11. Practice Paper 10 (Paper 3)_ans

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Practice Paper 7 Advo Education Group Pte Ltd

Suggested Solution Practice Paper 11

Organic chemistry and ionic eqm Adapted from 2014 AJC P3 Q3

1 (a) (i) Since Ka1 >>> Ka2, only the first dissociation makes an appreciable contribution to the pH of the solution. Thus, the H+ from Ka2 is ignored.

1.30 x 10–2

[H3O+] = 0.03606 mol dm–3

pH = –log[H3O+] = 1.44

(ii) 2NaOH + H2A A2– + 2Na+ + 2H2O

No. of moles of A2– formed = (25/1000) x 0.10 = 0.0025 mol

No. of moles of NaOH required for complete neutralisation = 0.0025 x 2 = 0.00500 mol

Volume of NaOH required = (0.0050 / 0.10) x 1000 = 50 cm3

[A2–] = 0.0333 mol dm –3

(iii) A2– + H2O HA– + OH–

Kb1 = Kw / Ka2

= 10–14 / (5.90 x 10–7) = 1.695 x 10–8 mol dm–3

1.695 x 10–8 (allow ecf marking)

[OH–] = 2.377 x 10–5 mol dm–3

pOH = –log[OH–] = 4.62

pH = 14 – 4.62 = 9.38

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1 (a) (iv)

[1m] for correct shape of curve until 50 cm3.

[1m] for correct indication of 2 equivalence points at 25.0 cm3 and 50.0 cm3 with corresponding pH at 9.38.

[1m] for correct indication of 2 pKa values (1.89 and 6.23) and corresponding volumes (12.5 cm3 and 37.5 cm3)

(v) Indicator: thymolphthalein

The working pH range of thymolphthalein lies within the range of rapid pH change at the end point.

or

The pH at the second equivalence point is within the working pH range of thymolphthalein.

(b) (i)

trans cis

[1m] for correct displayed formulae[1m] for correct labelling



volume of NaOH / cm325.0 50.0

pH

12.0

1.44

9.38

12.5 37.5

1.89

6.23

60.0


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1 (b) (ii) There is intramolecular hydrogen bonding involving H of –OH group with the neighbouring C=O group. Hence, there are fewer sites available for intermolecular hydrogen bonds to be formed between the molecules. Less energy is needed to overcome the less extensive intermolecular H–bonding during melting.

or

Cis–butenedioic acid results in kinks which cause the molecules to be less closely packed in the structure. Less energy is required to overcome the intermolecular forces.

(c)

(d)

geraniol

accept

or

N

P

Q

R




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[1m] eachReaction Type of reaction Deduction

1. 1 mole of geraniol decolourises 2 moles of Br2(aq)

electrophilic addition Presence of two C=C bonds in geraniol

2. Heating geraniol with excess conc. acidified KMnO4(aq) produces N, P and a colourless gas.

oxidative cleavage Presence of more than one C=C bond in geraniol

The colourless gas is CO2.

3. Both N and P react with 2,4–dinitrophenylhydrazine

condensation Presence of ketone group in N and P

4. P gives a yellow precipitate and compound Q on warming with alkaline aqueous iodine

tri–iodomethane reaction or oxidation

Presence of CH3CO– group in P

5. R can be produced from butenedioic acid on reaction with H2 in the presence of a catalyst

reduction / hydrogenation R is

[2m] for any correct 3 out of the 5 points [1m] for any correct 2 out of the 5 points




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Reaction Kinetics Adapted from 2014 PJC P3 Q1d

2 The kinetics of the uncatalysed reaction between peroxodisulfate ions and iodide ions can be investigated experimentally.

S2O82(aq) + 2I(aq) → 2SO4

2(aq) + I2(aq)

To find the rate equation: rate = k[S2O82(aq)]a[I(aq)]b for this reaction, a continuous method with

sampling is used.

In an experiment, 50.0 cm3 of 0.200 mol dm3 of aqueous sodium iodide was mixed with 50.0 cm3

of 2.00 mol dm3 aqueous sodium peroxodisulfate. At various time intervals, 10.0 cm3 of the reaction mixture was withdrawn and quenched with 50 cm3 of ice-cold water. The resultant mixture was titrated against 0.0250 mol dm3 aqueous potassium thiosulfate, K2S2O3, using starch as an indicator.

The reaction between thiosulfate and iodine is as follows:

2S2O32 + I2 → S4O6

2 + 2I

The results are shown below:

Time / min 0 2 4 6 12 16Volume of K2S2O3 (aq) / cm3 0 9.50 17.00 22.50 32.25 35.50

(a) Show that 40.00 cm3 of aqueous potassium thiosulfate is required to react with 10.0 cm3

of the reaction mixture when the reaction between peroxodisulfate and iodide ions is complete.

n(I) = (50.0/1000)(0.200) = 0.010 mol (limiting reagent) n(S2O8

2) = (50.0/1000)(2.00) = 0.10 mol

n(I2) produced in 100 cm3 of reaction mixture = 0.010 / 2 = 0.00500 mol n(I2) in 10.0 cm3 = 0.00500 / (100/10.0) = 0.000500 mol

n(S2O32) = 0.000500 x 2

= 0.00100 mol volume of S2O3

2 needed = 0.00100 / 0.0250 = 0.040 dm3

= 40.00 cm 3 (shown)

2 (b) By drawing a suitable graph, use it to show that the reaction is first order with respect to I-.




