Edexcel Jun 2015 (IAL) Paper 4 Q13 (with explained solutions) |

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Edexcel Jun 2015 (IAL) Paper 4 Q13

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SECTION B Answer ALL the questions. Write your answers in the spaces provided. 13 A student carried out a titration by adding 0.0540 mol dm3 hydrochloric acid to 25.0 cm3 of 0.0240 mol dm3 ammonia solution. A sketch graph of pH against volume of hydrochloric acid added is shown below. 14pH 70Vol of HCI / cm3 (a) *(i) Name the type of solution formed in the region ringed on the sketch graph and explain its chemical behaviour. (3)12 *P45075A01228*<br />
*(ii) Explain why the pH at the equivalence point of this titration is less than 7. Include an ionic equation in your answer. (3)(iii) By considering the amount of excess acid remaining, calculate the pH of the solution formed when 40.0 cm3 of 0.0540 mol dm3 hydrochloric acid has been added to 25.0 cm3 of 0.0240 mol dm3 ammonia solution. (4) *P45075A01328* 13 Turn over<br />
(b) (i) Show, using the data below, that the pH of water at 373 K is 6.13. (cid:116) H2O(l) (cid:85) H+(aq) + OH(aq) (cid:116) Kw = 5.501013 mol2 dm6 at 373 K (ii) At 373 K, is water neutral, acidic or alkaline? Explain your answer. (2) (2)(Total for Question 13 = 14 marks) 14 *P45075A01428*<br />
BLANK PAGE *P45075A01528* 15 Turn over<br />

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Mark +NH3 + H+ + present (at equivalence point) + react with water /hydrolysed + + H2O NH3 + H3O+ + + H2ONH4OH + H+ Acceptable Answers Notethe equations NH4 NH4 score all three marks Note the equation NH4 scores 2 marks, but if (aq) state symbols are given, scores 3 marks 1st mark: Ammonium ions /NH4 OR ammonium chloride/ammonium salt (1) 2nd mark Ammonium (ions) / NH4 by water /dissociate in water Ignore ammonium chloride reacts with water (1) 3rd mark NH4 OR NH4 Allow NH4 Note if no other mark awarded Just strong acidweak base (titration) / ammonium chloride is the salt of a strong acid and a weak base scores (1) only + + H2ONH4OH + H+ (1) +NH3 + H+ + + H2O NH3 + H3O+ Question Number 13(a) (ii)<br />
Question Number 13(a) (iii) Mark Acceptable Answers If final answer is 1.6(2), with correct working or without working, award 4 marks Mol of ammonia used = (25/1000 x 0.024) = 6x104 mol and Mol of acid added = (40/1000 x 0.054) = 2.16x103 (1) Mol of excess acid = 2.16x1036x104 = 1.56x103 mol (1) [H+] = 1.56x103/(65/1000) = 0.024 mol dm3 (1) pH = log [H+] = 1.6(2) (1) Ignore SF except 1 SF Allow TE for 2nd, 3rd marks Allow TE for 4th mark provided pH is less than 7 and it is based on some use of data in question Alternative method for 1st and 2nd marks Mol of ammonia used = (25/1000 x 0.024) = 6x104 mol and Volume of acid used = 6x104 x 1000 0.054 = 11.111 cm3 (1) Volume of acid left = 4011.111 = 28.889 cm3 Mol of excess acid = 28.889 x 0.054 1000 = 1.56x103 mol (1) Reject Mark 6.13 with no working Acceptable Answers Question Number 13(b)(i) EITHER [H+]2 = 5.5x1013 or [H+] = 5.5x1013 / 7.416x107 (1) (mol dm-3) pH = log5.5x1013 (= 6.12982 / 6.13) (1) OR pKw = 12.26 (1) pH =pKw (= 6.130) (1)<br />
Acceptable Answers Neutral (1) [H+] = [OH- ] /equal amounts of H+ and OH- ions OR Both [H+]and [OH-] have increased by the same amount (1) Reject Mark Acidic or alkaline for both marks Total for Question 13 = 14 marks Question Number 13(b) (ii)<br />

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