PRACTICAL QUESTIONS

TITRATION AND ENTHALPY

Answers

Dr Chris Clay http://drclays-alevelchemistry.com/

http://drclays-alevelchemistry.com/

Practical Questions

M1.(a) Any three from:

A method of weighing by difference / wash the solid from its weighing container into the beaker

If the nature of any washing is imprecise penalise once only.

Wash the (wet) rod into the flask / beaker after use

Do not allow a method where the solution is made up directly in the flask.

Wash the (wet) beaker into the flask after transfer

Ignore any instructions that refer to rinsing equipment (before use) or use of deionised water.

Wash the filter funnel (after transfer) into the flask

Use a teat pipette to make up to the mark on the volumetric flask

Ensure the bottom of the (liquid) meniscus is on the graduation mark

Mix / shake the final solution in the flask / invert flask Max 3

(b) Do (a) further titration(s)

Mark these points independently. 1

To obtain concordant results

Allow results with ± 0.1 1

[5]

M2.(a) As a droplet from the funnel could enter the burette / affect volume / readings / titre 1

(b) Air bubble in jet or wtte

Do not allow misreading burette or overshooting end point. 1

(c) Ensures all reagents are able to react / mix / come into contact

Accept no reagent is left unreacted on sides of flask

Do not allow any reference to ‘removal’ of the solution unless it is clear that it is added to the flask.

1

Practical Questions

(d) The added water does not affect the mols / amount of reagents / reactants / solution Z

Do not allow mols of solution or mols in the flask.

Allow water does not react with the reagents / water is not one of the reactants

Do not allow ‘water is not involved’ 1

[4]

M3.Pipette = 0.05 × 100 / 25.0 = 0.2%

Ignore precision 1

Burette = 0.15 × 100 / 24.25 cm3

Must show working

Allow one mark for two correct answers with no working 1

[2]

M4.Increase in volume

If a volume is quoted it must be less than 300 1

Smaller increase in T above room temperature Or increased contact between calorimeter and water Or smaller heat loss by evaporation / from the surface

1

[2]

M5.(a) (Q = mcΔT)

= 50 × 4.18 × 27.3

If incorrect (eg mass = 0.22 or 50.22 g) CE = 0 / 2 1

= 5706 J (accept 5700 and 5710)

Accept 5.7 kJ with correct unit. Ignore sign.

Practical Questions 1

(b) Mr of 2-methylpropan-2-ol = 74(.0)

For incorrect Mr, lose M1 but mark on. 1

Moles = mass / Mr

= 0.22 / 74(.0)

= 0.00297 moles 1

ΔH = –5706 / (0.002970 × 1000)

= –1921 (kJ mol–1)

If 0.22 is used in part (a), answer = –8.45 kJ mol–1 scores 3

(Allow –1920, –1919)

If uses the value given (5580 J), answer = –1879 kJ mol–1 scores 3

Answer without working scores M3 only.

Do not penalise precision.

Lack of negative sign loses M3 1

(c) ΔH = ΣΔH products – ΣΔH reactants OR a correct cycle

Correct answer with no working scores 1 mark only. 1

ΔH = −(−360) + (4 × −393) + (5 × −286)

M2 also implies M1 scored. 1

ΔH = –2642 (kJ mol–1) This answer only.

Allow 1 mark out of 3 for correct value with incorrect sign. 1

(d) (–2422 – part (b)) × 100 / –2422

Ignore negative sign.

Practical Questions Expect answers in region of 20.7

If error carried forward, 0.22 allow 99.7

If 5580 J used earlier, then allow 22.4 1

(e) Reduce the distance between the flame and the beaker / put a sleeve around the flame to protect from drafts / add a lid / use a copper calorimeter rather than a pyrex beaker / use a food calorimeter

Any reference to insulating material around the beaker must be on top.

Accept calibrate the equipment using an alcohol of known enthalpy of combustion.

1

(f) Incomplete combustion 1

[11]

M6.(a) Temperature on y-axis

If axes unlabelled use data to decide that temperature is on y-axis. 1

Uses sensible scales

Lose this mark if the plotted points do not cover half of the paper.

