4.3 ANSWERS TO EXAM QUESTIONS

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1 4. ANSWERS TO EXAM QUESTIONS. (a) (i) A proton donor () (ii) Fully ionised or fully dissociated () (iii) () mol dm 6 () 4 (b) (i) 50 0 /5 000 () = 0 06 mol dm () () (ii) Mol OH added = /000 = () Mol H used = 5 0 () Mol OH excess = () [OH ]= / = () [H ] = 0 4 /8 0 0 = 5 0 () ph = 9 () 8 (c) (i) 0 mol dm () (ii) [H] at ph = 0 7 is 0 mol dm () m v = m v 0 5 = 0 v () Hence v = 7 5 () Water added = = 5 () 5 [7]. (a) (i) K a = [H ][A [HA] ] () [H ] K a = () [HA] [H] = 0.8 =.54 0 mol dm () ( ).54 0 Ka = 0.5 = () mol dm () [] (ii) Decreases () Equilibrium shifts to right (endothermic process ) () [H ] as T () ph gets smaller 8

2 (b) (i) n () (ii) n n () = () (iii) Ka = [H ][X [HX] ] [H ] = () =. 0 4 mol dm () 5 (c) Weak acid/strong base ph at equivalence > 7 () methyl orange has colour change at ph < 7 () (d) Buffer can resist change in ph () on addition of small amounts of H (or OH ) () H (aq) X (aq) HX (aq) () [8]. (a) proton donor () substance formed when acid has lost proton / substance that becomes an acid by gaining a proton (not just proton acceptor) () (b) (i) acid: HBr base: Br (ii) acid: H SO 4 base: HSO 4 allow in (b) if both acids / bases are correct (ie give for a correct vertical pair) (c) (i) H O H OH / H O H O OH () (accept other types of arrow) [H ][OH ] (ii) K c = / expression based on H O equation () [H O] [H O] is (effectively) constant /concentration of H O is large / equilibrium in (i) is to left () (K c [H O]) = K w = [H ] [OH ] () (iii) [H] = [OH ] / [H ] = () [H] = = () ph = log 0 ( ) () = 6.77 () 4 (iv) endothermic and attempt at reason () more dissociation / ionization / H ions at higher temperature () if (iii) not completed, allow endothermic with sensible reason for mark if answer to (iii) is ph>7, allow mark for exothermic with attempt at reason [4]

3 4. (a) Definition of a base Proton acceptor () Essential feature Transfer of protons () Equation H OH H O OR H B BH () (b) only partially dissociated in solution () (c) K a = [H (aq)][ch [CH CH CH COO COOH(aq)] (aq)] () mol dm () (d) (i) resists change in ph () on addition of small amounts of strong acid or base () (ii) correct weak acid/co-base or correct weak base/co-acid () (iii) any suitable use () 4 [0] 5. (a) (i) 0..8 = M 5 () molarity = () (ii) K a = [H ][A [HA] ] () (iii) Volume of NaOH(aq) added.8/ = 5.9 cm () ph 4. to 4.5 () (iv) As [HA] = [A ] () K a = [H ] () ph = log 0 [H ] () hence K a = 0 4. = () Note: Mark K a consequentially to ph in a(iii) 9 (b) (i) The added OH reacts with HA or H () The equilibrium, HA H A, displacement to right or HA ionises () (ii) The added H reacts with A () The equilibrium, HA H A, displaced to left () 4 [] 6. (a) CO H HCO or Na CO HCl NaHCO NaCl () HCO H H O CO or NaHCO HCl NaCl CO H O () or H CO (b) 5 cm (c) Indicator Methyl orange (allow other correct indicators) () Explanation Methyl red changes colour over ph range. 4.4 (allow between and 7) ()

