NT Exampro 1

1. Propanoic acid is a weak acid; explain the term weak. incompletely / partially / slightly dissociated OR dissociation varies with concentration allow poor proton donor OR conjugate base strong

(1)

2. (i) Give an equation for the dissociation of propanoic acid and hence an expression for its dissociation constant, k2.

CH3CH2COOH CH3CH2COO- + H+

OR CH3CH2COOH + H2O CH3CH2COO- + H3O+ (1)

Ka = COOH]CH[CH

]][HCOOCH[CH

23

23+−

if H2O in equation: Ka = COOH]CH[CH

]O][HCOOCH[CH

23

323+−

Mark consequentially for second mark if acid is incorrect but MUST have H+ or H3O+.

(ii) At 25 °C ka for propanoic acid is 1.30 × 10-5 mol dm-3. Find the pH of a solution of propanoic acid of concentration 0.0100 mol dm-3. State any assumptions you make.

[H+] = 0.01101.30 5 ××= −cKa = 3.6 × 10-4 (1)

pH = –lg[H+] (1) = -lg(3.6 ×10-4 = 3.4(4) any rounding should be consistent (1) assumes dissociation of acid is small / concentration of acid does not change (1) if made no assumptions ie solved quadratic and correct answer (4) penalise arithmetic errors ONCE ONLY

(iii) Increasing the temperature of the propanoic acid solution causes the pH to decrease. what does this tell you about the enthalpy of dissociation? Justify your answer.

pH down means [H+] up (1) dissociation endothermic (1) equilibrium shifts in endothermic direction when temperature rises (1) mark consequentially

(9)

3. (i) pH = –lg[H+] (1) or lg ][

1+

H

(ii) Define Kw , the ionic product of water. Kw = [H+] [OH–] (1)

(Accept [H3O+] instead of [H

+])

(Total 2 marks)

www.chemactive.com

NT Exampro 2

4. (i) What is the principal property of a buffer solution?

maintains pH nearly constant (1) if small

(ii) The dissociation constant for ethanoic acid is 1.80 × 10–5 mol dm–3. Calculate the pH of a buffer solution which has a concentration of 0.0150 mol dm–3 with respect to ethanoic acid and 0.0550 mol dm–3 with respect to sodium ethanoate.

amounts of acid or base added (1) accept resists change to pH

Ka = 1.8 × 10–5 = )1(][

]][[

][

]][[

23

23

acid

saltH

HCOCH

COCHH +−+

=

[H+] = 1.8 × 10–5 × 0.015/0.055 = 4.91 × 10–6 (mol dm–3)

pH = –lg 4.91 × 10–6 = 5.3 (1)

Mark consequentially if salt and acid wrong way up Incorrect if used 0.015 twice but would get final mark consequentially

(Total 5 marks)

5. Phosphoric(V) acid, H3 PO4 , is a tribasic acid.

Write a formula of the potassium salts of this acid. KH2PO4

K2HPO4

K3PO4 (1)

do not allow if ions only

(Total 1 marks)

6. For the reaction

CaCO3 (s) CaO(s) +CO2 (g)

the equilibrium total pressure at 1200 K is 4 atm. Determine Kp. Kp = p(CO2) (1)

so Kp = 4 atm

(Total 3 marks)

4 (1) atm (1)

Do not allow square brackets

7. (a) State Raoult’s Law for an ideal binary liquid mixture. p(A) = po(A) × (A) (1) definition of symbols (1) (or words to that effect)

(2) www.chemactive.com

NT Exampro 3

(b) Liquid oxygen and liquid nitrogen form an ideal liquid mixture. At 70 K the vapour

pressures are 38.3 kPa for N2 and 6.40 kPa for O2. Find the composition of the vapour in

equilibrium with a liquid mixture which at equilibrium is an equimolar mixture of the two elements.

p(N ) = 38.3 x 0.5 = 19.15 kPa

p(O ) = 6.4 x 0.5 = 3.2 kPa

(N ) = 19.15/22.35 = 0.857

(O ) = 0.143 (by any method)

x

x∴

P = 22.35kPa OR recognition of ratio 6:1

(1)(1)

tot

OR 6/7 and 1/7 of final composition ∴ (1)

(1)

(1)

2

2

2

2

Answer must be quoted to 2 or more significant figures but penalise 6 or more significant figures once only on the paper

(3)

(c) Show by means of sketches of boiling point / composition diagrams and a brief comment why ideal mixtures of liquids of similar boiling temperature are more difficult to separate by fractional distillation than those with boiling temperatures more widely separated.