Lose this mark if the temperature axis starts at 0 °C. 1

Plots all of the points correctly ± one square

Lose this mark if the graph plot goes off the squared paper. 1

Draws two best-fit lines

Candidate must draw two correct lines.

Lose this mark if the candidate’s line is doubled or kinked. 1

Both extrapolations are correct to the 4th minute

Award this mark if the candidate’s extrapolations are within one

Practical Questions square of your extrapolations of the candidate’s best-fit lines at the 4th minute.

1

(b) 19.5 (°C)

Accept this answer only. 1

(c) 26.5 ± 0.2 (°C)

Do not penalise precision. 1

(d) (c) – (b)

Only award this mark if temperature rise is recorded to 1 d.p. 1

(e) Uses mcΔT equation

Allow use of this equation with symbols or values for M1 even if the mass is wrong.

1

Correct value using 25 × 4.18 × (d)

7.0 gives 732 J.

Correct answer with no working scores one mark only.

Do not penalise precision.

Allow answer in J or kJ.

Ignore sign of enthalpy change. 1

(f) 9.0(1) × 10–3

Do not allow 0.01

Allow 9 × 10–3 or 0.009 in this case. 1

(g) If answer to (e) in J, then (e) / (1000 × (f))

or

If answer to (e) in kJ, then (e) / (f)

7.0 and 9.01 × 10–3 gives 81.2 kJ mol–1

Practical Questions If answer to (e) is in J must convert to kJ mol–1 correctly to score mark.

1

Enthalpy change has negative sign

Award this mark independently, whatever the calculated value of the enthalpy change.

1

(h) The idea that this ensures that all of the solution is at the same temperature

Do not allow ‘to get an accurate reading’ without qualification. 1

(i) (i) Chlorine is toxic / poisonous / corrosive

Do not allow ‘harmful’. 1

(ii) Explosion risk / apparatus will fly apart / stopper will come out

Ignore ‘gas can’t escape’ or ‘gas can’t enter the tube’. 1

[16]

M7.(a) The value of the titre would be higher (than the true value.) 1

(b) It should have no effect. 1

The first titration can be ignored / subsequent titrations would be accurate

Allow references to the first titration being a ‘rough’ or ‘trial’ value. 1

[3]

M8.(a) M1 (q = mcΔT = 100 × 4.18 × 38(.0)) = 15 884 / 15 880 / 15 900 / 16 000 (J) (OR 15.884 / 15.88 / 15.9 / 16 (kJ))

Award full marks for correct answer

Mark is for value not expression (at least 2sf); penalise incorrect units here only if M1 is the only potential scoring point in M1-M3

1

Practical Questions M2 Moles (methanol = 1.65 / 32.0) = 0.0516 or 0.052

At least 2sf 1

M3 Heat change per moles = M1/M2 (15 884 / 0.0516 / 1000 = 308 (kJ mol−1) (allow 305 to 310)

At least 2sf; answer must be in kJ mol−1

1

M4 Answer = −308 (kJ mol−1) (allow −305 to −310)

This mark is for – sign (mark independently) 1

(b) Heating up copper / calorimeter / container / thermometer / heat capacity of copper / calorimeter / thermometer not taken into account OR Evaporation of alcohol/methanol OR Experiment not done under standard conditions

Not human errors (e.g. misreading scales)

Not impure methanol

Allow evaporation of water 1

(c) (100 × 0.5 / 38 =) 1.3 or 1.32 or 1.316% (minimum 2 sf)

Allow correct answer to at least 2sf;

Allow 1.31 or 1.315% 1

(d) Idea that heat loss is more significant issue OR Idea that temperature change/rise is (significantly / much) bigger than uncertainty

One of these two ideas only and each one must involve a comparison 1

(e) M1 Mass of ethanol = 500 × 0.789 (= 394.5 or 395 (g)) 1

M2 Moles of ethanol = M1 / 46.0 (= 8.576 or 8.58) 1

M3 Heat released = M2 × 1371 = 11800 (kJ) must be 3 sf 1

Correct answer to 3sf scores 3; correct value to 2sf or more than 3sf scores 2

Answers that are a factor of 10x out score 2 if given to 3sf or 1 if given to a different number of sf