4 (d) CO H H O CO Mol H = 0 0./000 =.6 0 () Mol CO =.8 0 = M 5/000 () = 0.07 M () (e) Volume of HCl(aq) added for first end-point 5 () Volume of HCl(aq) added for second end-point 45 () [0] 7. (a) A proton donor (b) Equation HCl(g) H O(l) H O (aq) Cl (aq) () Role of water base or proton acceptor () (c) Equation NH (g) H O(l) NH 4 (aq) OH (aq) () Role of water acid or proton donor () (d) Equation for formation H SO 4 HNO HSO 4 H NO () Role of nitric acid base or proton acceptor () H NO NO H O () (e) (i) only partially dissociated in aqueous solution () (ii) K a = [H ][A [HA] ] () (iii) not very big () (iv) strong () Although HX is not fully dissociated, the relative concentration of undissociated HX is very small () 5 [] 8. (a) NaOH HA NaA H O () or HA OH A H O (b) (i) Moles A = moles NaOH added () = = () Initial moles HA = = 0.05 () Allow Moles NaOH added = Moles HA remaining = () Allow ( ) 0 Mark conseq 4

5 (ii) moles A in (5 5) cm Hence [A ] = /40 = 0.75 () Allow and moles HA in 40 cm Hence [HA] = /40 = 0.58 () Allow and 0.5 Allow marks in (ii) conseq to answers in (i) (iii) K a = [H ] [A ] / [HA] = [H ] = / 0.75 () = () Allow (.9.44) 0 4 ph =.6 () Allow.6.64 and.6 Mark conseq to answers in (ii) 9 [0] 9. (a) only partially ionized / partially dissociated / not fully ionised () not not ionised at all (b) (i) K a = [C6H 5O ][HO [C H OH] 6 5 ] () accept [H] do not accept with [H O] included must include charges (ii) pk a = log K a () allow log (K a ) do not allow log [K a ] (iii) pk a = 0 (ignore units) () (iv) lower / smaller number () (c) (i) at end point ph = pk a = 9. () colour change detectable over range of ph units range = () (allow 8 0) (ii) (colourless to) pink / red () [In ] [HIn] / [In ] increases () not just equilibrium shifts to right (iii) equivalence point / end point of titration below ph 7 more acidic / lower than phenolphthalein range / is about ph 4 () not just the ph range is wrong [0] 5

6 0. (a) (i) ph = log (0) [H ] Note; (aq) not required; Not ln [H ] () (ii) K a = [H ] [X ] / [HX] Note; (aq) not required () Allow [A ] and [HA] Do NOT allow [H ] /[HX] (iii) K a = = [H ] /[HX] () 5 [H] = () () = not a conseq mark ph =.6 Mark conseq to [H ] above () or ph = pk a log 0 [HX] = =.6 () () () Note ph =.4 scores max 6 (b) (i) [H ] = = 0.75 () Allow and 0.4 ph = 0.6 () Allow Only allow ph mark if [H ] is correct (ii) [H ] = [Y ] = 0.75 (or a value from b(i)) () [HY] = = 0.05 () Allow K a = [H ] [Y ] / [HY] = (0.75) / 0.05 () K a = 4.5 () Allow Ignore units CE if [HY] is incorrect 6 []. (a) pk a log 0 K a (b) K a = () K a = [H ] /0.5 or [H ]=[X ] () [H ] = ( ) = () ph = log 0 [H ] =.00 () or ph = ½ pk a ½ log [HX] =.86 (0.4) =

7 (c) Ka [H ] [X ]/[HX] () [HX] = [X ] at half neutralisation () (d) Hence Ka= [H ] and pka = ph () There is no rapid/sharp/steep change in ph during a weak acid - weak base titration () Indicator need a sharp ph rise to change colour quickly () [0]. (a) (i) K a = [H ][CH [CH COO COOH] ] () [H ] (ii) () K a = () [CH COOH] () [H ] = () ph = -log 0 [H ] () can score independently =.96 0 () (4) ph =.7 () d.p. essential If forget can score () and () for ph = (b) (i) moles acid = 00 () = x = x = or = 6.7 (or 7) cm (or 6.6) () NOT 6 NOR 7.0 units must match (ii) Indicator: thymol blue () Explanation: weak acid strong base () equivalent at ph > 7 () or high ph 5 7

