1 In 1756, Baron Cronstedt, a Swedish geologist, was examining one of his country’s rare minerals. He heated it strongly and to his surprise it bubbled as if it were boiling and gave off clouds of steam. He decided to call this strange mineral a zeolite, which is Greek for ‘boiling stone’. Later research showed how the chemical structure of a zeolite gives it valuable catalytic properties.

About forty zeolites are known today, and the names and formulae of some of the commonest are shown below in Box 1.

The atoms inside the square brackets form the aluminosilicate framework of the mineral with the negatively charged oxygen. Cations are present to balance the negative charges.

Box 1 Some common natural zeolites

Analcime Na16[(AlO2)16(SiO2)32].16H2O

Chabazite Ca2[(AlO2)4(SiO2)8].13H2O

Clinoptilolite Na6[(AlO2)6(SiO2)30].24H2O

Stilbite NaCa4[(AlO2)9(SiO2)27].30H2O

(a) Examine Box 1 which shows the chemical formulae of five common zeolites.

(i) What is the relationship between the number of aluminium atoms and sodium ions as the only metal ions in a zeolite?

1:1 ratio.................................................................................................................................................

(ii) Using this relationship, deduce the formula of natrolite which has an aluminosilicate framework with the formula of [(AlO2)16(SiO2)24].16H2O.

Na16[(AlO2)16(SiO2)24].16H2O.................................................................................................................................................

(b) Deduce the formula of the resulting zeolite formed if the sodium ions of clinoptilolite are replaced with sufficient calcium ions to preserve overall electrical neutrality.

Ca3[(AlO2)6(SiO2)30].24H2O.................................................................................................................................................

(c)(i) Draw a ‘dot-and-cross’ diagram to show the bonding present in a silicate tetrahedron.

Class Reg Number

Candidate Name .......................................................................

Chemistry H2 9746Tutor TuteeRevision Exercise 16: Integrated Questions

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(ii) Show the ionic nature of the aluminate tetrahedron, assuming it contains aluminium and oxide ions.

AlO2- (Al3+ and O2-)

(iii) Hence, explain if it would be correct to describe a zeolite as consisting of a giant ionic lattice.

No. ions are hydrated and only held in place by relatively weak electrostatic forces involving the lone pair of the oxygen atom in the alumina-silicate framework. .................................................................................................................................................

(d)(i) Aluminium oxide reacts with sodium hydroxide and water to form sodium aluminate, NaAl(OH)4. This is how sodium aluminate is prepared for the laboratory synthesis of zeolites.

Write an equation to describe this reaction.

Al2O3 + 2NaOH + 3H2O 2NaAl(OH)4

.................................................................................................................................................

(ii) Sodium aluminate is also formed when aluminium is dissolved in sodium hydroxide solution. Water is involved in the reaction and hydrogen as is produced.

Write an equation to describe this reaction.

Al + NaOH + 3H2O NaAl(OH)4 + 3/2 H2

.................................................................................................................................................

(iii) Metals and their oxides generally react with dilute acids, but not with alkalis. Explain why it is unusual for aluminium and its oxides to be able to react with acids and alkalis.

Aluminium and its oxides are amphoteric due to Al3+ being small and highly charged and hence having high charge density and is able to polarise / distort the large anionic cloud on O2-, resulting in Al2O3 having ionic with some covalent character, allowing it to react with both acids and alkalis..................................................................................................................................................

(iv) Draw the structural formula for the silicate anion and deduce its molecular shape. Assume that the silicate ion is a simple ion.

Shape: trigonal planar

A Level Chemistry Notes and Questions Chapter 35

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2 In the late 19th century the two pioneers of the study of reaction kinetics, Vernon Harcourt and William Esson, studied the rate of the reaction between hydrogen peroxide and iodide ions in acidic solution.

H2O2 + 2I– + 2H+ 2H2O + I2

This reaction is considered to go by the following steps.

step 1 H2O2 + I– IO– + H2O

step 2 IO– + H+ HOI

step 3 HOI + H+ + I– I2 + H2O

The general form of the rate equation is as follows.

rate = k[H2O2]a[I–]b[H+]c

(a) Explain how the rate of reaction can be measured.

.................................................................................................................................................

.................................................................................................................................................

.................................................................................................................................................

(b) Suggest values for the orders a, b and c in the rate equation for each of the following cases.

