PRE-U STPM CHEMISTRY TERM 1

Chemistry Term 1 STPM

(iii) A catalyst decreases the activation energy for the 3
(a) (b) (c) (d)
forward and the reverse reactions by the same
amount. As a result, equilibrium is established Cl – OH – C H COO – SO 3 2–
6
5
in a shorter time.

–
4 HCO + H O  H CO + OH –
Energy 3 2 2 3
–

2–
E a = Activation energy HCO + H O  CO + H O +
E a 3 2 3 3
without catalyst
E a ' Quick Check 7.3
E a ' = Activation energy
with catalyst 1
Acid K /mol dm –3 pK
a a
Iodoethanoic acid 6.9  10 –4 3.16
Time –4
However, catalyst does not affect the yield or Nitrous acid 7.2  10 3.14
equilibrium constant for the reaction. It also Cynic acid 3.6  10 –4 3.44
does not alter the enthalpy change of reaction. –10
(b) The forward reaction is exothermic. Thus, it is Boric acid 5.8  10 9.23
favoured by low temperature. However, if the  – +
temperature is too low, the rate of reaction will be 2 (a) CH COOH + H O  CH COO + H O
3
3
2
3
+
–
slow and it takes a longer time for equilibrium to [CH COO ][H ]
3
establish. Thus, a compromise temperature of 450 C (b) K = [CH COOH]
o
a
3
is used where the yield is reasonable and the rate not (c) 1.80  10 mol dm –3
–5
too slow. In order to increase the rate further, finely (d) 4.74
divided iron is added as a catalyst. –6 –3
The forward reaction is accompanied by a decrease in 3 (a) 1.70  10 mol dm

+
the number of moles of gas. Hence, it is favoured by (b) N H + H O  N H + OH –
2
2
5
2
4
high pressure. However, if the pressure is too high, (c) 2.60  10 –3

the cost of production will also be high as the pipes 4 (a) HSO + H O  SO + H O +
–
2–
3
2
3
3
and storage tanks must be strong enough to withstand [H ][SO ]
+
2–
the high pressure. As a compromise, a pressure of (b) K = 3
–
500 atm is used. a [HSO ]
3
(c) 7.21
Chapter 7 Ionic Equilibria and Quick Check 7.4
Solubility Equilibria 1 (a) H CO + H O  H O + HCO –

+
2 3 2 3 3

–
HCO + H O  H O + CO 2–
+
Quick Check 7.1 3 2 3 3
[H ][HCO ]
–
+
(b) K = 3
(a) (b) (c) (d) (e) a, 1 [H CO ]
3
2
+
2–
Cu 2+ SO AlCl H + H + [H ][CO ]
3 3 3
K =
–
a, 2 [HCO ]
Quick Check 7.2 –17 3 2 –6
1 (c) 2.06  10 mol dm

+
Acid Base Conjugate Conjugate 2 (a) H PO  H + H PO 4 –
3
4
2
base acid
+
– 
H PO  H + HPO 2–
(a) NH + H O NH H O + 2 4 4
4 2 3 3 2–  + 3–
HPO  H + PO
(b) H O CN – OH – HCN 4 4
2 [H ][H PO ]
+
–
(c) [Cu(H O) ] 2+ H O [Cu(H O) OH] + H O + (b) K = 2 4
2 6 2 2 5 3 a, 1 [H PO ]
(d) HCOOH HCOOH HCOO – HCOOH + + 3 4 2–
2 [H ][HPO ]
4
K =
(e) H O + C H O – H O C H OH a, 2 –
3 2 5 2 2 5 [H PO ]
2
4
3–
+
(f) H O NH – OH – NH [H ][PO ]
2 2 3 4
K =
a, 3 [HPO ]
2–
4
2 –22 3 –9
(a) (b) (c) (d) (e) (f) (c) 1.99  10 mol dm
C H NH + H CO H SO NH OH + H O + CH CONH +
2 5 3 2 3 2 4 2 2 3 3 3
374
12 Answers.indd 374 3/26/18 4:06 PM

Chemistry Term 1 STPM
Quick Check 7.5 Quick Check 7.10
1 1 (a) 4.84
(b) 4.87
(a) (b) (c) (d) (e)
(c) The change is negligible because the initial solution is
pH = 2.47 pH = 0.097 pH = –0.38 pOH = 1.25 pOH = –0.45
a buffer solution.