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(c) The order of reaction with respect to S2O82 is reported to be one. You are required to

conduct a second experiment using the same experimental procedures to confirm the order of reaction.

Suggest suitable concentrations of aqueous sodium iodide and sodium peroxodisulfate to be used and explain how the data obtained could be used to confirm the order of reaction.

Mix 50.0 cm3 of 0.200 mol dm-3

of aqueous sodium iodide with 50.0 cm3 of 4.00 mol dm3

(or any other appropriate concentration) aqueous sodium peroxodisulfate and plot a volume of S2O3

2 needed against time, the gradient of the graph at time = 0 should be doubled compared to the first experiment if it is first order wrt S2O8

2 OR t1/2 is half.Or

Mix 50.0 cm3 of 0.200 mol dm3 of aqueous sodium peroxodisulfate with 50.0 cm3 of

2.00 mol dm3 aqueous sodium iodide and plot a volume of S2O32 needed against time,

the half life should be constant if it is first order wrt S2O82-. Note that in this case, I‒ is

added in excess. [Total: 7]




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Electrochemistry Adapted from 2011 AJC/P3/5

3 (a) (i) In the presence of water ligands, the partially filled 3d orbitals in vanadium ions are split into two levels (non–degenerate) with a small energy gap (correspond to visible light region)Energy is absorbed from the visible region when an electron promotes from a lower energy level d orbital to a vacant higher energy level d orbita l , i.e. d–d transition. The colour observed is the complement of the colour absorbed.

(ii) 2VO3– + 8H+ + Zn Zn2+ + 2VO2+ + 4H2O

Eocell = 1.00 – (– 0.76) = +1.76 V > 0

2VO2+ + 4H+ + Zn Zn2+ + 2V3+ + 2H2O Eo

cell = 0.34 – (– 0.76) = +1.10 V >0

2V3+ + Zn 2V2+ + Zn2+

Eocell = –0.26 – (– 0.76) = +0.50 V >0

Eocell = –1.20 – (– 0.76) = –0.44 V <0

From the Eocell calculations, VO3

– will be reduced by Zn to V2+, and the final colour of the solution will be violet.

(iii)n(NH4VO3) reacted =

n(SO2) = ½ (0.0257) = 0.01285 mol

Volume of SO2 needed = 0.01285 x 24000 = 308 cm 3

(b) (i) A homogeneous catalyst is a catalyst that is in the same phase as the reactants.

(ii) Step 1: Formation of intermediate

2VO2+ + 4H+ + 2I 2VO2+ + 2H2O + I2

Eocell = 1.00 – 0.54 = +0.46 V > 0

Step 2: Regeneration of the catalyst

2VO2+ + 2H2O + S2O82 2VO2

+ + 4H++ 2SO42

Eocell = 2.01 – 1.00 = +1.01 V > 0

(c) (i) cathode: VO2+ + 2H+ + e VO2+ + H2O

anode: V2+ V3+ + e

(ii) Eocell = 1.00 (0.26)

= +1.26 V

(iii) This helps to maintains electrical neutrality (prevent build up of charges in the two half cells) so that electrical energy can continue to flow.




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(d) Q = It = 14 x 20 x 3600 = 1.008 x 106 C

n(VO2+) = n(e–) = 1.008 x 106/96500 = 10.45 mol

[VO2+] initially = 10.45/5 = 2.09 mol dm 3

(e) (i) Any Eo value that is more negative than that of Eo

V3+/V2+ = 0.26V and its corresponding Eo

cell value correctly calculated.

(ii) Larger minimum external e.m.f needed to recharge the cell.

[Total: 20]

Electrochemistry Adapted from 2011 ACJC/P3/4

4 A chemist design an ion-specific probe for measuring [Ag +] in a NaCl solution saturated with AgCl using the following set- up.

(a) (i) Given the following standard half reactions.

Hg2Cl2(s) +2e 2Hg(s) + 2Cl-(aq) Eøred = +0.24 V

Ag+(aq) + e Ag(s) Eøred = +0.80 V

Obtain an overall balanced equation, including state symbols for the cell above and state the direction of the flow of electron in the cell.

[2]

2Ag+(aq) + 2Cl‒(aq) + 2 Hg(s) → Hg2Cl2(s) + 2Ag(s) [1] From Pt to Ag [1]

4 (a) (ii) An engineer wish to use the probe to analyse an ore sample. After pretreating the



Salt BridgePaste of Hg2Cl2 in Hg

Saturated KCl solution

Pt wire

NaCl solution saturated with AgCl

Ag wire

voltmeter


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sample, the chemist measured the cell voltage, Ecell as 0.53V.

