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OCR A Jan 2012 Paper 5 Q4
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![This question looks at acids, bases and buffer solutions. 10 (a) Nitric acid, HNO3, is a strong BrnstedLowry acid. Nitrous acid, HNO2, is a weak BrnstedLowry acid with a Ka value of 4.43104 mol dm3. (i) What is the difference between a strong acid and a weak acid?[1] (ii) What is the expression for the acid dissociation constant, Ka, of nitrous acid, HNO2? (iii) Calculate the pH of 0.375 mol dm3 nitrous acid, HNO2. Give your answer to two decimal places. [1] pH =[2] (iv) A student suggests that an acidbase equilibrium is set up when nitric acid is mixed with nitrous acid. Complete the equation for the equilibrium that would be set up and label the conjugate acidbase pairs. HNO3 + HNO2+[2]OCR 2012<br />
(b) Calcium hydroxide, Ca(OH)2, is a strong BrnstedLowry base. 11 (i) Explain what is meant by the term BrnstedLowry base.[1] (ii) Calculate the pH of 0.0400 mol dm3 Ca(OH)2. Give your answer to two decimal places. pH =[3] (c) Aqueous calcium hydroxide is added to nitrous acid, HNO2. Write the overall equation and the ionic equation for the reaction that takes place. overall:ionic:[2]OCR 2012 Turn over<br />
12 (d) Carbonic acid, H2CO3, is a weak BrnstedLowry acid formed when carbon dioxide dissolves in water. Healthy blood is buffered to a pH of 7.40. The most important buffer solution in blood is a mixture of carbonic acid and hydrogencarbonate ions, HCO3 . (i) Explain how the carbonic acidhydrogencarbonate mixture acts as a buffer in the control of blood pH. In your answer you should explain how equilibrium allows the buffer solution to control the pH.[5]OCR 2012<br />
13 (ii) Healthy blood at a pH of 7.40 has a hydrogencarbonate : carbonic acid ratio of 10.5 : 1. A patient is admitted to hospital. The patients blood pH is measured as 7.20. Calculate the hydrogencarbonate : carbonic acid ratio in the patients blood. [5] [Total: 22]OCR 2012 Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_5_QP_JAN12_question_4-0.jpg)
![F325 Mark Scheme January 2012 Question Expected answers 4 a i A strong acid completely dissociates AND a weak acid partially dissociates (cid:57) Marks Additional guidance ALLOW ionises for dissociates ii (Ka =) [H+] [NO2] [HNO2] (cid:57) iii FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer = 1.89 award 2 marks IF answer = 1.9 award 1 mark -------------------------------------------------------------------- pH = log 0.0129 = 1.89 (cid:57)(cid:57) OR pH = log 0.0129 = 1.9 (cid:57) not two decimal places iv HNO3 + HNO2NO3 + (cid:57) Acid 1 Base 2 Base 1 Acid 2 (cid:57)+ H2NO2 13 DO NOT ALLOW [H+]2 [HNO2] Square brackets are required IF there is an alternative answer to more decimal places, check calculator value ------------------------------------------------------------- Working to get to 0.0129 (mol dm3) Not required and no credit [H+] = Ka (cid:117) [HNO2] = 4.43 (cid:117)10(cid:16)4 (cid:117) 0.375 ALLOW 1 mark for an answer with more than 2 decimal places that rounds back to 1.89 State symbols NOT required 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 IF proton transfer is wrong way around ALLOW 2nd mark for idea of acidbase pairs, i.e.(cid:56) HNO3 + HNO2H2NO3 Base 2 Acid 1 Acid 2 Base 1 (cid:57) NOTE For the 2nd marking point (acidbase pairs), this is the ONLY acceptable ECF + + NO2<br />
F325 Mark Scheme January 2012 Question Expected answers Marks Additional guidance i Proton acceptor (cid:57) ii Marks are for correctly calculated values. Working shows how values have been derived. [OH] = 20.04(00) = 0.08(00) (mol dm3) (cid:57) [H+] = 1.00 (cid:117) 1014 0.08(00) OR 1.251013 (mol dm3) (cid:57) pH = log 1.251013 = 12.90 (cid:57) --------------------------------------------- pOH variation (also worth 3 marks) [OH] = 20.04(00) = 0.08(00) (mol dm3) (cid:57) pOH log 0.08(00) = 1.10 (cid:57) pH = 14.001.10 = 12.90 (cid:57) Ca(OH)2 + 2HNO2 (cid:111) Ca(NO2)2 + 2H2O (cid:57) H+ + OH (cid:127)(cid:111) H2O (cid:57) 4 b 14 i.e., NO ECF from impossible chemistry ALLOW H+ acceptor ALLOW by ECF 1.00 (cid:117)10(cid:16)14 calculated value of [OH(cid:16)] DO NOT ALLOW 12.9 not two decimal places --------------------------------------- COMMON ERRORS 12.60 12.6 12.30 12.3 1.40 (cid:57)(cid:57) no2 for [OH] (cid:57) (cid:57)(cid:57)2 [OH] (cid:57)2 [OH] AND 1 DP only NO marks no2 for [OH] AND 1 DP only ALLOW: 2H+ + 2OH (cid:111) 2H2O<br />
