JAWAPAN 20142015

SUGGESTED ANSWER PSPM 1
SESI 2014/2015

1 (a) Consider the following reaction.

2NO(g) + 2 CO(g) ⟶ N2(g) +2CO2(g)

If 300.00 g of CO has reacted, calculate the mass percentage, w/w, of CO2 in the
mixture of product.

ANSWER

No. of mol CO  300.00 g
28.0 g mol1

 10.714 mol

From equation, 2 mol CO  1 mol N2 x 1 mol N 2
10.714 mol CO  x mol N 2
x  10.714 mol CO
2 mol CO
 5.3571 mol N 2

From equation, 1 mol N 2  2 mol CO2

5.3571 mol N 2  x mol CO 2

x  5.3571 mol N 2 x 2 mol CO 2
1 mol N 2

 10.714 mol CO 2

Mass of CO2 = 10.714 mol X 44.0 g mol-1
= 471.42 g

Mass of N2 = 5.3571 mol x 28.0 g mol-1
= 150.00 g

%w w  mass of solute x 100
mass of solution

 471.42 g x 100

(471.42 150.00) g

 75.862%

40

(b) A 30.00 mL of 0.025 M sodium dichromate, Na2Cr2O7 solution is titrated with iron(II)

sulphate,Fe2SO4 solution in acidic condition, according to the following reaction:

Cr2O72- + Fe 2+ Cr3+ + Fe 3+

The titration requires 40.00 mL of Fe2SO4 solution to reach the end point.

(i) Balance the redox equation.

ANSWER

6e + 14H+ + Cr2O72- ⟶ 2Cr3+ + 7H2O
6 (Fe2+ ⟶ Fe3+ + e )

Cr2O72- + 6Fe2+ + 14H+ ⟶ 2Cr3+ +6Fe3+ +7H2O

(ii) Calculate the mass of Na2Cr2O7 needed to prepare a 0.025 M solution in a 50 mL
volumetric flask.

ANSWER

Molarity of Na 2Cr2O7  mol of solute

volume of solution L

mol Na 2Cr2O7  (0.025mol L1 )  50  L
 1000 

 1.3 x 103 mol

Mas s Na 2Cr2O7 needed  1.3 x103 mol x 262.0 g mol1
 0.33 g

(iii) Determine the molarity of the Fe2+ solution.

ANSWER

mol of Na 2Cr2O7  0.025 mol L1 x 30.00x103 L
 7.500 x 104 mol

From equation, 1 mol Cr2O7 2  6 mol Fe 2

7.500 x 104 mol Cr2O7 2  x mol

x  7.500 x104 mol Cr2O7 2 x 6mol Fe 2
1 mol Cr2O7 2

 4.500 x 103 mol

41

Molarity of Fe 2 solution  Mole of solute

Volume of solution (L)

 4.500 x 103 mol
40.00x 103 L

 0.1125 mol L1

2 (a) The formula of formic acid is HCO2H.One of the carbon-oxygen bond lengths in this
molecule is 1.36 Å while the other is 1.23 Å. Draw the Lewis structure of this
molecule and labels these bonds.

ANSWER

Formic acid: HCOOH

O

1.23 Å

HCOH

1.36 Å

(b)(i) Xenon can be covalently bonded to fluorine and oxygen to form xenon compounds,
XeF4 and XeO2F2. For both compounds,
(ii) determine the number of bonding electron pair(s) and lone electron pair(s) around
the central atom xenon.
(iii) state the molecular geometry.
(iv) determine the hybridisation of xenon atom.
predict their polarity.

ANSWER

F 3
F Xe F sp

F  P 3 
H sp
3
s sp

3
sp

sH

Hs 

No of bonding pairs XeF4 XeO2F2
electron 4 4
2 1
No.of lone pairs electron
Square planar Distorted tetrahedral @ see
Molecular shape saw

42

XeF4 ↑↓ ↑↓ ↑↓
5p
Xe: ↑↓
5s

Xe excited state: ↑↓ ↑↓ ↑↓ ↑↓ ↑↑
5s 5p 5d

Xe hybrid: ↑↓ ↑↓ ↑ ↑ ↑ ↑
Sp3d2

XeO2F2 ↑↓ ↑ ↑
5p
Xe2+ground state: ↑↓
5s

Xe2+ excited state: ↑↓ ↑ ↑↑ ↑↑
5s 5p 5d

Xe2+ hybrid: ↑↓ ↑ ↑↑ ↑
XeF4 Sp3d

XeO2F2

FF FO

Xe Xe

FF FO

μ = 0 (non polar) μ ≠ 0 (polar)

