Sydney_Grammar_2023_Chemistry_sol

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Svoxrv Gneunnen ScHoor 2023 TRIAL HIGHER SCHOOL CERTIFICATE EXAMINATION C { I a CANDIDATE NUMBER Chemistry Section I - Multiple Choice Select the alternative A, B, C or D that best answers the question. Fill in the response oval completely. Sample: (D) e 2+4= (A) 2 (B) 6 (C) 8 no no co nO If you think you have made a mistake, put a cross through the incoriBct answer and fill in the new answer. ao sI co Do If you change your mind and have crossed out what you consider to be the con.ect answer, then indicate the comect answer by writing the word cetv.l and drawing an arrow as follows. I urut / nI nl co no Start Here aO eO eO eO eO aO eO eO aO eO eO sO eO eO eO e! eO eO eO nlp cO cO cO cO cO cO cO cC cO cO oO uO oO oO oO oO oO oO oO nO 11. t2. 13. L4, 15. 16. 17. 18. 19. 20. eO eO eO aO eO erp eO eO eO eO eO eO sO eO eO eO sO sO eO eO cO cO cO cO cO cO cO cO cO cO oO nO nC oO oO oO oO nO oe pO )1. 2. 3. 4. f,. 6. 7. 8. 9. 10.

SECTION l: MULTIPLE CHOICE (20 marks) Attempt ALL Questions Use the Multiple-Choice Answer Sheet. 1 Which of the following would best enable 2,2 from octane? @ Mass spectrometry (B) Determination of molar mass using gravimetric analysis (c) Measuring volume of carbon dioxide produced when combusted (D) Addition of bromine water n13 .'- C ,4-trimethylpentane to be distinguished <-:[-: - I tsannot's ,hr ) is6wt{vs Sa0vt4 2 Separate 20.0 mL solutions of a weak monoprotic acid and a strong monoprotic acid of the same concentration are titrated with NaOH solution. Which of the following will be the same for these two titrations? (A) lnitial pH (B) pH at the equivalence Point @ Volume of NaOH required to reach the equivalence point (D) The conductivity of the initial acid solutions 3 Which of the following reagents would liberate carbon dioxide when mixed with a concentrated aqueous solution of sodium carbonate? q- y1uzd, an o-c''i ethanoic acid ethanamine ethanamide ethyl ethanoate 4 Which of the following conditions will maximise the yield of dinitrogen tetraoxide? 2 Noz(9) + Nzo+(g) LH #7 '2 kJ mol-1 ._ (A) Low temPerature, low Pressure rc) Low temperature, high Pressure l) (C) High temPerature, low Pressure (D) High temperature, high pressure d, (B) (c) (D) Page 3

lvdnpv ar School forrl Vl 0rentistry 2021 irial Exanrination J 5 10 mL of 0.01 mol L-1 nitric acid (HNO3) is diluted with 90 mL of water. What is the pH of the resulting solution? lo =_> lao PI + I 6 Which of the following hydrocarbons contains an atom with trigonal planar geometry? (A) (B) @ (D) (A) @ (c) (D) 1 2 3 4 propane propene propyne e--C 4 H u p1t 2-methylpropane 7 Ethene reacts with hydrogen gas in the presence of a Pd-C catalyst. Which of the following statements about this reaction is correct? (A) (B) (c) Ethanol is produced . X ehatne f *ol ueeA ,ry The reaction also produces a byproduct. The Pd-C is consumed in the reaction. X This is an addition reaction. S+ I What is the concentration of OH- ions (in mol L-1) in an aqueous solution in which [H+] = 2.0x10-3 mol L-1 at25'C? D - tp -tv 2.0x10-3 4.0x10-6 5.0x10-12 2.Ox1O-17 x 10 >x.lo @ (A) (B) lw nl Loft - Lo#- l j .21 X (D) Page 4

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lydney Grammar School lorm Vl Chemistry 202i irial Examination 9 The name 2-propyl-2-chloro-4,5-dibromopentane does not follow IUPAC conventions. What is the systematic name of this compound? ,1 (A) 4,5-dibromo-2-chloro-2-propylpentane (B) 2-chloro-4,5-dibromo-2-propylpentane (C) 4-chloro-6,7-dibromo-4-methylheptane (D) 1,2-dibromo-4-chloro-4-methylheptane 1O Hydrogen and iodine react at 500"C according to the equation Hz(g)+lr(g)=2Hl(g) The apparatus shown below is set-up. 6 0l 6{ + z Y'( 3 n o shLl'4 a;fi,t/,;^ "^3 Vessel X Vessel Y The tap between Vessels X and Y is opened and then the system is left_at 500'C until no further change occurs. Which of the following statements is true? X will contain more hydrogen than Y. X and Y will contain the same amount of Hl(g). X will contain less iodine than Y. Y will contain more Hl(g)than X. 11 An organic compound reacted with concentrated HCI and ZnClz to produce 2-chloro-2-methylpentane. What was the name of the original compound? (A) (c) (D) (A) 2-methylpentan-1-ol (B) 2-chloropentanal @ 2-methylpentan-2-ol (D) 2-methylpentanal - of sut*'|"']"; 2 molHl 1 1 mol mol He Iz Page 5 urt(k U

