TOPIC 19 ANSWERS & MARK SCHEMES QUESTIONSHEET 1. ph AND K W

Transcription

1 QUESTIONSHEET 1 a) ph - lg [H + ] / lg 1 [H + ] b) Water ionises slightly as follows: H 2 O(l) ž H + (aq) + OH - (aq) [H + (aq)][oh - (aq)] K c [H 2 O(l)] Since the ionisation is very slight, we regard [H 2 O] as constant K w [H + (aq)] [OH - (aq)] c) mol 2 dm -6 d) (i) Since for water [H + ] [OH - ]; K w [H + ] x mol 2 dm -6 [H + ] (5.14 x ) x 10-7 mol dm -3 (ii) ph -lg [H + ] -lg (2.267 x 10-7 ) 6.64 ph AND K W e) Endothermic At 373 K, K w is greater than at 298 K water is more dissociated Le Chatelier s Principle implies that endothermic reactions are favoured by increases in temperature the dissociation of water is endothermic f) (i) ph 7.47 [H + ] - antilg x 10-8 mol dm -3 [H + ] [OH - ] K w [H + ] x mol 2 dm -6 (ii) Yes [H + ] [OH - ] so water is neutral even though values have changed

2 QUESTIONSHEET 2 ph OF STRONG ACUDS AND BASES a) (i) 1.00 (ii) 2.40 (iii) 0.40 (iv) [H + ] / ph (v) [H + ] / ph b) (i) [H + ] (ii) ph c) (i) [H + ] mol dm -3 ; [HCl] mol dm -3 (ii) [H + ] mol dm -3 ; [H 2 SO 4 ] mol dm -3 (iii) [H + ] mol dm -3 ; [NaOH] mol dm -3

3 QUESTIONSHEET 3 STRENGTHS OF ACIDS ( AND p ) a) (i) One which is only partially dissociated in solution (ii) [C 2 H 5 COO - (aq)] [H + (aq)] [C 2 H 5 COOH(aq)] (iii) Since [H + ] [C 2 H 5 COO - ], [H + ] 2. [C 2 H 5 COOH] As there is only slight dissociation, assume that [C 2 H 5 COOH] at eqm original concentation (c) of acid [H + ] (. c) ( ) 1.64 x 10-3 mol dm -3 ph - lg ( ) 2.79 b) (i) p - lg where is the acid dissociation constant (ii) HS: mol dm -3 HT: mol dm -3 Delete for missing or incorrect units (iii) HS because it has the larger / smaller p c) (i) [NH 3 (aq)] [H + (aq)] [NH 4+ (aq)] (ii) p -lg -lg ( ) 9.25 d) Butanoic < 4-chlorobutanoic < 3-chlorobutanoic < 2-chlorobutanoic

4 QUESTIONSHEET 4 ph OF ACID OR BASE SOLUTIONS a) (i) HNO 3 (aq) H + (aq) + NO 3- (aq) [H + (aq)] [NO 3- (aq)] 0.2 mol dm -3 ph -lg [H + (aq)] -lg i.e (ii) C 3 H 7 COOH(aq) ¾ C 3 H 7 COO - (aq) + H + (aq) p mol dm -3 [H + ] 2 / [C 3 H 7 COOH] [H + ] (. c) [H + ] ( ) mol dm -3 ph -lg [H + ] -lg ( ) 2.76 Explanation Nitric acid is a strong acid / completely dissociated into ions in solution Butanoic acid is a weak acid / only partially dissociated and thus its [H + ] is lower b) (i) n (H + ) 10(10-3 ) mol [H + ] 10-2 mol dm -3 ph 2.0 (ii) n (OH - ) 20(10-3 ) mol [OH - ] mol dm -3 [H + ] / mol dm -3 ph 12.3 c) (i) HCOOH(aq) ¾ H + (aq) + HCOO - (aq) [H + (aq)] [HCOO - (aq)] [HCOOH(aq)] (ii) If ph 4.0, [H + ] 10-4 mol dm -3 [HCOO - ] [HCOOH] (10-4 ) 2 / ( ) mol dm -3 d) ph 3.0 so [H + ] mol dm -3 [Q - ] 0.2 mol in 500 cm 3 ò 0.4 mol dm -3 [HQ] [H + (aq)] [Q - (aq)] [HQ(aq)] ( ) 2 / mol dm -3

5 a) (i) HCl H + + Cl - [H + ] 0.05 mol dm -3 ph OF MIXED SOLUTIONS OF ACID AND BASE (ii) p x 10-5 mol dm -3 [H + ] (. c ) (1.738 x 10-5 x 0.05) 9.32 x 10-4 mol dm -3 (iii) n (HCl) 0.1(10-3 )(100) 1 x 10-2 mol n (KOH) 0.15(10-3 )(50) 0.75 x 10-2 mol n (H + ) 0.25 x 10-2 mol in ( ) 150 cm 3 [H + ] (0.25 x 10-2 )(10 3 ) / (150) mol dm -3 b) (i) n (H + ) 2(25.0)(10-3 )(0.22) x 10-3 mol (ii) n (OH - ) 20.0(10-3 )(0.25) 5.0 x 10-3 mol QUESTIONSHEET 5 (iii) n (H + ) used n (OH - ) added 5.0 x 10-3 mol n (H + ) unreacted ( ) x 10-3 mol (iv) Total volume cm 3 [H + ] (6.0)(10-3 )(10 3 ) / mol dm -3 ph -lg (0.1333) 0.875