M3 ignore units, but penalise incorrect units

Practical Questions M3 ignore sign

M2 and M3 – allow consequential marking [10]

M9.(a) Selects correct titres

If 3 or more titres used them MAX 1 for conseq M3 1

= 9.7(0) cm3

Calculates mean 1

mol HCL = 0.102 × 9.70/1000 = 9.89 × 10−4(allow 9.9 × 10-4 for M3 but check not via 4 titres in which case only 1 mark)

Calculates mol (working or result gains credit)

9.92 × 10−4 scores 1 if all 4 titres used

9.83 × 10−4 scores 1 if titres 1,2, and 3 used 1

(b) mol MHCO3 = ANS 3.1 × 10 (= 9.89 × 10−3)

Use ecf if wrong mean calculated above 1

1

Mr = 148 (3sf)

Allow ecf following wrong mass conversion 1

(c) Suggestion: Use a larger mass of solid OR use a more concentrated solution of MHCO3 OR less concentrated / more dilute solution of HCl OR more MHCO3

1

Cannot score justification mark unless suggestion correct, but suggestion could be after justification

Justification: So a larger titre/reading will be needed OR larger volume of HCl

Assume reference to the solution means the MHCO3 1

(d) This question is marked using levels of response.

Level 3 Must use volumetric flask to access level 3

Practical Questions Answer is communicated coherently and shows a logical progression from stage 1 to stage 2 then stage 3.

All stages are covered and the description of each stage is complete 6 marks

All stages are covered but up to 2 omissions/errors from different stages. If 2 omissions/errors from same stage only level 2 possible

5 marks

Level 2 Answer is mainly coherent and shows progression from stage 1 to stage 3

All stages are covered but 3 omissions/errors 4 marks

All stages are attempted 3 marks

Level 1 Answer includes isolated statements but these are not presented in a logical order or show confused reasoning.

2 stages attempted 2 marks

1 stage attempted 1 mark

Level 0 Insufficient correct chemistry to gain a mark.

0 marks

Indicative Chemistry content

Stage 1: transfers known mass of solid

a) Weigh the sample bottle containing the solid on a (2 dp) balance

b) Transfer to beaker* and reweigh sample bottle

c) Record the difference in mass

Or

d) Place beaker* on balance and tare

e) Transfer solid into beaker

f) Record mass

Or

g) Known mass provided

h) Transfers (known) mass into beaker*

i) Wash all remaining solid from sample bottle into beaker

Allow use of weighing boat

*Allow other suitable glassware including volumetric flask

Stage 2: Dissolves in water

Practical Questions a) Add distilled / deionised water

b) Stir (with a glass rod) or swirl

c) Until all solid has dissolved

Stage 3: Transfer, washing and agitation

a) Transfer to volumetric / graduated flask. Allow if a clear description/diagram given eg long necked flask with 250 cm3 mark

b) With washings

c) Make up to 250 cm3 / mark with water

d) Shakes/inverts/mixes 6

[14]

M10.(a) Mass of mineral on x-axis;

If axes unlabelled use data to decide if mass of mineral is on the x-axis.

1

Sensible continuous scales;

Lose this mark if the plotted points do not cover at least 9 squares by 7. Lose this mark if the graph plot goes off the squared paper. The graph does not have to start at the origin.

1

Plots points correctly ± one square;

Award this mark if the line is close to your line. 1

Draws a best fit straight line

Award this mark if best fit line is consistent with candidate’s plotted points. Lose this mark if line is kinked or doubled.

1

(b) 1.48 or 1.49 or 1.50 or 1.5 (g);

Accept these answers only Ignore precision of answer. Allow range 1.48 – 1.5

1

Practical Questions (c) 0.0124 (mol);

Accept 0.012, 0.0125. Allow answer without working.

1

(d) (1.49 / 0.0124) = 119.4 – 125.0;

Must divide answer to part (b) by answer to part (c) to score first mark. Allow consequential answer from part (b). Allow answer without working. Ignore precision of answer.