Sydney Grammar School form Yl Chemistry 2021 Trial [xamination 12 The following equilibrium exists in bromine water: grr(aq) + HzO(l) + Br(aq) + 2 H.(aq) + OBr(aq) (red-brown) (colourless) (colourless) Which of the following solutions could be added to the reaction mixture to cause the red-brown colour of bromine water to fade? (A) (B) @ (D) (4 (B) (c) HCr KBr AgNOs NaOBr @ o.2e mot (B) 0.40 mol (C) 0.60 mol (D) 0.71 mol Hr',t frL ClxUSlLl cl &r fea.,lto4 ----P ,1, ,o 13 which of the following salts has the highest molar solubility? calcium carbonate copper(ll) carbonate lead(ll) carbonate silver carbonate 33 Axlol a s,Pvto-5 n /,Q x to 1'* x t; rf V 'ld I n^av* 14 A exists in equilibrium with B according to the equation below: A(s) + B(s) lf 1.0 mole of A was allowed to reach equilibrium, how many moles of B would be formed if Ko is equal to 0.40. tql tl C '-2/ €, t -?L fb7 o +/, '/- )L l-x- o'r L, O'4 - A'L/'x' t= ffi= O,r-f Page 6

Sydrey 6ranrnrar \ch fonr Vl Ihenrirtry )021 Trial Exanrinrtion 15 Propan-2-ol is heated with concentrated sulfuric acid. Compared to propan-2-ol, the product of this reaction: oli (A) (B) (c) @ is more soluble in water. has a higher molar mass. has fewer signals in 13C NMR. has a lower boiling point. A section of a polymer is shown below. r"onL --.--2 ItrSOrl Quesflon 16 and 17 refer to the following information. ot"zo \"y'o 4 -H H I H-C I oo + I -H I H-C I H I H-C-H I H _H -H t"2o I c I H -H H I H-C I o H I H-C-H c I c I c I H H- H I c I H H I H I c-c-c I H I H 16 Which of the following shows the monomer used to produce the polymer shown above? (B) H I H_C-H I o H I H-C-H I o H tJ-U H c-=o \,H c c H c H H I H-C-H I (D) c I c I c I H H I H-C I H H c-o H o H H (c) H I H-C I o H I c I H H o 2 _H -H Page 7 H.

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Syd n ey 0iamnrar School lornr Vl Ihemistry 20ll Irial Exanrination 17 lf you are comparing this polymer to polyethylene, which of the following would be true? (A) This polymer is an addition polymer while polyethylene is a polymer. con;(sation High-density polyethylene would nave weftr intermolecular forces as the chains can pack into a more orderly solid. \ Being a polffter, this polymer is used to make clothing while polyethylene is used for carft€s. Both polymers could be made without the elimination of a small molecule' (B) (c) @ (A) (B) i8 Ammonia (NH3) is a weak base in aqueous solution with an ionisation constant K6. Which of the following represents the ionisation constant for the reaction: NH4+(aq) + HzO(l)= Nft(aq) + H3O+(aq) I Lr H,i iLort'J K k6 K K a a f 'ttls1 ' f4ro'^ltoil'1 f aro''JLurl.7 K K K* Kb Kb K* ku (D) fn,if..*n ] X kL kw k Page I

fonn Vl Ihenristry l02l lrial txamination Sydney 6:anrmar School 19 Which of the following plots correctly represents the conductometric titration of 0.05 mol L-1 HzSO+ with 0.1 mol L-1 of the weak organic base ethylamine? GO cl tr) a{ It {J {ll E : E o (J E} t{} 10 o $ 0 ?s Volume of eth$am:irn {mL} x) 30 Volume of ethylamine {mL} fi) :10 Volume of ethylamine {mL) fil iF in /I} fi ilo ro 40 tfr (B) {c} m 60 cl frl o{} IJ 5 :ill U E,'I 8ro o 60 90 !im UsCI E 6 Etr E Ero o 60 lo g1$ tl H30 n Hm E 8r0 0 o o 1I} 10 sl o uI Volume of ethylam'lne {mL} i$ Page 9

Svdnev 6ranrmar Sdool lorn Vl Ihenristry 20ll Irial Ixanrination 20 Solid calcium chloride is added to 200.0 mL of 0.12 mol L-j potassium sulfate solution at 298 K. What is the minimum mass of calcium chloride required to produce a precipitate? (A) 0.0033 s (B) 0.00e1 s (c) 0.228 g (D) 6.21 mg (^') k p = f u'nJf sor'J = +'73x); { ta'nJ - -<. 3Yio r /t,ll v lo I N) -r 4' /'L o, /a- z: 4,tl x lo-tx 0,J = t'&( Y /o''P ptt/ ^lo"c/r) . F, J(x to- r x/+o'o/ + axss' v) Page l0