8 .0 (c) () mol NaOH added = = () 40.0 If wrong M r : CE lose marks () and () then mark on consequentially max 4 () mol CH COOH left = = 0.70 () () mol CH COO formed = () (4) [H [acid] [ A ] ] = Ka OR ph = pka [salt] log [ HA ] If expression wrong no marks for 4 / 5 / 6 can score () to (4) in (5) etc () (5) [H ] = (0.70) (0.05) 0.05 OR ph = 4.76 log () 0.7 (6) ph = 4. () Correct answer gets Mark (5) is for use of correct values of (acid moles) and (salt moles) if one wrong allow ph conseq if both wrong, no further marks e.g. if candidate forgets substitution in () he loses () and (5) but can score () () (4) (6) = max 4 [acid] for ph = 4. if upside down; answer 5.9 scores [salt] for () () () 6 [6] 5. (a) moles HA = 0.50 =.75 0 () vol NaOH = = dm () 0.0 or 8.75 cm (b) (i) ph = log 0 [H ] () (ii) Value above 7 but below () (iii) phenol red / thymol blue / phenolphthalein / thymolphthalein i.e. indicator with 7 < pk in < 8

9 (c) (i) Only slightly dissociated () NOT not fully dissociated / ionised (ii) K a = NOT [ H ][ A ] [ HA] [ ] [ HA] H () (iii) For weak acid alone: [ H ] Ka = () [ HA] 5 [H ] = (.75 0 ) 0. 5 =.0 0 () ph =.69 () ph should be given to decimal places penalise answer to d.p. once in question 5 (d) moles OH added = = moles A = moles HA left () or [A ] = [HA] Ka = [H ] or ph = pk a () ph = 4.56 () [] 4. Penalise ph given to dp first time it would have scored only (a) (i) K w = [H ] [OH ] () (ii) ph = log [H ] () or in words or below unless contradiction (iii) Calculation: [H 4 ] = () = ph = 6.6 or 6.64 () Explanation: pure water [H ] = [OH ] () 5 (b) (i) [OH ] = 0.50 [H ] = 0 4 /0.5 = or poh = 0.8 (ii) ph =.8 () or ph=.7 moles OH = (5 0 ) 0.50 = () a moles H = (40 0 ) 0.0 = 4.8(0) 0 () b excess moles of OH = () c [H] = [OH ] = (4.5(0) 0 4 ) 000/75 d () e =.66 0 or poh =. ph =.78 () f or

10 [ ][ ] H X (c) (i) K a = [ HX] () (ii) [H ] = =.70 0 () [ ] H K a = [ HX] (.70 0 ) () = = () mol dm () (.70 0 ) or 0.47 = Notes (a) If K w includes H O allow 6.6 if seen otherwise no marks likely (b) (ii) If no vol, max 4 for a, b, c, f answr = 0.65 If wrong volume max 5 for a, b, c, e, f If no substraction max for a, b, d If missing 000 max 5 for a, b, c, d, f answer = 8.78 If uses excess as acid, max 4 for a, b, d, f answer =. If uses excess as acid and no volume, max for a, b answer =.5 (c) If wrong K a in (i) max in part (ii) for [H ] () and conseq units () but mark on fully from minor errors eg no [ ] or charges missing [8] 5. (a) Hydrogen bonding () between H O and NH () (b) (i) NH H O NH 4 OH () (ii) Ammonia is weak base () NOT partially ionised Equilibrium to left or incomplete reaction () (c) A proton donor () (d) Buffer solution: A solution which resists change in ph () when small amounts of acid or base added or on dilution () Reagent: NH 4 Cl () Allow a correct strong acid (e) (i) K a = [H ] [A ] / [HA] () = [H ] [0.5 4] () /.00 [H ] = / = () ph = log 0 [H ] = 4.47 () Allow ph conseq to [H ] if place decimals given (ii) H CH COO CH COOH () 5 [4] 0

