-
TOPIC 19 ANSWERS & MARK SCHEMESA2 Level
a) pH = (1)
b) Water ionises slightly as follows: H2O(l) ž H+(aq) + OH-(aq)
(1)
Since the ionisation is very slight, we regard [H2O] as constant
(1)
Kw = [H+(aq)] [OH-(aq)] (1)
c) mol2 dm-6 (1)
d) (i) Since for water [H+] = [OH-]; Kw = [H+]2 = 5.14 x 10-14
mol2 dm-6 (1)
∴ [H+] = √(5.14 x 10-14) = 2.267 x 10-7 mol dm-3 (1)
(ii) pH = -lg [H+] = -lg (2.267 x 10-7) = 6.64 (1)
e) Endothermic (1)At 373 K, K
w is greater than at 298 K ∴ water is more dissociated (1)
Le Chatelier’s Principle implies that endothermic reactions are
favoured by increases in temperature (1)∴ the dissociation of water
is endothermic (1)
f) (i) pH = 7.47 ∴ [H+] = - antilg 7.47 = 3.388 x 10-8 mol dm-3
(1)[H+] = [OH-]∴ K
w = [H+]2 = 1.15 x 10-15 mol2 dm-6 (1)
(ii) Yes (1)[H+] = [OH-] so water is neutral even though values
have changed (1)
pH AND KW
- lg [H+] / lg 1
[H+(aq)][OH-(aq)]
[H2O(l)]
Kc = (1)
[H+]
QUESTIONSHEET 1
-
TOPIC 19 ANSWERS & MARK SCHEMESA2 Level
pH OF STRONG ACUDS AND BASES
QUESTIONSHEET 2
a) (i) 1.00 (1)
(ii) 2.40 (1)
(iii) 0.40 (1)
(iv) [H+] = 1 × 10-14 / 0.005 = 2.0 × 10-12 (1) pH. = 11.7
(1)
(v) [H+] = 1 ×10-14 / 0.06 = 1.67 × 10-13 (1) pH. = 12.8 (1)
b) (i) [H+] = 0.200 (1)
(ii) pH = 0.700 (1)
c) (i) [H+] = 0.0398 mol dm-3 (1); [HCl] = 0.0398 mol dm-3
(1)
(ii) [H+] = 2.00 × 10-3 mol dm-3 (1); [H2SO
4] = 1.00 × 10-3 mol dm-3 (1)
(iii) [H+] = 5.01 × 10-12 mol dm-3 (1); [NaOH] = 2.00 × 10-3 mol
dm-3 (1)
-
TOPIC 19 ANSWERS & MARK SCHEMESA2 Level
STRENGTHS OF ACIDS (Ka AND pKa)
a) (i) One which is only partially dissociated in solution
(1)
(ii)
(iii) Since [H+] = [C2H
5COO-],
[H+]2 = Ka . [C
2H
5COOH] (1)
As there is only slight dissociation, assume that [C2H
5COOH] at eqm ≈ original concentation (c) of acid (1)
∴[H+] = √(Ka . c)
= √ (1.35 × 10-5 × 0.2) = 1.64 x 10-3 mol dm-3 (1)∴ pH = - lg
(1.64 × 10-3) = 2.79 (1)
b) (i) pKa = - lg K
a where K
a is the acid dissociation constant (1)
(ii) HS: Ka = 2.51 × 10-4 mol dm-3 (1)
HT: Ka = 7.94 × 10-5 mol dm-3 (1)
(iii) HS (1)because it has the larger K
a / smaller pK
a (1)
c) (i)
(ii) pKa = -lg K
a = -lg (5.62 × 10-10) = 9.25 (1)
d) Butanoic < 4-chlorobutanoic < 3-chlorobutanoic <
2-chlorobutanoic (1)
[C2H
5COO- (aq)] [H+(aq)]
Ka =
[C2H
5COOH(aq)]
(1)
[NH3(aq)] [H+(aq)]
Ka =
[NH4
+(aq)](1)
Delete (1) for missing or incorrect units
QUESTIONSHEET 3
-
TOPIC 19 ANSWERS & MARK SCHEMESA2 Level
pH OF ACID OR BASE SOLUTIONS
a) (i) HNO3(aq) → H+(aq) + NO