+
2 (a) H SO  2H + SO 4 2– 2 (a) 4.69
4
2
–3
(b) 0.50 mol dm (b) (i) 4.74
(c) 0.30 (ii) 4.65
–3
3 (a) 1.0 mol dm
(b) pOH = 0 Quick Check 7.11
–3
–3
4 (a) 1.34  10 mol dm 1
(b) 2.87 Salt BaSO Al(OH) Fe S Ca (PO )

5 (a) NH + H O  NH + OH – 4 3 2 3 3 4 2
+
3 2 4 Solubility/g dm –3 2.44  10 –3 2.26  10 –33 2.04 10 –16 1.22 10 –3
–
–3
(b) [OH ] = 5.1  10 mol dm –3
pOH = 2.29
Salt Silver sulphate Barium carbonate Silver iodide
Quick Check 7.6 K 5.00  10 M 3 8.16  10 M 2 1.50  10 –16 M 2
–7
–9
1 2.93 sp
2 11.1 Quick Check 7.12
–3
–3
3 (a) pH = 3.08; α = 4.12  10 1 3.55  10 g
(b) pH = 4.08; α = 4.12  10 –2 2 1.00  10 mol dm –6
–10
2
(c) Dilution increases both the pH and the degree of 3 No
dissociation.
–6
–3
4 (a) 7.87  10 mol dm Quick Check 7.13
(b) 8.90 1 The oxalate ions react with H to form undissociated
+
5 (a) (i) 2.87 H C O . 4
2
2
2–
(ii) 0.0134 C O + 2H → H C O
+
2
4
2
2
4
(b) (i) 2.96 The sulphate ion does not react with H .
+
(ii) 0.0164
2 BaCO reacts with the dilute HCl in the stomach to form
3
–4
6 (a) 4.30  10 soluble BaCl . 2
(b) 3.70  10 –7 BaCO (s) + 2HCl(aq) → BaCl (aq) + H O(l) + CO (g)
3
–3
7 (a) 3.12  10 mol dm –3 BaSO does not react with HCl. 2 2 2
4
(b) 3.21  10 mol dm –3
–12
(c) 11.49 STPM Practice 7
Objective Questions
Quick Check 7.7 1 A 2 C 3 C 4 A 5 C
1 6 B 7 B 8 D 9 C 10 C
(a) (b) (c) (d) (e) (f) (g)
11 A 12 C 13 D 14 D 15 D
Acid Neutral Acid Base Acid Base Base 16 A 17 A 18 D 19 A 20 B
21 C 22 A 23 B 24 C 25 D
Quick Check 7.8 26 C 27 A 28 C 29 C 30 C
31 D 32 B 33 B 34 C 35 D
1 (a) NaOH + ClCH COOH → ClCH COONa + H O
2
2
2
(b) 0.063 M Structured and Essay Questions
(c) 1.56 1 (a) The solubility product of a sparingly salt is defined
2 (a) 0.143 M as the product of the molar concentration of the
(b) NaOH + HXO → NaXO + H O constituent ions, in a saturated solution, each raised
4 4 2
(c) 0.200 mol dm –3 to the power of the stoichiometric coefficient.
(d) 100.5 (b) (i) Ca(OH) + 2HCl → CaCl + 2H O
2
2
2
(e) A = 35.5; X ≡ Chlorine M × 20 1
r
0.045 × 20.2 = 2
Quick Check 7.9 M = 0.023 mol dm –3
1 (d) and (f) only – –3
[OH ] = 2  0.023 = 0.046 mol dm
2 (a) 4.91 (c) 4.22 (e) 4.86 pOH = –log(0.046) = 1.34
(b) 4.91 (d) 9.67 (f) 4.47 ∴ pH = 14 – 1.34 = 12.66
375
12 Answers.indd 375 3/26/18 4:06 PM