11 6. (a) [H ][A ] K a = [HA] (All three sets of square brackets needed, penalise missing brackets or missing charge once in the question) (Don t penalise extra [H ] /[HA]) (b) K a = [H ] [HA] or [H ] = [A ] [H -4 ] = (.45 0 ) 0. 5 = ph =. (must be to dp) (allow 4th mark consequential on their [H ]) (c) (i) ph (almost) unchanged (Must be correct to score explanation) H removed by A forming HA or acid reacts with salt or more HA formed (ii) [H ] = 0.59 = or [A K ] = a [HA] [H ] 4 (.45 0 ] 0.5 = = 0.4 (mol dm ) (Allow 0.9 to 0.4 and allow 0.4) (If not used.59, to find [H ] can only score M for working) (If.59 used but [H ] is wrong, can score M for correct method and conseq M4) If wrong method and wrong expression, can only score M) (ii) Alternative scheme for first three marks of part (c)(ii) [HA] ph = pk a log [A ] pk a = =.84 log [A ] []

12 7. (a) (i) ph = log [H ] () [ ][ ] H X (ii) Expression for Ka: K a = [ HX] () Calculation: ph =.56 [H ] =.75 0 () H K a = [ ] [ HX] (.75 0 ) = 0. = () (mol dm ) or [H ] = [X ] () 5 depending on approximate made, values of K a = 0 5 using [HX] = using [HX] = using.8 and [HX] = using.8 and [HX] = upside down K a (b) (i) Expression for K w : K w = [H ] [OH ] () Value of K w : (.0 )0 4 (mol dm 6 ) () ignore units (ii) [H.0 0 ] = or poh =.5 () =. 0 ph =.65 () 4 must be dp in final answer (c) (i) H C O 4 OH HC O 4 H O () (ii) mol OH = (4.6 0 ) () =.87 0 mol H C O 4 = () [H C O 4 ] = /5 = () 4 if moles of H C O 4 not equal to half moles of OH, no further marks gained if mol OH =.9 0 ; hence mol H C O 4 = ; [H C O 4 ] = 0.08 [] 8. (a) (i) B; C; A; (ii) cresolphthalein OR thymolphthalein;

13 (b) (i) log[h ]; (ii) [H ] =.59 0 (or.6 or.) OR poh = 4 ph; [OH 4 0 ] = OR =.0;.58 0 = 7.9(4) 0 ; (if [H ] is wrong allow for [OH] = K W /[H ] or as numbers) (c) (i) K a = [H ] /[CH CH COOH] OR [H] /[HA] OR [H ] = [A ] etc; [H ] = l.5 l or expression without numbers; =.57 l0 ph =.90; (iii) K a = [H ] OR pk a = ph; ph = 4.87; (penalise dp once) [] 9. (a) log [H ] allow ecf if [ ] wrong and already penalised ignore units

14 [H ][X ] (b) (i) Ka = allow HA etc [HX] - [H ] not but mark on [HX] If expression wrong allow conseq units in (ii) but no other marks in (ii) (ii) [H ] ( ) = [HX] [0 50] If use K a =.4() 0 4 and pka =.85 = allow mol dm (iii) pk a =.86 Penalise dp of final answer < or > in ph once in paper (c) (i) = or.4(4) 0 Mark is for answer (M) (ii) = or 6. 0 Mark is for answer (M) (iii) (0.006) =.80 0 M is for (i) (ii) If x missed, CE i.e. lose M and the next mark gained (iv) = (0.08) 48 M4 is for answer If vol is not 48 0 (unless AE) lose M4 and next mark gained If multiply by 48 - this is AE - i.e. lose only M4 If multiply by 48 0 this is AE - i.e. lose only M4 (v) 0 4 / (0 4 / 0.08) M5 for K w /[OH ] (=.66 0 ) (=.6 0 ) or poh or poh =.46 (or poh =.40) If no attempt to use K w or poh lose both M5 and M6 ph =.57 (.58) M6 Allow M6 conseq on AE in M5 if method OK [] 4