3-(aq)
[H+(aq)] = [NO3-(aq)] = 0.2 mol dm-3 (1)
pH = -lg [H+(aq)] = -lg 0.2 = 0.699 i.e. 0.70 (1)
(ii) C3H
7COOH(aq) ¾ C
3H
7COO-(aq) + H+(aq)
pKa = 4.82 ⇒ K
a = 1.51 × 10-5 mol dm-3 (1)
Ka = [H+]2 / [C
3H
7COOH] ⇒ [H+] = √(K
a . c) (1)
[H+] = √(1.51 × 10-5 × 0.2) = 1.738 × 10-3 mol dm-3 (1)pH = -lg
[H+] = -lg (1.738 × 10-3) = 2.76 (1)
ExplanationNitric acid is a strong acid / completely dissociated
into ions in solution (1)Butanoic acid is a weak acid / only
partially dissociated and thus its [H+] is lower (1)
b) (i) n (H+) = 10(10-3)1.0 = 10-2 mol (1)[H+] = 10-2 mol
dm-3
∴ pH = 2.0 (1)
(ii) n (OH-) = 20(10-3)1.0 = 2 × 10-2 mol (1)[OH-] = 2 × 10-2
mol dm-3[H+] = 1 × 10-14/ 2 × 10-2 = 5 × 10-13 mol dm-3 (1)pH =
12.3 (1)
c) (i) HCOOH(aq) ¾ H+(aq) + HCOO-(aq)
(ii) If pH = 4.0, [H+] = 10-4 mol dm-3 = [HCOO-] (1)[HCOOH] =
(10-4)2 / (1.78 × 10-4) = 5.62 ××××× 10-5 mol dm-3 (1)
d) pH = 3.0 so [H+] = 1 × 10-3 mol dm-3 = [Q-] (1)0.2 mol in 500
cm3 ò 0.4 mol dm-3 = [HQ] (1)
[H+(aq)] [HCOO-(aq)]K
a =
[HCOOH(aq)](1)
QUESTIONSHEET 4
= (1 × 10-3)2/0.4 = 4 ××××× 10-5 mol dm-3 (1)[H+(aq)]
[Q-(aq)]
Ka =
[HQ(aq)]
-
TOPIC 19 ANSWERS & MARK SCHEMESA2 Level
pH OF MIXED SOLUTIONS OF ACID AND BASE
a) (i) HCl → H+ + Cl- ∴ [H+] = 0.05 mol dm-3 (1)
(ii) pKa = 4.76 ∴ K
a = 1.738 x 10-5 mol dm-3 (1)
[H+] = √(Ka . c ) = √(1.738 x 10-5 x 0.05) = 9.32 x 10-4 mol
dm-3 (1)
(iii) n (HCl) = 0.1(10-3)(100) = 1 x 10-2 mol (1)n (KOH) =
0.15(10-3)(50) = 0.75 x 10-2 mol (1)∴ n (H+) = 0.25 x 10-2 mol in
(100 + 50) = 150 cm3 (1)∴ [H+] = (0.25 x 10-2)(103) / (150) =
0.0167 mol dm-3 (1)
b) (i) n (H+) = 2(25.0)(10-3)(0.22) = 0.011 = 11.0 x 10-3 mol
(1)
(ii) n (OH-) = 20.0(10-3)(0.25) = 5.0 x 10-3 mol (1)
(iii) n (H+)used = n (OH-)added = 5.0 x 10
-3 mol (1)∴ n (H+)
unreacted = (11.0 – 5.0)10-3 = 6.0 x 10-3 mol (1)
(iv) Total volume = 20.0 + 25.0 = 45.0 cm3 (1)∴ [H+] =
(6.0)(10-3)(103) / 45.0 = 0.1333 mol dm-3 (1)∴ pH = -lg (0.1333) =
0.875 (1)
QUESTIONSHEET 5
-
TOPIC 19 ANSWERS & MARK SCHEMESA2 Level
BRØNSTED – LOWRY THEORY
a) (i) Acid A species that can donate a proton to a base (1)
Base A species (with a lone pair of electrons) that can accept a
proton from an acid (1)
(ii) Conjugate acid The species formed when a base has accepted
a proton from an acid (1)
Conjugate base The species left when an acid has donated a
proton to a base (1)
Two or more sentences with correct spelling, punctuation and
grammar in which the meaning is clear.