F325 Mark Scheme January 2012 Question Expected answers 4 d Equilibrium H2CO3H+ + HCO3(cid:57) Action of buffer Added alkali H2CO3 reacts with added alkali OR H2CO3 + OH (cid:111) OR added alkali reacts with H+ OR H+ + OH (cid:111) (cid:57) Equilibrium (cid:111) right OR equilibrium shifts forming H+ OR HCO3(cid:57) 15 Marks Additional guidance 2 have been ANNOTATIONS MUST BE USED Equilibrium sign is required IGNORE HAH+ + A DO NOT ALLOW H2CO32H+ + CO3 2 DO NOT ALLOW NaHCO3Na+ + HCO3IGNORE H2O + CO2H2CO3 IF HAH+ + A OR H2CO32H+ + CO3 used above: ALLOW all marks that meet marking alternatives as written NOTE The 1st added acid mark cannot then be accessed Equilibrium responses must refer back to a written equilibrium BUT IF H2CO3 (cid:111) H+ + HCO3shown above, assume that any equilibrium comments apply to the correct equilibrium IF more than one equilibrium shown, it must be clear which equilibrium is being referred to ALLOW added alkali reacts with weak acid Quality of Written Communication Mark is for linking the action of the buffer in controlling added alkali and hence pH<br />
F325 Mark Scheme January 2012 Question Expected answers Marks Additional guidancereacts with added acid (cid:57) Added acid HCO3 Equilibrium (cid:111) left OR equilibrium shifts forming H2CO3 (cid:57) ii FIRST, CHECK THE ANSWER ON ANSWER LINE OR 1 : 0.15: H2CO3 and award 5 marks. IF answer = 6.6 : 1 CHECK ratio is HCO3 IF answer = 0.15 : 1 , CHECK ratio is H2CO3 : HCO3and award 4 marks -------------------------------------------------------------------- In blood at pH 7.40, [H+] = 10pH = 107.40 = 3.98108 (mol dm3) (cid:57) Ka = [H+] [HCO3] [H2CO3 ] = 3.98 (cid:117) 108 (cid:117) 10.5 OR Ka = 4.18107 (mol dm3) (cid:57) (cid:32) Ka [H+] OR 4.18 (cid:117) 10(cid:16)7 In blood at pH 7.20, [H+] = 10pH = 107.20 = 6.31108 (mol dm3) (cid:57) [HCO3 ] [H2CO3] = 6.6 1 ALLOW any answer with > 1 decimal place that rounds back to 6.62 OR 6.63 OR 6.6 : 1 (cid:57) (up to calc. value, see below) 6.31(cid:117) 108 (cid:57)is required for this mark BUTHCO3 ALLOW added acid reacts with conjugate base ONLY if HCO3 DO NOT ALLOW salt reacts with added acidis present in equilibrium with H2CO3 IF there is an alternative answer, check to see if there is any ECF credit possible using working below ----------------------------------------------------------- ANNOTATIONS MUST BE USED FOR ALTERNATIVE using HendersonHasselbalch equation below -------------------------------------------------------------- ALLOW 3.98108 up to calculator value of 3.981071706108 correctly rounded ALLOW 6.31108 up to calculator value of 6.309573445108 correctly rounded ----------------------------------------------------------------------------- Common errors 0.15 : 1 (cid:57)(cid:57)(cid:57)(cid:57) Inverse ratio of H2CO3 : HCO3 16.6 : 1 OR 0.06 : 1 (cid:57)(cid:57)(cid:57)(cid:57) 10.5/1 swapped over in 2nd mark giving Ka value of 3.79 x 109 ALLOW answer with > 1 decimal place that rounds back to 16.64 OR 16.65 4 d ALTERNATIVE approach for concentrations using HendersonHasselbalch equation (5 marks) pH = pKa + log pKa = pHlog [HCO3 ] [H2CO3] [HCO3 ] [H2CO3] OR logKa + log = 7.40log 10.5 1 [HCO3 ] [H2CO3] (cid:57) = 6.38 (cid:57) (subsumes previous mark) Calculator: 6.378810701 16<br />
F325 Mark Scheme January 2012 Question Expected answers Marks Additional guidance At pH = 7.20, log [HCO3 ] [H2CO3] [HCO3 ] [H2CO3] = 100.82 (cid:57) = pHpKa = 7.206.38 = 0.82 (cid:57) (subsumes previous mark) = 6.6 1 OR 6.6 : 1 (cid:57) Total 22 17<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_5_QP_JAN12_answer_4-0.jpg)