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CF.l($ - lXS CANDIDATE NUMBER Question 21 (3 marks) The table provides thermodynamic data about two bromide salts. A.orH (kJ.mol-1) A"orS (J Kn mol-1) LiBr -48.8 21.5 KBr 19.9 89.0 Compare and explain the solubilities of the two bromide salts at 300 K. You should include calculations in your answer. I . LtB. Xr : -*?'?: KU Mr=17 .1 - , .... X ..... 1. n 4r k ...Q..*...b etL.. ..sk t / ^h e n :... ..aa. r K..*.t. J,.. 3... G*: #):::l-'?f fIvr Gier %)z -(: QQ*:I -l -t o Page I I

G ramnrar Iornr Vl ?0ll lrial Ixrru;riri,l I Question 22 (7 marks) Marks Sulfur trioxide decomposition reaches equilibrium at 2OO'C according to the equation: 2 SOs(g) r:2 SOz(g) + Oz(g) (a) Use Collision Theory to state and explain the effect, if any, of an increase in the overall pressure. 3 I ln €.K.a,2.n4 .... F.r.*aee(*).1n.1f€.4,F..A...fb..n. a a #Y.ffiiJ a //. n p. t :... ..b--fo.*..a...!m ?.n.kn,. (b) ldentify the effect, if any, on the value of K"q, it the overall pressure of the system is increased. Velue "f Kq t2 (nxfblr't (c) A 0.40 molsample of sos(g) is placed in a2..0 t'vesseland allowed to reach equilibrium. Given that K"n = 1.30 x 10-e at this temperature, calculate the equilibrium concentration of SOz(g). QJ ee CoJ afi5wet utrlLt uutLt n0 2 ttu"[-s €"r o x a. :Q. k.::fr ..**...1.. *- 3 l.30 rlb 3Dx/D )s +o. """""""""tl(""" o, t..&x +z Q.:2.: .... H ... ......... 4*... x- K l-rc xtb 2 z 9' 7 --s - I s nonk5 Fot ;;,Afr;;'n/+6; l*z' o. 3 7' 3 x o.olx I I taarb Fo'-. ane ' 5t1nfr,Zf,efu,f fto8=7*:. +7t/2t'k H ? 1 e.tf o ( *= /}tn39 v lD

iornr V! 2021 Trial Ixarnination Sydney I r School Question 23 (3 marks) Complete the reactions below by drawing the structure of all organic reactant(s) and/or organic product(s). (a) HCI(sr) (b) conc HaSO - ..... : o - heat +Heo ? Marks I H o il c H I c I H H I c I H H I c I H H 2 t 1 c.L 1r H- c- L-O-H Pl PI lrl t't E Page I 3

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Sydrrtv (rranrnrrr -cirool Iornr Vl ftrenristry l0ll Trial lr:antinatitrn Question 24 (6 marks) Lactic acid CH3CH(OH)COOH is a weak monoprotic acid. (pK4 = O.B5) ar (a) Write an equation for the reaction of lactic acid with water. PHla tt kA*t*; H,a 3 -*>et 6 4 %j tr6 (b) ldentify a conjugate acid/base pair from this reaction. Acid: ..... €Wr*U.bU.) ttr a: Marks 1 1 conjusate base:... 4. ltgAfl (a A). m.1. 4a (c) State the equilibrium constant expression, K4, tor kut cU(an) a'J t (d) Calculate the pH of a 0.2Q mol L-1 solution of lactic acid ..Ko. eHl Ko: f Q...2..-'K x..sm/{..*.. aS *11 a'Z 4?/') /"e """"""2"""""' z=). 82.5.x. -G la r '; 3; r.4 .......... {:3! 5 x f H= ( s'et€ rrd 27 3 marQg €o- a,rreef arrPFtc/ent Z ma*Ls $r arrec one e-o"rdf 1 3 t I "{ oJor t;g Page I 4

Sydney Granrnrar School Ioinr Vl llrernistrv )021 lrial Ixaminattol Question 25 (3 marks) You have 0.1 mol L-1 solutions of each of NaNOs and Na2COs. Predict whether these two solutions are acidic, neutral or basic, explaining your reasoning with chemical equation(s), where relevant. t..5 .... txd.!.rc,| ..... *.s .... b..a.th .... No+..s)a( I ( MNAe NQs bas..a 3{.e ..... H..*i. ' l .. aor L .. #r .. krIre.f.. tt* ;( aa// se eYz/a """/""""" .... mrJ ..... m. il/.,n.. re J. Ng6s ts ba ar'c /4e/ 6of fif * tle 43 Na ldl-Lj *utq J, alt?I ro,// fitercE" r4a,,L F.. Ls bos Art "/- U;- v-rs s €ran 4 frlaL g'** l/t o/f at& s o 4 fi^ ?b l ra 3 I marl €' r ;Lou ,U . j tlt/tf Yl,* 7ro ./ \1 _ I UU' 13 bastc u_7- t €tl c+ +of q HA d '1 L Page l5 7

Question 26 (5 marks) Marks Compound X has the molecular formula C 4 H 10 O and is highly soluble in water. It does not react when heated with acidified KMnO 4 nor acidified K 2 Cr 2 O 7 . (a) Draw the structure and name compound X . 2 Name: ........................................................................................................................... (b) Predict the number of signals that compound X will show in 13 C NMR. 1 ........................................................................................................................................... (c) Draw two isomers of compound X that contain the same functional group as compound X . 2