A study was carried out in which both [H2O2] and [H+] were kept constant at 0.05 mol dm-3, and [I–] was plotted against time. The following curve was obtained.

(c) Calculate the initial rate of this reaction, indicating its units.

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rate = 6.5–7.5 × 10–6 mol dm–3 s–1

(d) Use half-life data calculated from the graph to show that the reaction is first order with respect to [I–].

half-life measured and quoted as ≅ 90–94 s

evidence of two half-lives measured

(e) Use the following data to deduce the orders with respect to [H2O2] and [H+], explaining your reasoning.

lines 1 and 2: as [H2O2] increases by 0.07/0.05 = 1.4, so does rateso order w.r.t. [H2O2] = 1

lines 1 and 3: increase in rate (1.8) is also the increase in [H2O2],so rate is independent of [H+] (or zero order)

(f) From your results, deduce which of the three steps is the slowest (rate determining) step.

the first step/or the relevant equation.................................................................................................................................................

9701_w08_qp4 Q23(a) (4-aminophenyl)ethanoic acid (4-APEA) and its derivatives are being investigated as

possible drugs to treat chronic inflammation of the intestines. The synthesis of 4-APEA from methylbenzene is shown in the following scheme.

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(i) Draw the structures of the compounds G and H in the boxes above.

G is 4-nitromethylbenzeneH is 4-nitrophenylethanoic acid

(ii) Suggest reagents and conditions for the following steps.

step II: Cl2, uv light

step III: KCN (in ethanol) + heat / reflux

step V: Sn or Fe + HCl (+ heat)

(b) Silk from silkworms, used as a fabric shows a different secondary structure to that produced by spiders.

(i) What sort of bonding would you expect to occur between adjacent parts of the protein chains in each form of silk?

silkworm – hydrogen bonds spider – van der Waals’ OR hydrogen bonds

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(ii) Suggest two differences in properties that these forms of silk could have. Explain your answer.

Spider silk is more elastic/flexible/less rigid than silkworm silk/has a lower density

Silkworm silk absorbs water more easily.................................................................................................................................................

(iii) Spider dragline silk contains large amounts of the amino acid glycine. How does this affect the properties of the silk?

This increases the elasticity/hydrophobic nature of the silk.................................................................................................................................................

9701_w08_qp4 Q5, 104(a) Explain why most compounds containing transition metals are coloured, whereas

compounds of non-transition metals are usually colourless.

Colour is due to the absorption of visible lightAtom needs vacancy(ies) in the d-orbitalsThe d-orbitals are split into two energy levels by ligands (d-d splitting)Energy is used to promote electrons from lower to upper d-orbitals (d-d transition)OR Energy gap in non-transition metals does not lie in visible range.................................................................................................................................................

(b) When the neutral V(II) complex V(H2O)4Cl2 is added to water, the green complex dissolves to form a violet solution. Explain why this change of colour takes place.

Ligand exchange between chloride and water occurs

d-orbital energy gap with Cl– ligands is different to that with H2O ligands.................................................................................................................................................

(c) State the colour and formulae of the aqueous ions of vanadium in oxidation states III and IV.

V(III) is V3+ (or [V(H2O)6]3+) and is greenV(IV) is VO2+(aq) and is blue NOT V4+

.................................................................................................................................................

(d) Titration of a sample containing 0.0010 mol of an oxychloride of vanadium, VOClx, required 20.0 cm3 of 0.020 mol dm–3 KMnO4(aq) for its complete oxidation in acidic solution.

(i) Use E data from the Data Booklet to predict the final oxidation state of the vanadium after the titration.

MnO4–/Mn2+ is +1,52V, higher than VO2

+/VO2+ so final state is 5.................................................................................................................................................

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(ii) Use the titration data given above to calculate the change in oxidation number undergone by the vanadium during the titration.

moles of e– = 0.02 x 5 x 20/1000 = 0.002Hence 2 moles of electrons are used per mole of vanadiumChange is from V(III) to V(V).................................................................................................................................................

(iii) Hence deduce a value for x in VOClx.

x is 1, hence VOCl.................................................................................................................................................

9701_w03_qp6 Q95 Suggest explanations for the following observations, writing relevant equations where

possible.

(a) Adding aqueous barium chloride to an orange solution of sodium dichromate(VI) causes a yellow solid to be precipitated, and an acidic solution to remain.

Equilibrium shifts to the right as CrO42– ions are removed and hence

the solution becomes more acidic.................................................................................................................................................