3 (a) Limestone,CaCO3, decomposed to solid calcium oxide and carbon dioxide gas when
heated at high temperature. At 30 oC, a volume of 107.3 mL of the gas was collected
by displacement of water with a total pressure of 1 atm. Calculate;

the number of moles of carbon dioxide produced.
(i)

ANSWER

CaCO3(s)  CaO(s) + CO2 (g)

43

PH2O  31.8 mmHg
 0.0418 atm

PT  PCO2  PH2O
PCO2  1.0  0.0418

 0.9582 atm

PCO2 V  n CO2 RT

n CO2  (0.9582)(0.1073)
(0.08206) (303.15)

 0.0041 mol

(ii) the mass of limestone decomposed.
[Vapour pressure of water at 30oC is 31.8 mmHg]

ANSWER

1 mol CO2 ≡ 1 mol CaCO3
0.0041 mol CO2 ≡ 0.0041 mol CaCO3

 mass CaCO3 = 0.0041 X 100.1
= 0.4137g

(b) Briefly describe three types of crystalline solids in terms of their interparticle forces.

ANSWER
3 type of crystalline solid

 Metallic crystal/metallic solid
- Electrostatic forces between positively metal ion and the sea of valence
electron.
- Metallic bond @ metal cations with a cloud of delocalized electrons.

 Ionic crystal/ionic solid
- Electrostatic forces between positively ion and negatively ion
- Ionic bond

 Molecular covalent/simple molecular/simple molecular covalent
- Weak van der Waals forces

 Giant covalent
- Covalent bond between atoms.

44

4 (a) HA is a weak acid with an acid dissociation constant,Ka=2.95x10-8.If the
concentration of the acid is 1.12 M, calculate

(i) pH of the solution.
ANSWER

HA(aq) ⇌ H+(aq) + A-(aq)

[ ]o 1.12 0 0
∆[ ]
-x +x +x
[ ]f
1.12-x x x

Ka  [H ][A ]
[HA ]

2.95 x108  (x)(x)
1.12  x

2.95 x108  x 2
1.12

[H ]  x  1.82 x 104 M
pH   log[H ]
  log1.82 x 104
 3.74

(ii) Percent dissociation.

ANSWER

% dissociation  [ ]dissociated x 100%
[ ]initial

 1.82 x 104
x 100%
1.12

 0.0163%

(b) The solubility product constant,Ksp of calcium phosphate,Ca3(PO4)2, in pure water is
1.2 x 10-26 at 25°C.

(i) Write the expression for the solubility product constant.

ANSWER

Ca3(PO 4)2(s) 3Ca2+ + 2PO 3 - (aq)
4

Ksp  [Ca 2 ]3[PO43 ]2

45

(ii) Calculate the molar solubility of calcium phosphate and concentration of each ion.
ANSWER

Ca3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO43-(aq)
x 3x 2x

K sp  [Ca2 ]3 [PO 4 3 ]2

1.2 x 1026  (3x)2 2x2

1.2 x 1026  108x5
x5  1.1111x 1028

 2.57 x 106 M
 molar solubility of Ca 3 (PO 4 )2  2.57 x 106 M
[Ca2 ]  7.70 x 106 M
[PO 4 2 ]  5.14 x 106 M

5. The ion of atom X has 8 outermost electrons and 10 inner electrons with a charge of
-1. Discuss all the rule(s) and principle(s) used to fill the electrons in the orbital of
atom X. Explain the change in the radius of atom X as it changes from a neutral
atom to a negatively charged ion.

ANSWER

Aufbau principle:
 electron occupy orbitals in the order of increasing energy level of the
orbitals.
 Orbitals with the lowest energy are always occupied first.

Pauli’s exclusion principle:
 Only two electrons may occupy the same orbital and that these two
electrons must have opposite spins.

Hund’s rule:
 When electrons are placed in a set of orbitals with equal or degenerate
energies, the electrons must occupy them singly with parallel spins before
they occupy the orbitals in pairs.

The extra electron is added to the same shell (n=3). Extra repulsion is produced
by the incoming electron causing the atom to expand. As a result X- ion is
bigger than atom X.