15 0. (a) (i) [H ][OH ] log [H ] (ii) [H ] = [OH ] (iii) (.0 0 ) 0.5 =.0 0 (iv) [H] = ( = ) ph = 0.40 (b) (i) Ka = [H][CH CH COO] [CH CH COOH] = [H] [CH CH COOH] [H] = ( ) 0.5 ( =.0 0 ) ph =.89 (c) (i) ( ) 0.5 = (ii) (6.5 0 ) (.0 0) = (iii) mol salt formed =.0 0 (H) = Ka [CH CH COOH] [CH CH COO) = ( ) (5.5 0 (.0 0 ) / V ) / V ( = ) ph = 4.5 [6]. (a) Concentration of acid: m v = m v hence 5 m = OR moles NaOH =.7 0 ; m = /5= 0.09; (b) (i) K a =[H ][A ]/[HA] not K a = [H ] / [HA]; (ii) pk a = logk a ; (iii) [A ] = [HA]; hence K a = [H ] [A ] / [HA] = [H ] and logk a = log[h ]; (c) ratio [A ] : [HA] remains constant; hence as [H ] = K a [HA] / [A ]; [H ] remains constant; 5

16 (d) (i) ph of 0.50 mol dm- HCl = 0.60 and ph of 0.50 mol dm- HCl = 0.8; ph change = 0.; (ii) moles HCl = = v = OR v = / =50; water added = 50 0 = 0 cm ; []. (a) proton or H donor () (b) (i) partially ionised or dissociated () not fully (ii) NH () not NH 4 OH but allow in equation not H O but allow in equation if both weak acid and base stated NH H O NH 4 OH () (c) (i) HCOOH(aq) H O(l) HCOO (aq) H O (aq) () allow H O(aq) (ii) H O or water () HCOO or methanoate ion () (iii) [H ][HCOO ] K a = () [HCOOH] allow [OH ] 4 (d) (i) addition of small amounts of acid () addition of small amounts of base () allow volumes, allow alkali, penalise missing small once only not weak dilution () (ii) sodium methanoate or sodium hydroxide () allow salt of methanoic (or this) acid not just an ion (methanoate) (iii) OH added () or base H reacts with OH () or forming water More HCOOH dissociates to restore equilibrium () allow equilibrium moves to right must only describe addition of base; if both base and acid addition given, MAX ex 7 [5] 6

17 . (a) HCl(g) H O(l) H O (aq) Cl (aq) () allow H O(aq) (b) ph = log[h ] () = 0.0 (± 0.0) () (c) ph = 7 () neutral solution or [H ] = [OH ] () (d) (i) moles H =.6 95 = moles () range 0.0 ± 0.00 (ii) moles OH 45 =.7 = 0. moles () 000 range 0. ± 0.00 (iii) XS OH = =.6 0 moles () range 0.00 to Volume = = 40 cm () [OH ] = = M () 40 range to K w = [H ] [OH ] = 0 4 mol dm 6 () [H ] = 4 K w 0 = [OH ] =.89 0 M () range (0.8 to.5) 0 ph =.4 () must show dp range.87 to.55 8 [] If no for Ba(OH) then H is in XS MAX 4 ex 6 If no volume used then MAX 4 ex 6 If no 000 for molarity then MAX 4 ex 6 Combinations of TWO of these MAX ex 6 All THREE ZERO 7

18 4. (a) only partially dissociated in water (b) (i) Added H reacts with A () Equilibrium moves left & HA forms, restoring [H ] and ph () (ii) Added OH reacts with H forming H O () Equilibrium moves right & HA dissociates, restoring [H ] and ph () 4 (c) (i) K a = [H ][A [HA] ] (ii) Equation for [H] [H ] = K a [HA] () [A ] [HA] and [A ] () are altered to same extent maintaining [H ] and ph () [9] 5. (a) (i) pka = log 0 Ka () (ii) () mol dm () (iii) [H ] = [A ] or Ka = [H ] /[HA] () Hence = [H ] /0.80 [H ] = ( ) = () ph = log 0 [H ] () ph =.4 () Marked consequentially to a(ii) 7 (b) (i) Mol NaOH = mol X () mv/000 = /000 = () (ii) Mol HX remaining = original mol HX mol NaOH added () (5 0.9/000) = () (iii) Concentration of X /(5 0.5) = 0.7 () Concentration of HX /5.5 = 0.4 () ph of solution K a = [H ] [X ]/[HX] () [H ] = /0.7 () ph = 4.04 () Marked consequentially to b(iii) 9 (c) Change in ph Very small fall or slight change () Explanation H X HX () Equilibrium restored HX H X OR [9] 8

