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OCR A Jun 2010 Paper 5 Q3
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![The chemicals that we call acids have been known for thousands of years. However, modern theories of acids have been developed comparatively recently. It wasnt until the early 1900s that the concept of dissociation became accepted by the scientific community and the concept of pH was introduced. A student carried out a series of experiments with acids and alkalis. (a) Propanoic acid, CH3CH2COOH, is a naturally occurring weak acid. The equation for the dissociation of propanoic acid is shown below. CH3CH2COOH(aq) H+(aq) + CH3CH2COO(aq) The student wanted to prove that propanoic acid is a weak acid. The student had access to a pH meter and 0.100 mol dm3 propanoic acid. Explain how the student could prove that propanoic acid is a weak acid by taking a single pH measurement. Show how the student could then calculate the acid dissociation constant, Ka, for propanoic acid.[4] (b) The student measured the pH of a solution of sodium hydroxide at 25 C. The measured pH was 13.46. Calculate the concentration of the aqueous sodium hydroxide.OCR 2010 concentration =mol dm3 [2]<br />
(c) A student made a buffer solution by mixing an excess of propanoic acid to an aqueous solution of sodium hydroxide at 25 C. This buffer solution contains an equilibrium system that minimises changes in pH when small amounts of acids and alkalis are added. Explain why a buffer solution formed when an excess of propanoic acid was mixed with aqueous sodium hydroxide. Explain how this buffer solution controls pH when 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.[7]OCR 2010 Turn over<br />
(d) A student added nitric acid to propanoic acid. A reaction took place to form an equilibrium mixture containing two acidbase pairs. Complete the equilibrium below and label the two conjugate acidbase pairs. HNO3 + CH3CH2COOH+[2] (e) Finally, the student reacted an aqueous solution of propanoic acid with a reactive metal and with a carbonate. (i) Write an equation for the reaction of aqueous propanoic acid with magnesium.[1] (ii) Write an ionic equation for the reaction of aqueous propanoic acid with aqueous sodium carbonate.[1] [Total: 17]OCR 2010<br />
BLANK PAGE QUESTION 4 STARTS ON PAGE 10 PLEASE DO NOT WRITE ON THIS PAGEOCR 2010 Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_5_QP_JUN10_question_3-0.jpg)
![F325 Question 3 a Marks Expected Answers measured pH > 1 OR [H+] < 0.1 (mol dm3) (cid:57) [H+] = 10pH (cid:57) Ka = [H+][CH3CH2COO] OR [H+]2 (cid:57) [CH3CH2COOH] [CH3CH2COOH] Calculate Ka from (cid:57) [H+]2 0.100 Mark Scheme June 2010 Additional Guidance ALLOW C2H5 throughout question ALLOW [H+] < [CH3CH2COOH] OR [H+] < [HA] ALLOW measured pH is higher than expected ALLOW measured pH is not as acidic as expected ALLOW a quoted pH value or range > 1 and < 7 OR between 1 and 7 ALLOW [H+] = antilog pH OR [H+] = inverse log pH ALLOW [H+][A] OR [H+]2 [HA] [HA] IF Ka is NOT given and Ka = [H+]2 0.100 is shown, award mark for Ka also is automatically awarded the last 2 marks) [H+]2 (i.e. Ka = 0.100 ALLOW 3.4673685051014 and correct rounding to 3.51014 ALLOW 0.28840315 and correct rounding to 0.29, i.e. ALLOW 0.288 ALLOW alternative approach using pOH: pOH = 1413.46 = 0.54 (cid:57) [OH] = 100.54 = 0.29 (mol dm3) (cid:57) Correct answer gets BOTH marks Marks are for correctly calculated values. Working shows how values have been derived. [H+] = 1013.46 = 3.47 x 1014 (mol dm3) (cid:57) [OH] = 1.01014 = 0.29 (mol dm3) (cid:57) 3.471014<br />
F325 Question Mark Scheme June 2010 Marks Expected Answers Propanoic acid reacts with sodium hydroxide forming propanoate ions/sodium propanoate OR CH3CH2COOH + NaOH (cid:111) CH3CH2COONa + H2O (cid:57) Some propanoic acid remains OR propanoic acid AND propanoate (ions) / sodium propanoate present (cid:57) equilibrium: CH3CH2COOHH+ + CH3CH2COO(cid:57) Added alkali CH3CH2COOH reacts with added alkali OR CH3CH2COOH + OH (cid:111) OR added alkali reacts with H+ OR H+ + OH (cid:111) (cid:57) (cid:111) CH3CH2COO OR Equilibrium (cid:111) right (cid:57) Added acid CH3CH2COO reacts with added acid OR [H+] increases (cid:57) (cid:111) CH3CH2COOH OR Equilibrium (cid:111) left (cid:57) Additional Guidance ANNOTATIONS MUST BE USED ALLOW C2H5 throughout question ALLOW Adding NaOH forms propanoate ions/sodium propanoate (imples that the NaOH is added to the propanoic acid) ALLOW: weak acid AND its conjugate base/salt present Throughout, do not penalise comments that imply that pH is constant in presence of buffer DO NOT ALLOW HA and A in this equilibrium expression For description of action of buffer below, ALLOW HA for CH3CH2COOH; ALLOW A for CH3CH2COO Equilibrium responses must refer back to a written equilibrium. IF no equilibrium shown, use the equilibrium as written in expected answers (which is also written on page 6 of the paper) ALLOW weak acid reacts with added alkali ALLOW conjugate base reacts with added acid DO NOT ALLOW salt reacts with added acid<br />
F325 Mark Scheme June 2010 Question Expected Answers HNO3 + CH3CH2COOHCH3CH2COOH2 + + NO3(cid:57) acid 1 base 2 acid 2 base 1 (cid:57) Marks i 2CH3CH2COOH + Mg (cid:111) (CH3CH2COO)2Mg + H2 (cid:57) Additional 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 acidbase pairs are. IF proton transfer is wrong way around then ALLOW 2nd mark for idea of acidbase pairs, i.e. + (cid:56) HNO3 + CH3CH2COOHCH3CH2COO + H2NO3 base 2 acid 1 base 1 acid 2 (cid:57) IGNORE state symbols ALLOW ionic equation: 2H+ + Mg (cid:111) Mg2+ + H2 IGNORE any random charges in formula of (CH3CH2COO)2Mg as long as the charges are correct (charges are treated as working) i.e. (CH3COO)2 Mg OR (CH3COO)2 However, Mg2+ instead of Mg on the left side of equation is obviously wrong State symbols NOT requiredMg should not be penalised ii 2H+ + CO3 OR 2H+ + CO3 OR H+ + CO3 2 (cid:127)(cid:111) H2O + CO2 2 (cid:127)(cid:111) H2CO3 2 (cid:127)(cid:111) HCO3 (cid:57) Total 17<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_5_QP_JUN10_answer_3-0.jpg)