46

Atoms X, Y and Z are in periods n, n+1 and n+2, respectively. These atoms are
also in the same group. Discuss the trend in electronegativity exhibited by these
atoms

Answer

X, Y and Z are located in periods 3, 4 and 5 respectively. As we go down the
group from X to Z, new shell (n) increase and inner electrons are added than
shielding effect increases. Furthermore valence electrons being far away from
the nucleus, so the attractive forces of electrons become smaller. Therefore, the
electronegativities of atom will decrease.

6 Draw and explain the structures of BH3, NH3 and PH3 using valence shell electron
pair repulsion theory. Also, show the overlapping of orbitals in the PH3 molecule.

The boiling points of BH3, NH3 and PH3 are 173 K, 240 K and 185 K, respectively.
Explain why the boiling point of PH3 is greater than BH3 but lower than NH3.

ANSWER

VSEPR theory state that the repulsion of bonding pair-bonding pair < bonding

pair – lone pair < lone pair-lone pair. H

BH3 HB
 Has 3 bonding pairs. H

 Molecular geometry : trigonal planar

 Based on VSEPR theory, the electron pairs are arranged as far as possible

in order to minimize the repulsion. The repulsion of bonding pair-bonding

pair repulsion is equal.

 Bond angle: 120o

NH3 N

 Have 3 bonding pairs and 1 lone pair. HH H
 Molecular geometry: trigonal pyramidal

 Based on VSEPR theory, the electron pairs are arranged as far as possible

in order to minimize the repulsion. The repulsion of lone pair-bonding pair

> bonding pair-bonding pair.

 Bond angle: 107.5o

47

P

PH3 HH H
 Have 3 bonding pairs and 1 lone pair.

 Molecular geometry: trigonal pyramidal

 Based on VSEPR theory, the electron pairs are arranged as far as possible
in order to minimize the repulsion. The repulsion of lone pair-bonding pair

> bonding pair-bonding pair.
 Bond angle: < 109.5o

P(ground state):

3s 3p 3

P(excited state): sp

3s 3p  P sp3 
Hs sH
P(hybrid): sp3
sp3
3
Hs 
sp

 PH3 is a polar molecule, presence of dipole-dipole interaction.
 BH3 is nonpolar molecule, presence of London forces.
 Strength of dipole-dipole interaction > London forces.
 NH3 has hydrogen bonding and has stronger forces than PH3.

7 (a) The melting points of C2H6, CH3OH, NaCl and Si increase in the order as shown
below. Explain the trend.
C2H6 < CH3OH < NaCl < Si

ANSWER

Melting point increases in the order of attraction between molecules or atoms.
 C2H6 is non-polar compound and can form weak Van der Waals forces
between molecules.
 CH3OH has hydroxyl group and can form hydrogen bond between
molecules.
 NaCl is ionic compound and all atoms in the compound are held together
by electrovalent electrostatic forces.
 Si has gigantic covalent network in which Si atom is bonded tetrahedrally
to each other, thus form infinite amount of covalent bond and required
high energy to break all bonds.

48

(b) A 10.00-L vessel contains 0.0681 mol phosphorous trichloride,PCl3 and 0.2056 mol
chlorine,Cl2 at 250 °C. If 0.0316 mol phosphorous pentachloride, PCl5, is produced at
equilibrium, calculate Kp for this reaction.

Discuss four factors that can increase the amount of PCl5 produced.

ANSWER

PCl3(g) + Cl2(g) → PCl5(g)

ni 0.0681 0.2056 0

nc -x -x +x

nf 0.0681-x 0.2056-x x = 0.0316

= 0.0365 =0.174

[ ] 0.00365 M 0.0174 M 0.00316 M

Kc = PCl5   =  0.00316 = 49.756
PCl3 Cl2 0.00365  0.0174 

Kp = Kc(RT)∆n = 49.756 (0.08206×523.15)−1 = 1.16

Le Chatelier state that when a system at equilibrium is subjected to change,
the system will adjust to counteract the effect.
 Increase pressure: Reaction will moves forward because the number of

moles of reactants is more than that of the product.
 Increase temperature: Reaction will moves forward in order to counteract

with the added heat by undergo endothermic reaction.
 Add reactants: Reaction will moves forward in order to decrease the

concentration of reactants.
 Remove product: Reaction will moves forward in order to increase the

concentration of product.

49

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