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Edexcel Jan 2016 (IAL) Paper 4 Q19
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![A E R A S I H T N I E T I R W T O N O D A E R A S I H T N I E T I R W T O N O D 19 Propanoic acid is a weak Brnsted-Lowry acid. (a) Define the term Brnsted-Lowry acid. (1)(b) What is the pH of a 0.100 mol dm3 solution of hydrochloric acid? How would the pH of any weak acid of the same concentration differ from the pH of hydrochloric acid? (1)(c) Methanoic acid was mixed with propanoic acid. (i) Use your Data Booklet to decide which acid is stronger. Justify your answer. (1)(ii) Hence complete the following equation. HCOOH + C2H5COOH (cid:85)+(d) Calculate the pH of a solution of 0.050 mol dm3 sodium hydroxide. [Kw = 1.01014 mol2 dm6] (1) (2) A E R A S I H T N I E T I R W T O N O D *P46940A01928* 19 Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/l_smalljpgJanuary_2016_IAL_QP_-_Unit_4_Edexcel_Chemistry_A-level_19.jpg)
![Reject Reject Mark Mark Different (from 1) Reject Mark Acceptable Answers Proton/ H+ donor Acceptable Answers pH of HCl = 1 and pH of weak acid is greater /higher than 1 Allow any number >1 and <7 Acceptable Answers HCOOH/ methanoic acid is stronger because its Ka is bigger/higher OR its pKa is smaller / lower (The data: Ka pKa Methanoic acid 1.6 x 10-4 3.8 Propanoic acid 1.3 x 10-5 4.9 ) IGNORE Discussion of inductive effect Acceptable Answers (HCOOH + C2H5COOH ) HCOO + C2H5COOH2 ALLOW TE for equation with propanoic acid as proton donor giving HCOOH2 HCOOH is stated to be weaker + and C2H5COO- if Reject Mark COOH C2H6COOH+ Question Number 19a Question Number 19b Question Number 19c(i) Question Number 19c(ii)<br />
Acceptable Answers Reject Mark [H+ ] = (1 x 1014 /[OH]) = 2 x 1013 (mol dm3) (1) pH = 12.7 (1) OR pOH / -log 0.05 = 1.3 (1) pH = (14 -1.3 =) 12.7 (1) Correct answer with no working scores 2 provided at least 3 SF Allow TE on first mark provided answer >7 13 13 Reject Acceptable Answers C2H5COOH + NaOHC2H5COO() Na(+) + H2O ALLOWforC2H5COO + Na+ for C2H5COO() Na(+) IGNORE State symbols even if incorrect Mark Question Number 19d Question Number 19e(i)<br />
Acceptable Answers Question Number 19e(ii) Allow salt/ C2H5COONa/ propanoate ion/ C2H5COO/ base for A Allow propanoic acid/ C2H5COOH for HA First mark Ka = [H+][A] [HA] OR log Ka = log[H+] + log [A]/[HA] OR pH = pKalog [HA]/[ A] ALLOW any of these equations re-arranged or used correctly (1) Next four marks Mol NaOH before mixing = (20 x 0.05/1000) = 0.001 and mol propanoic acid before mixing = (20 x 0.25/1000) = 0.005 (1) Mol propanoate in mixture = 0.001 OR [propanoate] = (0.001/40 x1000) = 0.025 (mol dm3) (1) Mol propanoic acid in mixture = 0.004 OR [propanoic acid] = (0.004/40 x 1000) = 0.1(mol dm3) (1) [H+] = (1.3 x 105 )(0.1) 0.025 pH = 4.28/ 4.3 (1) Correct pH with no working scores last 4 marks ALLOW Other methods leading to 4.28 e.g. based on equal volumes being mixed so mol propanoate are in double the volume and so concentration is 0.025 mol dm3 Reject Mark<br />
Reject Mark Acceptable Answers First mark The mixture contains a large amount/ reservoir of a (weak) acid/propanoic acid and its conjugate base/ propanoate ions /salt (1) Second mark Only awarded if at least one equation given Added OH combines with H+ (H+ + OHH2O) from propanoic acid followed by dissociation of more propanoic acid C2H5COOHC2H5COO + H+ OR Added OH combines with propanoic acid OH + C2H5COOHC2H5COO + H2O (1) Third mark (pH is unchanged because added OH is removed) change in concentration of C2H5COO and C2H5COOH is small / ratio [salt]/[acid] hardly changes (1) (Total for Question 19 = 15 marks) Question Number 19e(iii)<br />](https://learnah.org/wp-content/uploads/2018/08/January_2016_IAL_MS_-_Unit_4_Edexcel_Chemistry_A-level_19-0.jpg)
