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OCR A Jun 2015 Paper 5 Q4
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![11 4 A student is supplied with 0.500 mol dm3 potassium hydroxide, KOH, and 0.480 mol dm3 propanoic acid, C2H5COOH. The acid dissociation constant, Ka, for C2H5COOH is 1.35105 mol dm3. (a) C2H5COOH is a weak BrnstedLowry acid. What is meant by a weak acid and BrnstedLowry acid ?[1] (b) Calculate the pH of 0.500 mol dm3 potassium hydroxide. (c) The student dilutes 25.0 cm3 0.480 mol dm3 C2H5COOH by adding water until the total volume is 100.0 cm3. pH =[2] (i) Write the expression for Ka for C2H5COOH. (ii) Calculate the pH of the diluted solution. [1]OCR 2015 pH =[3] Turn over<br />
(d) Aqueous propanoic acid, C2H5COOH, reacts with carbonates and alkalis. 12 (i) Write the full equation for the reaction of aqueous propanoic acid with sodium carbonate.[1] (ii) Write the ionic equation for the reaction of aqueous propanoic acid with aqueous potassium hydroxide.[1] (e) A student prepares a buffer solution containing propanoic acid C2H5COOH and propanoate ions, C2H5COO. The concentrations of C2H5COOH and C2H5COO are both 1.00 mol dm3. The following equilibrium is set up. C2H5COOH(aq) C2H5COO(aq) + H+(aq) The acid dissociation constant, Ka, for C2H5COOH is 1.35105 mol dm3. (i) Calculate the pH of this buffer solution. Give your answer to two decimal places. pH =[1] (ii) A small amount of aqueous ammonia, NH3(aq), is added to the buffer solution. Explain, in terms of equilibrium, how the buffer solution would respond to the added NH3(aq).[2]OCR 2015<br />
(iii) The student adds 6.075 g Mg to 1.00 dm3 of this buffer solution. 13 Calculate the pH of the new buffer solution. Give your answer to two decimal places pH =[4] [Total: 16]OCR 2015 Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_5_QP_JUN15_question_4-0.jpg)
![Mark Scheme June 2015 Answer Marks Guidance (calculator 0.301029995) For pOH method:, ALLOW pOH = log[OH] = 0.3(0) (cid:57) ALLOW pH = 140.3 = 13.7 (cid:57) ALLOW 13.7 up to calculator value of 13.69897 correctly rounded. ALLOW ECF from incorrect [H+(aq)] provided that pH >7 IGNORE [H+ ]2 OR [H+ ] [A] [HA] [C2H5COOH] ALLOW [H3O+] for [H+] IGNORE state symbols F325 Question 4 (a) (b) Proton/H+ donor AND Partially dissociates/ionises (cid:57) FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer = 13.7(0), award 2 marks[H+] = 1.00 1014 0.5(00) pH = log 2(.00)1014 = 13.7(0) (cid:57) OR 2(.00)1014 (mol dm3) (cid:57) (c) (i) (Ka =) [H+ ] [C2H5COO ] [C2H5COOH] (cid:57) 16<br />
F325 Question 4 (c) Mark Scheme June 2015 Answer Marks Guidance (ii) FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer = 2.9(0), award 3 marks[C 2H5COOH] = 0.12(0) mol dm3 (cid:57) [H+] = K a[C2H5COOH] = 1.35 1050.12(0) OR 1.27103 (mol dm3) (cid:57) pH = log 1.27103 = 2.9(0) (cid:57) NOTE: The final two marks are ONLY available from attempted use of Ka AND [C2H5COOH] ALLOW HA for C2H5COOH and A for C2H5COO ALLOW ECF from incorrectly calculated [C2H5COOH] ALLOW 1.27103 to calculator value of 1.272792206103 correctly