The Nernst equation can be used to measure the concentration of silver ions using the probe,

Ecell = Eøcell – log10 K

Where n is the no of moles of electrons transferred in the overall reaction and K is the equilibrium constant for the overall equation in a(i).

Assuming the concentration of Cl‒ is so high that it is essentially constant, use the Nernst equation to calculate the concentration of silver ions in the ore sample [3]

K = 1/ [Ag+]2 since [Cl-] is high and effectively constant) [1] n = 2 [1]

0.53 = 0.56 – log10 1/ [Ag+]2

[Ag+] = 0.311 mol dm‒3 [1]

(iii) Given that, AgCl(s) + e Ag(s) + Cl‒(aq) Eø

red = +0.22 V

Using the half equation above and any other relevant data from the Data Booklet, derive the Eø

cell for the overall reaction: Ag+(aq) + Cl‒(aq) AgCl(s) and use the equation given in a(ii), calculate a value of the Ksp for silver chloride at equilibrium, if Ecell of any process at equilibrium is 0V. [3]

Ag+(aq) + Cl‒(aq) AgCl(s) Eøcell = +0.80 – 0.22 = 0.58V [1]

Given that for a process at equilibrium, the Ecell = 0V K = 1 / Ksp(AgCl) and n = 1 [1]

0 = 0.58 ‒ log10 1/Ksp

1/Ksp = 6.27 x 109

Ksp = 1.59 x10‒10 mol2dm‒6 [1]

4 (b) (i) To prepare the saturated solution of AgCl and NaCl. AgCl is added to a 200 cm3 solution of NaCl of concentration of 1 x 10‒3 mol dm-3. Using the Ksp value obtained in




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a(iii), predict the maximum mass of AgCl to be added to obtain a saturated solution. If you had not been able to obtain a value for the Ksp in a(iii), assume a value of 1.59 x 10‒10 mol2 dm‒6 [2]Ksp = [Ag+][Cl–] = 1.59 x 10–10

Let s be the solubility of AgCl in the given NaCl solution.(s )(s+ 1 x 10-3) = 1.59 x 10–10 Assume s << 1.0 x 10–3,s = 1.59 x 10–7 mol dm–3 [1]Maximum mass of AgCl that could dissolve = 1.59 x 10–7 x 0.200 x 143.5 = 4.56 x 10–6 g [1]

(ii) The concentration of chloride ions from silver chloride in (b)(i) is less than the square root of the Ksp value of silver chloride. Explain why this is so. [1]

AgCl(s) ⇌ Ag+(aq) + Cl–(aq)Cl‒(aq) from NaClcaused the above equilibrium to shift left by Le Chatelier’s Principle. Thus, the solubility of AgCl is suppressed due to common ion effect. [1]

(iii) Suggest why when aqueous sodium thiosulfate was added to the resultingsolution in (b)(i), more silver chloride was able to dissolve. [3] AgCl(s) ⇌ Ag+(aq) + Cl–(aq) S2O3

2– forms a complex with Ag+. [1]This decreases [Ag+]. [1]

Hence, the above equilibrium shifts right which causes more AgCl todissolve. [1]

(c) In 1999, researchers in Israel reported a new type of alkaline battery, called the “superiron” battery. This battery used the same anode reaction as an ordinary alkaline battery.

The overall equation for the cell was found to be:

2 FeO42‒(aq) + 8H2O(l) + 3 Zn(s) → 3Zn(OH)2(s) + 2 Fe(OH)3(s) + 4 OH‒(aq)

(i)Construct Construct the half-equations, with state symbols, for each electrode reaction in alkaline conditions. [2]

Anode: Zn(s)+2OH(aq) → Zn(OH)2(s) + 2e

Cathode: FeO42(aq) + 4H2O (l) + 3e → Fe(OH)3 (s) + 5OH(aq)

4 (c) (ii) A “super-iron” battery should last longer than an ordinary alkaline battery of the same size and weight.

Calculate the quantity of charge released by the reduction of 10.0 g of K2FeO4 to A is Easy with ADVO!



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Fe(OH)3.

[3]

Amount of FeO42 = = 0.0505 mol

Amount of e = 3(0.0505) mol Q = 3(0.0505) 96500 = C (missing units 1)

(iii) Suggest a reason why the “super-iron” battery can last longer, given that for anormal alkaline battery, the reaction at the cathode is:

2MnO2(s) + 2NH4+(aq) + 2e → 2NH3(aq) + 2MnO(OH) (s) [1]

Storage capacity of alkaline batteries are cathode limited and hencethe cathode in the “super iron” battery accepts 3 moles of electrons/ more electrons than the normal alkaline battery. 1 FeO4

2 Ξ 3e vs 1 MnO2 Ξ 1e




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11. Practice Paper 10 (Paper 3)_ans