P.1
Suggested Solution of AL-CHEMISTRY Paper II 1998
Suggested Solution
1.
a)
i)
When HCl is added to A and B, the pH of A drops to a greater extent
than that of B.
(1)
It is because B is a buffer solution.
(0.5)
The H+ added will react with the lactate to form lactic acid and thus the
drop in pH is relatively small.
(0.5)
iii)
At the equivalence point, all the lactic acid is converted into lactate.
The pH at the equivalence point is the same as the pH of an aqueous
solution of the salt.
(1)
For the two solutions, the concentations of lactate formed at the
equivalence points are the same. Therefore, the pHs of the two
solutions are the same.
(1)
Marks
For solution A,
HA
Initial
At eqm
1.38x10-4 =
ii)


0.15
0.15 -x
H+ +
A-
0
x
0
x
x2
0.15  x
(1)
Assume x<<0.15
x2 = 2.07 x 10-5
[H+] = 4.55 x 10-3
pH = 2.34 (accept 2.3 to 2.4)
Let V be the total volume at equivalence point.
For solution A, all the 50.0ml 0.15M lactic acid is converted to lactate
at the equivalence point. Thus, the concentration of lactate in A is
(50.0 x 0.15)/V.
For solution B, the concentration of lactate is 0.05M initially.
At the equivalence point, total concentration of lactate in B is
((0.05+0.10) x 50.0)/V.
Therefore, the concentration of lactate at the equivalence point are
equal.
(1)
For solution B,
Initial
Final
HA +
0.30/2
0.15
0.10
HA
Initial
At eqm

OH- 
0.10/2
0.05
0


0.10
0.10 -y
A- + H2O
0
0
0.05 0.05
H+ +
A-
0
y
0.05
0.05 + y
y (0.05  y )
0.10  y
Assume y<<0.05
[H+] = 2.76 x 10-4
pH = 3.56 (accept 3.5 to 3.6)
1.38  10 4 
For mixing two solutions, the new concentrations of them must be
calculated first. If there is reaction between the two solutions such
as lactic acid and sodium hydroxide in this case, the new
concentration should also be determined before the calculation of the
equilibrium.
(1)
(1)
(1)
b)
PV = nRT
PV = m RT
M
mRT
M=
PV

RT
M=
P
(
1
.15 gdm 3 )(8.31JK 1mol 1 )( 298K )
M=
95.3kPa
3
3
1
1
M = (1.15  10 gm )(8.31JK mol )( 298K )
3
2
95.3  10 Nm
M = 29.9g mol-1
Let CnH2n+2 be the formula of compound D.
(1)
(0.5)
(1.5)
P.2
(12.01  n)  (1.008  (2n  2))  29.9
n=2
Therefore, the molecular formula of compound D is C2H6.
between C atoms are restricted.
(0.5, 05)
Therefore, diamond is hard.
In graphite, C-C bonds within each layer are strong covalent bond. (0.5)
However, the weak Van der Waals' force between layers
(0.5)
allow the layers to slip over each other.
(0.5)
Thus, graphite is soft and can be used as lubricant.
(1)
Unit should be given for molar mass while there is no unit for relative
atomic mass.
b)
i)
t1 
2
ln 2
k
1.27  10 
c)
F
i)
(1)
k=5.46x10-10yr-1
N
ln o  kt
Nt
ln 1  9 =(5.46x10-10yr-1)t
1
t = 4.2x109yrs
B
F
ii)
F
(the dotted can also be neglected)
The vacant orbital on boron in BF3
overlaps with the orbital holding the lone pair electrons
on nitrogen in NH3 to form a dative bond.
H
F
B
F
F
(0.5)
(0.5, 0.5)
(0.5)
H
a)
i)
ii)
(1)
(1)
(0.5)
(1.5)
All the radioactive reactions are first order reaction. The following
N
ln 2
equations can be used:
and
t1 
ln o  kt
2
k
Nt
(1)
N
H
ii)
The description of the vacant orbital accept the lone pair electrons can also
be accept.
2.
ln 2
k
In diamond, all the carbon atoms form single covalent bond with four
neighbouring carbon atoms.
(0.5)
However, in graphite each carbon atom form  bond with three other
carbons and the sideway overlap of the unhybridized p-orbital is allowed.
(0.5)
This causes the partial double bond character.
(0.5)
The stronger bond between carbons in the layers of graphite makes the
C-C bond distance shorter.
(0.5)
The carbons between each layers are not covalently bond
(0.5)
The layers are attracted by the weak Van der Waals' forces, so the the
C-C distance between each layer is the longest.
(0.5)
In diamond, all the carbon atoms are tetrahedrally bonded with strong
covalent bond.
(0.5)
This makes it have strong directional character and the relative motion
The assumption is that there is no serious escape of 40Ar from the sample.
(1)
The assumption of constant temperature is not accepted because the
activation energy in radioaction reaction is zero and the rate of the
reaction is unrelated to the temperature.
c)
i)
For hydrogen-oxygen mixture,
H=  242  1.34  104 kJkg1
0.018
For methanol-oxygen mixture,
Hc[CH3OH] = 2Hf[H2O]+Hf[CO2]-Hf[CH3OH]
= 2(-242) + (-394) -(-239)
= -639kJmol-1

