0.08 M HNO2 was added to 40.0 mL of water, 40.0 mL of 0.03
M NaOH and 40.0 mL of 0.03 M NaNO2 , respectively. Determine the pH of the solution in
each beaker.
[Ka for HNO2 = 4.5 X 10 -4]

Dilution of weak acid
i) HNO2(aq) + H2O (l) → NO2- (aq) + H3O + (aq)

Neutralisation
ii) HNO2 (aq) + NaOH (aq) → NaNO2 (aq) + H2O (l)

Formation of buffer solution
iii) HNO2 (aq) + NaNO2 (aq) → acidic buffer solution

HNO2(aq) + H2O (l) → NO2-(aq) + H3O + (aq)

ni 2x10-2x0.08 - 00
= 0.0016
- +x +x
nΔ -x - x x
nf 0.0016 –x x/0.06 x/0.06
[ ]f (0.0016-x)/ 0.06

Ka = (x/0.06)(x/0.06)

( 0.0016-x)/0.06
4.5x 10-4 = (x/0.06)(x/0.06)

( 0.0016-x)/0.06
Since Ka is larger than 1 X 10-5,
x2 + 2.7 x 10-5 x – 4.32 x 10-8 = 0
x= 1.9478 x 10-4 mol
[H3O+] = 3.2463 x 10-3 M

78

pH = -log [H3O+]
= -log (3.2463 x 10-3)
= 2.49

HNO2(aq) + NaOH(aq) → NaNO2(aq) + H2O (aq)

ni 0.0016 0.0012 00
nΔ -0.0012
-0.0012 +0.0012 +0.0012
nf 0.0004
0 0.0012 0.0012
[ ]f 0.0004/ 0.06
= 6.6667x10-3 - 0.0012/0.06 0.0012/0.06

= 0.02 =0.02

pH  -log Ka  log [NO - ]
2

[HNO 2 ]

 - log 4.5x10-4  log 0.02
6.6667 x 10 -3

 3.82

[ NO2-] = 1.2x10-3 M [HNO2]= 0.02667M

pH  -log Ka  log [NO - ]
2

[HNO 2 ]

 - log 4.5x10-4  log 1.2x10-3
0.02667

 2.0

79

1. a) SUGGESTED ANSWER PSPM 1
SESI 2017/2018

A 25.0 mL of 0.050 M silver nitrate solution is mixed with 25.0 mL of 0.050 M of
calcium bromide solution to give 0.105 g of solid silver bromide.

i. Write the balanced chemical equation for this reaction.
2AgNO3 (aq) + CaBr2 (aq) → 2AgBr (s) + Ca(NO3)2 (s)

ii. Determine the limiting reactant.

n AgNO3 = n CaBr2  0.05 x 25
1000

 1.25 x103 mol

From equation,
2 mol AgNO3 ≡ 1 mol CaBr2

1.25 x 10-3 mol AgNO3 ≡ 1.25 x 103 mol AgNO 3 x 1 mol CaBr 2
2 mol AgNO 3

= 6.25 x 10-4 mol CaBr2
Number of mole of CaBr2 needed (6.25 x 10-4 mol) < number of mol CaBr2
available (1.25 x 10-3 mol)
CaBr2 is excess reacant and limiting reactant is AgNO3

iii. Calculate the percentage yield of silver bromide.

From equation,
2 mol AgNO3 ≡ 2 mol AgBr

1.25 x 10-3 mol AgNO3 ≡ 1.25 x 10-3 mol AgBr
mass AgBr = 1.25 x 10-3 mol x 187.8 gmol-1
= 0.2348 g

% yield  actualyield x 100
theoritical yield

 0.105 x 100
0.2348

 44.72%

80

b) An organic compound contains only carbon, hydrogen and oxygen with a mass
composition of 41.40% C, 3.47% H and 55.13% O. If 0.05 mol of this compound
weighs 5.80 g, determine its

i. empirical formula

% mass C H O
Moles 3.47 55.13
ratio 41.40
3.47  3.47 55.13  3.45
41.40  3.45 1 16
12
1 1
1

Empirical formula is CHO

ii. molecular formula

(CHO)n = molar mass

(12n  n  16n)  5.80 g
0.05 mol

29n 116
n 4

Molecular formula is C4H4O4

2. a) The azide ion, N3 - , exists in several resonance forms.
i. Draw possible resonance structures for the azide ion.