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Question 27 (7 marks) Marks Equal volumes of hydrogen and bromine are introduced at T = 0 minutes to a sealed vessel and allowed to reach equilibrium at 60˚C. H 2 (g) + Br 2 (g) ⇌ 2 HBr(g) ∆H = -103 kJ mol -1 (a) Calculate K eq for this reaction at 60˚C. 3 ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Concentration x 10 -3 (M) Time (minutes) [H 2 ] [HBr]

Question cont. Marks (b) At 12 minutes, the temperature was changed . Deduce whether the temperature was increased or decreased and explain the change in concentration of H 2 (g) and HBr(g) in terms of activation energy of the opposing reactions. 4 .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ..........................................................................................................................................

Sample answer Q27(b) As [H 2 ] increases and [HBr] decreases, the equilibrium has shifted to the reactants side. The forward reaction is exothermic (∆H<0) so the reverse reaction is endothermic. As the endothermic reaction has been favoured, the temperature must have been increased. The endothermic direction has a higher activation energy than the exothermic direction, so a T increase means that the proportion of particles with kinetic energy > activation energy increases more for the endothermic reaction than the exothermic. Both reactions have an increase in successful collisions, but the increase is more significant for the reverse reaction than for the forward reaction, so the rate of the reverse reaction increases more. This ultimately results in a new equilibrium position with higher [H 2 ] and lower [HBr].

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Question 28 (4 marks) Butan-2-one can be produced from 2-bromobutane in two steps. Complete the reaction scheme below to show how 2-bromobutane can be converted to butan-2-one. Include the reaction conditions for each step and diagrams for the intermediate compound and butan-2-one. intermediate compound 2-bromobutane butan-2-one

Question 29 (5 marks) Marks 0.561 g of ethanol undergoes complete combustion using the equipment shown below. The initial temperature of the water was 20.0 o C. (a) Write a balanced chemical equation for the complete combustion of ethanol. 1 ........................................................................................................................................... (b) The enthalpy of combustion of ethanol is -1367 kJ mol -1 . Assuming half of the energy released from the burner is transferred into the water, calculate the final temperature of the water. 4 ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...........................................................................................................................................

Question 30 (7 marks) Marks Alpha-linolenic acid (ALA) is an omega-3, essential fatty acid. It is found in seeds and oil, and when extracted is a colourless liquid with a density of 0.91 g/mL. Its molar mass is 278.4 g mol -1 . With a formula of C 18 H 30 O 2 , ALA’s structure is shown in the diagram below: (a) On the diagram above, circle the functional groups of this molecule. 1 Marking criteria Marks Circles alkanoic acid/carboxyl group (-COOH) AND alkene groups (C=C) only 1 Markers Note: • Students should circle all the alkene groups. (Circling just one example of this functional group was accepted) • The carbonyl and hydroxy groups should not have been circled separately. • C-C single bonds are not a functional group. (b) Predict whether this molecule would be water-soluble, explaining your reasoning. 2 Marking criteria Marks Correct prediction of solubility with a thorough explanation of insolubility with reference to: - dispersion forces and hydrogen bonding - polarity of water - non-polar fatty acid (dominance of non-polar tail in fatty acid) OR Correct prediction of solubility with a thorough explanation of insolubility with reference to: - enthalpic considerations - entropic considerations 2 Correct prediction of solubility and a sound explanation that lacks depth and detail. 1 Sample answer: This molecule will not be soluble in water since this large molecule is dominated by a long non-polar hydrocarbon tail. This hydrophobic (non-polar) tail exhibits dispersion forces that are not strong enough to overcome the strong hydrogen bonds that exist between the polar water molecules and hence will not dissolve. Even though the fatty acid has a polar acid functional group that could form some hydrogen bonding with the water molecules, this interaction will be outweighed by the dispersion forces that exist between the long non-polar tails of the fatty acid, that form the majority of the molecule.

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(c) 2 mL of ALA is mixed with 2 mL of bromine water, shaken and then left to stand for a few minutes. Identify and explain two observations you would make. You may find a diagram to be helpful. 2 Marking criteria Marks Identifies and explains TWO observations 2 Identifies and explains ONE observation. OR Identifies TWO observations 1 Sample answer: - The orange-brown bromine water is decolourised due to the addition reaction of bromine water across the C=C double bonds in the ALA forming a colourless product. - The non-polar ALA will separate and float on top of the aqueous layer since ALA is less dense than water and insoluble in water. Markers note: It was not enough for an explanation to simply say the bromine water would decolourise due to presence of C=C. Students needed to explain how the reaction occurs. Note: Clear is not a colour (i.e., bromine water goes clear did not score marks for observation as the bromine water was clear to start with (i.e., clear brown/orange at star)t, also it is not the ALA that goes colourless as this was colourless at the start. (d) When used, ALA is often partially hydrogenated, so it is an unhealthy trans-fat that has a single carbon-carbon double bond remaining. Calculate the volume of hydrogen gas at 100 kPa and 25 o C required to convert 1.0 g of ALA to its equivalent trans-fat. 2 Marking criteria Marks Correct answer 2 One step correct 1 Sample answer: 2H 2 needed to convert 2 C = C in ALA n ALA = m/MM= 1/278.4 = 3.59 x 10 -3 mol n H2 = 2 x n ALA = 7.18 x 10 -3 mol H 2 needed. At 25 o C and 100KPa, v H2 = n x 24.79 = 0.178L = 0.18 L (2 sf) Markers note: Responses need to show working clearly and use some words to show what they are calculating. Many responses failed to recognise that 2 moles of H 2 were needed in the reaction.