b) Acid Conjugate acid(i) CH
3COOH H
2O (1)
(ii) H2SO
4H
2NO
3+ (1)
(iii) HSO4
- H2O (1)
c) (i) Ethylamine has a lone pair of electrons on the nitrogen
atom of the amino group (1)which can accept a proton from an acid
(1)
(ii) CH3CH
2NH
3+ (1)
(iii) Either pH = 11.5 so pOH = 14 -11.5 = 2.5 (1)[OH-] =
-antilg 2.5 = 3.16 x 10-3 mol dm-3 (1)
Or Or pH = 11.5 so [H+] = 3.162 × 10-12 mol dm-3 (1)K
w = [H+] [OH-] = 10-14 mol2 dm-6
[OH-] = 10-14 / 3.162 × 10-12 = 3.16 ××××× 10-3 mol dm-3 (1)
d) As an acid: HSO4-(aq) + H
2O(l) ¾ H
3O+(aq) + SO
42-(aq) (1)
Or HSO4-(aq) + OH-(aq) ¾ H
2O(l) + SO
42-(aq) (1)
As a base: HSO4-(aq) + H+(aq) ¾ H
2SO
4(aq) (1)
e) Base Br− (1)Conjugate base H
2PO
4-(1)
QUESTIONSHEET 6
-
TOPIC 19 ANSWERS & MARK SCHEMESA2 Level
THEORY OF INDICATORS
a) Soluble in water (1)Two distinct colours in acid and base
forms (1)Intensely coloured so as to be seen when only a few drops
are used (1)
b) (i) HIn(aq) ž H+(aq) + In-(aq) (1)
(ii) The pH range over which the colour changes (1)
(c) (Approximately) pH 4.0 – 6.0 (i.e. pKIn ± 1) (1)
c) (i) Thymolphthalein (1)Titration of a weak acid with a strong
base (1)End-point likely to be in region pH 9-10 i.e. ≈ pK
in (1)
(ii) Methyl orange (1)Titration of a weak base with a strong
acid (1)End-point likely to be in region pH 3-5 i.e. ≈ pK
In (1)
(iii) No indicator is suitable (1)Titration of weak acid with
weak base (1)No sharp end-point / pH changes gradually in the
region of the equivalence point (1)
[H+(aq)] [In-(aq)]K
in =
[HIn(aq)](1)
QUESTIONSHEET 7
-
TOPIC 19 ANSWERS & MARK SCHEMESA2 Level
14
12
10
8
6
(5) 4.87
4(3) 2.93
2
010 20 30 40
7
Vol NaOH added
pHpH
NaOH curve: start at pH 13(1) shape (1) ends > pH 1 (1)NH
3 curve: start ≈ pH 11 (1) shape (1) ends > pH 1 (1)
end-points the same (1) buffer region correct (1)mid-point
buffer region at pH approximately 9 (1)axes labelled correctly
(1)Divide by 2, round up /5 total
b) (i) n (OH-) = 0.2(18.0)10-3 = 3.6 x 10-3 mol∴ n (HA) = 3.6 x
10-3 mol in 25.0 cm3 (1)[HA] = (3.6)(10-3)(103) / 25.0 = 0.144 mol
dm-3 (1)
(ii) When HA has been half converted to Na+ A-, [HA] = [A-] so
that Ka = [H+] (1)
∴ pKa = pH when half the neutralising alkali has been added =
3.58
∴ Ka = - antilg 3.58 = 2.63 x 10-4 mol dm-3 (1)
(iii) [H+] = √(Ka . c)
= √(2.63 x 10-4 x 0.144)= 6.15 x 10-3 mol dm-3 (1)
∴ pH = -lg (6.15 x 10-3) = 2.21 (1)
c)
TITRATION CURVES
a) In original acid, [H+] = √(Ka . c) = √(1.35 x 10-5 x 0.1) =
1.162 x 10-3 mol dm-3
hence original pH = -lg (1.162 x 10-3) = 2.93 (1 – if shown on
graph)Since solutions are equimolar, end-point will be at 20.0 cm3
alkali added (1 – if shown on graph)pH at the end-point ≈ 9 (1- if
shown on graph)
14
Shape of graph (2)
Volume 0.1 M HNO3 added
0.1 M NH3
0.1 M NaOH
○
○
○
○
○
○
○
○
○
○
○
Equivalence point
/cm3
QUESTIONSHEET 8
Buffer range
0
7
14
pH
-
TOPIC 19 ANSWERS & MARK SCHEMESA2 Level
ACID-BASE STRENGTH AND ENTHALPY OF NEUTRALISATION
a) (i) The enthalpy change / enthalpy released when 1 mol water
is formed (1)by the reaction between acid and base in dilute