Chemistry Term 1 STPM
–
2+
–
(ii) Ca(OH) (s)   Ca (aq) + 2OH (aq) H + + CH CH COO (aq) : CH CH COOH(aq)
2 (added) 3 2 3 2
2+
– 2
K = [Ca ][OH ] When a little base is added to the mixture, the added
sp
2
–5
3
–
(iii) K = (0.023)(0.046) = 4.87  10 mol dm –9 OH would react with the undissociated
sp
–
–
(iv) Ammonia is a weak base. The [OH ] in aqueous CH CH COOH molecules to form CH COO ions.
3
3
2
ammonia is not sufficient to cause OH – (added) + CH COOH(aq) :
3
precipitation of calcium hydroxide. CH COO (aq) + H O(l)
–
2
3
2.33 × 10 –3
(v) No. of moles of Ca(OH) = As a result, the pH of the buffer solution remains
2 74.1 constant.
–5
= 3.14  10 mol (c) Using the expression for acidic buffer solutions:
Solubility of Ca(OH) = 0.023 mol dm –3 [Propanoic acid]
2
3.14 × 10 –5 pH = pK – log
a
Volume of water needed = [Sodium propanoate]
0.023 0.050
–3
= 1.36  10 dm 3 5.9 = 4.87 – log x
+
[RCOO ][H O ] –3
–
2 (a) (i) K = 3 x = 0.54 mol dm
a [RCOOH] ∴ Mass of sodium propanoate to be added to
(ii) Using: 500 cm propanoic acid
3
ABBBB
+
[H ] = K C 1
a = (0.54 × 74)g
ABBBBBBBBBBBB
–4
= (7.5  10 )(0.18) 2
–2
= 1.16  10 mol dm –3 = 20.0 g
pH = 1.94 4 (a) pK = –log K
a a
(b) (i) A solution whose pH does not change signifi- (b) C H B(OH) + 2H O  C H B(OH) + H O +
–

5
6
2
3
3
5
6
2
cantly on the addition of a little acid or base. C H COOH + H O  C H COO + H O +
–

–
(ii) H + + RCOO → RCOOH 6 5 2 –9 6 5 3
(1.59 × 10 )(0.12)
+
(added) (c) [H ] = ABBBBBBBBBBBBB
–
OH – + RCOOH → RCOO + H O –5 –3
(added) 2 = 1.38  10 mol dm
(iii) Using the formula: pH = 4.86
[Acid] (d) Lower. Because benzoic acid is a stronger acid than
pH = pK – log
a [Salt] phenylboronic acid.
0.18 (e) (i)
= 3.12 – log pH
0.35
= 3.41
(c) On addition of sodium salt which provides the
–
RCOO ions, the following equilibrium will shift to
the left: 8.8 X
–

RCOOH  RCOO + H +
+
The [H ] concentration decreases causing the pH to 4.2 X
increase.
3 (a) Bronsted-Lowry defines an acid as a proton donor. 6 12 16.7 21.4 Volume of NaOH/cm 3
When propanoic acid dissolves in water, the
following equilibrium is set up. (ii) For benzoic acid:
CH CH COOH(aq) + H O(l)  25.0  M = 0.10  12

3 2 2
CH CH COO (aq) + H O (aq) M = 4.80  10 mol dm –3
+
–
–2
3 2 3
In the above equilibrium, the CH CH COOH For phenylboronic acid:
3
2
molecule donates a proton to H O molecule. As a 25.0  M = 0.10  9.4
2
–2
result, CH CH COOH is a Bronsted-Lowry acid. M = 3.76  10 mol dm
–3
2
3
(b) CH CH COOH(aq)  CH CH COO (aq) + H (aq) 5 (a) The ammonium ion is a Bronsted-Lowry acid.
–
+

3 2 3 2
100% NH (aq) + H O(l)  NH (aq) + H O (aq)

+
+
CH CH COONa(s) + aq 9: 4 2 3 3
3 2 (b) The O–H covalent bond in benzoic acid is strong.
+
–
CH CH COO (aq) + Na (aq)
3 2 (c) Formation of a water-soluble complex with
In a mixture of propanoic acid and sodium concentrated HCl.
propanoate, there will be a large amount of –  2–
undissociated CH CH COOH molecules and PbCl (s) + 2Cl (aq)  [PbCl ] (aq)
2
4

