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1 M.(a) [H + ] = or = Allow ( ) M = If [HX] / [X - ] or. upside down, or any addition or subtraction lose M & ph = 4.5 (correct answer scores 3) Can score for correct ph conseq to their [H + ], so ph = 5.0 scores one Must be to 2 dp Alternative using Henderson Hasselbach Equation ph = pka log[hx] / X ] = log( ) log( ) Allow ( ) M pka = If [HX] / [X ] or upside down, can only score ph = = 4.5 so ph = 5.0 Must be to 2 dp Page 2

3 M mol salt after addition = = 0.25 For X if no subtraction or error in subtraction (other than AE) (or subsequent extra add or sub) MAX 3 ph = (pka log[hx] / [X ]) = log( ) log(0.266 / 0.25) If errors above in both addition AND subtraction can only score for insertion of their numbers except if addition and subtraction reversed then ph = 4.58 scores 2 ph = M4 ph = = 4.43 If [HX] / [X ] upside down, lose & M4 (or next two marks) but can score M5 for correct ph conseq to their working, so if M & correct, ph = 5.09 scores 3. Allow more but not fewer than 2dp here. M5 [8].(a) HCO 3 CO H + or H 2 O + HCO 3 CO H 3 O + Must have equilibrium sign but mark on to (b) Ignore state symbols (b) Acid: Increase in concentration of H + ions, equilibrium moves to the left. Allow H + ions react with carbonate ions (to form HCO 3 - ) Alkali: OH - reacts with H + ions, equilibrium moves to the right (to replace the H + ions) Concentration of H + remains (almost) constant [4] Page 4

4 .(a) Z Mark independently. The idea that the solution contains both HA and A (b) ph [HA] = [A ] Accept solution half neutralised. ph = pk a Accept [H + ] = K a [5] M4.(a) (only) slightly or partially dissociated / ionised Ignore not fully dissociated. Allow low tendency to dissociate or to lose / donate a proton. Allow shown equilibrium well to the left. Otherwise ignore equations. (b) 2CH 3CH 2COOH + Na 2CO 3 2CH 3CH 2COONa + H 2O + CO 2 OR 2CH 3CH 2COOH + CO 3 2 2CH 3CH 2COO + H 2O + CO 2 OR Page 5

5 CH 3CH 2COOH + Na 2CO 3 CH 3CH 2COONa + NaHCO 3 OR CH 3CH 2COOH 2 + CO 3 CH 3CH 2COO + HCO 3 Must be propanoic acid, allow C 2H 5COOH. Not molecular formulae. Allow multiples. Ignore reversible sign. Not H 2CO 3. (c) [OH ] = = M Correct answer for ph with or without working scores 3. [H + ] = = OR poh =.62 If 2 missed or used wrongly can only score for correct calculation of ph from their [H + ]. ph = 2.38 Lose if not 2 decimal places: 2.4 scores scores (missing 2) ; 2. scores scores (dividing by 2).8 scores 0. (d) (i) K a = Ignore ( ) here but brackets must be present. Must be correct acid and salt. If wrong, mark part (ii) independently. (ii) M K a = OR with numbers Page 6

6 Correct answer for ph with or without working scores 3. Allow HX, HA and ignore ( ) here. May score M in part (i). [H + ] = ( ) or (K a [C 6H 5COOH]) (= ( = ) ph = 6.2 may score 2 if correct working shown and they show the square root but fail to take it. But if no working shown or wrong K a = used which also leads to 6.2, then zero scored. ph = 3.06 Must be 2 decimal places ie 3. loses. (iii) M [H + ] = = Correct answer for mass with or without working scores 5. Allow 0 4. [X ] = Ignore ( ) here. If [HX] / [X ] upside down, can score M plus M4 for = And M5 for g. Page 7

7 M4 = M5 Mass (C 6H 5COONa) = =.09 g or. g Wrong method, eg using [H + ] 2 may only score M and M5 for correct multiplication of their M4 by 44 (provided not of obviously wrong substance). (e) M CO 2 Allow NO x and SO 2. ph (It) falls / decreases If M wrong, no further marks. mark & independently acidic (gas) OR reacts with alkali(ne solution) / OH OR CO 2 + 2OH CO H 2O OR CO 2 + OH HCO 3 Not forms H 2CO 3 H 2SO 3 H 2SO 4 etc OR H + ions. [7] M5.(a) This question is marked using levels of response. Refer to the Mark Scheme Instructions for Examiners for guidance on how to mark this question. All stages are covered and the explanation of each stage is generally correct and virtually complete. Answer is communicated coherently and shows a logical progression from stage and stage 2 to stage 3. Steps in stage 3 must be complete, ordered and include a comparison. Level marks Page 8

8 All stages are covered but the explanation of each stage may be incomplete or may contain inaccuracies OR two stages are covered and the explanations are generally correct and virtually complete. Answer is mainly coherent and shows a progression from stage and stage 2 to stage 3. Level marks Two stages are covered but the explanation of each stage may be incomplete or may contain inaccuracies, OR only one stage is covered but the explanation is generally correct and virtually complete. Answer includes some isolated statements, but these are not presented in a logical order or show confused reasoning. Level 2 marks Insufficient correct Chemistry to warrant a mark. Level 0 0 marks Indicative Chemistry content Stage : difference in structure of the two acids The acids are of the form RCOOH but in ethanoic acid R = CH 3 whilst in ethanedioic acid R = COOH Stage 2: the inductive effect The unionised COOH group contains two very electronegative oxygen atoms therefore has a negative inductive (electron withdrawing)effect The CH 3 group has a positive inductive (electron pushing) effect Stage 3: how the polarity of OH affects acid strength The O H bond in the ethanedioic acid is more polarised / H becomes more δ + More dissociation into H + ions Ethanedioic acid is stronger than ethanoic acid 6 (b) Moles of NaOH = Moles of HOOCCOO formed = Extended response Moles of HOOCCOOH remaining = = Page 9

9 K a = [H + ][A ] / [HA] [H + ] = K a [HA] / [A ] [H + ] = ( / V) / ( / V) = ph = log 0( ) =.406 =.4 Answer must be given to this precision (c) 5H 2C 2O 4 + 6H + + 2MnO 4 2Mn CO 2 + 8H 2O OR 5C 2O H + + 2MnO 4 2Mn CO 2 + 8H 2O Moles of KMnO 4 = / 000 = Moles of H 2C 2O 4 = 5 / = Concentration = moles / volume (in dm 3 ) = / 25 = (mol dm 3 ) If : ratio or incorrect ratio used, and M4 can be scored [5] Page 0

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