diagrams to show wider difference between b.p. (1) if b.p's close ∴ v.p's each

component close ∴ effective separation more difficult ∴more stages in the fractionating column necessary (or words to that effect) (2)

(3)

(d) The vapour pressure / composition diagram at constant temperature for a mixture of two liquids A and B which shows a strong deviation from Raoult’s Law is shown below. (x(A) = mole fraction of A.

x( ) = 1A

vapourpressure

x( ) = 0Amole fraction

www.chemactive.com

NT Exampro 4

(i) Sketch on the axes below the boiling point / composition diagram for this system,

showing the curves for both liquid and vapour.

x x(A)=1 composition (A)=0

θ θb

bb

b

θ θ (A) < (B) azeotrope inmore or less correct % position azeotrope is minimum liquid/vapour lines correct and labelled (labels need only show vap. and liq. once)

(1)(1)

(1)(1)

(ii) consider a mixture for which the mole fraction of A is initially 0.75. Use your diagram to explain what happens when this mixture is fractionally distilled, making clear the nature of the distillate and of the residue. distillate is azeotrope (1) residue is A (1) suitable discussion (1) referred to diagram (1) (eg use of tie lines)

(8) (Total 16 marks)

8. State Raoult’s Law for an ideal binary liquid mixture. p(A) = po(A)x(A) (1) definition of symbols (1) (or words to that effect)

(Total 2 marks)

9. Liquid oxygen and liquid nitrogen form an ideal liquid mixture. At 70 K the vapour pressures are 38.3 kPa for N2 and 6.40 kPa for O2. Find the composition of the vapour in equilibrium with a liquid mixture which at equilibrium is an equimolar mixture of the two elements.

p(N ) = 38.3 x 0.5 = 19.15 kPa

p(O ) = 6.4 x 0.5 = 3.2 kPa

(N ) = 19.15/22.35 = 0.857

(O ) = 0.143 (by any method)

x

x∴

P = 22.35kPa OR recognition of ratio 6:1

(1)(1)

tot

OR 6/7 and 1/7 of final composition ∴ (1)

(1)

(1)

2

2

2

2

Answer must be quoted to 2 or more significant figures but penalise 6 or more significant figures once only on the paper

(Total 3 marks)

www.chemactive.com

NT Exampro 5

10. Show by means of sketches of boiling point / composition diagrams and a brief comment why

ideal mixtures of liquids of similar boiling temperature are more difficult to separate by fractional distillation than those with boiling temperatures more widely separated.

diagrams to show wider difference between b.p. (1) if b.p's close ∴ v.p's each

component close ∴effective separation more difficult ∴more stages in the fractionating column necessary (or words to that effect) (2)

(Total 3 marks)

11. The vapour pressure / composition diagram at constant temperature for a mixture of two liquids A and B which shows a strong deviation from Raoult’s Law is shown below. (x(A) = mole fraction of A.

x( ) = 1A

vapourpressure

x( ) = 0Amole fraction

(i) Sketch on the axes below the boiling point / composition diagram for this system, showing the curves for both liquid and vapour.

x x(A)=1 composition (A)=0

θ θb

bb

b

θ θ (A) < (B) azeotrope inmore or less correct % position azeotrope is minimum liquid/vapour lines correct and labelled (labels need only show vap. and liq. once)

(1)(1)

(1)(1)

(ii) Consider a mixture for which the mole fraction of A is initially 0.75. Use your diagram to explain what happens when this mixture is fractionally distilled, making clear the nature of the distillate and of the residue.

distillate is azeotrope (1) residue is A (1) suitable discussion (1) referred to diagram (1) (eg use of tie lines)

(Total 8 marks)

www.chemactive.com

NT Exampro 6

12. (a) Ammonia is manufactured by direct synthesis in the Harber Process:

N2(g) + 3H2(g) = 2NH3(G) ∆H = 92Kj mol–1

(i) Write an expression for the equilibrium constant, Kc, for this reaction and give its units.

KNH

N Hmust be square bracketsc =

[ ]

[ ][ ]( )3

2

2 23

1

mol–2 dm6 (1) units marked consequentially on above Kc ( or Kp if used)

the unit mark can be obtained if the correct units are mentioned in (ii)

(ii) When 3 mol of hydrogen and 1 mol of nitrogen were allowed to reach equilibrium in a vessel of 1 dm3 capacity at 500°C and 1000 atm pressure, the equilibrium mixture contained 0.27 mol of N2, 0.81 mol of H2 and 1.46 mol of NH3.