3
2
2+
CH CH COO ions. (d) Mg (aq) + 2NH (aq) + 2H O(l) 
–
3
2
3
+
2
When a little acid is added to the mixture, the H + Mg(OH) (s) + 2NH (aq)
4
2
added would react with the CH CH COO ions to Addition of NH Cl shifts the equilibrium to the left
–
4
2
3
form undissociated CH CH COOH molecules. causing the Mg(OH) to dissolve.
3 2 2
376
12 Answers.indd 376 3/26/18 4:06 PM

Chemistry Term 1 STPM

+
6 (a) pH = –log[H ] (b) (i) 50 kPa
(b) pH = 3.10 (ii) 0.58 mole fraction of P
–log[H ] = 3.10 (c) 0.56 mole fraction of P
+
[H ] = 7.94 × 10 mol dm
–4
+
–3
(c) (i) Phenolphthalein Quick Check 8.5
(ii) Let the concentration of HA = M mol dm 1 (a) (i)
–3
22.50 × 0.100 = M × 25.00
46
M = 0.090 mol dm
–3
(d) HA is a weak acid that undergoes partial dissociation l
in aqueous solution. Pressure/kPa
ABBB
(e) Using: [H ] = K C V 35
+
a
(7.94 × 10 ) = 0.090K a
–4 2
K = 7.00 × 10 mol dm
–6
–3
(f) pH a 0 Mole fraction of X 100
(ii)
Temperature/°C Composition l Boiling
7 V
(b)(ii)
point
of
vapour
(b)(i)
V
22.5
50
Chapter 8 Phase Equilibria 0 Mole fraction of X 100

o
2 (a) Temperature/ C
Quick Check 8.1
1 0.479
2 (a) 0.162 (b) 973.7 g 90
V
3 148.5 g
80
Quick Check 8.2 (b)(i)
70
1 37.68 kPa
2 (a) A solution that obeys Raoult’s law 60
(b) (i) 38.56 kPa (b)(ii)
(ii) 34.90 kPa 0 0.2 0.4 0.6 0.8 1.0
(iii) 27.67 kPa Mole fraction of X
Quick Check 8.3 (b) (i) 71.0 °C (ii) 0.72
1 (a) 0.38 mole fraction of C H 6 Quick Check 8.6
6
(b) 0.75 mole fraction of C H
6 6 1 (a) P. P has a lower boiling point than Q.
2 (a) 26.1 kPa (b) (i) 77 °C (ii) Azeotrope (iii) Pure Q
(b) Mole fraction of hexane = 0.684 (c) The mixture with a constant boiling point. The
Mole fraction of heptane = 0.316 composition of the vapour and the liquid mixture
remains the same even on fractional distillation.
Quick Check 8.4
(a) (i) & (ii) 2 (a) Positive. The azeotrope has the lowest boiling point.
(b) The intermolecular forces between water and
Pressure/kPa
ethanol molecules in the mixture are weaker than
those between the molecules in their pure liquids.
(c) (i)
70
(a)(i)
Vapour pressure
60
(b)(i) (a)(ii)
50
40
(c) (b)(ii) 0 95 100
% by mass of ethanol
30
0 Mole fraction of P 1
377


12 Answers.indd 377 3/26/18 4:06 PM

Chemistry Term 1 STPM

(ii) (d)
Temperature/°C 100 78.5 Temperature/°C 126





78 98
0 0.11 95 100 0 Mole fraction of octane 1
% by mass of ethanol
(e) Fractional distillation
(d) (i) Azeotrope 2 (a) A mixture with a constant boiling point. Its
(ii) Water
composition remains constant on distillation.
3 (a) 54.75 kPa (b)
(b) The mixture shows negative deviation. The bonds 122
between the molecules in the mixture are stronger V
than those between molecules in the pure liquids. Temperature/°C V
4 (a) 0.11 100 l l
(b) (i) 87
Vapour pressure 0 30 65 1