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Question 31 (8 marks) Marks This question is about buffers. (a) State what is meant by the term buffer and describe the chemical composition of an acid buffer solution in general terms. 2 Marking criteria Marks • Describes a buffer system as helping to maintain pH/resist changes to pH • Correctly describes the chemical composition and equimolar concentrations of an acid buffer 2 One of the above 1 Sample answer: A buffer resists changes in pH when acid or base is added to a system. An acid buffer is composed of a 50:50 ratio (or equimolar or approx. equal amounts) of a WEAK acid with its conjugate base e.g., CH 3 COOH/CH 3 COONa Markers note: Responses for the composition of the buffer often lacked detail such as: - the equivalent amounts of weak acid : conjugate base - recognising that the acid used needed to be weak. Some responses did not address the question re the composition of an acid buffer and students generically wrote the components of all buffers . eg weak acid/base with conjugate base/acid. 50.0 mL of 0.10 mol L -1 ammonia solution is mixed with 50.0 mL of 0.060 mol L -1 hydrochloric acid and the resulting mixture forms a buffer. The K b of NH 3 is 1.78 x 10 -5 . (b) Describe what would happen if an additional small amount of acid solution was added to this buffer. Use an equation to support your explanation. Marking criteria Marks Thorough description involving shift of equilibrium to counteract change with a correct equation showing equilibrium arrows 2 Sound description relating to a suitable equation OR a correct equilibrium equation. 1 Sample answer: The buffer system formed is NH 3 (aq) + H 2 O (l) ⇌ NH 4 + (aq) + OH - (aq) If H 3 O + added to this buffer system, it would remove OH - from the equilibrium system due to the reaction of H 3 O + + OH - → 2H 2 O . The equilibrium shifts to the RHS (LCP), therefore resisting change to pH, since H 3 O + has been removed and pH = - log [ H 3 O + ]. The buffer reestablishes equilibrium, and pH is minimally affected. OR The buffer system formed is NH 3 (aq) + H 3 O + (aq ) ⇌ NH 4 + (aq) + H 2 O(l)

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If H 3 O + added to this buffer system, the equilibrium shifts to the RHS (LCP), therefore resisting change to pH, since H 3 O + has been removed and pH = - log [ H 3 O + ]. The buffer reestablishes equilibrium, and pH is minimally affected. Markers note: Many responses did not demonstrate a thorough understanding of buffers. Responses for the buffer needed an equilibrium arrow and a description of the shift in equilibrium when an acid added to the system. (c) Calculate the pH of the buffer produced when the two solutions were mixed. 4 Marking criteria Marks Correct answer with all working 4 One error in calculation but all correct steps. 3 Two correct steps 2 Any correct step eg calculating moles ammonia and HCl, identifying limiting reagent, calculating [ NH 3 ], [ NH 4 + ] , expression for K b , pOH or pH calculation 1 Sample answer: n(NH 3 ) = 0.10 x 0.05 = 0.005 mol n(HCl) = 0.060 x 0.05 = 0.003 mol HCl is limiting. Therefore, in buffer n(NH 3 ) = 0.002 (0.005 – 0.003) and n(NH 4 + ) = n(HCl) = 0.003 This is in 100 mL, therefore : [ NH 3 ] = 0.002/.1 = 0.02M, [ NH 4 + ] = 0.00 3 /.1 = 0.0 3 M NH 3 + H 2 O ⇌ NH 4 + + OH - NH 3 H 2 O NH 4 + OH- I 0.020 0.030 0 C - x + x + x E 0.020 - x 0.030 + x x K b NH 3 = 1.78 x 10 -5 (x is negligible compared to concentrations of 0.02 M and 0.03 M) 1.78 x 10 -5 = [ NH 4 + ] [ OH - ] /[ NH 3 ] = 0.030 x/0.020 x = [ OH - ] = (1.78 x 10 -5 x 0.020)/0.030 = 1.186666 x 10 -5 M pOH = - log [ OH - ] = 4.926 pH = 14 – pOH = 9.07

