ANSWERS TO EXERCISES AND REVISION QUESTIONSequals its initial concentration, and applying the
equilibrium law,Ka [HCO^3- (aq)][H 3 O+(aq)]
[CO 2 (aq)]
If the only acidity is due to dissolved carbon dioxide,
[HCO 3 – (aq)][H 3 O+(aq)], giving
Ka [H^3 O+(aq)] 2
[CO 2 (aq)]
(ii)[H 3 O+(aq)] (4.3 10 –^7 4.0 10 –^2 )
1.3 10 –^4 mol dm–^3
(corresponding to 0.3% ionization).
pHlog (1.3 10 –^4 )3.89
16.8 (i)WB–SA, therefore acidic (ii)WA–WB,KaKb
so approximately neutral (iii)WB–SA, therefore acidic
(iv)SA–SB, therefore neutral (v)SB–WA, so basic.
16.9M(C 6 H 5 COOH)122 g mol–^1 ,
M(C 6 H 5 COONa)144 g mol–^1
Number of moles of C 6 H 5 COOH0.100/122
8.20 10 –^4.
Number of moles of C 6 H 5 COONa0.080/144
5.6 10 –^4.
Concentration of (C 6 H 5 COOH ) 8.2 10 –^4 / 0.100
8.2 10 –^3 mol dm–^3.
Concentration of (C 6 H 5 COONa ) 5.6 10 –^4 / 0.100
5.6 10 –^3 mol dm–^3.[H 3 O+(aq) CAKa(T) 8.20^10– (^3) 6.5 10 – 5
Cs 5.6 10 –^3
9.5 10 –^5 mol dm–^3
pHlog (9.5 10 –^5 )4.02
C 6 H 5 COOH(aq)H 2 O(l) C 6 H 5 COO–(aq)H 3 O+(aq)
Explanation of buffer action is as follows. (a) Addition of
H 3 O+(aq) causes association of the C 6 H 5 COO–(aq) and
H 3 O+(aq), so restoring [H 3 O+(aq)]. (b) Addition of OH–(aq)
neutralizes H 3 O+(aq), and more benzoic acid ionizes to
compensate.
16.10Choice of indicators from Table 16.4 for a weak
base–strong acid titration are bromocresol green and
methyl red. The remaining indicators begin to change
colour before or after the pH at the equivalence point
( pH 4–8).
16.11The ingested hydrogencarbonate increases the
buffer capacity of the athlete.
Unit 17
Exercises
17A
(i)
(ii)
(iii)C 12 H 26
17B
17C
(i) (ii)
(iii)
17D
(i)2-methylpropane
(ii)4-ethylheptane
(iii)2,2-dimethylpropane
(iv)4-ethyl-3,5-dimethyloctane
17E
(i)
(ii)
(iii)
(iv)
17F
(i)n-butane (ii)399 K
17G
(i)CnH 2 n
(ii)The ring is far less ‘strained’ – the bond angle is
larger than in cyclopropane.
(iii)
17H
(i)670, 660, 660, 620, 660; yes, usually around
660 kJ mol–^1
(ii)654 kJ mol–^1
(iii)–CH 2
(iv)4814 kJ mol–^1.
17I
(i)See Box 17.6 (ii)No, unlike ethane.
17J
(i)hex-1-ene
(ii)but-2-ene
(iii)4,4-dimethylpent-2-ene
(iv)3-bromoprop-1-ene
17K
(i) (ii)
(iii) (iv)
(v)
17L
(i)
(ii)geometric (iii)structural
(iv) 1istrans-1,2-dibromoethene, 2 iscis-1,2-
dibromoethene, 3 is 1,1-dibromoethene.
17M
(i)CH 2 CH 2 HBrCH 3 – CH 2 Br (bromoethane)
(ii)CH 2 CH 2 H 2 SO 4 CH 3 – CH 2 – OSO 3 H (ethyl
hydrogensulfate)
(iii)CH 3 – CHCH 2 Br 2 CH 3 – CHBr–CH 2 Br (1,2-
dibromopropane)
(iv)CH 3 – CHOH–CH 2 OH
(v)n(CH 2 CHCN)···CH 2 – CHCN–CH 2 – CHCN–CH 2 –
CHCN–CH 2 – CHCN–CH 2 – CHCN···.
17N
(i)CH 3 – CH 2 CCH
(ii)propene, CH 3 – CHCH 2 and propane, CH 3 CH 2 CH 3
(iii)CHBr 2 – CHBr 2 1,1,2,2-tetrabromoethane.
17O
(i)
mass of C present 21.99^1244 6.00 mg
percentage C 7.02^6 100 85.5%
mass of H present 8.95^2
18 1 mg
percentage H 7.02^1 100 14.3%
Empirical formula:
CH
no of moles 85.5 14.3 12 1
ratio 7.1 : 14.3
1 : 2
Therefore, the empirical formula is CH 2.
(ii)molecular formula X C 3 H 6 , structure CH 3 – CHCH 2
Y is CH 3 CH 2 CH 3
Z is CH 3 CHBrCH 2 Br
17P
(i)
(ii) 3 119 357 kJ mol–^1
(iii)Benzene has a lower enthalpy of formation than that
suggested by the Kekulé structure.
- 3 H 2
3 = C
Br
BrH
H
H
C
CH 3 CH 2 CH 2 CHBrCH CH 2
CH 2 ClC CH 2
CH 3
CH 3 CH 2 CH 2 C CCH 3
CH 3 CH 3
CH 3 CCH 2
CH 3 CH 2 CH CHCH 2 CH 3 CH 3
C 6 H 12
Cyclohexane
CHCH 22
CH 2
CH 2
H 2 C CH 2
H 2 C
C 6 H 14
Hexane
CH 3 CH 2 CH 2 CH 2 CH 2 CH 3
Boiling point / K
Number of carbon atoms
0
100
300
600
510
HH
H
C
H
HH
H
HCC
C
H
HH
CHH
H
C C
HH
H
C
H
H
C
H
H
CH
HH
H
C
H
HH
HCC
C
H
HH
CHH
H
HH
CC
H
H
C
H
H
C
H
H
CH
H H
H
C
C
H
H
HH
H
HCC
C
H
HH
HH
C
H
H
C
H
C
H
H
C
H
H
C
H
H
CH
HH HH
H
C
H
H
HH
H
HHCC C
H
H
C C
H H
H
C
C
H
H
HH
H
HHCC
HH
C
HHHH
H
C
H
H
HH
H
HHCCCC
HHHHH
HHHHH
HHCCCCC
C C
H H
H H
H H
HHHHHH H
HHHHHH H
C
C 8 H 18
HHCCCCC
H
H
CC
H
H
HHC
H
H
C
H
H
HHC
H
H
C
H
H
C
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