(c) (i) Water % HNO 3
(ii) The azeotrope
(d) (i) Raoult’s law states that the partial vapour
0 0.11 1
Mole fraction of HCl pressure of liquid A in a solution is equal to the
product of the vapour pressure of pure A and
the mole fraction of A in the mixture, at the
(ii) same temperature.
Temperature/°C 108.6 (ii) The intermolecular forces in the mixture
(ion-dipole attraction) are stronger than
100
those between molecules of the pure liquids
(hydrogen bonds).
3 (a) Raoult’s law states that the partial vapour pressure of
– 84
liquid A in a solution is equal to the product of the
0 0.11 1 vapour pressure of pure A and the mole fraction of
Mole fraction of HCl
A in the mixture, at the same temperature.
(c) HCl reacts with water to form ions. (b) (i) The intermolecular forces in the mixture (van
+
HCl + H O → H O + Cl – der Waals forces) are of the same type and
2 3
The intermolecular forces in the mixture is stronger strength as those present between molecules of
than those between HCl molecules and between H O the pure liquids.
2
molecules. (ii) (iii)
(d) Hydrogen chloride
STPM Practice 8 Pressure Pressure Vapour
Objective Questions Temperature
1 B 2 A 3 C 4 C 5 D Liquid
6 A 7 D 8 D 9 B 10 D
11 C 12 A 13 C 14 C 15 D
0 1 0 0 1 1
Structured and Essay Questions Mole fraction of benzene Mole fraction of benzene
Mole Fraction of 1-propanol
1 (a) van der Waals forces (iv) (I): Benzene (because it has a lower boiling
(b) A solution that obeys Raoult’s law point)
(c) The intermolecular forces between molecules of
pure liquid (van der Waals forces) are of the same (II): Methylbenzene
nature and strength as those between molecules in
the mixture.




378






12 Answers.indd 378 3/26/18 4:06 PM

Chemistry Term 1 STPM

4 (a) A mixture with a constant boiling point. Its 6 (a) (i) (ii)
composition remains constant on distillation.
(b) Pressure
Pressure Temperature 82


78

0 1 0 1
Mole fraction of ethanol Mole fraction of ethanol
0 % HBr 100
(b) The intermolecular forces between ethanol and
A mixture of water and HBr shows negative cyclohexane molecules in the mixture are weaker
deviation from Raoult’s law because the forces than those between ethanol molecules, and between
holding the H O and HBr molecules together in
2 cyclohexane molecules in the pure liquids. Hence,
the mixture is the strong ion-dipole attraction, as more heat is absorbed to break the bonds and less
compared to the weaker hydrogen bonds between heat is released when new bonds are form. The
water molecules and the van der Waals forces mixing process is endothermic.
between HBr molecules.
(c) 126 7 (a) Positive
(b) Azeotrope is a mixture with constant boiling point.
Temperature (°C) 100 l V (c) Composition: 0.6 mole fraction of P and 0.4 mole
V The composition of the solution and the vapour are
the same at all temperatures.
l
–67
Boiling point: 60 °C
fraction of Q
(d) (i) 77 °C
(ii) First distillate: The azeotrope (60% P and 40%
0 47 60 100
% HBr Q)
Last distillate: Pure Q (100% Q)
(d) (i) HBr (ii) The azeotrope
5 (a)
Temperature/°C STPM Model Paper (962/1)
82
80 SECTION A
78 1 C 2 D 3 D 4 C 5 A
6 D 7 B 8 D 9 D 10 B
76
11 A 12 C 13 C 14 C 15 C
74 l V
V
72 (d)(i) SECTION B
l
70
16 (a) NH + H : NH 4 +
+
3
+
68 NH : H + NH –
3 2
66 (b) + +
(d)(ii) H H – –
0 0.2 0.4 0.6 0.8 1.0
Mole fraction of X H N H H N H H N H N
(b) Positive deviation. (Because the boiling point/ H H H H
composition curve shows a minimum).
+
+
(c) Boiling point = 67.0 °C (c) NH . The N atom in NH is surrounded by 4
4
4
–
Azeotrope = 0.72 mole fraction of X bonding pairs while in the NH , it surrounded by 2
2
(d) (i) 71.4 °C lone-pairs and 2 bond pairs. The repulsion between
(ii) 0.54 mole fraction of X lone pairs is stronger thus pushing the N9H bonds
–
(iii) The azeotrope in NH closer to one another.
2
(iv) Pure X 17 (a) pH = –log[H ]
+
(b) (i) pH = 3.1
∴ [H ] = 7.9 × 10 mol dm –3
–4
+
[HCl] = 7.9 × 10 mol dm –2
–4
379
12 Answers.indd 379 3/26/18 4:06 PM