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Markers note: If solve with H-H: pH = pK a + log [A - ] [ H A ] pK a = 14 – pK b = 14 – (-log 1.78 x 10 -5 ) = 9.25 (Or find via K a = K w /K b = 1 x 10 - 14 /1.78 x 10 -5 = 5.61798 x 10 - 10 , pKa = - log 5.61798 x 10 - 10 ) In buffer n(NH 3 ) = 0.002 (0.005 – 0.003) and n(NH 4 + )= 0.003 [ NH 3 ] = 0.002/.1 = 0.02M, [ NH 4 + ] = 0.003/.1 = 0.03M pH = pK a + log [A - ] [ H A ] pH = 9.25 + log (0.02/0.03) = 9.07 OR pOH = pK b + log [N H 4 + ] / [N H 3 ] = - log 1.78 x 10 - 5 + log (0.03/0.02) = 4.9256…. pH = 14 – 4.9256… = 9.07 Question 32 (3 marks) Determine the maximum mass of calcium hydroxide that will dissolve in 230 mL of 0.040 M barium hydroxide solution at 25˚C. Sample answer: K sp of Ca(OH) 2 = 5.02 x 10 -6 K sp = [Ca 2+ ][OH - ] 2 [Ba(OH) 2 ] = 0.04 M, therefore [OH - ] = 0.08M (hydroxide ion will have negligible change in concentration with dissolution of calcium hydroxide) K sp = [Ca 2+ ][OH - ] 2 5.02 x 10 -6 = x . 0.08 2 x = 7.843 x 10 -4 mol L -1 n Ca2+ = c x v = 7.843 x 10 -4 x 0.230 = 1.804 x 10 -4 mol L -1 n Ca2+ = n Ca(OH)2 mass calcium hydroxide = n x MM = 1.804 x 10 -4 x 74.096 = 0.0134 g Markers note: • Students are advised to show their working and logic clearly. Using some words to show what is being calculated is strongly advised! (A bunch of numbers on a page is not very helpful.) Marking criteria Marks Correct answer with calculations 3 One error in calculation 2 One correct step. eg correct [OH - ], correct molar mass of Ca(OH) 2 1

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Question 33 (3 marks) Deduce whether a precipitate will form when 5.0 mL of 0.010 mol L -1 magnesium sulfate solution is added to 10.0 mL of 0.020 mol L -1 sodium carbonate solution at 25˚C. Marking criteria Marks Correct • calculations for [ Mg 2+ ] and [CO 3 2- ], • calculation for Q sp • relationship between Q sp and K sp and deduction of ppt forming 3 One minor error in above but must show calculation of Qsp 2 One correct step. 1 Sample answer: [ Mg 2+ ] = (0.01 x 0.005)/0.015 = 3.33 x 10 -3 mol L -1 [ CO 3 2- ] = Vc = (0.020 x 0.01)/0.015 = 1.33 x 10 -2 mol L -1 Q sp = [ Mg 2+ ] x [CO 3 2- ] = 4.44 x 10 -5 K sp (MgCO 3 )= 6.82 x 10 -6 Q sp > K sp , therefore a precipitate will form. Markers note: • Students are advised to show their working clearly. • Some responses showed calculations for a limiting reagent. i.e., Students identified that number of moles of Mg 2+ was limiting reagent and tried calculating Qsp using [Mg 2+ ] = [CO 3 2- ] = 3.33 x 10 -3 mol L -1 . If correct steps, Qsp = 1.11 x 10 -5 and Qsp > 6.82 x 10 -6 , therefore a ppt forms, followed this answer, students could score a maximum of 2 marks.

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Sydney Grammar School Form VI Chemistry 2023 Trial Examination Question 34 (9 marks) Two organic compounds, A and B , are isomers with a chemical composition by mass of carbon 54.5%; hydrogen 9.2%; and oxygen 36.3%. A is soluble in water, while B is a pleasant-smelling liquid. The mass spectrums of both A and B have the M + peak at 88 but are otherwise not helpful in distinguishing between the isomers as they both show peaks at an m/z ratio of 15, 29, 43 and 73. A’s carbon-13 NMR has three peaks (one each at 184, 35 and 19 ppm), while B ’s equivalent has four peaks (171, 60, 21 and 17 ppm). The IR and proton NMR spectra are shown on the following pages, along with proton NMR shift data. Determine the structure of each of the isomers. Draw and name the isomers in the boxes provided on the following pages. Justify your choices based on the information provided, making sure to reference ALL spectral types. Marked holistically Marks Criteria 9 • Draws and names Compound A as 2-methylpropanic acid • Draws and names Compound B as ethyl ethanoate • Clear, logical justification with detailed reference to all four spectra and other supplied information • Eliminates all possible alternatives 8 Missing one item from 9 – often incorrect name OR alternative elimination 7 Missing two items from 9 5-6 Correctly identified compounds but gives limited justification 3-4 Gives a list of features with limited justification AND correctly identifies at least one of the compounds 1-2 Any new information of relevance Marking notes: MP = methyl propanoate; EE = ethyl ethanoate Identification of peaks on graphs were also marked when justification was insufficient in written response. Generally, well done. The following items were NOT marked down BUT: • Good practice to use units (e.g., 184 ppm ) • Good practice to refer to datasheet when making claims about peaks • Good practice to read IR peak as value at the bottom of the peak.