rounded ALLOW 2.9(0)103 to calculator value of 2.895242493 correctly rounded ALLOW use of quadratic equation which gives same answer of 2.90 from 0.120 mol dm3COMMON ERRORS (MUST be to AT LEAST 2 DP unless 2nd decimal place is 0) pH = 2.59 2 marks pH = 5.79 2 marks pH = 5.19 1 mark pH = 4.87 0 marks log (1.35 x 1050.480) Original conc, no (cid:165) log(cid:165) (1.35 x 1050.480) Original conc log(1.35 x 1050.120) log(1.35 x 105) = 4.87 log Ka No (cid:165) 17<br />
F325 Question 4 (d) (i) 2C2H5COOH + Na2CO32C2H5COONa + CO2 + H2O (cid:57) (d) (ii) H+ + OHH2O (cid:57) (e) (i) pH = log 1.35105 = 4.87 (cid:57) (e) (ii) Added ammonia C2H5COOH removes added NH3/alkali/base OR C2H5COOH + NH3 / OHOR NH3/alkali reacts with/accepts H+ OR H+ + NH3OR H+ + OH(cid:57) EqulibriumC2H5COO OR Equilibriumright (cid:57) Mark Scheme June 2015 Answer Marks Guidance (Question asks for 2 DP) IGNORE state symbols and use of equilibrium sign FOR CO2 + H2O ALLOW H2CO3 ALLOW C2H5COONa+ OR C2H5COO + Na+ BUT BOTH + andcharges must be shown ALLOW NaC2H5COO ALLOW C2H5COOH + OHC2H5COO + H2O IGNORE state symbols ONLY correct answer DO NOT ALLOW 4.9 ALLOW use of HA/weak acid/acid for C2H5COOH; ALLOW use of NH4OH for NH3 ALLOW A for C2H5COO ASSUME that equilibrium applies to that supplied in the question, i.e. IGNORE any other equilibria 18<br />
F325 Question 4 (e) (iii) CHECK WORKING CAREFULLY AS CORRECT NUMERICAL Mark Scheme June 2015 Answer Marks Guidance ANSWER IS POSSIBLE FROM WRONG VALUES ================================================== ALLOW HA and A throughout (1 mark) Amount of Mg 6.075 n(Mg) = 24.3 = 0.25(0) mol (cid:57)Moles/concentrations (2 marks) n(C2H5COOH) = 1.00(20.25) = 0.50 (mol) (cid:57) (C2H5COO) = 1.00 + (20.25) = 1.50 (mol) (cid:57)n [H+] and pH +] = 1.35105(1 mark) 0.50 1.50 OR 4.5106 (mol dm3) [H pH = log 4.5106 = 5.35 2 dp required (cid:57) NOTE: IF there is no prior working, ALLOW 4 MARKS for [H+] =1.35105IF the ONLY response is pH = 5.35, award 1 mark ONLY Award a maximum of 1 mark (for n(Mg) = 0.25 mol) for: 0.50 1.50 AND pH = 5.35 pH value from Ka square root approach (weak acid pH) pH value from Kw /1014 approach (strong base pH) FULL ANNOTATIONS MUST BE USEDFor n(Mg), 1 mark ALLOW ECF for ALL marks below from incorrect n(Mg) ECF ONLY available from concentrations that havesubtracted 0.50 OR 0.25 from 1 for [C2H5COOH]added 0.50 OR 0.25 to 1 for [C2H5COO] i.e. For moles/concentration 1 mark (1 mark lost) 1. n (C2H5COOH) = 0.75 AND n(C2H5COO) = 1.25 2. n(C2H5COOH) = 0.50 AND n(C2H5COO) = 1.25 3. n(C2H5COOH) = 0.75 AND n(C2H5COO) = 1.50ALLOW ECF ONLY for the following giving 1 additional mark and a total of 3 marks 0.75 1.25 pH = log 8.1106 = 5.09 1. [H +] = 1.351052. [H +] = 1.351053. [H +] = 1.351050.50 1.25 pH = log 5.4106 = 5.27 0.75 1.50 pH = log 6.75106 = 5.17ALLOW alternative approach based on HendersonHasselbalch equation for final 1 mark ALLOW log Ka for pKa pH = pKa + log 1.5 OR pKalog 0.5 0.5 1.5 pH = 4.87 + 0.48 = 5.35 (cid:57) Total 16 19<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_5_QP_JUN15_answer_4-0.jpg)