(b) Adding aqueous ammonia to a solution of copper(II) sulphate produces first a pale blue precipitate, then a deep blue solution.

NH3 + H2O NH4+ + OH– (i.e. ammonia solution contains OH– ions)

Cu2+ + 2OH– Cu(OH)2 (pale blue ppt) Then 4NH3 + Cu2+(aq) = [Cu(NH3)4]2+ (deep blue solution) NH3 is a stronger ligand than H2O and displaces it.Reference to IP < Ksp due to equilibrium position shifting right as [Cu2+] decreases to favour dissolution of Cu(OH)2 due to formation of deep blue solution. Reference to LCP..................................................................................................................................................

(c) Chromium(III) chloride crystallises in three forms, all having the formula CrCl3(H2O)6. One form is violet, and another is green. Adding aqueous silver nitrate to a solution of the violet form precipitates all the chloride it contains, but only two thirds of the chloride from the green form is precipitated.

violet – [Cr(H2O)6]3+ 3Cl– green – [Cr(H2O)5Cl–]2+ 2Cl–.H2O.................................................................................................................................................

9701_w03_qp6 Q106 The diagram shows a laboratory illustration of a simple hydrogen-oxygen fuel

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cell.

(a) Write the half equation for the reaction occurring at the left hand (oxygen) electrode when the cell operates.

O2 + 4H+ + 4e- 2H2O.................................................................................................................................................

(b) State the polarity (+ or –) of the left hand (oxygen) electrode. ⊕

(c) Use the Data Booklet to calculate the voltage produced by this cell.

Ecell = +1.23 V

.................................................................................................................................................

(d) Only a very small current can be drawn from this laboratory cell. Suggest one way in which it could be modified to enable a larger current to be drawn from it.

a better/larger salt bridge or a diaphragm or larger (area of) electrodes or increase concentrations/pressure.................................................................................................................................................

(e) A fuel cell in an orbiting satellite is required to produce a current of 0.010 A for 400 days.Calculate the mass of hydrogen that will be needed.

time = 400 x 24 x 60 x 60 = 34 560 000 secondscharge = current x time = 0.01 x 34 560 000 = 345 600 C moles of H = 345 600/96 500 = 3.6 mol mass of H = 3.6 g∴

(f) State one advantage, and one disadvantage of using fuel cells to power road vehicles compared to hydrocarbon fuels such as petrol.

advantages: less pollution/CO2/NOx etc. or cleaner by-productsless dependence on fossil fuels/finite resources any one

disadvantages: more expensive (to develop or to run)takes up more spacepoor power-to-volume ratiohydrogen is difficult to store or to transport any one

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NOT hydrogen is explosive/flammable.................................................................................................................................................9701_w04_qp4 Q2

7 The water which runs from waste heaps associated with copper mines contains low concentrations of dissolved copper ions. Companies have begun removing Cu2+ ions from the water by the process of ligand exchange solvent extraction.

A compound which is a good ligand for copper ions is dissolved in an organic solvent that is immiscible with water. When this solution is shaken with the water containing the copper ions, the following reaction takes place:

Cu2+(aq) + 2LH(organic) CuL2(organic) + 2H+(aq) Equation 7.1

(where L represents the ligand).

(a) Explain why the process represented in Equation 7.1 is classed as a ligand exchange reaction.

Initially, Cu2+(aq) exists as Cu(H2O)62+, where the 6 water molecules are ligands. From

equation 7.1, L- molecules now replace the water molecules in Cu(H2O)62+, forming

CuL2, thereby constituting a ligand-exchange reaction..................................................................................................................................................

(b) The effect of the process is to remove a low concentration of copper ions from water and to transfer them, at much higher concentration, to the organic solvent. The process can be reversed by then shaking the organic solution with moderately concentrated acid. This pushes Cu2+ ions back into aqueous solution and, again, an increase in concentration can be achieved.

(i) Use Equation 7.1 to explain why extraction of Cu2+ ions into the organic solvent and then their reversal back into the aqueous phase is pH dependent.

From Equation 7.1, H+ is the product of the forward reaction and the reaction is pH dependent since by LCP, a high pH (i.e. low [H+]) will cause equilibrium to shift right to favour the forward reaction to counteract the decrease in [H+]..................................................................................................................................................

(ii) Suggest how an increase in concentration of copper ions is achieved at each stage of the extraction.

Small volume of extracting solvent used.................................................................................................................................................