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tydney Grammar School GompoundA L2 10 ,eqzlfr* form Vl Chemistry 2021 Trial Ixamination IE Gompound B Singlet integral 1.5 ->f D Triplet: integral '1.5 4 I 5 0 -7 t1 Quartet: integral 1 ,+ 109876543210 Doublet integral 6 ----f Singtet: integral 1 Muhiplet integral 'l *__f ppm HSF-03-S34 Page 28 ppm

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!viinr:v Grannr;rr I chool lornr Vl (henrstry l0)l lriai Ir:anrination Justification: ................. Hq-ss Sper: _M+ N-sn*"k Cr. {s ' 86 r3 Ccatst\taJ\t .?^ t< motcw ,.nq 3 eF ts .c9.1t1 .... 9.*:tr:*: Lt*t^ "lo C1g,;vtv\e^ c- "l* "F lrt ..eI .... f$p..tx... ***1...h... * .... &r*.0. H .d*+r .... fslx sr*hq!L...q*-.V{ ..... Su+sSenur.!...$99,.f.2 .... :*!*es* ......... .B/ss6l $rpy-e foLgn .h*... .:y.*...* .... s.*r-o{..,...*.p.f.*y tEX*(,t cc,rl conct'gr'cqr\ +, A """u""""" . . . gLf 9,). . . . . *nel-I. . .$s)/.a&.f r. . . . ..e^e(. . . :kx d. (br C-o q'a,l C=O bo^c,LS N\z'"A h'.1 o^l sl" a^q_ Orr. 4&t tsn .**') ..... 1 +*B LuLr, ..".4 ........ ::.e.nd::7 liC..X.X.q.: ..... I].ts:.rx .... :'.i.:i;Jr1::it,.t:*:y.ty.,.\]...r.:;:::!.{.11::.:1::...:.=$:-".:i...H.:.)i...:: ..... ir,,q,,(,r, "!r r., f. :.. : .*xr...t *F1.. . k*.*.1i..::f...*y-..s..h...rys.rgf: envr\onmcnl .**.'1... #yf r.. *.**h {!.d .... 4x. .?.. ff!: .jfffY **.4..5.x.3. , .... r\-\o -t-- t6O * \8 .4v. :**xrt...! ..... ef Jrcas ( u ) Suit"," $".. B'g Q,A cr.bon -envrYonroQA,+i Q anv\trna^f cR pucr-k olo^ruL.$teol +o lrrN?.( So *topf*" cr$cul3*^\{rz eF g*hLigo't't'en th C-O b;'{ ( f"r a*+" SL'et) ?l^. sap- nz,{{- YT +" +-* j'*{i'*l*n Page 30

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l'ornr V1 [hentrttr;' ?lill lrirl Lxrntiiulioti \yrir ey 6r;lnrnrlt !chot,l Structure of A OH l\ '" n"t , no-C c---c-ll a b )*'H H '? 'p{ h n { IR spo,. ftr A(B 4aN Compound A: Name 2 - c^4*h^l o"ooanoiC e- one/ "''.''Y"1""t"".' compound B: Name ?-to.?\ ett^anoa*e Structure of B H I t4-C- Ar h o It c b \l nt fr-c*c-l-\ .c r dr I 1 vl ll1i"":..rr,,-< l..etv. i.;r,rrr.-rir.1ii-:,,r {-r'tl l"g.':.,- t !r--,'>\.:,:1rr_i S"*:ia ' tc.rga broo,J Feel( o/'ou\r\af Sooo c/"-r o'""t'13\ w.'fi c-t't geccKs. 'f,".\ s <rLcro.aA'o-rtr+'c ol +e O*H bc*'o( '\ cr-c-.cts-r 'or-tr.otqic'.sk-etlc-o'nSr\tetuFxl7r^acacbogt'lceccdcevrcLu*t"nlA' b(){'^ARBc-c+,tet^rn(lta*yega.,th4}eA^4{94|5l9..,n.1alptt<.tYr -h4 et.-.o^crq..'lBfit WV, J} C. e*rt-"\Xtr be^r,t L-- O ( tt" SO - I?SO c''n' ' , b0 clc4e' $r'd') o.6t,:hh7rcrl,..r6,nq.r^-k1 gt*ks qrts't'k +* Ne,t't A n.'r B con'tr=.fn "j Einf <zu-.6o o"kto &* &k( 9'2"'k-9 R,. r'g' i )(a-.,*, t r>-p-{r- ('1t, 1,.i fr,.,r1 ..r-)"{ '..--tt ;! { "i 11 r';r- l! 'r( ':i( ( ' I }' J f" r' ,'(' s; e \arynn pea,AS or4 <nrftocwre\+t' te-tr' otr tn*egre-l a1 s\(e w7)rc4lu'''t Ccn-\t\tork w)A B "g^'+'ltgofOn a, lf\ i+&t @ cnvirtnmqnf il ^ *\ L ccoH $'aS' g-"ter\aAb^ &t cooH 3*"p I ts ctc.^lts[ Sh!3 -+o .ctP** 12 ePn ;) cL l So Cects l34e F "0 ld NI,(R Aa*"-\ (Q 'o * rB"o mP# l*-^ gtwt t"r tl[e\3 o. t €-ssy ,rsek-+ ^ot'&'t< ,l+( x>ie- Co.4p^co'r ''' 6 \uA?^\ '1- .1.^< +o shgirtv f"yolrcgnn '^"tf (\ stt69 rv\G{ Xcc^1.{SL$+ec( pen{cnrt9 +' 'F- dPJL\e t"Iqg C'col-{ 3r'oc'P ' lr.onnnnruf . ll r: a do"6(e1 ' @ rS c< nnut+'Plc & Le.rr^t\eqlq QQ d'" +o 9 Vr '\-'\ qarl @ I 6 , cor.-sis+e^+ 'J 6 utV ,n+ t ru(e t^e,-b ..dT '' f3', "H NMR ; -s+,$,tct"{rbn . p leaAe ge-n J\.- 'l-t Nvq c[.*".t:cd st"\D+ o{cia VV

