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OCR A Jun 2014 Paper 5 Q6
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![6 Ethanoic acid, CH3COOH, is a weak BrnstedLowry acid. 13 (a) An acidbase equilibrium is set up when ethanoic acid is added to water. Write the equation for the equilibrium that would be set up and label the two conjugate acidbase pairs.++[2] (b) An aqueous solution of CH3COOH has a pH of 3.060. This solution contains both hydrogen ions and hydroxide ions. (i) How can an aqueous solution of an acid contain hydroxide ions?[1] (ii) Calculate the concentration of hydroxide ions in this solution of ethanoic acid. concentration of hydroxide ions =mol dm3 [2]OCR 2014 Turn over<br />
(c) A student adds an excess of aqueous ethanoic acid to solid calcium carbonate. The resulting solution is able to act as a buffer solution. 14 (i) Write a full equation for the reaction between ethanoic acid and solid calcium carbonate.[1] (ii) Explain why this buffer solution has formed.[1] (iii) Explain how this buffer solution controls pH when either an acid or an alkali is added. In your answer you should explain how the equilibrium system allows the buffer solution to control the pH.[5]OCR 2014<br />
15 (d) A biochemist plans to make up a buffer solution with a pH of 5.000. The biochemist adds solid sodium ethanoate, CH3COONa, to 400 cm3 of 0.200 mol dm3 ethanoic acid. Ka for ethanoic acid = 1.75105 mol dm3 Calculate the mass of sodium ethanoate that the biochemist needs to dissolve in the ethanoic acid to prepare this buffer solution. Assume that the volume of the solution remains constant at 400 cm3 on dissolving the sodium ethanoate.OCR 2014 [5] [Total: 17] Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_5_QP_JUN14_question_6-0.jpg)
![Mark Scheme June 2014 Guidance IGNORE state symbols (even if incorrect) ALLOW 1 AND 2 labels the other way around. ALLOW just acid and base labels if linked by lines so that it is clear what the acidbase pairs are ALLOW A and B for acid and base IF proton transfer is wrong way around ALLOW 2nd mark for idea of acidbase pairs, i.e. + + OH (cid:56) CH3COOH + H2OCH3COOH2 Base 2 Acid 1 Acid 2 Base 1 (cid:57) NOTE For the 2nd marking point (acidbase pairs), this is the ONLY acceptable ECF i.e., NO ECF from impossible chemistry ALLOW Kw = [H+] [OH] OR [H+] [OH] = 1014 (mol2 dm6) IGNORE breaking for dissociation IGNORE water contains H+ and OH IGNORE H2OH+ + OH i.e. no equilibrium sign IGNORE 2H2OH3O+ + OH i.e. no equilibrium sign F325 Question (a) Answer CH3COOH + H2OH3O+ + CH3COO (cid:57) Acid 1 Base 2 Acid 2 Base 1 (cid:57) Marks (b) (i) Water dissociates/ionises OR H2OH+ + OH OR 2H2OH3O+ + OH (cid:57) 19<br />
F325 (b) (ii) FIRST, CHECK THE ANSWER ON ANSWER LINE Mark Scheme June 2014 IF there is an alternative answer, check to see if there is any ECF credit possible using working below. ------------------------------------------------------------- ALLOW 2 SF: 8.7 (cid:117) 104 up to calculator value of 8.7096359 (cid:117) 104 correctly rounded ALLOW alternative approach using pOH: pOH = 143.06 = 10.94 (cid:57) [OH] = 1010.94 = 1.15 (cid:117) 1011 (mol dm3) (cid:57) IGNORE state symbols ALLOWprovided that reactants on LHS For CO2 + H2O, ALLOW H2CO3 ALLOW Ca(CH3COO)2 ALLOW (CH3COO)2Ca2+ BUT DO NOT ALLOW if either charge is missing or incorrect IF answer = 1.15 (cid:117) 1011, award 2 marks -------------------------------------------------------------------- [H+] = 103.06 = 8.71 (cid:117) 104 (mol dm3) (cid:57) [OH] = 1.00 (cid:117) 1014 8.71 (cid:117) 104 = 1.15 (cid:117) 1011 (mol dm3) (cid:57) ALLOW answer to two or more significant figures 2SF: 1.1 (cid:117) 1011; 4SF: 1.148 (cid:117) 1011; calculator 1.148153621 (cid:117) 1011 2CH3COOH + CaCO3(CH3COO)2Ca + CO2 + H2O (cid:57) (c) (i) 20<br />