Chemistry Term 1 STPM
(ii) Using the expression: (b) The order of reaction with respect to a given substance
[H ] = K C is the power to which the molar concentration of the
+
b
(7.9 × 10 ) = (1.8 × 10 )C substance is raised in an experimentally determined
–5
–4 2
C = 3.5 × 10 mol dm –3 rate equation.
–2
[CH COOH] = 3.5 × 10 mol dm –3 (c) (i) Rate = k[NO Cl]
–2
3
(c) CH COOH is a weak acid that dissociates partially in (ii) Mechanism: 2
3
water. As a result, more of it must be present to NO Cl : NO + Cl Slow (Rate-determining
provide the same H concentration. 2 2 step)
+
(d) (i) H9C≡C9H + NH : H9C≡C + NH 3
–
–
2
(ii) In the above reaction, ethyne acts a proton NO Cl + Cl : NO + Cl Fast
2
2
2
donor (Bronsted-Lowry acid) while the NH 2 – (iii) During the reaction, the colour of the mixture
ion acts as a proton acceptor (Bronsted-Lowry changes from yellow (NO Cl) to brown (NO ).
2
2
base). 19 (a) A buffer solution is a solution that resists changes in
pH on the addition of a little acid or base.

–
+
SECTION C (b) CH COOH(aq)  CH COO (aq) + H (aq)
3
3
CH COONa(aq) : CH COO (aq) + Na (aq)
–
+
18 (a) (i) Change in temperature: 3 3
The Boltzmann distribution curve for a Due to common ion effect, the mixture would
chemical reaction at two different temperatures contain a large amount/resoviour of undissociated
–
is shown below: CH COOH molecules and the CH COO ions.
3
3
On the addition of a little acid, the added H would
+
Fraction of molecules react with the CH COO ions to form undisoociated
–
T >T 1 2 with energy, E ≥ E at T 2 CH COOH molecules according to the equation:
3
1
a
Fraction of molecules T 2 a 1 On the addition of a little base, the added OH would –
Fraction of molecules
3
T
with energy, E ≥ E at T
H
+ CH COO (aq) : CH COOH(aq)
+
–
3
3
(added)
–
react with CH COOH molecules to form CH COO
3
3
ions according to the equation:
OH
–
(addad) + CH COOH(aq) :
3
CH COO (aq) + H O(l)
–
E Energy 3 2
–
+
a In both cases, the added H or OH are ‘destroyed’
For a reaction to occur, the reactant molecules and the pH of the buffer solution remains unchanged.
must collide with one another with energy (c) Using the expression for a buffer solution:
greater than the activation energy E . When the [CH COOH]
a
3
temperature increases (T > T ), the fraction of pH = pK – [CH COO ]
a
–
2
1
molecules having energy greater than the 3
activation energy increases and also the Since the molarity of CH COOH and CH COONa
3
3
collisions are now more energetic. This results are the same and the volume of the mixtures is
3
in an increase in the rate of reaction. The constant (total = 50.0 cm ), the concentration of the
reverse is true when the temperature is lowered, species can be equated to their respective volumes in
and the rate of reaction decreases. the mixture. Hence,
(ii) Presence of a catalyst: pH = pK – [V(CH COOH)]
3
A catalyst provides an alternative pathway, with a [V(CH COONa)]
lower activation energy for the reaction to Rearranging: 3
occur.
[V(CH COOH)]
3
pH = –log [V(CH COONa)] + pK a
Fraction of molecules with catalyst Hence, a graph of pH against log [V(CH COONa)]
3
Activation energy
[V(CH COOH)]
3
Activation energy
3
without catalyst
would give a straight line with a negetive slope, and
the intersect at the pH axis gives the value of pK for
Energy ethanoic acid. a
pH 3.92 4.28 4.70
As can be seen from the Boltzmann distribution [Methanoic acid]
curve above, in the presence of a catalyst, more log [Sodium methanoate] 0.95 0.6 0.18
molecules have energy greater than the new
(lower) activation energy. This would result in
more ‘effective collision’ leading to an increase
in the rate of reaction.
380
12 Answers.indd 380 3/26/18 4:06 PM