(-1) (-1) (0) (-2) (-2) (0)
N N+ N2- N2-N+ N
N- N+ N-
(+1) (+1)
(+1)

Structure Structure Structure
1 2 3

ii. Use formal charges to select the most stable structure.
Structure 1.

81

iii. Using Lewis structure, explain why N3 – ion exists, whereas trifluoride ion, F3 - ,
does not exist.

F F F0

F is in period 2, central atom cannot form expanded octet, so
F3- ion does not exist.
N is in period 2, central atom obey octet rule, so N3- ion is exists.

[9 marks]

b) Chloroform, CHCl3, is a common organic solvent. If H in CHCl3 is replaced
by Cl, it becomes CCl4, a toxic solvent. For each of CHCl3 and CCl4
compound.
i. draw the molecular shape.

CHCl3
No. of valence electron :

C =4
H= 1
3 Cl = 21
Total = 26

Lewis structure : H
H
C
Cl C Cl Cl Cl Cl
molecular shape: tetrahedral
Cl

CCl4
No. of valence electron :

C= 4
4 Cl = 28
Total = 32

Lewis structure : Cl
Cl C
Cl Cl Cl
Cl C Cl molecular geometry : tetrahedral
Cl

82

ii. show the bond polarity, H
CHCl3 C
Cl Cl Cl
CCl4
Cl
C
Cl Cl Cl

iii. predict the polarity.

The magnitude of C-H dipole moment is not equal to C-Cl dipole
moment. The four dipole moments do not cancel out each other,
  0. Therefore CHCl3 is an asymmetry and a polar molecule.

All C-Cl dipole moments are equal in magnitude. The four dipole
moments are arranged in tetrahedral shape causes them to cancel out
each other,  = 0. Therefore CCl4 molecule is a symmetry and a non-polar
molecule.

3. A 3.0 L flask at 298 K contains a mixture of 2.0 mol helium gas and 3.0 mol xenon gas. The

van der Waals constants for helium and xenon are given in TABLE 3.

TABLE 3
Gas a (L2 atm mol-2) b (L mol-1)

Helium 0.03421 0.02370

Xenon 4.194 0.05105

Given the van der Waals equation: ( )( )

a) State two (2) postulates of the kinetic molecular theory of an ideal gas in relationship
to parameters a and b as in TABLE 3.

Kinetic molecular theory of an ideal gas in relationship to:

parameter a -The attractive and repulsive forces between gas molecules are
negligible.
@
The intermolecular forces are negligible.

parameter b -The combined volume of all molecules of the gas is negligible
relative to volume of the gas container.

@
The total volume of all gas molecules is negligible compared to the volume in
which the gas is contained.

83

b) Which gas is expected to behave ideally and which gas exhibits a marked deviation
from ideal behavior?

Helium is expected to behave ideally while xenon exhibits a marked deviation
from ideal behaviour.

c) Explain your answer in 3(b) by referring to the significance of a and b values in TABLE
3.

Helium has smaller value of a which means the attractive and repulsive
forces between helium atoms are very weak and almost negligible. Helium
also has a smaller value of b which means the combined volume of all
helium atoms is negligible compared to the volume of the gas container.
Therefore, helium behaves almost like an ideal gas.

Whereas, xenon has a bigger value of a which means the attractive and
repulsive forces between xenon atoms are stronger and significant. A
bigger value of b means that the combined volume of all xenon atoms is
significant compared to the volume of the gas container. Therefore, xenon
deviates from ideal gas behaviour.

d) Calculate total pressure of the gas in the mixture using an ideal gas equation.

PT = n T RT
V

(2.0  3.0) mol  0.08206 L atm mol1 K1  298 K

=

3.0 L

= 41 atm

@

PHe = nHeRT
V

2.0 mol  0.08206 L atm mol1 K1  298 K

=

3.0 L

= 16.3 atm

PXe = nXeRT
V

3.0 mol  0.08206 L atm mol1 K 1  298 K

=

3.0 L

= 24.5 atm

84

PT = 16.3 atm + 24.5 atm
= 41 atm

e) Using the van der Waals equation, the total pressure of the system is calculated to be
38.1 atm. Explain the reasons behind the differences between this value and the
pressure calculated in 3(d).