(c) Research chemists have carried out experiments to find equilibrium constant, Kc, values for reactions like the one in Equation 7.1. Aqueous solutions of Cu2+ ions were shaken with solutions of a ligand in an organic solvent and allowed to reach equilibrium. In one experiment, which was maintained at pH = 2 (i.e. [H+] = 1.0 x 10–2 mol dm–3) and 298 K throughout, the equilibrium mixture was analysed and found to correspond to the following concentrations:

[Cu2+(aq)] = 0.0020 mol dm–3

[CuL2(organic)] = 0.045 mol dm–3

[LH(organic)] = 0.10 mol dm–3

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(i) Write an expression, in terms of concentrations, for Kc for the reaction in Equation 7.1.

(ii) Calculate the ratio of the concentration of Cu2+ in the organic phase to the concentration of Cu2+ in the aqueous phase, and hence the percentage of copper extracted in the experiment.

(iii) Calculate a value for Kc at 298 K from the results of this experiment.

Kc = 0.223

OCR L_A_Level Chemistry Salters SAM8 It has recently been reported that an incident occurred at the Dounreay fast-reactor nuclear

plant in Scotland in May 1977 in which about 2 kg of sodium were dumped down a shaft which had earlier been used for the disposal of radioactive waste. (Liquid sodium is used as a coolant in this type of reactor.) The shaft was partially flooded with seawater, and the violent reaction between sodium and water led to an explosion which scattered radioactive material over the nearby area.

(a) A number of highly exothermic reactions occur when sodium comes into contact with water. The principal reaction is

2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g) Equation 8.1

(i) Suggest why the reaction of sodium with water, in the restricted situation of the old mine shaft, gave rise to an explosion.

The reaction of Na with water produces H2 gas, which upon accumulation in a sealed vessel causes the build up of pressure giving rise to an explosion..................................................................................................................................................

(ii) Calculate the standard enthalpy change for the reaction in Equation 8.1 using the standard enthalpy change of formation values which follow.

Hf /kJ mol–1: H2O(l) = –286, NaOH(aq) = –470

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(b) Sodium and magnesium are neighbours in the Periodic Table. Sodium hydroxide is considerably more soluble in water than magnesium hydroxide, Mg(OH)2. In part, solubility is controlled by the enthalpy change of solution (Hsoln). This is itself determined by other enthalpy changes: for example, the enthalpies of hydration (Hhyd) of the cations and anions. Some data for the cations, Na+ and Mg2+, are given in the table below.

(i) Explain why water molecules are able to interact with both cations and anions.

.................................................................................................................................................

(ii) Explain, in terms of the three quantities: charge, ionic radius and extent of hydration, why Hhyd for Mg2+ is more negative than Hhyd for Na+.

.................................................................................................................................................

(c) Solubility is also controlled by the entropy change (S) which accompanies solution.

(i) In terms of the number of ions per mole of each compound, explain why this entropy change would be expected to be more positive for Mg(OH)2 than for NaOH.

.................................................................................................................................................

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(ii) In terms of the arrangement of water molecules, explain why this entropy change would be expected to be more negative for Mg(OH)2 than for NaOH.

.................................................................................................................................................

(iii) Name another entropy change which contributes to the total entropy change accompanying solution, and explain how it arises.

.................................................................................................................................................

OCR L A Level Salters SAM9 New Zealand has no oilfields of its own and until recently relied completely on imported oil

to meet its need for liquid fuels. The country does, however, have large reserves of natural gas (which is largely methane) and, since 1985, much of the petrol needed in New Zealand has been produced by chemical conversion of methane into liquid, hydrocarbon fuel.

The first stage in this process involves production of methanol from methane using the reactions in Equations 3.1 and 3.2. Data about these reactions are shown in the table below.

CH4(g) + H2O(l) CO(g) + 3H2(g) Equation 3.1CO(g) + 2H2(g) CH3OH(g) Equation 3.2

(a) Catalysts play a key role in increasing the rates of Reactions 3.1 and 3.2.

Briefly describe two other ways in which the conditions above are chosen to increase the rates of Reactions 3.1 and 3.2.

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.................................................................................................................................................

(b) The conditions used for Reactions 3.1 and 3.2 are chosen to give optimum yields at equilibrium from these reactions. For each reaction, explain why the conditions of temperature and pressure chosen give an optimum yield of products.

.................................................................................................................................................