1

(e) Answer to part (e) close to 120.3;

Allow consequential answer from part (d).

Allow correct calculation of x 1

(f) x must be a whole number; 1

(g) Good / straight line so results good / reliable;

Allow consequential answers from candidate’s graph Do not allow ‘so results are accurate’.

1

Anomaly at 1.34 g;

Allow anomaly clearly indicated on the graph. 1

(h) Ensure reaction / decomposition goes to completion;

Do not allow ‘to make fair test’ or ‘improve reliability’ Accept to ‘remove all carbon dioxide and water’.

1

(i) (i) Percentage errors too high / errors in weighing too high;

Do not allow ‘to make fair test’ or ‘improve reliability’ Do not allow ‘errors’ on its own.

1

Practical Questions (ii) Incomplete decomposition or words to that effect;

Do not allow ‘to make fair test’ or ‘improve reliability’ Do not allow ‘takes too long’ or ‘wastes chemicals’ Do not allow ‘not all of the water removed’.

1

(j) 39.05 / 18 = 2.170 and 60.95 / 84.3 = 0.723;

Allow Mr of MgCO3.H2O = 138.3 1

MgCO3.3H2O;

54 / 138.3 + 39.05% MgCO3.3H2O without working scores 1 mark.

1

(k) Atom economy for Reaction 1 is (40.3 / 84.3) x 100 = 47.8%

Maximum 1 mark if no working. Ignore precision of answers.

1

Atom economy for Reaction 2 is (40.3 / 58.3) x 100 = 69.1% 1

(l) No gas produced in stomach / won’t cause wind;

Do not allow ‘gas produced’ on its own. 1

[19]

M11.(a) (i) Volume of crater-lake solution on x-axis

Do not penalise missing axes labels.

If axes unlabelled use data to decide.

Lose this mark if axes mis-labelled. 1

Sensible scales

Lose this mark if plotted points do not cover at least half the paper or plot goes off the squared paper.

1

Practical Questions

All points plotted correctly +/– one square 1

(ii) Draws appropriate line of best fit, omitting point at 20 cm3 / 15 cm3

Lose this mark if the line deviated towards the anomalous result.

Lose this mark if the candidate’s line is doubled or kinked.

Candidate does not have to extrapolate to the origin. 1

(iii) 16.5 cm3 +/– 0.5 cm3

Accept this answer only.

Do not mark consequentially on candidate’s graph. 1

(iv) Value corresponding to 10 cm3 crater-lake solution / 6.00 cm3

Must have correct identity for explanation mark.

Accept results aren’t concordant. 1

Greatest % error from use of burette

Accept difficult to be accurate with small volumes (owtte). 1

(b) (i) pV = nRT

Accept any correct rearrangement.

Ignore case. 1

(ii) V = 81.0 × 10–6 or 8.1 × 10–5

1

n = (1 × 105 × 81.0 × 10–6) / (8.31 × 298)

Mark consequentially on candidate’s volume. 1

n = 3.27 × 10–3 (mol)

Practical Questions Correct answer without working scores one mark only.

Allow consequential mark using incorrect conversion.

Incorrect units lose this mark. 1

(iii) Mr CaCO3 = 100.1 (M1)

Accept 100 (can score this mark in calculation for M2 and M3). 1

Moles CaCO3 = (3.27 × 10–3 × 10) = 3.27 ×10–2 (M2)

Do not penalise lack of units.

Allow b(ii) × 10

Allow 1.25 × 10–3 × 10 1

Mass CaCO3 = M1 × M2 (= 3.27 g)

Correct mass without working scores one mark only.

Allow 1.25 × 10–2 × 10 × 100.1= 12.5 g 1

(iv) (3.27 / 95) × 100

Accept (b(iii) / 95) × 100.

Do not penalise precision. 1

3.44 g

Do not penalise lack of units.

Using 12.5 g gives 13.2 g

Correct answer without working scores 2 marks. 1

(v) Abundant / readily available

Accept not caustic or alkaline.

Non-corrosive

Accept insoluble so safe to add in excess (owtte). 1

[17]

Practical Questions