Calculate Kc at this temperature.

Kc =( . )

( . )( . )( )

146

0 27 0811

2

3

= 14.86 or 14.9 or 15 (1)

penalise more than 4 sig figs (lose (1))

(4)

(b) Predict and explain the effect of an increase in temperature on:

(i) the value of Kc: Kc decreases (1)

since reaction exothermic / reverse reaction endothermic (1) (increase in temperature) moves equilibrium to left / causes endothermic reaction to occur (1)

(ii) the rate of forward reaction. increases (1) energies / speed of molecules increased (1) more have energy > EAct (1) or sufficient energy to react instead of EAct

reference to EAct or sufficient energy to react essential.

(6) (Total 10 marks)

13. When ammonium salts are dissolved in water, the following reaction occurs.

NH4 + (aq) + H2O(1) NH3(aq) + H3O+ (aq)

(a) Identify the acid/ base conjugate pairs in this reaction by writing appropriate symbols under each of the species in the equation above.

A1 B2 B1 A2 (1) must show related pairs

www.chemactive.com

NT Exampro 7

(b) Write an expression for the dissociation constant, K2, for NH 4 + (aq)

Ka = [ ][ ]

[ ]

NH H

NH

3 3

4

0+

+ allow H+, water must not

(c) Calculate the pH of a solution of ammonium chloride 0.100 mol dm–3 at 298 K, the Ka

value for NH4+ being 5.62 × 10–10 mol dm–3 at this temperature.

be included

5.62 × 10–10 = [Total H30+ marks]2 /0.1 (1)

[Total H3O+ marks] = √(5.62 × 10–11)

= 7.497 × 10–6 (1)

∴ pH = 5.1 (1) if H+ concentration not correct but correct pH calculated from it (1) correct answer only (1)

(Total 5 marks)

14. Benzene and methylbenzene may be separated by fractional distillation. Sketch the general form of the boiling point/composition diagram for such a mixture and use it to explain the basis on which fractional distillation rests.

0 1

b.p. of benzene

temperature/ºC

b.p. of methylbenzene

mole fraction of methylbenzene vapour line (1) liquid line (1) tie line (1) vapour richer in more volatile component (1) must be a verbal statement repeat tie lines (1)

∴ distillate is benzene (½) and residue is methylbenzene (½) must be a verbal statement

(6) (Total 6 marks)

15. (a) When ethanoic acid is dissolved in water, the following equilibrium is established:

CH3CO2H + H2O CH3CO2– + H3O+ I

When hydrogen chloride dissolves in ethanoic acid, the equilibrium established is:

CH3CO2H +HCI CH3CO2H2 + Cl– II

Comment on the role of the ethanoic acid in:

www.chemactive.com

NT Exampro 8

(i) I (acid) –proton/H+ donor (1) if just acid (½).

(1) (ii) II (base) – proton/H+ acceptor/remover (1) if just base (½)

(1)

(b) What is the relationship between the species CH3CO2H2+ and CH3CO2H? conjugate

(1)

(acid/base) pair (1)

(c) The value of Ka for ethanoic acid at 298 K is 1.74 × 10–5 mol dm–3 and for methanoic

acid, HCO2H, it is 1.60 × 10–4 mol dm–3 at the same temperature.

(i) Write an expression for Ka for XH3CO2H.

KH CH CO

CH CO Ha

[ ][ ]

[ ]( )

+ −3 2

3 2

1

if [H2O] included (0) accept [H3O+]

(1)

(ii) Hence calculate the pH of a 0.100 mol dm–3 solution of CH3CO2H at 298 K.

1.74 × 10–5 = [ ]

.

H+ 2

01or [H+] = 1.32 × 10–3 (1)

pH = 2.88 / 2.9 / 2.90(1)

(2)

(d) The pH of a 0.050 mol dm–3 solution of HCO2H is 2.55.