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lydney Granrnar 5clool I'otnr Vl Ihenrirtrv 20?l lrial Exanrinatron lH NMR chemical shift data Type of proton si(cH3)4 [lMS) R-CHs R-CHrR R-CHRz R-C=C-H (alkyne) -CO-CHz- (aldehydes, ketones or esters) R-CHrNHz R-CHz-X (X = F, Cl, Br, l) -CHz-O- (alcohols, ethers or esters) R-OH R-NHz RzC=CHR (alkene) R-COONH-R (amide) Ar-H (aromatic) R-CHO (aldehyde) R-COOH 6/ppm 0 0.9-1.0 1.2-1.5 1.5-2.O 2.0-3.1 2.1-2.7 2.4-3.O 3.0-4.5 3.3-4.8 1-6 1-5 4.5-7.0 5-9 6.9-9.0 9.4-10.0 9.0-13.0 7v PHQ ,| B' - 3 LUc .Tf ge--ks a^i etvlroncnerv*r M,ta3 Fq*fto^ c{o.hc. G} @,691:qp= tt!'S'1-S2"2"V srypo4r *@ 4" z-,3,3 o( tt$ o[*. fi4{ re tfr +-{d.U"l-A-a^t 0Y\ €o-ct. BC +{a{- <nutVon6U/'} {e-F , C-o11\r\.1€nl ut\ J* n+l r.r(e ,.rit^ ar{rt.u,toy hfd^g^ q,\r1\,'ot$'ilrg i3 cto*,.sL.$t<l *. agprox t+gP*, ct nr.$*r+ ^d, k 33 ^S'ts pf- rwry g,k o.['o."q -f,*' oc\t""c"r13 1o e- O bc,^rl M & eNFq/ S l[ -i\ a. i.r.Xe-t ("rt'*t+.]tic,b k) I cuasistcltl v&n *q vrtl rulq "t-o a'oQ5ac<z'tca +o 3 \3d^oa4^I M enu,,w^'*c^a @ " - @ o a .$igleb , cJs,Vi.,tp wla .i{ n+ \ }''(r wt?r^ +d'aal*5 +D 2 tyu?u l\ anu.\on"r*^+ @ Page 29

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Question 35 (7 marks) Marks The following graph shows how the pH changes during the titration of 25.00 mL of a solution of a weak monoprotic acid (HA) with NaOH. (a) Using the graph, identify the pH at the equivalence point. 1 Paid range 8.4-9.2 (b) Bromocresol purple is an indicator that changes from yellow to violet over the range 5.2 – 6.6. If bromocresol purple was used to detect the end point of this titration, what effect, if any, would this have on the calculated concentration of HA compared to its actual concentration? 1 Less than actual ( Explanation - V decrease from graph; so n(OH) decrease; so n(H + ) decrease; so c(HA) decreases ) - not required in answer

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(c) Calculate the pK a value of the HA acid. Give your answer to 2 decimal places. Marks Criteria 5 • Reads volume at equivalence points as 8.2 mL • Chooses a suitable point on graph to use in calculations • Takes any equilibrium shift into account in calculation • Calculates pKa as 4.50 to 2 d.p. 4 As above less one point 3 As above less two points 2 Any two correct calculations – with working shown 1 Any relevant information • Note it was possible to calculate 4.50 but not get full marks if errors or working not shown. • Many marks were not lost by not showing working. • Do not round too soon. Codes used in marking: Code Meaning IPC inflection point chosen when it is impossible to read graph accurately when pH changing so quickly (usually ended up with a pKa = 5.61) E Errors Sample Answer – but other points can be chosen From titration curve, V(NaOH at equivalence point) = 8.2 mL n(NaOH) = n(OH – ) = Vc = 0.0082 x 0.1 = 8.2 x 10 -4 mol = n(H + ) = n(HA) Initial [HA] = n/V = 8.2 x 10 -4 / 0.025 = 3.28 x 10 -2 M Initial [H+] = 10 -pH = 1 x 10 -3 M (since curve starts at pH = 3) HA ∏ H + + A – I 3.28 x 10 -2 0 0 C -0.001 +0.001 +0.001 E 3.18 x 10 -2 0.001 0.001 K a = [H + ][A – ] / [HA] = (0.001) 2 / 3.18 x 10 -2 = 3.14 x 10 -5 p K a = -log K a = 4.50

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Sydney_Grammar_2023_Chemistry_sol

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