June 2014 ethanoate for CH3COO ALLOW names: ethanoic acid for CH3COOH ALLOW calcium ethanoate OR (CH3COO)2Ca for CH3COO IGNORE acid, salt, conjugate base; responses must identify the acid and conjugate base as ethanoic acid and ethanoate IGNORE ethanoic acid is in excess (in question) BUT DO ALLOW some ethanoic acid is left over/present/some ethanoic acid has reacted IGNORE equilibrium: CH3COOHH+ + CH3COO Dissociation of ethanoic acid only F325 (c) (ii) solution contains CH3COOH AND CH3COO (cid:57) Mark Scheme 21<br />
F325 (c) (iii) Quality of written communication, QWC Mark Scheme June 2014 2 marks are available for explaining how the equilibrium system allows the buffer solution to control the pH on addition of H+ and OH- (see below) ------------------------------------------------------ CH3COOHH+ + CH3COO (cid:57) ----------------------------------------------------- CH3COOH reacts with added alkali OR CH3COOH + OH (cid:111) OR added alkali reacts with H+ OR H+ + OH (cid:111) (cid:57) Equilibrium (cid:111) right OR Equilibrium (cid:111) CH3COO (cid:57) (QWC) CH3COO reacts with added acid (cid:57) Equilibrium (cid:111) left OR Equilibrium (cid:111) CH3COOH (cid:57) (QWC) FULL ANNOTATIONS MUST BE USED -------------------------------------------------------------------------------- Note: If there is no equilibrium equation then the two subsequent equilibrium marks are not available: max 2 DO NOT ALLOW HAH+ + A DO NOT ALLOW more than one equilibrium equation. ----------------------------------------------------- ALLOW response in terms of H+, A and HA IF more than one equilibrium shown, it must be clear which one is being referred to by labeling the equilibria. ALLOW weak acid reacts with added alkali DO NOT ALLOW acid reacts with added alkali ALLOW conjugate base reacts with added acid DO NOT ALLOW salt/base reacts with added acid 22<br />
F325 (d) FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer = 11.48 OR 11.5 (g), award 5 marks -------------------------------------------------------------------- [H+] = 105 (mol dm3) (cid:57) -------------------------------------------------------------------------- [CH3COO] = 1.75 (cid:117)10(cid:16)5 (cid:57) (cid:117) 0.200 = 0.350 mol dm3 (cid:57) 105 n(CH3COONa/CH3COO) in 400 cm3 400 1000 = 0.350 (cid:117) = 0.14(0) (mol) (cid:57) ------------------------------------------------------------------------ mass CH3COONa = 0.140 (cid:117) 82.0 = 11.48 OR 11.5 (g) (cid:57) For ECF, n(CH3COONa/CH3COO) must have been calculated in step before Mark Scheme June 2014 FULL ANNOTATIONS MUST BE USED -------------------------------------------------------------------------------- IF there is an alternative answer, check to see if there is any ECF credit possible. 105 (cid:57) (cid:117) 0.08 = 0.14(0) (mol) (cid:57)(cid:57) Incorrect use of [H+] = ( [CH3COOH] (cid:117) Ka ) scores zero BUT IGNORE if an alternative successful method is present Incorrect use of Kw, 1 max for [H+] = 105 (mol dm3) BUT IGNORE if an alternative successful method is present ------------------------------------------------------------------------ ALLOW n(CH3COONa/CH3COO) = 1.75 (cid:117)10(cid:16)5 Note: There is no mark just for n(CH3COOH) in 400 cm3 = 0.200 (cid:117) 400 1000 ----------------------------------------------------------- As alternative for the 4th and 5th marks, ALLOW: mass of CH3COONa in 1 dm3 = 0.350 (cid:117) 82.0 = 28.7 g (cid:57) mass of CH3COONa in 400 cm3 = 28.7 (cid:117) 400 1000 ------------------------------------------------------------- COMMON ECF 4.592 OR 4.6 g AWARD 4 marks use of 400/1000 twice = 0.08 (mol) = 11.48 g (cid:57) 23<br />
F325 Mark Scheme June 2014 ALLOW variants of HendersonHasselbalch equation. pKa = log(1.75 x 105) = 4.757 (cid:57) Calc: 4.75696 log [CH3COO(cid:16) ] [CH3COOH] = pHpKa = 54.757 = 0.243 = 100.243 = 1.75 (cid:57) [CH3COO(cid:16) ] [CH3COOH] [CH3COO] = 1.75 (cid:117) 0.200 = 0.350 mol dm3 (cid:57) n(CH3COONa/CH3COO) in 400 cm3 ------------------------------------------------------------------------ mass CH3COONa = 0.140 (cid:117) 82.0 = 11.48 OR 11.5 (g) (cid:57) = 0.350 (cid:117) 400 1000 = 0.14(0) (mol) (cid:57) Total 17 24<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_5_QP_JUN14_answer_6-0.jpg)
