Chemistry 9701/34 — May/June 2016
Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme
Topics Presentation of Data and Observations · Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation · Qualitative Analysis
Borax is an alkali which has many uses. In this experiment you will determine in the chemical formula of borax, , by titration with hydrochloric acid.
FB 1 is a solution containing of borax, .
FB 2 is hydrochloric acid, .
methyl orange indicator
Method
Dilution of FB 2
- Pipette of FB 2 into the volumetric flask.
- Make the solution up to using distilled water.
- Shake the solution in the volumetric flask thoroughly.
- This diluted solution of hydrochloric acid is FB 3. Label the volumetric flask FB 3.
Titration
- Fill the burette with FB 3.
- Pipette of FB 1 into a conical flask.
- Add several drops of methyl orange.
- Perform a rough titration and record your burette readings in the space below.
The rough titre is ..................... .
- Carry out as many accurate titrations as you think necessary to obtain consistent results.
- Make sure any recorded results show the precision of your practical work.
- Record in a suitable form below all of your burette readings and the volume of FB 3 added in each accurate titration.
Answer
Rough titre: 26.4 cm^3.
Accurate titrations (all burette readings to the nearest 0.05 cm^3):
| Burette reading / cm^3 | 1 | 2 | 3 |
|---|---|---|---|
| initial | 0.00 | 1.00 | 2.00 |
| final | 25.35 | 26.40 | 27.45 |
| titre (volume of FB 3 added) | 25.35 | 25.40 | 25.45 |
Three accurate titres obtained; 25.35, 25.40 and 25.45 cm^3 are concordant, as all lie within 0.10 cm^3 of one another.
See working — representative accurate titres: 25.35, 25.40, 25.45 cm^3
Background Concept
In an acid–base titration the volume of one solution needed to react completely with a fixed volume of another is found by adding it from a burette until the indicator changes colour. The reliability of the result depends on the quality of the burette readings: readings must be taken to the nearest 0.05 cm^3 (half a small division), the titre is final reading minus initial reading, and several accurate titres must agree closely so that a trustworthy mean can be taken. A rough titration first locates the approximate end point; accurate titrations then bracket it.
Understanding the Question
This part asks you to carry out the whole titration and record the evidence on the paper. You are not expected to state one value — you are expected to show the method, the rough titre, and a proper results table with initial and final burette readings and the derived titre for each accurate titration. The mark scheme rewards the conventions: correct headings with units, readings to 0.05 cm^3, at least two titres within 0.10 cm^3 of each other, and no inconsistent readings.
Approach
Work in a fixed order: rough titration first; then repeat accurately. Each accurate run is a fresh filling of the burette, so record initial and final readings each time and subtract to get the titre. Use a table with clear headings (quantity, unit, and column for each titration). Keep your eyes on the methyl orange colour change (orange to red/pink for acid in the burette added to the alkaline FB 1).
Step-by-Step Reasoning
The marks come from the record, not from any single reading: (1) show a rough titre; (2) show a table whose headings are initial/start reading, final/end reading and titre/volume of FB 3 added, each with cm^3 as unit; (3) give every accurate reading to the nearest 0.05 cm^3 — never to a whole cm^3 and never use 50.00 as an initial reading; (4) make sure at least two accurate titres fall within 0.10 cm^3 of each other; (5) avoid doing an extra titration that breaks the concordance. The representative titres above (25.35, 25.40, 25.45) satisfy these requirements, with a spread of only 0.10 cm^3.
Key Takeaways
The examiner is testing your practical discipline, not the chemistry: consistent precision, sensible headings, and concordant titres. These habits — reading to half a division and checking agreement — carry marks in every titration question on Paper 3.
Common Mistakes
Recording readings only to whole cm^3 (loses the 0.05 cm^3 precision mark); using 50.00 as the initial reading (expressly disallowed); performing an accurate titration that falls more than 0.10 cm^3 from the others after two concordant ones have already been done; leaving out units from table headings.
Things to Be Careful About
The titre is always the difference between final and initial readings. If you refill the burette, your initial reading changes, so record every initial reading, not just the first. Quote readings to the same precision throughout, and keep the concordant set together so the mean used in part (b) is defensible.
