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OCR A Jun 2013 Paper 5 Q4
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![This question looks at pH values and reactions of acids, bases and buffers. (a) 0.14 mol dm3 solutions of hydrochloric acid, HCl, and chloric(I) acid, HClO (pKa = 7.43), have different pH values. Explain why the pH values are different and calculate the pH of 0.14 mol dm3 solutions of HCl and HClO to two decimal places. Show any working in calculations.[5] (b) Aluminium powder is added to aqueous ethanoic acid, CH3COOH. Write full and ionic equations for the reaction that takes place. full equationionic equation[2]OCR 2013<br />
(c) Calculate the pH of a 0.40 mol dm3 solution of NaOH. (d) In biochemistry, buffer solutions based on methanoic acid can be used in the analysis of urine samples. (i) Explain what is meant by the term buffer solution. Describe how a buffer solution based on methanoic acid can act as a buffer. In your answer you should explain how the equilibrium system allows the buffer solution to control the pH. [2][7]OCR 2013 Turn over<br />
10 (ii) A chemist prepares a buffer solution by mixing together the following: 200 cm3 of 3.20 mol dm3 HCOOH (Ka = 1.70104 mol dm3) and 800 cm3 of 0.500 mol dm3 NaOH. The volume of the buffer solution is 1.00 dm3. Explain why a buffer solution is formed when these two solutions are mixed together.Calculate the pH of this buffer solution. Give your answer to two decimal places.OCR 2013 [6] [Total: 22]<br />
11 BLANK PAGE PLEASE DO NOT WRITE ON THIS PAGEOCR 2013 Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_5_QP_JUN13_question_4-0.jpg)
![F325 Mark Scheme June 2013 Question 4 (a) Answer HCl is a strong acid AND HClO is a weak acid (cid:57) HCl: pH = log 0.14 = 0.85 (2 DP required) (cid:57) HClO: CHECK THE ANSWER ON ANSWER LINE IF answer = 4.14, award all three calculation marks -------------------------------------------------------------------- Ka = 107.43 OR 3.7 x 108 (mol dm3) (cid:57) [H+] = Ka (cid:117) [HClO] OR Ka (cid:117) [HA] OR Ka (cid:117) 0.14 OR 3.7 x 10(cid:16)8 x 0.14 (cid:57) pH = 4.14 (2 DP required) (cid:57) Marks Guidance ANNOTATE WITH TICKS AND CROSSES, etc ALLOW HCl completely dissociates AND HClO partially dissociates ALLOW HCl (cid:111) H+ + Cl AND HClOH+ + ClO IGNORE HCl is a stronger acid than HClO IGNORE HCl produces more H+ IF there is an alternative answer, check to see if there is any ECF credit possible using working below ------------------------------------------------------------- ALLOW 2 SF to calculator value: 3.715352291 x 108, correctly rounded IGNORE HCl if it is clear that it is a slip Always ALLOW calculator value irrespective of working as number may have been kept in calculator. Note: pH = 4.14 is obtained from all three values above From no square root, pH = 8.28. Worth Ka mark only 10<br />
F325 Question 4 (b) (c) Mark Scheme Marks Answer 2Al + 6CH3COOH (cid:127)(cid:111) 2(CH3COO)3Al + 3H2 (cid:57) 2Al + 6H+ (cid:127)(cid:111) 2Al3+ + 3H2 (cid:57) FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer = 13.6(0), award 2 marks -------------------------------------------------------------------- [H+] = OR 1.0 (cid:117) 10(cid:16)14 [OH(cid:16) ] OR 2.5 x 1014 (mol dm3) (cid:57) Correctly calculates pH = log 2.5 x 1014 = 13.6(0) (cid:57) Kw [OH(cid:16) ] OR 1.0 (cid:117) 10(cid:16)14 0.4(0) June 2013 Guidance IGNORE state symbols ALLOW correct multiples, e.g.: Al + 3CH3COOH (cid:127)(cid:111) (CH3COO)3Al + 1.5H2 ALLOW any unambiguous formula for (CH3COO)3Al, i.e. (CH3CO2)3Al, Al(CH3CO2)3, (CH3COO)3Al3+, etc. Note: IF charges are shown, they must be correct with bothand 3+ shown ALLOW multiples, e.g.: Al + 3H+ (cid:127)(cid:111) Al3+ + 1.5H2 ALLOW alternative approach using pOH: pOH = 0.4(0) (cid:57) pH = 140.40 = 13.6(0) (cid:57) ALLOW ECF from [H+] derived using Kw and [OH] BUT DO NOT ALLOW an acid pH. ALLOW one or more decimal places 11<br />
