2019 EJC Prelim H2 Chem P3 (solutions with examiner s comments)

© EJC [Turn Over EUNOIA JUNIOR COLLEGE JC2 Preliminary Examination 2019 General Certificate of Education Advanced Level Higher 2 CHEMISTRY Paper 3 Free Response 9729/03 20 September 2019 2 hours Candidates answer on answer booklet. Additional Materials: Answer Booklet Data Booklet READ THESE INSTRUCTIONS FIRST Write your name, civics group and registration number on the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staplers, paper clips, glue or correction fluid. Section A Answer all questions. Section B Answer one question. Begin each question on a fresh page of the answer booklet. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 13 printed pages and 1 blank page.

2 © EJC 9729/03/J2PE/19 Section A Answer all the questions from this section. 1 Phosphoric(V) acid , H3PO4, is widely used in the production of fertilizers and the acidification of foods and beverages such as soft drinks. It behaves as a tribasic acid. (a) Phosphoric(V) acid can be formed from the reaction of phosphorous( V) oxide, P4O10, or phosphorous(V) chloride, PCl5, with water. (i) Write an equation for the reaction that occurs when a small amount of PCl5 is dissolved in water. State the expected observations and the colour of the resultant solution when a few drops of Universal Indicator is added. [2] PCl5(s)+ 4H2O(l)  H3PO4(aq) + 5HCl(aq) White fumes observed, red solution (ii) Phosphorus(V) chloride dissolves in some inert pol ar solvents to form solutions which conduct electricity. This is due to the presence of the two ions, [PCl4]+ and [PCl6]. Draw structure of the [PCl6] ion, indicating the types of bonding within the ion, and state its shape. [2] octahedral Comments:  The equation was generally well -attempted, although some students gave the equation for the reaction with limited water (instead of excess water). This was not accepted as the question stated “small amount of PCl5”.  Many students missed out the expected observations, or assumed that effervescence would be given out. Comments:  The most common error for the structure omitted the use of one dative bond. Some others used 6 dative bonds which would result in several electron-deficient chlorine atoms.  The shape around the central atom was generally well-attempted.  Candidates are advised to learn how to sp ell specific technical terms (including “octahedral”).

3 © EJC 9729/03/J2PE/19 [Turn Over (b) In an experiment, 25.0 cm 3 of 0.080 mol dm –3 phosphoric(V) acid is titrated against 0.100 mol dm–3 NaOH. The initial pH of the solution was 1.69. Assume that the initial pH is due to the first dissociation of H3PO4 only. (i) Show, by calculations, that H3PO4 is a weak acid. [1] If H3PO4 is a strong acid,   3H 0.080 mol dm 1.69   pH lg 0.080 1.10   Hence H3PO4 is a weak acid. or   3 340.080 mol dm H PO   + 1.69 3H 10 0.02042 mol dm Hence H3PO4 is a weak acid. (ii) The numerical values of the second and third acid dissociation constants of H3PO4 are as follows.      3 4 2 4H PO aq H PO aq H aq  Ka1       2 2 4 4H PO aq HPO aq H aq    Ka2 = 6.17  10–8       23 44HPO aq PO aq H aq    Ka3 = 4.79  10–13 Calculate the equilibrium concentration of H 3PO4 after the first dissociation and hence, a value of Ka1. [2] Eqm (1)  34H PO aq   24H PO aq  +  H aq Initial conc / mol dm–3 0.080 0 0 change in conc / mol dm–3 –0.02042 +0.02042 +0.02042 eqm conc / mol dm–3 0.05958 0.02042 10–1.69 = 0.02042 equilibrium concentration of H3PO4 = 0.0596 mol dm–3   2 2 a1 34 H 0.02042 H PO 0.05958K   337.00 10 mol dm Comments:  Calculations alone without any brief comment are not credited.  Students should not use Ka to show as the use of Ka assumes that H3PO4 is a weak acid (i.e. a circular argument). Comments: Note: the amt of dissociated H3PO4 cannot be ignored as the acid is 25% dissociated.  Some students did not answer to the question about equilibrium concentration of H 3PO4. Some students have the misconception that equilibrium concentration of H 3PO4 = concentration of H+.

4 © EJC 9729/03/J2PE/19 (iii) Calculate the first equivalence volume of NaOH for the titration. [1] 34 3 amount of NaOH required initial amount of H PO 25.0 0.0801000 2.00 10 mol      32.00 10first equivalence volume of NaOH 0. 3 s.f. or 2 d p 1 .. 0 320.0 cm (iv) Calculate the pH after the addition of 10.00 cm 3 of NaOH. Give your answer to 2 decimal places. [1]   3 a1At half equivalence point, pH p lg 7.00 10K      2.15 (v) Calculate the pH of the resultant solution if 0.20 cm3 of 0.080 mol dm–3 HCl was accidentally added to the solution in (b)(iv). [2] 50.20amount of HC added 0.080 1.60 10 mol1000    l 3 5 3 34 1new amount of H PO 2.00 10 1.60 10 1.016 10 mol2          3 5 3 24 1new amount of H PO 2.00 10 1.60 10 0.984 10 mol2               3 3 24 3 a1 3 34 1.016 107.00 10H H PO H 7.2276 10H PO 0.984 10 VK V                         pH lg H    2.14 Comments:  This question was surprisingly poorly attempted.  Many candidates have the misconception that since H 3PO4 does not fully dissociate in water, it would not fully react with NaOH either.  Correct concept: When H+ from the dissociation of H3PO4 reacts with NaOH, the position of equilibrium (1) shifts to the right, so all H3PO4 would react. Comments:  Many students did not recognise the solution as a buffer.

5 © EJC 9729/03/J2PE/19 [Turn Over (vi) The pH at the second equivalence point of the titration is 9. 77. Suggest an equation to account for the basic equivalence point. [1] 2 4 2 2 4HPO H O H PO OH   (vii) In total, 50.00 cm3 of NaOH was added during the titration . Using the information given in (b), and your answers in (b)(ii) to (b)(iv), sketch the pH –volume added curve for the titration. Label the following key points on the curve.  Initial pH  pH at second equivalence point  pH at the half-equivalence points [3]  Label initial point (0.00, 1.62)  Label max. buffer capacity (10.00, 2.15)  Label second equivalence point (40.00, 9.77)  Label all pKa2 and pKa3 values  Correct shape (buffer region should have gentle gradient)  Correct axes with units Comments:  Many students did not recognise the solution as a buffer.  Among stude nts who did, some forgot to include the volume in the calculation of concentration. Some forgot that both the concentrations of the weak acid and conjugate base would be affected. Comments:  Many candidates incorrectly used a non-reversible arrow.  Some candidates chose the wrong species as the reactant. pKa1 = 2.15 pH pKa2 = 7.21 1.69 pKa3 = 12.32 VNaOH / cm3 9.77 0 10 20 30 40 50

6 © EJC 9729/03/J2PE/19 (c) Consumption of an excessive amount of soft drinks can cause tooth decay as the acid present in the beverage reacts with Ca10(PO4)6(OH)2 present in tooth enamel.               22 10 4 4 2 62Ca PO OH s 8H aq 10Ca aq 6HPO aq 2H O      l The Ca2+ ions lost from enamel can form a precipitate with the 3 4PO  ions present in the soft drink.         23 4 3 4 23Ca aq 2PO aq Ca PO s  Ksp = 1.3  1026 (i) Write an expression for the solubility product, Ksp, of Ca3(PO4)2, indicating its units. [2] 3223 sp 4 Ca POK 