Pressure of a gas is the result of the collisions between the gas molecules with the
wall of their container. The higher the frequency and impact of the collisions, the
higher the gas pressure.

Helium and xenon are real gases with weak attractive forces between atoms.
These forces lessen the impact of collision of a given atom with the wall of the
container. Therefore, the actual pressure calculated using the van der Waals
equation is lower than the pressure calculated using the ideal gas equation.
The combined volume of gas molecules is NOT the reason in this case because
bigger gas molecules reduces the unoccupied volume and therefore increases the
gas pressure.

4. a) In a series of acids of the halogen group, hydroiodic acid, HI, is the strongest acid
while hydrofluoric acid, HF, is the weakest.
i. Explain what is meant by strong and weak acids.

Strong acid is an acid that dissociates completely, while weak acid is an
acid that dissociates partially.

ii. Briefly explain the main factor in determining the strength of the acid.

The polarity of the bond formed between atom F and H. The weaker the
bond, the easily it can dissociates to produce high concentration of H3O+.

iii. Calculate the degree of dissociation,  of 0.20 M HF solution.
[Ka = 6.8 x 10-4 at 25C]

[ ]/M HF (aq) + H2O (l) F- (aq) + H3O+ (aq)
I --
C 0.20 +x +x
E xx
-x
0.20 – x

[ ][ ]
[]
()

()

x1 = 0.01133 , x2 = -0.01200 (rejected)

85

degree of dissociation,  =

=
= 0.057

b) Acetic acid, CH3COOH, is a weak acid with Ka = 1.8 x 10-5. It dissociates in water

according to the following equation:
CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)

i. Rewrite the above equation and label the conjugate acid and conjugate base.

CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)
acid base conjugate base conjugate acid

ii. Determine the pH of 0.10 M.

[ ]/M CH3COOH (aq) + H2O (l) CH3COO- (aq) +H3O+ (aq)
I 0.10 00
C -x +x +x
E xx
0.10 – x

[ ][ ]
[]
()
()

Since Ka is very small, assume 0.10 – x  0.10

()

x = 1.34 x 10-3 M = [H3O+]
pH = - log [H3O+]

= - log 1.34 x 10-3
= 2.87

iii. Predict the percentage of dissociation when the acid concentration is
increased.
Percent of dissociation remain unchanged when the acid concentration
is increased.

86

5. a) Sketch the energy level diagram to show the electronic transitions which give
rise to the first five lines in the Lyman series of the hydrogen atom. Explain.

energy

n=6
n=5

n=4

n=3

n=2

n=1

Line spectrum of Lyman series.

When energy is supplied to an atom, electron at the ground state will absorb the
energy. This electron will excite to higher energy level and the electron is now at
its excited state. The electron is unstable and will fall back to lower energy level,
n = 1 by releasing specific amount of energy with specific wavelength in the form
of photon. The photon released is passed through a prism and fall on the
photograph plate to be recorded as line spectrum.

5. b) Element Y is in Period 3 of the Periodic Table. The first six successive
ionization energies of element Y are given in Table 5.

Ionization First TABLE 5 Fourth Fifth Sixth
1011 Second Third 4963 6274 21267
energy
(kJ mol-1) 1907 2914

Determine the group for Y and explain your answer. Write the valence electronic
configuration and the valence orbital diagram of Y. Explain how the Pauli exclusion

87

principle, Aufbau principle and Hund’s rule are applied in drawing the valence orbital
diagram of element Y.

• Sharp increase occur at IE6 (IE6>>IE5).
• The sixth electron is removed from inner shell.
• Element Y has 5 valence electron.
• Element Y in group 15
• Valence electronic configuration of Y : 3s2 3p3

Valence orbital diagram :

• Pauli exclusion principle state that no two electrons can have same set of four
quantum number. Hence, only 2 electrons are located at same orientation
with different spin.

• Aufbau principle state that in the ground state of an atom or ion, electrons fill
atomic orbitals of the lowest energy level before higher energy level. Energy
level 3s orbital is lower than 3p orbital.

• Hund’s rule state that when electron are filled on degenerate orbital (such as
3p orbital), electron are filled singly with same spin before its paired.

6. a) Sodium bicarbonate, NaHCO3 is used in baking as raising agent. When dissolved in

water, the salt dissociates into ions. Write the Lewis structure for the bicarbonate
ion, HOCO2- , and draw its resonance structures.