6 BRØNSTED LOWRY THEORY a) (i) Acid A species that can donate a proton to a base Base A species (with a lone pair of electrons) that can accept a proton from an acid (ii) Conjugate acid The species formed when a base has accepted a proton from an acid Conjugate base The species left when an acid has donated a proton to a base Two or more sentences with correct spelling, punctuation and grammar in which the meaning is clear. b) Acid Conjugate acid (i) CH 3 COOH H 2 O (ii) H 2 SO 4 + H 2 NO 3 (iii) - HSO 4 H 2 O c) (i) Ethylamine has a lone pair of electrons on the nitrogen atom of the amino group which can accept a proton from an acid (ii) CH 3 CH 2 NH 3 + (iii) Either ph 11.5 so poh [OH - ] -antilg x 10-3 mol dm -3 Or Or ph 11.5 so [H + ] mol dm -3 K w [H + ] [OH - ] mol 2 dm -6 [OH - ] / mol dm -3 d) As an acid: HSO 4- (aq) + H 2 O(l) ¾ H 3 O + (aq) + SO 4 2- (aq) Or HSO 4- (aq) + OH - (aq) ¾ H 2 O(l) + SO 4 2- (aq) As a base: HSO 4- (aq) + H + (aq) ¾ H 2 SO 4 (aq) e) Base Br Conjugate base H 2 PO 4- QUESTIONSHEET 6

7 THEORY OF INDICATORS a) Soluble in water Two distinct colours in acid and base forms Intensely coloured so as to be seen when only a few drops are used b) (i) HIn(aq) ž H + (aq) + In - (aq) K in [H + (aq)] [In - (aq)] [HIn(aq)] QUESTIONSHEET 7 (ii) The ph range over which the colour changes (c) (Approximately) ph (i.e. pk In ± 1) c) (i) Thymolphthalein Titration of a weak acid with a strong base End-point likely to be in region ph 9-10 i.e. pk in (ii) Methyl orange Titration of a weak base with a strong acid End-point likely to be in region ph 3-5 i.e. pk In (iii) No indicator is suitable Titration of weak acid with weak base No sharp end-point / ph changes gradually in the region of the equivalence point

8 A2 Level QUESTIONSHEET 8 TITRATION CURVES a) In original acid, [H + ] (. c) (1.35 x 10-5 x 0.1) 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 cm 3 alkali added (1 if shown on graph) ph at the end-point 9 (1- if shown on graph) ph ph (5) (3) Equivalence point Shape of graph (2) Vol NaOH added /cm 3 b) (i) n (OH - ) 0.2(18.0) x 10-3 mol n (HA) 3.6 x 10-3 mol in 25.0 cm 3 [HA] (3.6)(10-3 )(10 3 ) / mol dm -3 (ii) When HA has been half converted to Na + A -, [HA] [A - ] so that [H + ] p ph when half the neutralising alkali has been added antilg x 10-4 mol dm -3 (iii) [H + ] (. c) (2.63 x 10-4 x 0.144) 6.15 x 10-3 mol dm -3 ph -lg (6.15 x 10-3 ) 2.21 c) ph M NaOH 0.1 M NH 3 7 Buffer range 0 Volume 0.1 M HNO 3 added NaOH curve: start at ph 13 shape ends > ph 1 NH 3 curve: start ph 11 shape ends > ph 1 end-points the same buffer region correct mid-point buffer region at ph approximately 9 axes labelled correctly Divide by 2, round up /5 total

9 QUESTIONSHEET 9 ACID-BASE STRENGTH AND ENTHALPY OF NEUTRALISATION a) (i) The enthalpy change / enthalpy released when 1 mol water is formed by the reaction between acid and base in dilute aqueous solution in the mole ratio required by the equation (ii) Chemical equations ½ H 2 SO 4 (aq) + NaOH(aq) ½ Na 2 SO 4 (aq) + H 2 O(l) ; H -57 kj mol -1 CH 3 COOH(aq) + NaOH(aq) CH 3 COONa (aq) + H 2 O(l); H -54 kj mol -1 Explanation of difference The essential reaction is H + (aq) + OH - (aq) H 2 O(l) Ethanoic acid is a weak acid / a solution contains relatively few H + (aq) ions Ionisation requires the breaking of covalent bonds which is an endothermic process This reduces the overall energy released in the neutralisation Maximum 3 marks b) (i) As volumes / concentrations are the same, and HCN is also a monoprotic acid H -57.8(1.0) / kj mol -1 (ii) HCN is a very weak acid / the H _ C bond is strong and ionisation thus requires a considerable amount of energy NH 3 is a weak base and its ionisation also requires energy These two factors reduce the enthalpy of neutralisation by some 52 kj mol -1 Maximum 4 marks (iii) As sodium hydroxide is a strong base, no energy will be required for its dissociation The result obtained will lie between the other two values