From your accurate titration results, obtain a suitable value for the volume of FB 3 to be used in your calculations.
Show clearly how you obtained this value.
of FB 1 required ..................... of FB 3.
Working
Tick the concordant titres 25.35, 25.40 and 25.45 cm^3.
Mean = (25.35 + 25.40 + 25.45) / 3 = 25.40 cm^3
Answer
25.0 cm^3 of FB 1 required 25.40 cm^3 of FB 3.
25.40 cm^3
Background Concept
The mean titre is the single best estimate of the end-point volume. Only concordant values — those within a total spread of 0.20 cm^3 — should be averaged; an anomaly must be ignored because it shows a spurious reading. The mark scheme requires working to be visible (ticks next to the chosen values or the sum shown).
Understanding the Question
You have three accurate titres from part (a). You must show which you have selected and then state the mean titre to 2 decimal places. The paper then uses this value in all subsequent calculations.
Approach
Check the spread of the three values; here 25.35, 25.40 and 25.45 cm^3 span 0.10 cm^3, well within 0.20 cm^3, so all three are used. Add and divide by 3. Round to 2 decimal places.
Step-by-Step Reasoning
The sum is 25.35 + 25.40 + 25.45 = 76.20 cm^3; dividing by 3 gives 25.40 cm^3 exactly. This is reported with two decimal places, as required for a mean titre. If the mean had come out at 25.425, you would need to round to 25.43 (to the nearest 0.01), unless the value were exactly 25.425, in which case the note in the mark scheme permits 25.425 (three decimal places) for halving quirks.
Key Takeaways
A mean is only as good as the set it is calculated from; always justify the selection by showing which titre values were used. Report the mean to 0.01 cm^3.
Common Mistakes
Averaging an outlier along with the concordant values; failing to show the working or ticks; giving the mean to one decimal place when the readings justify two.
Things to Be Careful About
The mean feeds the entire calculation chain (c)(i)–(v), so a sloppy mean propagates forward. Round only at the end of each step, and keep the unit cm^3 attached to the value.
Calculations
Show your working and appropriate significant figures in the final answer to each step of your calculations.
Calculate the number of moles of hydrochloric acid present in the volume of FB 3 calculated in (b).
Working
FB 2 is diluted from 10.0 cm^3 to 250 cm^3, a 25-fold dilution:
[FB 3] = 2.00 / 25 = 0.0800 mol dm^-3
n(HCl) = (0.0800 × 25.40) / 1000 = 0.002032 mol
Answer
moles of HCl = 2.03 × 10^-3 mol
2.03 x 10^-3 mol
Background Concept
Moles are found from molar concentration and volume: n = cV, with volume in dm^3. The dilution of FB 2 (10.0 cm^3 made up to 250 cm^3) reduces its concentration by a factor of 250/10 = 25, so the diluted acid FB 3 has concentration 2.00/25 = 0.0800 mol dm^-3. The titrated volume of FB 3 is the mean from part (b).
Understanding the Question
The titration uses FB 3, not the original FB 2, so you first need the concentration of FB 3. You are then asked how many moles of HCl sit in the mean titre volume of 25.40 cm^3.
Approach
Convert the mean volume from cm^3 to dm^3 by dividing by 1000, then multiply by the diluted concentration 0.0800 mol dm^-3.
Step-by-Step Reasoning
n(HCl) = c × V = 0.0800 mol dm^-3 × (25.40/1000) dm^3 = 0.0800 × 0.02540 = 0.002032 mol. Because the values used have three significant figures, the answer should be given to three significant figures: 2.03 × 10^-3 mol.
Key Takeaways
Always handle a dilution by scaling the concentration by the volume ratio before using cV. Converting cm^3 to dm^3 (÷1000) is the step that most commonly introduces a factor-of-ten error.
Common Mistakes
Using 2.00 mol dm^-3 (undiluted) instead of 0.0800; forgetting to divide by 1000; quoting an excessive number of decimal places without rounding to significant figures.