(c) In the second stage of the process, methanol is converted into a mixture of hydrocarbons by Reactions 3.3 and 3.4. (In Equation 3.4, the mixture of hydrocarbons (petrol) is represented by octene, C8H16.) The reactions take place at 600 K in the presence of a zeolite catalyst.

2CH3OH(g) CH3OCH3(g) + H2O(g) Equation 3.34CH3OCH3(g) C8H16(g) + 4H2O(g) Equation 3.4

(i) In the space below, write an expression for Kp for the equilibrium in Equation 3.3 in terms of the partial pressures of the gases involved.

(ii) Under the conditions used in the industrial process, Reaction 3.3 reaches equilibrium. Calculate the equilibrium partial pressure of methoxymethane (CH3OCH3) when the partial pressure of methanol at equilibrium is 0.142 atm. (Kp = 9.00)

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(d) Zeolites are crystalline aluminosilicate materials with structures containing a network of linked channels through which molecules can pass. The channels restrict the size of the hydrocarbon molecules produced and their passage out of the zeolite.

In the zeolite used as a catalyst for Reactions 3.3 and 3.4, only molecules with up to 12 carbons atoms can be formed and pass through the channels. Despite this, up to 200 hydrocarbon compounds are present in the reaction product. This large number is due to the fact that, for most hydrocarbons, there are several ways in which the carbon and hydrogen atoms can be arranged for any given formula.

(i) Give three ways in which the arrangement of carbon and hydrogen atoms can give rise to different molecules of the same formula.

.................................................................................................................................................

(ii) Describe another industrial use for zeolites.

.................................................................................................................................................

OCR SAM Salters10 This question is regarding transition metal chemistry.

(a) What do you understand by the term ligand?

An atom, ion or molecule that has a lone pair of electrons that can form a dative bond to the metal ion..................................................................................................................................................

(b) Choose two neutral molecules and one ion that can act as ligands. Draw a diagram to show clearly how one of these ligands combines with a metal ion such as Cr3+.

.................................................................................................................................................

(c) The salt tetrasodium ethylenediaminetetra-acetate, EDTA, is a polydentate ligand that forms six bonds to a metal.

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With many metal ions EDTA forms strong complexes that are usually very soluble; consequently it has been used to counteract the effects of poisoning by heavy metals such as cadmium. The complexed cadmium can then be excreted in urine.

One disadvantage of its therapeutic use is that it also forms soluble complexes with cations such as Ca2+ and Zn2+ that occur naturally in the body and which are essential for health.

(i) On the above formula of EDTA, circle six atoms that can form bonds to a metal ion.

(ii) Use the following data to explain how EDTA alleviates cadmium poisoning. Explain whether calcium or zinc ions might also be complexed at similar concentrations and suggest how these problems could be overcome.

Ca2+ + edta4- [Ca(edta)]2- Kc = 5 x 1010dm3mol-1

Cd2+ + edta4- [Cd(edta)]2- Kc = 4 x 1016dm3mol-1

Zn2+ + edta4- [Zn(edta)]2- Kc = 3 x 1016dm3mol-1

.................................................................................................................................................

(d) The ‘crown ether’ A and the ‘crown thioether’ B are polydentate ligands that are used to remove harmful metals from the environment.

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(i) Explain what is meant by a polydentate ligand.

Polydentate – can form more than one (dative) bond per molecule of ligand.................................................................................................................................................

(ii) What features of both compound A and compound B make them suitable to act as ligands?

They contain lone pairs of electrons on oxygen and sulphur..................................................................................................................................................

(iii) Suggest the co-ordination number in, and the shape of, the complex formed between A and Sr2+ ions.

6, octahedral.................................................................................................................................................

(iv) Suggest the co-ordination number in, and the shape of, the complex formed between B and Ni2+ ions.

4, square planar or tetrahedral.................................................................................................................................................

9701_s04_qp6 Q9, 9701_w06_qp6 Q1011 This question is about the properties and reactions of the oxides of some elements in their

+4 oxidation state.(a) Chlorine dioxide, ClO2, is an important industrial chemical, used to bleach wood pulp for

making paper, and to kill bacteria in water supplies. However, it is unstable and decomposes into its elements as follows.

2ClO2(g) Cl2(g) + 2O2(g)

Compound B complexes well with ions of metals that are towards the right of the Periodic Table, and can be used to extract nickel, copper and lead ions from solutions of industrial waste.