F325 Question (i) 4 (d) Mark Scheme June 2013 Answer A buffer solution minimises pH changes (cid:57) on addition of small amounts of acid/H+ or alkali/OH/base (cid:57) -------------------------------------------------------------------------------- HCOOHH+ + HCOO (cid:57) Equilibrium sign essential For effect of acid and alkali, ALLOW wrong carboxylic acid (e.g. CH3COOH) OR HA; ALLOW CHOOH for acid (effectively ECF) ALLOW COOH for base ALLOW responses based on COOHH+ + COO DO NOT ALLOW other incorrect formula, e.g. CH3OOH Added alkali HCOOH reacts with added alkali/base/OH OR added alkali/OH reacts with H+ (cid:57) QWC: Equilibrium shifts forming HCOO OR H+ OR (HCOOH) Equilibrium (cid:111) right (cid:57) Added acid HCOO reacts with added acid/H+ (cid:57) QWC: Equilibrium shifts forming HCOOH OR (HCOOH) Equilibrium (cid:111) left (cid:57) 12 Marks Guidance ANNOTATE WITH TICKS AND CROSSES, etc ALLOW resists pH changes ALLOW buffer solutions maintains a nearly/virtually constant pH DO NOT ALLOW a response that implies that the pH is actually constant, e.g. does not change pH; maintains pH -------------------------------------------------------------------------- DO NOT ALLOW COOH OR CHOOH OR COOH DO NOT ALLOW HAH+ + A Quality of written communication, QWC 2 marks are for explaining how the equilibrium system allows the buffer solution to control the pH on addition of H+ and OH ALLOW HA OR weak acid reacts with added alkali DO NOT ALLOW this mark if there is no equilibrium system shown, e.g. HCOOHH+ + HCOO is absent ALLOW A OR conjugate base reacts with added acid IGNORE salt reacts with added acid DO NOT ALLOW this mark if there is no equilibrium system shown, e.g. HCOOHH+ + HCOO is absent<br />
F325 Mark Scheme June 2013 Question 4 (d) (ii) HCOOH reacts with NaOH forming HCOO/HCOONa Answer OR HCOOH + NaOH (cid:111) HCOONa + H2O (cid:57) Equilibrium sign allowed (Some) HCOOH/(weak) acid remains OR HCOOH/(weak) acid is in excess (cid:57) Calculation CHECK THE ANSWER n(HCOOH) OR [HCOOH] = 0.24(0) (mol / mol dm3) (cid:57) n(HCOO) OR [HCOO] OR [HCOONa] = 0.4(00) (mol / mol dm3) (cid:57) [H+] = Ka (cid:117) (cid:57) [HCOOH] [HCOO ] ) = 3.99 (cid:57) 0.24 0.4 pH = log [H+] = log(1.70 (cid:117)104 (cid:117) ---------------------------------------------------------- OR use of HendersonHasselbalch equation: pH = pKa + log [HCOO(cid:16)] [HCOOH] OR pH = logKa + log [HCOO(cid:16)] [HCOOH] (cid:57) = 3.77 + 0.22 = 3.99 (cid:57) IF answer = 3.99, award all four calculation marks Marks Guidance ANNOTATE WITH TICKS AND CROSSES, etc DO NOT ALLOW just methanoate/HCOO forms formulae or names of reactants also required ALLOW HCOOH + OH (cid:111) HCOO + H2O (cid:57) IGNORE conjugate base/salt forms IGNORE HCOOH has been partially neutralised Note: There must be a clear statement that 0.24 and 0.4 apply to moles or concentrations of HCOOH and HCOO. DO NOT ALLOW these values if unlabelled ALLOW HA/acid and A/salt for HCOOH and HCOO DO NOT ALLOW ECF for this mark: 3.99 is the ONLY correct answer --------------------------------------------------------------- ALLOW HA/acid and A/salt for HCOOH and HCOO ALLOW pH = pKalog [HCOOH] [HCOO(cid:16)] OR pH = logKalog [HCOOH] [HCOO(cid:16)] ALLOW = 3.77(0.22) = 3.99 DO NOT ALLOW ECF for this mark: 3.99 is the ONLY correct answer Total 22 13<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_5_QP_JUN13_answer_4-0.jpg)