Things to Be Careful About
Keep track of the identity of the solution: FB 3 is the dilute acid used in the burette. The dilution factor is exactly 25 because 10.0 cm^3 was placed in a 250 cm^3 flask and made up to the mark.
1 mole of borax is neutralised by 2 moles of hydrochloric acid.
Calculate the number of moles of borax that react with the hydrochloric acid in (i).
Working
1 mol borax : 2 mol HCl
n(borax) = 0.5 × 0.002032 = 0.001016 mol
Answer
moles of borax = 1.02 × 10^-3 mol
1.02 x 10^-3 mol
Background Concept
The balanced neutralisation of borax by hydrochloric acid consumes 2 mol of HCl per 1 mol of borax — the ratio is stated in the question. Stoichiometry converts the moles of the reactant you measured (HCl) into moles of the substance you are analysing (borax) by multiplying by the mole ratio.
Understanding the Question
You know from part (i) how many moles of HCl were neutralised. You must now find the moles of borax that reacted, using the given 1 : 2 ratio.
Approach
Divide the moles of HCl by 2, because there are two moles of HCl for every mole of borax.
Step-by-Step Reasoning
n(borax) = n(HCl) ÷ 2 = 0.002032 ÷ 2 = 0.001016 mol, which is 1.02 × 10^-3 mol to three significant figures. This value corresponds to the moles of borax present in the 25.0 cm^3 sample of FB 1 taken in the pipette.
Key Takeaways
The stoichiometric ratio is the bridge between measured reagent and analysed substance — always write the ratio explicitly and decide whether to multiply or divide by it.
Common Mistakes
Multiplying by 2 instead of dividing (which would double the moles of borax); forgetting that the stated ratio already gives borax : HCl as 1 : 2, not the reverse.
Things to Be Careful About
This result concerns the 25.0 cm^3 aliquot only — it is not yet the amount in 1.00 dm^3, which is handled in part (iii).
Use your answer to (ii) to calculate the number of moles of borax in of FB 1.
Working
Scaling from 25.0 cm^3 to 1.00 dm^3 is a ×40 factor (1000/25).
n(borax in 1.00 dm^3) = 40 × 0.001016 = 0.04064 mol
Answer
0.0406 mol dm^-3
0.0406 mol dm^-3
Background Concept
The 25.0 cm^3 aliquot is only a fraction of 1.00 dm^3 of FB 1. To convert the amount in the aliquot to the amount in 1.00 dm^3, multiply by the ratio 1000/25 = 40. Because the concentration of FB 1 is given as 15.5 g dm^-3, knowing the molar amount in 1.00 dm^3 lets us connect mass concentration to molar concentration.
Understanding the Question
Part (ii) gave moles of borax in 25.0 cm^3 of FB 1. This part asks for the moles in 1.00 dm^3 — i.e. the molar concentration of borax in FB 1.
Approach
Multiply the result of (ii) by 40. Do not round intermediate numbers before multiplying; round only at the reported step.
Step-by-Step Reasoning
n(borax in 1.00 dm^3) = 40 × 0.001016 = 0.04064 mol dm^-3. Reported to three significant figures this is 0.0406 mol dm^-3 — the molar concentration of borax in FB 1.
Key Takeaways
Unit conversion of an aliquot to a full dm^3 uses the volume ratio 1000/volume-of-aliquot. This 40× factor appears commonly in titration calculations and is worth recognising at a glance.
Common Mistakes
Using 0.04 (or worse 0.4) as the factor instead of 40; mixing up the direction of the scaling; rounding 0.001016 to 0.001 before multiplying, which distorts the final Mr.
Things to Be Careful About
The mark for (ii) and (iii) is a single combined mark for correct use of factors, so both applications must be right. Keep three significant figures through the chain so the Mr in (iv) is reliable.
Use your answer to (iii) and the information on page 2 to calculate the relative formula mass, , of borax.
Working
FB 1 contains 15.5 g of borax per dm^3, which is 0.04064 mol per dm^3.