aqueous solution (1)in the mole ratio required by the equation
(1)
(ii) Chemical equations½ H
2SO
4(aq) + NaOH(aq) → ½ Na
2SO
4(aq) + H
2O(l) ; ∆H = -57 kJ mol-1 (1)
CH3COOH(aq) + NaOH(aq) → CH
3COONa (aq) + H
2O(l); ∆H = -54 kJ mol-1 (1)
Explanation of differenceThe essential reaction is H+(aq) +
OH-(aq) → H
2O(l) (1)
Ethanoic acid is a weak acid / a solution contains relatively
few H+(aq) ions (1)Ionisation requires the breaking of covalent
bonds which is an endothermic process (1)This reduces the overall
energy released in the neutralisation (1)Maximum 3 marks
b) (i) As volumes / concentrations are the same, and HCN is also
a monoprotic acid∆H = -57.8(1.0) / 10.8 = -5.35 kJ mol-1 (1)
(ii) HCN is a very weak acid / the H_C bond is strong (1)and
ionisation thus requires a considerable amount of energy (1)NH
3 is a weak base (1)
and its ionisation also requires energy (1)These two factors
reduce the enthalpy of neutralisation by some 52 kJ mol-1
(1)Maximum 4 marks
(iii) As sodium hydroxide is a strong base, no energy will be
required for its dissociation (1)The result obtained will lie
between the other two values (1)
QUESTIONSHEET 9
-
TOPIC 19 ANSWERS & MARK SCHEMESA2 Level
BUFFER SOLUTIONS
a) (i) A buffer solution is one whose pH changes only slightly
(1)on the addition of moderate quantities of acid or base (1)
(ii) Acid buffer Ethanoic acid and sodium ethanoate (1)(or any
other weak acid and its salt with a strong base)
Alkaline buffer Aqueous ammonia and ammonium chloride (1)(or any
other weak base and its salt with a strong acid)
Biological buffer Carbonic acid and sodium hydrogencarbonate in
blood (1)(or any amino acid, or specified correct protein)
b) (i) The quantity of H+/OH- ions it can absorb without a
significant change in its pH (1)
(ii) The concentration of the buffer / the amounts of reactants
available to counteract the addition of acid or base (1)The ratio
of acid (or base) to its salt (1)
c) (i) Reaction with H+ ions H+ (aq) + CH3COO- (aq) → CH
3COOH (aq) (1)
Reaction with OH- ions OH- (aq) + CH3COOH (aq) → CH
3COO- + H
2O (l) (1)
(ii) Species present H3N+ CH
2COOH(aq) (1) and H
3N+ CH
2COO- (aq) (1)
Reaction with H+ ions H3N+ CH
2COO- (aq) + H+ (aq) → H
3N+ CH
2COOH (aq) (1)
Reaction with OH- ions H3N+ CH
2COOH (aq) + OH- (aq) → H
3N+ CH
2COO- (aq) + H
2O (l) (1)
d) Initially, since ethanoic acid is a weak acid, there is a
relatively low concentration of hydrogen ions (1)As these react
with the alkali, pH changes rapidly (1)At the same time, however,
sodium ethanoate is formed (1)Once a moderate amount of this is
present, the solution acts as a buffer (1)until such time as its
buffering capacity is exhausted (1)which is near to the end-point
of the titration (1)Maximum 5 marks
QUESTIONSHEET 10
-
TOPIC 19 ANSWERS & MARK SCHEMESA2 Level
pH OF BUFFER SOLUTIONS
a) (i)
[H+(aq)] = 1.70 × 10-5 × 0.1 / 0.2 = 8.5 × 10-6 (1)pH = 5.07
(1)
(ii)
[H+(aq)] = 1.35 × 10-5 × 0.1 / 0.2 = 6.75 × 10-4 (1)pH = 3.17
(1)
(iii)
