Skip to content
OCR A Specimen Paper 1 Q20
Answers available below
Show answer
Question
16
17
18
19
20
21
22
23
24
![22 20 A student investigates the reactions of two weak monobasic acids: 2-hydroxypropanoic acid, CH3CH(OH)COOH, and butanoic acid, CH3CH2CH2COOH. (a) The student wants to prepare a standard solution of 2-hydroxypropanoic acid that has a pH of 2.19. Plan how the student could prepare 250 cm3 of this standard solution from solid 2-hydroxypropanoic acid. In your answer you should provide detail of the practical procedure that would be carried out, including appropriate quantities and necessary calculations. Ka for 2-hydroxypropanoic acid is 1.38104 mol dm3 at 25 C. Specimen . . . . . . [8]OCR 2014 H432/01<br />
23 (b) 2-Hydroxypropanoic acid is a slightly stronger acid than butanoic acid. The two acids are mixed together and an acidbase equilibrium is set up. Suggest the equilibrium equation and identify the conjugate acidbase pairs. CH3CH(OH)COOH + CH3CH2CH2COOH(c) To prepare a buffer solution, 75.0 cm3 of 0.220 mol dm3 butanoic acid is reacted with 50.0 cm3 of 0.185 mol dm3 sodium hydroxide. Ka for butanoic acid is 1.5105 mol dm3 at 25 C. (i) Calculate the pH of 0.185 mol dm3 sodium hydroxide at 25 C. Give your answer to two decimal places. Specimen (ii) Calculate the pH of the buffer solution at 25 C. Give your answer to two decimal places. Show all your working. pH =[2] [2]OCR 2014 H432/01 Turn over pH =[4]<br />](https://learnah.org/wp-content/uploads/2018/08/Specimen_QP_-_Paper_1_OCR_Chemistry_A_A-Level_20-0.jpg)
![H432/01 Question 20 (a) Mark Scheme June 20XX Guidance ALLOW 5 marks for 6.80 g through any calculation. ALLOW ECF for incorrect calculation of mass. Mass used must be linked to calculation. Answer Marks (cid:57) [H+]2 Ka (6.46 (cid:117) 103)2 = 1.38 (cid:117) 104 1000 [H+] = 10pH = 102.19 = 6.46 (cid:117) 103 (mol dm3) (cid:57) [CH3CH(OH)COOH] = = 0.0302 (mol dm3) (cid:57) n(CH3CH(OH)COOH) = 0.302250 = 0.0755 mol (cid:57) Mass of CH3CH(OH)COOH = 0.075590 = 6.80 g (cid:57) Dissolve 6.80 g of the solid in distilled water (less than 250 cm3) in a beaker (cid:57) (then) transfer the solution to a 250 cm3 volumetric flask AND ensure that all solution is washed out of beaker (washings transferred to volumetric flask) (cid:57) (then) make solution up to 250 cm3 with distilled water AND ensure thorough mixing by inverting the flask several times (cid:57) Specimen CH3CH(OH)COOH AND CH3CH(OH)COO CH3CH2CH2COOH AND CH3CH2CH2COOH2 Both pairs identified (cid:57) (b) CH3CH(OH)COO + CH3CH2CH2COOH2 + (cid:57) State symbols NOT required ALLOW labels acid 1, base 1 etc. ALLOW ECF for second mark 16 P M T<br />
H432/01 Question (i) (c) Answer [H+] = 11014 = 5.4051014 (Use of Kw) (cid:57) 0.185 pH = log(5.4051014) = 13.27 (cid:57) Marks Mark Scheme June 20XX (ii) n(A) = 9.25103 (mol) (cid:57) Guidance ALLOW 5.4054054051014 and correct rounding to 5.41014 ALLOW alternative approach using pOH: pOH = log(0.185) = 0.73 pH = 140.73 = 13.27 Correct answer scores BOTH marks ALLOW 13.267 ALLOW HA/acid and A/salt throughout for butanoate and butanoic acid ALLOW pKa = log Ka OR log 1.5103 OR 4.82 ALLOW ECF from incorrect values of n(A) or n(HA) ALLOW pH = log(1.51054.93 25.7 25.9 10 10 ) = (cid:117) (cid:117) 3Specimen (cid:117) 125 (cid:117) 125 ) = 4.93 (cid:57) 25.9 25.7 10 10 3n(HA) = 0.01659.25103 = 7.25103 (mol) (cid:57) [H+] = Ka[HA] (cid:57) [A] pH = log(1.51050.058) = 4.93 0.074 OR pH = log(1.51051000 1000 (cid:117) (cid:117) Final mark also via HendersonHasselbalch equation: pH = pKalog = 4.82(0.11) = 4.93 OR pH = pKa + log = 4.82 + 0.11 = 4.93 (cid:57) Total 16 17 P M T<br />](https://learnah.org/wp-content/uploads/2018/08/Specimen_MS_-_Paper_1_OCR_Chemistry_A_A-Level_20-0.jpg)