Lewis structure:

-

O
H OCO

Resonance Structure: - -

O O
H OCO H OCO

Determine the shape of the HOCO2- ion and the hybridization of central atoms. Draw
and label the overlapping of the orbitals to show the formation of the covalent
bonds.

Molecular Geometry for the HOCO2- ion = Trigonal planar

88

Type of hybridisation:

2 1 -
HO
O1 3
O-
C

i. O1 = sp2
ii. O2 = sp3
iii. C = sp2
Valencce Orbital diagram:

H= O2 2p
1s
GS=
O3 = 2s
2s
2p ES=

sp2 2p

O1 C 2p

GS= 2p GS= 2p
2s 2p 2s
sp3
ES= ES= 2p
2s 2s

ES= ES=

sp2

Orbital Overlaping:

sp2 sp2
2p
O

sp2

 
sp2
O
sp3 sp3 2p C
sp2 2p
O sp2



 3 sp3

sp

H 1s

89

(b) Metallic compounds have some physical properties which are different from covalent
compounds. State four (4) properties and relate these properties to metallic bonding.
[8 marks]
Four properties of metallic compound:
i) High melting and boiling point
Metallic compounds have strong electrostatic attraction between
positively charged metal ion and negatively charge electron sea. More
energy needed to break this attraction.

ii) Able to conduct electricity
Metallic compounds have free moving electrons.

iii) Malleable
Delocalized electrons enable the metal atoms to roll over each other.
Metal atoms only slide to each other.

iv) Luster
Photons of light do not penetrate very far into the surface of a metal and
are typically reflected on the metallic surface.

7 (a) Carbon dioxide can be used to extract caffeine under supercritical condition. Some of
the physical properties of carbon dioxide are shown below.

Triple point temperature -51°C
Triple point pressure 5 atm
Critical temperature 31°C
Critical pressure 73 atm

What is meant by triple point and critical point?

Triple point: Temperature and pressure at which solid, liquid and gas
Critical point: simultaneously exist in equilibrium.
Temperature and pressure at which gaseous phase and liquid
phase are indistinguishable.

90

Using the data given above, sketch a labeled phase diagram for carbon dioxide. Show
the supercritical point in the diagram.

Phase diagram of carbon dioxide

Pressure (atm)

73 point

5
1

-78.5 -51 31
Temperature (°C)

Using the phase diagram, under what conditions can dry ice be formed to act as
coolants?

Dry ice can be formed to act as coolants under atmospheric pressure (1 atm) and
the temperature of -78.5°C. Under these conditions, dry ice breaks down and turns
directly into carbon dioxide gas rather than a liquid. The super-cold temperature
and the sublimation feature make dry ice great for refrigeration or as coolants. For
example, if we want to send something frozen across the country, we can pack it in
dry ice. It will be frozen when it reaches its destination, and there will be no messy
liquid left over like we would have with normal ice.

b) In an experiment, 0.10 mol of N2O4 and 0.55 mol of NO2 are mixed in a closed vessel at
350 K with a total pressure of 2.0 atm. At this temperature, the equilibrium constant,
Kp is 3.89. The reaction is given as follows:
N2O4 (g) 2NO2 (g)

In which direction will the reaction proceed to reach equilibrium?

91

PN2O4  X N2O4  Ptotal

 n N2O4  Ptotal
n total

 0.1 mol   2.0 atm
0.10  0.55mol

 0.3077 atm

PNO2  X NO2  Ptotal

  1  X N2O4  Ptotal

 1  0.1538 2.0 atm

 1.6923 atm

 QPP2  1.69232  9.31
NO2 0.3077

PN 2O 4

Since QP > KP:
 The system is not at equilibrium
 Initially, there are more NO2 in the reaction mixture
 To reach equilibrium QP = KP, the reaction will shift backward

If the above experiment is carried out at 500 K, the new KP is 1700. Is the reaction
exothermic or endothermic? Explain.

 At higher temperature, the value of KP increases
 The system consumes more N2O4 and produces more NO2
 This indicates that at higher temperature, the equilibrium system is

disturbed
 According to Le Chatelier’s principle, equilibrium position will be shifted to

the right as to lower the temperature of the system by absorbing the added
heat
 Therefore, the forward reaction is endothermic

92

8. a) Hypochlorous acid, HOCl, is a monoprotic acid. An aqueous solution of
0.028 M HOCl has a pH of 4.5. Calculate the Ka value for this acid.
A buffer solution is prepared by mixing 100 mL of 0.04 M HOCl with 100 mL of 0.02
M NaOCl. Calculate the pH of the solution. Explain how a mixture of HOCl and
sodium hypochlorite, NaOCl, solution behaves as a buffer solution when a small
amount of strong acid is added.

