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