639
H=
 7.99  103 kJkg1
(44  2  18)  10 3
(1)
(1)
(1)
P.3
ii)
The effectiveness for the hydrogen-oxygen mixture,
 1.34  104
 744
18
The effectiveness for methanol-oxygen mixture,
 7.99  103
 300
1
(44  2  18)
3
Slope = -Ea/R
Ea = -R (slope) = -(8.31JK-1mol-1)(-11646K)
= 96.8kJmol-1
(1)
Rememeber to label the axis with units.
There is no unit for lnk.
(1)
ii)
Based on the above data, hydrogen is the more effective fuel in powering
rockets.
(1)
3.
a)
i)
1 graph is plotted.
A lnk against
T
k (s-1)
2.46x10-5
4.75x10-4
5.76x10-3
5.48x10-2
lnk
-10.61
-7.65
-5.16
-2.90
T (K)
273
293
313
333
(1.5)
From the graph
at 353K lnk = -0.92
thus k = 0.40s-1 (accept 036 to 0.41s-1)
(1.5)
Calculate the value of k by using two sets of data is not accepted.
1/T (K-1)
366x10-3
341 x10-3
319 x10-3
3.00 x10-3
k = ln 2
t1
t1
iii)
or
3.
b)
i)
2
=1.73s
(1.56 to 1.90s)
(1.5)
2
The progress of the reaction can be monitored by measuring the volume
of carbon dioxide.
(1)
monitored by titrating the concentration of the dicarboxylic acid using a
standard solution of an alkali.
(1)
P.4
ii)
At a higher temperature (T2), more particles have higher kinetic energy
refer to the curves in (i).
(0.5)
Therefore, there would be more particles having sufficient energy higher
than the activation energy.
(0.5)
Thus, the number of effective collision increase and the rate of reaction
increase.
(0.5)
becomes less negative, while that of the cathodic reaction
becomes less positive.
II)
Or
iii)
b)
i)
(1)
The cell is rechargeable because the products of the
electrochemical reactions remain in the system.
(1)
the cathodic and anodic reactions are reversible with no side
products. Therefore, the reverse reaction can occur upon
recharging.
(1)
KD = [CH 3COOH ] H 2O
[CH 3COOH ]alcohol
0.5  x
)
3.05 = (
100
x
(
)
200
x = 0.0198 (accept 0.19 to 0.20)
The efficiency of ethanoic acid extraction
= 0.198  100% =39.6%
(accept 39.0 to 40.0%)
(1)
(1)
(0.5)
0.5
As the partition coefficient is a ratio, no need to convert the concentration
into molarity. e.g.
3.05 = (
In reaction (1), the catalyst provides an alternative reaction pathway with
a lower activation energy for the reaction to occur.
(0.5)
By lowering the kinetic barrier, the rate of reaction increase.
(0.5)
4.
a)
i)
I)
II)
ii)
or
iii)
At anode,
Pb(s) + SO42-(aq)  PbSO4(s) + 2e(0.5)
At cathode,
PbO2(s) + SO42-(aq) + 4H+(aq) + 2e-  PbSO4(s) + 2H2O(l) (0.5)
Pb(s) + PbO2(s) + 2SO42-(aq) + 4H+(aq)  2PbSO4(s) + 2H2O(l)
(0.5)
e.m.f. = +1.69 - (-0.36) = +2.05V
(1.5)
Pb(s)∣PbSO4(s)∣H2SO4(aq)∣PbO2(s)∣PbO2(s) Eo=+2.05V
[Pb(s) + SO42-(aq)]∣PbSO4(s)∣H2SO4(aq)∣[PbO2(s)
+ SO42-(aq) + 4H+(aq)]∣[PbSO4(s) + 2H2O(l)]
Eo=+2.05V
(2)
I)
(1)
During discharge, H2SO4 are removed from the solution.
The reduction potential of the anodic reaction
(2)
ii)
0.50M  0.1dm3  x
)
0.1dm3
x
(
)
0.2dm3
For the first extraction,
0.5  y
)
3.05 = (
100
y
(
)
100
y = 0.123 (accept 0.12 to 0.13)
(1)
For the second extraction,
0.5  0.123  z
)
3.05 = (
100
z
(
)
100
z = 0.093 (accept 0.09 to 0.1)
(1)
P.5
The efficiency of ethanoic acid extraction
= (0.123  0.093)  100%
= 43.3%
0.5
(accept 43.0 to 44.0%)
(0.5)
There is a surplus of 3.7% extraction efficiency by using two 100ml portions of
the alcohol instead of using 200ml in a single extraction.
(0.5)
Therefore, (ii) is a better extraction method.
(0.5)