[H+(aq)] = 1.35 × 10-3 × 0.01 / 0.2 = 6.75 × 10-5 (1)pH = 4.17
(1)
b) (i)
(ii) Since equal volumes are mixed, the equilibrium
concentration of sodium propanoate will be half its
originalconcentration, i.e. 0.100 mol dm-3 (1)
EitherK
a (CH
3CH
2COOH) = - antilg 4.87 = 1.349 × 10-5 mol dm-3 (1)
From b) (i), [acid] in buffer = [salt] [H+]/Ka (1)
= 0.1 × 10-5/1.349 × 10-5 = 0.074 mol dm-3 (1)
OrpH = pKa + lg([salt]/[acid]) (1)5.0 = 4.87 + lg (0.1/[acid])
(1)0.13 = lg (0.1/[acid])1.349 = 0.1/[acid][acid] in buffer =
0.1/1.349 = 0.074 mol dm-3 (1)
So originally the concentration of propanoic acid was 2 × 0.074
= 0.148 mol dm-3 (1)
QUESTIONSHEET 11
[CH3COO-(aq)] [H+(aq)]
Ka =
[CH3COOH(aq)]
(1)
[BrCH2COO-(aq)] [H+(aq)]
Ka =
[BrCH2COOH(aq)]
(1)
[BrCH2COO-(aq)] [H+(aq)]
Ka =
[BrCH2COOH(aq)]
(1)
[CH3CH
2COO-(aq)] [H+(aq)]
Ka =
[CH3CH
2COOH(aq)]
(1)
-
TOPIC 19 ANSWERS & MARK SCHEMESA2 Level
TEST QUESTION I
a) S 26.89% O 13.45% Cl 59.66%
Moles % = 26.89/32 13.45/16 59.66/35.5 (1)= 0.8403 0.8406
1.6806
Ratio: 1 1 2 (1)
Empirical formula is SOCl2 (1)
b) Ag+(aq) + Cl-(aq) → AgCl(s)n (Cl-) = n (Ag+) = 25.0(10-3)0.1
= 2.5 x 10-3 mol Cl- in 25.0 cm3 (1)∴ c (Cl-) = 2.5(10-3)(103) /
25.0 = 0.100 mol dm-3 (1)i.e. 2 mol Cl- per mol compound, as per
empirical formula.Molecular formula is also SOCl2(1)
c) The white precipitate is barium sulfite / BaSO3 (1)since it
dissolves in acid to release SO
2 (1)
To confirm: n (SOCl2) in 25.0 cm3 = 0.05 (25.0)/103 = 1.25 x
10-3 mol
This should give 1.25 x 10-3 mol BaSO3 ⇒ (1.25 x 10-3) 217 =
0.271 g BaSO
3 (1)
The weak acid HQ is therefore sulfurous acid / H2SO3 (1)and the
strong acid HP is hydrochloric acid / HCl (1)
Equation: SOCl2(l) + 2H
2O(l) → H
2SO
3(aq) + 2HCl(aq) (1)
d) (i) pH = 0.975 ⇒ [H+] = 0.10593 mol dm-3 (1)[NB: a
subtraction is to follow, which will reduce the number of
significant figures and cause inaccuracy in thecalculation if
“rounding up” is carried out here]
(ii) Since 0.05 mol SOCl2 gives 0.10 mol H+ from HCl (1)
then (0.10593 – 0.10) = 0.00593 mol H+ from H2SO3 (1)
(iii) Percentage from weak acid = 0.00593(100) / 0.10593 = 5.60%
(1)
QUESTIONSHEET 12
-
TOPIC 19 ANSWERS & MARK SCHEMESA2 Level
TEST QUESTION II
a) Refer to graph:axes labelled (1)plotting of points (1)2
straight lines (1)intersect at 17.0 cm3 (1)readings shown (1) total
(5)
b) From graph, 17.0 cm3 1.0 M NaOH required (1) (Award this mark
for correct reading from graph)
n (HZ) = n (NaOH) = 17.0(10-3)(1.0) = 1.7 x 10-2 mol in 20 cm3
(1)∴ c (HZ) = 1.7(10-2)(103) / 20 = 0.85 mol dm-3 (1)
c) pH = 2.85 ∴ [H+] = 1.413 x 10-3 mol dm-3 (1)K
a= [H+] [Z-] / [HZ]= (1.413 x 10-3)2 / [0.85 – (1.413 x 10-3)]
(1)
[Note: original acid has dissociated partially]= 2.35 x 10-6 mol
dm-3 (1)
d) For end-point at 17.0 cm3, ∆T = 5.25 (1) (Award this mark for
correct reading from graph)q = m.s.∆T = (20.0 + 17.0)(4.2)(5.25) =
815.85 J (1)∆H = -q/n
= -815.85/(17.0 x 10-3) = -47991 J mol-1 = -48.0 kJ mol-1
(1)
Volume of NaOH added /cm3
Max
imum
tem
pera
ture
ris
e / 0
C
17.0
QUESTIONSHEET 13
5.25
5
4
3
2
1
05 10 15 20 25 30 35