19 6. (a) Monoprotic acid An acid which gives only one proton () Example HCl etc () (b) (i) log 0 [H ] or in words () (ii).58 M (allow.6) () (c) Mol H = () =.75 0 Mol OH = () = Excess OH = () [OH ] = /60 () = [H ] = 0 4 /7.5 0 () =. 0 () ph = log 0. 0 =.9 () NB Consequential marking if [OH ] not calculated to maximum of 5 7 [] 7. (a) Equation for HCl(g) HCl(g) H (aq) Cl (aq) () Equation for KOH(s) KOH(s) K (aq) OH (aq) () (b) K w = [H ] [OH ] (c) strong base, fully dissociated () or [OH ] = 0.06 M () [H ] = K w [OH ] () = 0 4 = M () () 0.06 ph = log 0 [H ] () ph =. () neutral solution, [H ] = [OH ] () ph = 7 () 7 (d) (i) 755 cm of 0.0 M acid contain mol H = moles () 000 moles OH used for neutralisation = () (ii) ph =.6 [H ] = 0.6 =.5 0 M () [OH ] = = K [H w ] () [OH ] =.98 0 M () () in 755 cm there are =.0 0 mol () (iii) Total moles = (9.06.0) 0 = 0.0 mol () (iv) M r = 9 6 = 56 m = = 0.68g () [8] 9

20 8. (a) HF H O H O F () HCl H O H O Cl () for HF, must have reversible arrow allow (aq) in HCl equation (b) (i) ph = log0[h ] or equivalent word definition () allow -log[h O ] or -log[h (aq)] (ii) [H]= mol dm ph =.(0) () if correct definition demonstrated in (ii), but word definition in (i) wrong, allow mark transfer from (ii) to (i) (c) (i) K a = [H ][F [HF] ] () do not expression allow with [H O]; allow [H O ] allow consequential mark from wrong equation in (b) providing [H ] present (ii) [H ] K a = or [H ] = K a [HF] () [HF] 4 [H ] = = () ph =. /.8 () allow mark for correct ph from wrong [H ] (d) hydrogen fluoride or HF () donates a proton (to the nitric acid) () conjugate base F (this mark dependent on correct identification of acid) () [] 9. (a) (i) proton donor () (ii) partially dissociated (into ions) (not weakly dissociated) () (b) (i) K a = [CH [CH COO ][H COOH] ] () allow either [H O ] or [H ] in expression must include charges if [H O] included then no mark (ii) pk a = log(l ) = 4.77 / 4.8/4.80 () 0

21 (c) (i) indicator: phenolphthalein () explanation: weak acid-strong base / ph change above 7.0 () link between ph change and indicator range () (ii) colourless to red / pink / purple () if methyl orange named as indicator - wrong but allow second explanation mark and colour change mark in (ii) (ie pink / red yellow / orange) (iii) NaOH CH COOH CH COONa H O () if charges shown they must be correct accept H OH H O (d) excess acid NaOH / acid NaOH = NaEt () in : ratio of volumes / : mol / : acid: salt () (e) (i) OH / C=O bonds are polar (words or on diagram) due to different electronegativities of O and H (or O and C) () lone pairs of electrons on O atoms () attraction of δ H atom in OH for δ O atom / lone pair in C=O between different molecules () Max 0. (a) Weak acid An acid which only partially ionises () Example Ethanoic, carbonic etc () [ ][ ] H A (b) Expression Ka = [ HA] () Units mol dm (or mol l ) () (c) (i) The dissociation of water is an endothermic process () Less dissociation on cooling (or equilibrium moves to water or Kw decreases) () less H (or[h ] lower) () (ii) Because [H ] = [OH - ] () 4 (d) Resists change in ph () on addition of small quantities of acid or base () [4] [0]