Chemistry Term 1 STPM

pH Since the ionic quotient is larger than the
solubility product (2.00 × 10 ). Precipitation
–13
will occur.
6.0
20 (a) (i)
5.0
4.5
H N H H N H H N H H H H
4.0 O C O O C O O C O CI CI CI
H H H
3.5
3
(ii) NH : 1 lone pair + 3 bond pair.
3
Shape: trigonal pyramidal
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
log [CH 3 COOH] N
[CH 3 COOH – ]
[CH COOH] H H
When log 3 = 0
–
[CH COO ]
3 H
pH = 4.85 CO : 2 double bond pair.
∴ pK of ethanoic acid = 4.85 Shape: Linear
2
a
–log K = 4.85 O C O
a
∴ K = 1.42 × 10 mol dm –3 HCl: 1 bond pair + 3 lone pair
–5
a
(d) (i) Solubility product of a sparingly soluble salt is Shape: Linear
the products of the molar concentration H CI
aqueous ion (raised to their respective
stoichiometric coefficient) produced in a (iii) All three molecules exist as simple covalent
saturated solution of the salt at constant molecules with covalent bonds holding the
temperature. atoms together in the molecules.
(ii) The common ion effect refers to the reduction The intermolecular forces between CO and
2
in the solubility of a sparingly soluble when a HCl molecules are the weak van der Waals
soluble compound containing one of the ions of forces. However, the van der Waal’s force in
the salt is added to the saturated solution. CO is stronger because the molecule has more
2
Silver chloride is sparingly soluble in water. In a electrons (22) than HCl (18). This accounts for
saturated solution of silver chloride, the the higher boiling point of CO compared with
2
following equilibrium is established. HCl.
AgCl(s) + aq  Ag (aq) + Cl (aq) Although NH has only 10 electrons, the
+
–

3
If a little Cl ions (e.g. from dilute HCl) is added intermolecular force is the stronger hydrogen
–
to the saturated solution, then according to Le bond that needs more energy to break. This
Chatelier’s Principle, the equilibrium would accounts for NH having the highest boiling
3
shift to the left-hand side and as a result, less point.
silver chloride will dissolved. The same effect is (b) (i) Hydrogen bond.
observed when a little Ag (aq) ions (e.g. from
+
AgNO ). Thus the addition of either Ag or Cl – H CH OH
+
2
3
(the common ions) would decrease the solubility O
of silver chloride in water. O H H H
(iii) Mn (aq) + 2OH (aq) 9: Mn(OH) (s) H
–
2+
2
K of Mn(OH) = [Mn (aq)][OH (aq)] 2 OH H
2+
–
sp 2 HO OH
In the mixture:
1 H OH
[MnSO ] = (0.020)
4
2 (ii) Let the number of water molecules bonded to
= 0.010 M glucose in honey be x.
1 Molecular formula: C H O xH O
[NH ] = (0.050) = 0.025 M 6 12 6 2
3
2 % by weight of H O in honey
2
To calculate the OH of the NH solution: 18x
–
3 = × 100%
[OH ] = K C 180 + 18x
–
b
= (1.80 × 10 )(0.025) = 28.6 %
–5
= 6.71 × 10 M ∴ x = 4
–4
The number of H O molecules bonded to each
The ionic quotient of [Mn ][OH ] glucose molecule in honey is 4.
2
2+
– 2
= (0.010)(6.71 × 10 )
–4 2
= 4.50 × 10 mol dm –9
3
–9
381
12 Answers.indd 381 3/26/18 4:06 PM