HOCl(aq) + H2O(l) OCl- (aq) + H3O+(aq)
- 0 0
[ ]i 0.028 - +x
+x
[ ]c -x - x x
[ ]f 0.028 – x

pH = - log [H3O+]

4.5 = - log x
x = 3.16 x 10-5 M

[HOCl] = 0.02797M
[OCl-] = [H3O+] = 3.16 x 10-5 M

 Ka  OCl - H3O
[HOCl]

   3.16 x 105 3.16 x 105 0.040.1 = 4 x 10-3 mol
0.02797

 3.57 x 108 n HOCl =

[HOCl] = 4 x 103  0.02 M
0.2

n NaOCl = 0.020.1 = 2 x 10-3 mol
[NaOCl] =
2 x 103  0.01 M
0.2

OCl  
 pH = - log Ka + log

HOCl
0.01
= - log 3.57x 10-8 + log 0.02

= 7.15

Buffer solution contain HOCl act as an acid and OCl- that acts as a base.
When small amount of strong acid is added to it, OCl- will neutralized it.

Thus, the pH is not much affected.

OCl- (aq) + H3O+(aq) → HOCl(aq)

93

b) The solubility product, Ksp for calcium sulphate, CaSO4, is 2.0 x 10-5 at 250C.
Calculate the molar solubility of CaSO4 in water and in 0.10 M sodium sulphate,
Na2SO4. State common ion effect on the solubility of CaSO4.

CaSO4(s) Ca2+(aq) + SO42-(aq)

xx

Ksp = [Ca2+][SO42-]

2  105 = x2

x = 4.47 x 10-3 M

Molar solubility in water is 4.47 x 10-3 M

CaSO4(s) Ca2+(aq) + SO42-(aq)

y y + 0.10 M

Ksp = [Ca2+][SO42-]

2  105 = (y)( y + 0.10) ; assume x<< 0.10, y+ 0.10 = 0.10

2  105 = (y)( 0.10)

y = 2 x 10-4 M

Molar solubility in Na2SO4 is 2 x 10-4 M
 Solubility of CaSO4 in pure water higher than in 0.10 M Na2SO4 solution, SO42-

(common ion) is present.
 The equilibrium position shifts backward.
 Solubility of CaSO4 decreases.
 The present of common ion, SO42- reduces the solubility of CaSO4.

.

94

SUGGESTED ANSWER PSPM 1
SESI 2018/2019

1. a) Bromine has proton number of 35. FIGURE below shows a mass spectrum of
bromine.

i. Write the notations for all isotopes of bromine.

79 Br and 81 Br
35 35

ii. Calculate the relative atomic mass of bromine.

Average atomic mass = (isotopic mass abundance)
 abundance

= 79 51   81 49
49  51

= 79.98
∴ relative atomic mass = 79.98

b) A reagent bottle contains a stock solution of 0.90% by mass of sodium chloride, NaCl.
The density of the solution is 1.00 g cm-3. Calculate
i. The mole fraction of NaCl.
Assume mass of solution = 100 g
Mass of NaCl = 0.90 g

Mole of NaCl = 0.90 g = 0.0154 mol
58.5 g mol

Mass of H2O = 100 g – 0.90 g = 99.1 g

Mole of H2O = 99.1 g = 5.506 mol
18 g mol

95

∴Mole fraction of NaCl = 0.0154 mol mol

0.0154  5.506

= 2.79×10-3

ii. The molality of NaCl solution.

Molality = Mole of NaCl (mol)
mass of solvent (kg)

= 0.0154 mol
0.0991 kg

= 0.155 m

iii. The volume of the stock solution required to prepare 100 mL of 0.01 M NaCl

solution.