22 . (a) (i) Strong = fully dissociated ) Weak = partially dissociated ) in aqueous solution both needed () (ii) K a [H] [HX].74 (0 ) = 0.6 = () mol dm (not M) () Explanation: HX barely dissociates OR [H ] very small () so [HX] eqm = [HX] original () and [H ] = [X ] in an aqueous solution of weak acid () (b) (i) moles H in 8 cm = =.88 0 () moles OH in V eqv cm = () (V eqv 0. 0 ) V eqv = =.0 cm () 0. Start Half equivalence Equivalence Double equivalence Volume/cm Ba(OH) solution added ph for titration of S ph for titration of W (ii) Strong acid S Weak acid W Half-equivalence Volume of base = 6 cm equivalence Volume of base = 4 cm 9 cm of strong acid S () in (8 6) cm = 4 cm () [H ] = 9 0.6/4 = 0.06 M () ph =. () cm of base B in excess () in (8 4) cm = 4 cm () [OH ] = 0./4 = M () [H ]= K w /[OH ] () =.46 0 M () ph =.84 () [HX] = [X ] (given) ph = pk a or [H ] = K a () = log ( ) = 4.68 () Same as for strong acid S ph =.84 () [ Max] 0 (iii) Graph: Sensibly scaled volume axis () 4 points in text correctly plotted ( for each error) () vertical portion at equivalence on graph for S () single high ph curve for both S and W () Weak acid (W) has marked ph rise at start ()

23 ph Volume Ba(OH) /cm 5 (c) (i) Buffer Properties resists change in ph () on adding small amounts of acid or base () W as a Buffer Plenty of X present to mop up H () Plenty of HX present to mop up OH () OR equations showing same e.g. H X HX, OH HX X H O (ii) acid buffers act at low ph, basic buffers act at high ph () halfneutralised W is an acid buffer () basic buffer: mix weak base with the salt of its coacid () OR correct specific example 6 [0]

24 . (a) ph = log[h ] K [H ] a = or [H ] = [CH COOH] [A ] [H ]= (or.6 0 ) ph =.79 (penalise dp or more than dp once in the qu) (b) (i) Solution which resists change in ph /maintains ph despite the addition of (small amounts of) acid/base (or dilution) (c) (ii) CH COO H CH COOH must show an equation full or ionic in which ethanoate ions are converted to ethanoic acid (i) K a [CH COOH] [H ] = [CH COO ] if rearrangement incorrect, no further marks = (= ) ph = 4.58 (ii) Ml moles H added = = 0.0 M moles ethanoic acid after addition = = 0.6 M moles ethanoate ions after addition = = 0.09 M4 [H ] = K a [CH COOH] =.74 0 [CH COO ] 5 0.6/V 0.09/V ( = ) M5 ph = 4.5 The essential part of this calculation is addition/subtraction of 0.0 moles to gain marks M and M. If both of these are missing, only mark Ml is available. Thereafter treat each mark independently, except if the expression in M4 is wrong, in which case both M4 and M5 are lost. alternative scheme for part (c)(i) [CH COOH] ph pk log [CH COO ] a pk a = ph = (4.76 log ) = alternative for penultimate mark of part (c)(ii) 0.6 ph = 4.76 log 0.09 [5] 4

M1. (a) ph = log[h + ] 1. ph = 2.79 (penalise 1 dp or more than 2dp once in the qu) despite the addition of (small amounts of) acid/base (or dilution)

M1. (a) ph = log[h + ] 1. ph = 2.79 (penalise 1 dp or more than 2dp once in the qu) despite the addition of (small amounts of) acid/base (or dilution) M. (a) ph = log[h + ] [H + ]=.74 0 5 0.5 (or.62 0 3 ) ph = 2.79 (penalise dp or more than 2dp once in the qu) (b) (i) Solution which resists change in ph /maintains ph despite the addition of (small amounts

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