Compound A complexes with ions of metals on the left of the Periodic Table, and can be used to extract radioactive strontium ions from groundwater. (Complexes formed by the ions of Group II metals are similar to the complexes formed by transition metal ions.)

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(i) The chlorine atom is in the middle of the ClO2 molecule. Using the chlorine-oxygen bond energy as 278 kJ mol–1, and other values from the Data Booklet, calculate H for the above reaction.

ΔH = 4 × 278 – 244 – 2 × 496 = –124 kJ mol–1

(ii) Assuming the Cl–O bonds in chlorine dioxide are double bonds, predict the shape of the ClO2 molecule. Explain your answer.

shape is bent/V-shaped/non-linear (or diagram)due to (one) lone pair and/or (1) odd/unpaired electron (or shown on diag) (assume electrons are on chlorine unless explicitly stated otherwise, in which case award no mark)

(iii) Cl O2 can be made in the laboratory by reacting KClO3 with concentrated H2SO4. Other products are K2SO4, KClO4 and H2O.

Construct a balanced equation for this reaction.

3KClO3 + H2SO4 K2SO4 + KClO4 + H2O + 2ClO2

.................................................................................................................................................

(b) All the oxides of the elements in Group IV in their +4 oxidation state are high melting point solids except CO2.

(i) Explain this observation by describing the bonding in CO2, SiO2 and SnO2.

CO2: simple + molecular/covalent or weak intermolecular forcesSiO2: giant/macro + molecular/covalentSnO2: ionic/electrovalent (ignore “giant”).................................................................................................................................................

(ii) State the difference in the thermal stabilities of SnO2 and PbO2. Illustrate your answer with an equation.

SnO2 is stable, PbO2 is not or SnO2 is the more stable PbO2 PbO + ½ O2

.................................................................................................................................................

CO2 dissolves in water to form a weakly acidic solution containing the hydrogencarbonate ion.

(iii) Write an equation for the reaction of CO2 with water, and write an expression for the equilibrium constant, Kc.

H2O + CO2 ( ) H⇌ + + HCO3¯

Kc = [H+][HCO3¯]/[H2O][CO2] or = [H+][HCO3¯]/[CO2]

(iv) Explain the role of the hydrogencarbonate ion in controlling the pH of blood, illustrating your answer with relevant equations.

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HCO3¯ + H+ H2CO3 or H2O + CO2 (or equation with H3O+)

HCO3¯ + OH¯ CO3

2- + H2O (NB NOT H2CO3 + OH¯ ).................................................................................................................................................

9701_s08_qp412 The following account describes the preparation of Péligot’s salt, named after the 19th

century French chemist who first made it.

Place 6.0 g of potassium dichromate(VI) in a 100 cm3 beaker and add 8.0 g of concentrated hydrochloric acid and 1.0 cm3 water. Warm the mixture gently; if carefully done the dichromate(VI) will dissolve without the evolution of chlorine. On cooling the beaker in an ice bath the solution will deposit long orange-red crystals of Péligotʼs salt.

An analysis of Péligot’s salt showed that it contained the following percentages by mass:

K, 22.4%; Cr, 29.8%; Cl, 20.3%; O, 27.5%.

(a) Calculate the empirical formula of Péligot’s salt.

(b) Suggest a balanced equation for the formation of Péligot’s salt.

K2Cr2O7 + 2HCl 2KCrClO3 + H2O.................................................................................................................................................

(c) The instructions suggest that strong heating might cause chlorine to be evolved.

(i) What type of reaction would produce chlorine in this system?

oxidation.................................................................................................................................................

(ii) Use the Data Booklet to identify relevant half equations and E values for the production of chlorine from the reaction between K2Cr2O7 and HCl.

.................................................................................................................................................

Use these equations to write the overall full ionic equation for this reaction.

.................................................................................................................................................

(iii) The use of dilute HCl (aq) does not result in the production of chlorine. Suggest why this is so.

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(dilution will) lower Eo for Cr2O72–/Cr3+ or raise Eo for Cl2/Cl–

or lower [Cl–] or [H+] will shift equilibrium in eqn to the left hand side.................................................................................................................................................

(iv) Use the Data Booklet to suggest a reason why it is not possible to prepare the bromine analogue of Péligot’s salt by using HBr(aq) instead of HCl (aq).

Br2/Br– = +1.07 V, so Cr(VI) would oxidise Br– (easily).................................................................................................................................................

9701_w07_qp4 Q3End of Paper

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