10 QUESTIONSHEET 10 BUFFER SOLUTIONS a) (i) A buffer solution is one whose ph changes only slightly on the addition of moderate quantities of acid or base (ii) Acid buffer Ethanoic acid and sodium ethanoate (or any other weak acid and its salt with a strong base) Alkaline buffer Aqueous ammonia and ammonium chloride (or any other weak base and its salt with a strong acid) Biological buffer Carbonic acid and sodium hydrogencarbonate in blood (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 (ii) The concentration of the buffer / the amounts of reactants available to counteract the addition of acid or base The ratio of acid (or base) to its salt c) (i) Reaction with H + ions H + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) Reaction with OH - ions OH - (aq) + CH 3 COOH (aq) CH 3 COO - + H 2 O (l) (ii) Species present H 3 N + CH 2 COOH(aq) and H 3 N + CH 2 COO - (aq) Reaction with H + ions H 3 N + CH 2 COO - (aq) + H + (aq) H 3 N + CH 2 COOH (aq) Reaction with OH - ions H 3 N + CH 2 COOH (aq) + OH - (aq) H 3 N + CH 2 COO - (aq) + H 2 O (l) d) Initially, since ethanoic acid is a weak acid, there is a relatively low concentration of hydrogen ions As these react with the alkali, ph changes rapidly At the same time, however, sodium ethanoate is formed Once a moderate amount of this is present, the solution acts as a buffer until such time as its buffering capacity is exhausted which is near to the end-point of the titration Maximum 5 marks

11 QUESTIONSHEET 11 ph OF BUFFER SOLUTIONS a) (i) [CH 3 COO - (aq)] [H + (aq)] [CH 3 COOH(aq)] [H + (aq)] / ph 5.07 (ii) [BrCH 2 COO - (aq)] [H + (aq)] [BrCH 2 COOH(aq)] [H + (aq)] / ph 3.17 (iii) [BrCH 2 COO - (aq)] [H + (aq)] [BrCH 2 COOH(aq)] [H + (aq)] / ph 4.17 b) (i) [CH 3 CH 2 COO - (aq)] [H + (aq)] [CH 3 CH 2 COOH(aq)] (ii) Since equal volumes are mixed, the equilibrium concentration of sodium propanoate will be half its original concentration, i.e mol dm -3 Either (CH 3 CH 2 COOH) - antilg mol dm -3 From b) (i), [acid] in buffer [salt] [H + ]/ / mol dm -3 Or ph pka + lg([salt]/[acid]) lg (0.1/[acid]) 0.13 lg (0.1/[acid]) /[acid] [acid] in buffer 0.1/ mol dm -3 So originally the concentration of propanoic acid was mol dm -3

12 TEST QUESTION I a) S 26.89% O 13.45% Cl 59.66% Moles % 26.89/ / / Ratio: Empirical formula is SOCl 2 QUESTIONSHEET 12 b) Ag + (aq) + Cl - (aq) AgCl(s) n (Cl - ) n (Ag + ) 25.0(10-3 ) x 10-3 mol Cl - in 25.0 cm 3 c (Cl - ) 2.5(10-3 )(10 3 ) / mol dm -3 i.e. 2 mol Cl - per mol compound, as per empirical formula. Molecular formula is also SOCl 2 c) The white precipitate is barium sulfite / BaSO 3 since it dissolves in acid to release SO 2 To confirm: n (SOCl 2 ) in 25.0 cm (25.0)/ x 10-3 mol This should give 1.25 x 10-3 mol BaSO 3 (1.25 x 10-3 ) g BaSO 3 The weak acid HQ is therefore sulfurous acid / H 2 SO 3 and the strong acid HP is hydrochloric acid / HCl Equation: SOCl 2 (l) + 2H 2 O(l) H 2 SO 3 (aq) + 2HCl(aq) d) (i) ph [H + ] mol dm -3 [NB: a subtraction is to follow, which will reduce the number of significant figures and cause inaccuracy in the calculation if rounding up is carried out here] (ii) Since 0.05 mol SOCl 2 gives 0.10 mol H + from HCl then ( ) mol H + from H 2 SO 3 (iii) Percentage from weak acid (100) / %

13 QUESTIONSHEET 13 TEST QUESTION II a) Refer to graph: axes labelled plotting of points 2 straight lines intersect at 17.0 cm 3 readings shown total (5) Maximum temperature rise / 0 C Volume of NaOH added /cm 3 b) From graph, 17.0 cm M NaOH required (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 cm 3 c (HZ) 1.7(10-2 )(10 3 ) / mol dm -3 c) ph 2.85 [H + ] x 10-3 mol dm -3 [H + ] [Z - ] / [HZ] (1.413 x 10-3 ) 2 / [0.85 (1.413 x 10-3 )] [Note: original acid has dissociated partially] 2.35 x 10-6 mol dm -3 d) For end-point at 17.0 cm 3, T 5.25 (Award this mark for correct reading from graph) q m.s. T ( )(4.2)(5.25) J H -q/n /(17.0 x 10-3 ) J mol kj mol -1