Skip to content
OCR A Jun 2012 Paper 5 Q3
Answers available below
Show answer
Question
1
2
3
4
5
6
7
![3 Butanoic acid, CH3(CH2)2COOH, is the butter acid, formed when butter turns rancid and tastes sour. A student prepares an aqueous solution of butanoic acid with a concentration of 0.250 mol dm3. The Ka of butanoic acid is 1.51105 mol dm3. (a) (i) Write the expression for the acid dissociation constant of butanoic acid. (ii) Calculate the pKa of butanoic acid. [1] (iii) Calculate the pH of the 0.250 mol dm3 butanoic acid. Give your answer to two decimal places. pKa =[1] pH =[3] (b) The student adds aqueous butanoic acid to magnesium. The student then adds aqueous butanoic acid to aqueous sodium carbonate. (i) Write the ionic equation for the reaction between aqueous butanoic acid and magnesium.[1] (ii) Write the ionic equation for the reaction between aqueous butanoic acid and aqueous sodium carbonate.[1]OCR 2012<br />
(c) The student adds 50.0 cm3 of 0.250 mol dm3 butanoic acid to 50.0 cm3 of 0.0500 mol dm3 sodium hydroxide. A buffer solution forms. (i) Explain why a buffer solution forms.[2] (ii) Calculate the pH of the buffer solution. The Ka of butanoic acid is 1.51105 mol dm3. Give your answer to two decimal places. pH =[5] (d) The student adds methanoic acid, HCOOH (Ka = 1.82104 mol dm3), to butanoic acid. A reaction takes place to form an equilibrium mixture containing two acidbase pairs. Complete the equilibrium below and label the conjugate acidbase pairs. HCOOH + CH3(CH2)2COOH+[2] [Total: 16]OCR 2012 Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_5_QP_JUN12_question_3-0.jpg)
![F325 Mark Scheme June 2012 Question (a) (i) Answer (Ka =) [H+][CH3(CH2)2COO ] [CH3(CH2)2COOH] (cid:57) Marks Guidance ALLOW CH3CH2CH2COOH OR C3H7COOH in expression DO NOT ALLOW use of HA and A in this part. DO NOT ALLOW: [H+][CH3(CH2)2COO ] [CH3(CH2)2COOH] [CH3(CH2)2COOH] : CON [H+]2 ALLOW 4.82 up to calculator value of 4.821023053 DO NOT ALLOW 4.8 IF alternative answer to more or fewer decimal places, check calculator value and working for 1st and 2nd marks ------------------------------------------------------------ ALLOW use of HA and A in this part Calculator: 1.942935923 x 103 ALLOW use of calculated Ka value, either calculator value or rounded on script. pH must be to 2 decimal places ALLOW ECF from incorrectly calculated [H+] and pH ONLY when values for both Ka AND [CH3CH2CH2COOH] have been used, i.e. 1.5 x 105 AND 0.250. e.g.: pH = 5.42 2 marks pH = 2.11 2 marks log(1.51 x 105 x 0.250) Nolog( 1.51(cid:117)10(cid:16)5 0.250 ) log( 1.51(cid:117)10(cid:16)5 0.250 pH = 4.22 1 mark DO NOT ALLOW just log(1.51 x 105) = 4.82 NO MARKS ) No(ii) pKa = logKa = 4.82 (cid:57) (iii) FIRST, CHECK THE ANSWER ON ANSWER LINE [Ka][CH3(CH2)2COOH] OR 1.51(cid:117) 10(cid:16)5 (cid:117) 0.250 IF answer = 2.71 award 3 marks -------------------------------------------------------------------- [H+] = (cid:57) [H+] = 1.94 x 103 (mol dm3) (cid:57) pH = log[H+] = 2.71 (cid:57) 10<br />