Using this, together with the data in (c) and your answer to (c)(i):

(i) state which of the two acids is the stronger; methanoic (1)

(1)

(ii) comment on the relative pH values of the two acids. pH less for methanoic (½) despite it being more dilute (½) strength depends on Ka / degree of dissociation (1)

pH depends on concentration as well as strength / pH is measure of concentration of hydrogen ions sensible comment on relative electron release effect of the methyl group (1)

(3)

(e) (i) Sketch with reasonable accuracy on the axes below, how the pH changes during the titration of 20.0 cm3 of a 0.100 mol dm–3 solution of methanoic acid with 0.050 mol dm–3 sodium hydroxide solution. correct start pH 2 – 3 (1) line must be rising for this mark to be awarded

correct place for vertical line at 40 cm3 (1)

correct range for vertical portion (or slightly off vertical)

ie start at 6/7 end at 10/11 (1)

(3)

www.chemactive.com

NT Exampro 9

(ii) Select using the data below a suitable indicator for this titration. Give a brief

reason for your choice based on the curve drawn in (e)(i).

Indicator pH Range Bromocresol green 3.5 – 5.4 Bromothymol blue 6.0 – 7.6 Phenol red 6.8 – 8.4 Phenol red or bromothymol blue (1)

pH range for colour change in correct pH range for vertical portion of graph (1)

note: the choice of indicator is consequential on the straight vertical portion of the graph drawn

(2) (Total 15 marks)

16. This question concerns redox behaviour and the following data will be found useful.

E /V

Fe3+ (aq) + e– Fe2+ (aq +0.77 ½Cl2(g) + e– Cl– (aq) +1.36

mNo4– (aq) + 8H+ (aq) + 5E– Mn2+ (aq) + 4H2O(I) + +1.52

(a) In potassium manganate(VII) titrations, the solutions are acidified with dilute sulphuric acid.

(i) Using the data above, explain why dilute hydrochloric acid is not used for this purpose. CI– oxidised (by manganate(VII) / CI– ions form CI2 (1)

quantitative relevance of redox potentials (1)

but calculations not

hence volume of manganate(VII) unreliable / too large (1)

required

(3)

(ii) Why is potassium manganate(VII) usually placed in the burette, despite the difficulties it presents in reading the burette? colour change would be colourless to pink/purple (1)

not

which is very easy to see / very sharp end-point (1)

clear instead of colourless

or

(2)

inverse argument

(b) A test for Mn2+ ions in solution is to react them with sodium bismuthate(V), NaBiO3, in the presence of nitric acid. A purple colour will develop owing to the formation of MnO4– ions in the solution. The ionic half equation for the reduction of BiO3– ion is:

BiO3– + 6H+ + 2e– Bi3+ + 3H2O

(i) Use this half equation and the data above to write an ionic equation for the oxidation of the Mn2+ ion.

5BiO3– + 14H+ + 2Mn2+ → 5Bi3+ + 2MnO4

– + 7H2O

multiplication 5 to 2 (1)

correct overall equation (1)

(2)

(ii) Suggest qualitatively, how the E value for BiO3– | Bi3+ compares with that for

MnO4–| Mn2+

www.chemactive.com

NT Exampro 10

E for bismuthate must be more positive/greater/bigger/higher than that for

manganate(VII) (1)

(1)

(c) An alloy contains iron and manganese only. One arming with dilute nitric acid 2.30 g of this alloy gave a solution containing iron(III) ions and manganese (II) ions. Treatment of this solution with excess sodium bismuthate(V) completely oxidised all the Mn2+ ions present to MnO4– ions.

The excess bismuthate(V) ions were then completely destroyed and the solution made up to 250cm3 with distilled water and thoroughly shaken.

Titration of 25.0cm3 portions of this solution required 25.0cm3 of standard 0.100 mol dm–3 iron (II) sulphate solution.

(i) Write the equation for the reaction occurring during the titration.

MnO4– + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O

multiplication 5 to 1 (1)

correct overall equation (1)

(2)

(ii) Calculate the percentage of manganese present in the alloy.

moles Fe2+ = 0.1 × 25

1000 (1) = 2.5 × 10–3

moles manganate(VII) = moles Mn2+ = 2 5 10

5

3. × −

(1) = 5 × 10–4

moles Mn2+ in 250 cm3 = 10 × (0.5 × 10–3) (1) = 5 × 10–3

mass Mn in alloy = 55 × (5 × 10–3) (1) = 0.275g

% Mn = (55 × 5 × 10–3) × 100

2 3. (1)

= 12%

marks are for the processes but if the final answer is incorrect

ie ≠ 12% to 2 – 4 (inc.) significant figures then penalise (–1)

if answer is correct but not given to 2–4 (inc.) significant

figures then penalise (–1)

correct answer with some working (5)

correct answer and no working (1)

if equation is incorrect in (c)(i) but the processes are

correct for that equation then (max 4)

(5) (Total 15 marks)

www.chemactive.com