(a) (K a =) [H 3 O + (aq)] [HS (aq)] [H 2 S(aq)] [H 3 O + ] 2 numerator (2) (K a2 =) [H 3 O + (aq)][ S 2 (aq)] [HS (aq)] [H 3 O + ] 2 numerator Allow H + (aq) for H 3 O + (aq) Ignore missing / incorrect state symbols (b)(i) (H 2 S + H 2 O H 3O + HS (2) Initially 0.00 0 0 At eqm 0.00 x x K a = x 2 ) 0.00 M: x 2 = 8.9 x 0-9 (mol 2 dm -6 ) M2: (x = 9.4393 x 0-5 ) [HS ] = 9.44 x 0-5 / 0.0000944 (mol dm -3 ) For M2, answer must be to 3 sf Correct answer without working scores (2) (b)(ii) ([H + ] = ( 8.9 x 0-9 =) 9.439 x 0-5 (mol dm -3 )) ph = (-log 9.439 x 0-5 ) = 4.025/4.025 /4.03/4.0 4/4.02 TE on answer to (b)(i) provided ph <7
*(b)(iii) Any THREE from: (3) Assumption [H 2 S] equilibrium = [H 2 S] initial The dissociation of H 2 S is negligible 0.0000944 is very small compared to the initial concentration of H 2 S/ 0.00 (hence a valid assumption) Assumption 2 [H 3 O + ] = [HS ] / [H + ] = [HS ] Ignore any H + from (the dissociation of) water / H + only from H 2 S Assumption 3 Ignored ionization of HS / HS doesn t (significantly) dissociate further K a2 very much smaller than K a Assumption 4 Measurements at 298 K / standard temperature IGNE References to the concentration of water References just to standard conditions
(c)(i) M: General shape of an acid-base curve with the ph increasing and either one or two steep / vertical sections shown NOTE Penalise a ph range for a single vertical with a range of eight or more ph units (as this is the typical range for a strong monobasic acid with a strong base titration curve) M2: Vertical / steep section at 25 cm 3 (5) M3: Vertical / steep section at 50 cm 3 M4: Either equivalence point labelled anywhere on vertical section or x-axis M5: Initial ph =.5 and a recognisable plateau in the ph range of 2 to 3 (c)(ii) The ph when 2.5 cm 3 of NaOH has been added the ph at half-equivalence (for the first equivalence point) ALLOW ph at half neutralisation Allow TE from an incorrect graph
2(a)(i) HC 2 O 4 (aq) + H 2 O(l) C 2 O 4 2 (aq) + H 3 O + (aq) (or ) ALLOW H 2 O(aq) Equation states ALLOW for mark HC 2 O 4 (aq) C 2 O 2 4 (aq) + H + (aq) States mark is not stand alone but can be awarded if the equation has a minor error e.g. an incorrect charge 2 2(a)(ii) K a = [C 2 O 2 4 ] [H 3 O + ] / [HC 2 O 4 ] K a = K a = [C 2 O 4 2 ] [H + ] / [HC 2 O 4 ] No TE on incorrect equation in (a)(i) Penalise incorrect charges in (i) and (ii) once only [H + ] 2 / [HC 2 O 4 ] [H + ][A ]/ [HA] 2 (a) (i No TE on (a)(ii) 3 K a = 0 4.28 5.24807 x 0-5 (mol dm 3 ) K a = [H + ] 2 / [HC 2 O 4 ] K a = [H + ] 2 / 0.050 [H + ] = (0.05 x 0 4.28 ) =.6988 x0 3 (mol dm 3 ) TE on incorrect K a value ph = log.6988 x 0 3 = 2.7905 = 2.8 For final mark TE on algebraic / arithmetical errors providing ph.3 Correct answer with no working scores 3 Ignore SF except SF
2(b)(i) IGNE explanations First mark: HC 2 O 4 /hydrogenethanedioate ion ionization negligible ALLOW Acid for HC 2 O 4 Slight / partial / incomplete / does not dissociate for negligible [HC 2 O 4 ] equilibrium = [HC 2 O 4 ] initial / 0.050 (mol dm 3 ) Second mark: [H + ] due to ionization of water negligible auto ionization of water negligible Use of NaHC 2 O 4 for HC 2 O 4 sodium hydrogenethanedioate for hydrogenethanedioate ion throughout this item 2 [H + ] only due to ionization of HC 2 O 4 /acid [C 2 O 4 2 ] = [H + ] IGNE references to temperature and to HA and A Penalize omission of [] in discussion once only
2(b)(ii) Ethanedioic acid is a (much) stronger acid (than hydrogenethanedioate ion / sodium hydrogenethanedioate) Ethanedioic acid has a (much) smaller pk a (than hydrogenethanedioate) Ionization / dissociation of ethanedioic acid is (much) greater (than hydrogenethanedioate) Reverse arguments Ethanedioic acid is a strong acid / fully dissociated 2 IGNE NaHC 2 O 4 ionization negligible Approximation of negligible ionization invalid / incorrect [H 2 C 2 O 4 ] equilibrium not equal to [H 2 C 2 O 4 ] initial Just approximation invalid No TE on 8(a)(iii) IGNE Second ionization occurs 2(c)(i) Start ph at 2.8 ALLOW 3 2 4 Vertical section at 25 cm 3 within ph range 6- and 2.5-4 units long end ph (approaching) value in range 2-3 (asymptotically) deviation from vertical maximum before final ph