F325 Mark Scheme June 2012 Marks Guidance 2Mg2+ 2 + 2 CH3(CH2)2COOH (cid:127)(cid:111) IGNORE state symbols ALLOW Mg + 2 CH3(CH2)2COOH (cid:127)(cid:111) 2CH3(CH2)2COO + Mg2+ + H2 DO NOT ALLOW on RHS: (CH3(CH2)2COO Ions must be shown separately IGNORE state symbols ALLOW CO3 2 CH3(CH2)2COO + H2O + CO2 ALLOW as product H2CO3 ALLOW names throughout ALLOW sodium salt of butanoic acid ALLOW CH3(CH2 2COOH + NaOH (cid:111) CH3(CH2)2COONa + H2O DO NOT ALLOW just forms a sal i.e. identity of product is required t/conjugate base Question Answer (b) (i) Mg + 2H+ (cid:127)(cid:111) Mg2+ + H2 (cid:57) (ii) CO3 2 + 2H+ (cid:127)(cid:111) H2O + CO2 (cid:57) (i) CH3(CH2)2COONa OR CH3(CH2)2COO forms OR CH3(CH2)2COOH + OH (cid:111) CH3(CH2)2COO + H2O (cid:57) CH3(CH2)2COOH is in excess OR acid is in excess OR some acid remains (cid:57) (c) 11<br />
F325 Mark Scheme June 2012 Question Answer (c) (ii) Moles (2 marks) amount CH3(CH2)2COOH = 0.0100 (mol) (cid:57) amount CH3(CH2)2COO = 0.0025 (mol) (cid:57) Concentration (1 mark) [CH3(CH2)2COOH] = 0.100 mol dm3 AND [CH3(CH2)2COO] = 0.025 mol dm3 (cid:57) [H+] and pH (2 marks) 0.100 0.025 [H+] = 1.51(cid:117)10(cid:16)5 (cid:117) (cid:57) pH = log 6.04 x 10-5 = 4.22 (cid:57) = 6.04 x 10-5 (mol dm3) pH to 2 DP 0.025 0.100 OR pKalog 0.100 pH = pKa + log 0.025 (cid:57) TAKE CARE with awarding marks for pH = 4.22 There is a mark for the concentration stage. If this has been omitted, the ratio for the last 2 marks will be 0.0100 and 0.0025. 4 marks max. Common errors pH = 5.42 inverted. As above for 4.22 but with acid/base ratio Award 4 OR 3 marks Award zero marks for: 4.12 from no working or random values pH value from Ka square root approach (weak acid pH) pH value from Kw /1014 approach (strong base pH) Marks Guidance ANNOTATIONS MUST BE USED -------------------------------------------------------------------------------- ALLOW HA and A throughout Mark by ECF throughout ONLY award final 2 marks via a correct pH calculation via Ka (cid:117) question (i.e. not just made up values) [CH3(CH2)2COOH] [CH3(CH2)2COO(cid:16)] using data derived from that in the pH = 4.820.60 = 4.22 (cid:57) ALLOW logKa for pKa Common errors pH = 4.12 use of initial concentrations: 0.250 and 0.050 given in question. Award last 3 marks for: 0.250/2 AND 0.050/2 = 0.125 AND 0.025 (cid:57) 1.51(cid:117)10(cid:16)5 (cid:117) pH = log[H+] = 4.12 (cid:57) Award last 2 marks for: 0.125 0.025 = 7.55 x 10-5 (mol dm3) (cid:57) 1.51(cid:117)10(cid:16)5 (cid:117) pH = log[H+] = 4.12 (cid:57) 0.250 0.050 = 7.55 x 10-5 (mol dm3) (cid:57) pH = 5.52 As above for 4.12 but with acid/base ratio inverted. Award 2 OR 1 marks as outlined for 4.12 above 12 ALLOW alternative approach based on HendersonHasselbalch equation for final 2 marks<br />
Mark Scheme June 2012 Marks Guidance State symbols NOT required ALLOW 1 and 2 labels the other way around. ALLOW just acid and base labels throughout if linked by lines so that it is clear what the acid-base pairs are For 1st mark, DO NOT ALLOW COOH (i.e. H at end rather than start) but within 2nd mark ALLOW COOH by ECF IF proton transfer is wrong way around then ALLOW 2nd mark for idea of acidbase pairs, i.e. HCOOH + CH3(CH2)2COOHHCOOH2 base 2 acid 1 acid 2 base 1 (cid:57) For H2COOH+ shown with wrong proton transfer, DO NOT ALLOW an ECF mark for acidbase pairs + + CH3(CH2)2COO (cid:56) F325 Question (d) Answer HCOOH + CH3(CH2)2COOHHCOO + CH3(CH2)2COOH2 (cid:57) acid 1 base 2 base 1 acid 2 (cid:57) CARE: Both + andcharges are required for the products in the equilibrium DO NOT AWARD the 2nd mark from an equilibrium expression that omits either charge Total 16 13<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_5_QP_JUN12_answer_3-0.jpg)