Density of solution = Mass of solution (g)
volume of solution (cm3)

1.00 g cm-3 = 100 g
volume of solution (cm3 )

Volume of solution = 100 cm3

Molarity = Mole of NaCl (mol)
Volume of solution (L)

= 0.0154 mol
0.1 L

= 0.154 M

Mole of diluted NaCl solution = (0.1 L)(0.01 M)
= 1×10-3 mol

∴Volume NaCl solution needed = 1×10-3 mol
0.154 M

= 6.49×10-3 L

c) Silicon tetrachloride, SiCl4 can be prepared by heating silicon in excess chlorine gas.

Si(s)  2Cl2(g)  SiCl4(l)

i. Calculate the mass of silicon needed to produce 400 g SiCl4 if the percentage
yield is 42.5%.

42.5% = 400 g 100

theoretical yield

96

Theoretical yield (mass of SiCl4) = 941.18 g

Mole of SiCl4 = 941.18 g = 5.533 mol
170.1 g mol

1 mol SiCl4 ≡ 1 mol Si
5.533 mol SiCl4 ≡ 5.533 mol Si

∴Mass of Si = (5.533 mol)(28.1 g/mol) = 155.5 g

ii. If 15 mol of chlorine is used, determine the amount (mole) of unreacted

chlorine.
1 mol SiCl4 ≡ 2 mol Cl2

5.533 mol SiCl4 ≡ 11.066 mol Cl2

∴ unreacted Cl2 = (15 - 11.066) mol = 3.934 mol

2. a) The wavelength that produces a line, B in Brackett series is 2165.6 nm.

i. Determine the transition that forms the B line.

1  RH  1  1   n2
  n12 n22  , n1
 

1  1.097 107 m1  1  1 
2.1656106 m  42 n22 
 

n1 = 7
∴n =7 to n=4

ii. Calculate the energy emitted for the transition.

E  RH  1  1 
 ni2 n2f 

= 2.181018 J 1  1 
 72 42 

= - 9.18×10-20 J

iii. Another line, C was form with wavelength of 1817.5 nm. Explain qualitatively,
whether line B or line C, has higher energy emitted.
 Line C has higher energy emitted than line B.
 Energy is inversely proportional to wavelength.
 Line C has smaller wavelength than line B.

97

b) The electronic configuration of element D is 1s2 2s2 2p6 3s2 3p3.
i. Give a set of quantum number for the 9th electron.
(n=2, l=1, m=0, s=+1/2)

ii. Draw the orbital diagram of the valence electrons.

3s 3p

iii. Draw and label the 3D shape of orbitals occupied by the valence electrons.

zz

xy xy
3s 3px
z z

x yx y

3py 3pz

iv. Explain the filling of the valence electrons in 2(b)(iii) according to the
appropriate rule(s)/principle(s).
 Aufbau principle.
Electrons added to the lower energy orbital first. The valence electrons
are fill the 3s orbital before 3p orbital. 3s orbital has lower energy than
3p orbital.

 Hund’s rule.
When electrons are, add into degenerate orbitals, each orbitals are,
filled singly with electron of the same spin before it is pair. 3p orbitals
are degenerate orbitals. Therefore, the remaining 3 valence electrons is
filled singly in each orbital with the same spin.

98

 Pauli’s Exclusion Principle.
No two electrons can have same set of quantum number. Only 2
electrons can occupy the same atomic orbitals, thus 2 electrons must be
filled in 3s orbital with opposite spin.

3. a) Ammonia, NH3 and boron triflouride, BF3 are covalent compounds. NH3 and BF3
react to form H3NBF3 molecule.
i. Explain why NH3 obeys octet rule but BF3 does not.

HNH HBH

HH

 N in NH3 is surrounded by 8 valence electrons while B in BH3 is
surrounded by only 6 valence electrons.

 BH3 is incomplete octet.

ii. Show the formation of H3NBF3 molecule using Lewis dot symbol and label the

bond formed.

HH HH

+H N BH HN BH

HH HH

Dative bond [5 marks]

b) Oxygen difluoride, OF2 is a strongly oxidizing colourless gas.
i. Determine the molecular geometry of this molecule.

FO

F

 Number of electron pair around oxygen atom: 4
 Electron pair arrangement: tetrahedral
 2 bonding pair and 2 lone pair.
 Based on VSEPR theory, the electrons pair will be located as far as

possible to minimize the repulsion among them. The lone pair-lone pair
repulsion > lone pair-bonding pair repulsion > bonding pair-bonding pair
repulsion.
 Molecular geometry: V-shape

FO

F

99