2(c)(ii) ) F mark: Methyl yellow range = 2.9 4 and the phenolphthalein range = 8.2 0 ALLOW pk in (methyl yellow) = 3.5 and pk in (phenolphthalein) = 9.3 2 Second mark: (The volumes are different) because ethanedioic acid is dibasic / diprotic / has two replaceable/acidic hydrogen atoms ALLOW dicarboxylic (acid) (therefore there are two stages to the neutralization) Methyl yellow range coincides with neutralization of first proton and phenolphthalein range coincides with neutralization of second proton
3 (a)(i) Ka = [CH 3 CO 2 ] [H + ]/[CH 3 CO 2 H] Ka = [CH 3 CO 2 ] [H 3 O + ]/[CH 3 CO 2 H] Use of [CH 3 COO ] instead of [CH 3 CO 2 ] and [CH 3 COOH] instead of [CH 3 CO 2 H] IGNE state symbols even if wrong Numerator as [H + ] 2 Expressions in terms of HA alone Round/curved brackets () Any other carboxylic acid 3(a)(ii).7 x 0-5 = [H + ] 2 / 0.5 [H + ] =.7 x 0-5 x 0.5 / 2.95(476) x 0 3 ph = (-log[h + ]) = 2.53529 = 2.54 4.77 or 4.8 from using ph = -log Ka loses both marks 2 = 2.5 ALLOW TE for second mark from any hydrogen ion concentration as long as ph less than 7 Correct answer alone scores (2) ALLOW ph = 2.53 if [H + ] is rounded to 2.92 x 0 3 IGNE sf except 3(a)(iii) 20 (cm 3 ) IGNE units 0.02 dm 3
3(a)(iv) Moles of excess NaOH = 0/000 x 0.50 4 = 5 x 0-3 So [NaOH/OH - ] = 5 x 0-3 x 000/50 = 0.0 mol dm -3 EITHER Kw route: [H + ] x 0. = x 0-4 So ph = - log x 0-4 / 0. = 3 poh route: poh = So ph = (4 ) = 3 ALLOW TE throughout Correct final answer scores (4)
3(a)(v) Starting at ph 2-3 AND finishing at ph between 2 and 3.7 inclusive 3 Vertical section at 20 cm 3 S-shaped curve, with gradual rise and vertical section within the ph range 5.5 and.5 and of 3 to 5 units in length These are stand alone marks
3(b)(i) EITHER 4 [base] = Ka [acid]/[h + ] Or [H+] = (0 -ph4.70 ) =.995 x 0-5 [base] =.7 x 0-5 x /(.995 x 0-5 ) = 0.852 moles base = 0.852 x 0.5 = 0.426 (mol) mass base = 0.426 x 82 = 34.9 g IGNE sf except Correct answer, with or without working (4) ph = pka log[acid]/[base] 4.70 = 4.8 log [/[base]] Log[/[base]] = 0. [base] = 0.794(328) (mol dm -3 ) So in 500 cm 3 Moles = 0.794 x 0.5 = 0.397 mol Mass = 0.397 x 82 = 32.554/32.6 g (ALLOW using pka = 4.77)
3(b)(ii) i Acceptable Answers First mark Buffer has large amount/ excess/ reservoir of CH 3 COOH (and CH 3 COO ) Second mark OH ions added react with CH 3 COOH CH 3 COOH + OH CH 3 COO + H 2 O OH + H + H 2 O and CH 3 COOH CH 3 COO + + Reject Mark 3 Equations described in words Third mark Ratio / values of [CH 3 COOH] to [CH 3 COO ] remains (almost) unchanged IGNE concentration of hydrogen ions remains constant ALLOW answers in terms of HA and A
4 (a) K a = (0-0.64 ) = 2.3 x 0 - / 2.2909 x 0 - (mol dm -3 ) Ignore sf except 4 (b)(i) K a = [HCOO - ][H + ] [HCOOH] written as HCO 2 - and HCO 2 H with H 3 O + instead of H + K a = [H + ] 2 [HCOOH] without also giving full expression Allow K a = [A - ][H + ] [HA] if formula of HA and A - given as HCOOH and HCOO - 4 (b)(ii).6 x 0-4 = [H + ] 2 3 0.50 [H + ] =.6 x 0-4 x0.5 (= 8x0-5 = 8.94 x 0-3 ) ph = (2.048455) = 2.05 / 2.0 Correct answer with no working (3) ph =2 ph =2. TE for third mark if [H + ] calculated incorrectly No TE from incorrect K a expression Ignore sf except 4 (b)(iii) All H + comes from acid / none from water / [H + ] = [HCOO - ] [H + ] = [A - ] Dissociation of acid is negligible / very small [HA] initial =[HA] equilibrium K a is measured at 298K Just dissociation of acid is partial
4 (c)(i) HCOOH + CH 3 COOH 2 both correct 4 (c)(ii) (HIO + CH 3 COOH ) H 2 IO + + CH 3 COO - / (HIO + CH 3 COOH ) HIOH + + CH 3 COO - Ignore position of positive charges
4 (d) (ph = 4.9) so [H + ] = (.2589254 x 0-5 ) =.259 x 0-5 2 ( K a = [HCOO - ] [H + ] [HCOOH] =.6 x 0-4.259 x 0-5 ) = 2.7 (:) / 3(:) (HCOO - per HCOOH or base:acid) (2.709252 from unrounded [H + ] 2.708499 from [H + ] rounded to.259 x0-5 2.3 from [H + ] rounded to.3 x0-5 ) TE from error in [H + ] Allow 800:63 Correct answer scores 2 Accept (0.0786828) = 0.079 HCOOH per HCOO - for acid:base ratio (0.0786874) = 0.079 from rounded ph pk a = -log K a = 3.79 3.79 = 4.9 log [base] [acid] log [base] =. [acid] [base] = (2.882496) = 2.9 (:) [acid] Correct answer scores 2 Accept 0.0776/ 0.078 HCOOH per HCOO - for acid:base ratio (0.0776247) TE from error in pk a Ignore sf except