M_r = 15.5 / 0.04064 = 381.4
Answer
M_r of borax = 381
381
Background Concept
The relative formula mass links the two concentrations: molar concentration (mol dm^-3) equals mass concentration (g dm^-3) divided by Mr. Rearranging, Mr = mass in 1 dm^3 ÷ moles in 1 dm^3.
Understanding the Question
You know from part (iii) that 1.00 dm^3 of FB 1 contains 0.04064 mol of borax, and the page-2 information states that the same volume contains 15.5 g of borax. Dividing the mass by the moles gives the Mr.
Approach
Use Mr = 15.5 / (iii). Divide 15.5 by 0.04064, without rounding the denominator, then round the quotation to three or four significant figures.
Step-by-Step Reasoning
M_r = 15.5 / 0.04064 = 381.4 (to three significant figures, 381). Because subsequent parts use small differences, keep the unrounded 381.4 in your head for part (v) rather than the rounded 381, to avoid rounding drift.
Key Takeaways
Mass concentration ÷ molar concentration = Mr is a fundamental relation for interpreting a standard solution's label. Keep unrounded values moving forward when a later step subtracts comparable numbers.
Common Mistakes
Rounding the denominator (e.g. using 0.041 instead of 0.04064) which shifts Mr by several units; dividing incorrectly (15.5 × 0.04064 instead of ÷); dropping the unit (Mr is dimensionless).
Things to Be Careful About
The value 15.5 g dm^-3 is exact data given on the paper, so treat it as having ample precision; the significant-figure limit is set by the measured titration values.
Calculate x in the formula of borax, .
Use data from the Periodic Table on page 12.
Working
M_r(Na_2B_xO_7·10H_2O) = 2(23) + 10.8x + 7(16) + 10(18) = 338 + 10.8x
381 = 338 + 10.8x
x = (381 − 338) / 10.8 = 4.02
Answer
x = 4
4
Background Concept
Every formula has a relative formula mass composed of the relative atomic masses of its elements. For Na2BxO7·10H2O, the known part is 2 Na (2 × 23) + 7 O (7 × 16) + 10 water molecules (10 × 18) = 338, leaving the B atoms to contribute 10.8x. Setting the total equal to the experimental Mr from part (iv) lets you solve for x.
Understanding the Question
The measured Mr from (iv) must equal the theoretical sum of atomic masses. Only x is unknown, so isolate it: x = (Mr − 338) / 10.8. Your arithmetic should land very close to an integer, which is the chemical constraint — x counts whole boron atoms.
Approach
Write the full Mr expression, subtract the fixed 338, divide by 10.8, and round the decimal result to the nearest integer. Confirm the value satisfies the significant-figure rule (3 or 4 s.f.) and is a whole number.
Step-by-Step Reasoning
Known atoms: 2 Na = 46; 7 O = 112; 10 H2O = 180 → total known = 46 + 112 + 180 = 338. The B atoms contribute 10.8x. Thus 338 + 10.8x = 381.4 (using the unrounded Mr), so x = (381.4 − 338) / 10.8 = 43.4 / 10.8 = 4.02 → x = 4. A whole number is chemically sensible because you cannot have a fraction of a boron atom; 4.02 rounds cleanly to 4, confirming the experiment.
Key Takeaways
Determining a molecular parameter by titration combines stoichiometry, unit handling, and formula arithmetic in one chain. The integer check is a powerful internal validation: if x came out at, say, 3.7, you would re-check your mean titre and arithmetic because x must be whole.
Common Mistakes
Forgetting one of the fixed contributions (missing the 10H2O is the classic error — it changes Mr by 180 and spoils x); rounding Mr to 381 before subtracting, which gives x = (381 − 338)/10.8 = 3.98 → 4 (still lands on 4, but less comfortably); quoting x as 4.02 instead of the integer 4; giving an answer with wrong significant figures.
Things to Be Careful About
The question asks for x, an integer, so the final answer must be a whole number even though your calculation produces a decimal. Use the unrounded Mr from part (iv) in the subtraction to keep the arithmetic tight, and quote all reported values to 3 or 4 significant figures.
The rest of this paper
2 more questions- Q2Presentation of Data and Observations · Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation14M
- Q3Qualitative Analysis · Presentation of Data and Observations13M