Chemistry 5070/32 — October/November 2015

Walkthrough

This part is a real titration. The burette is filled with P, and a pipette is used to put $25.0$ cm³ of Q into the conical flask. An indicator is added. The colour change at the end point tells you when the reaction is complete. Record the burette reading before and after running the acid in; the titre is the difference.

You must repeat the titration until you get two or three results that are close together. Tick the two best, concordant titres, and find only their mean. The difference between two readings of each titration and the precision of the burette matter. In the actual Paper 3 marks are awarded for accuracy against the supervisor’s value, for concordance, and for a correct average. There is therefore no single printed value; the answer has to be made from the candidate’s own readings.

Key Takeaways

• A titre is a difference between final and initial burette readings.
• Repeat until titres agree closely (concordant).
• Only the ticked, best results are averaged.
• The average volume of P is then used in later mole calculations.

Common Mistakes

• Averaging all titrations instead of only the concordant ones.
• Recording burette readings without the required decimal places, for example writing 20 rather than 20.0.
• Using final reading as the volume instead of final minus initial.
• Miscopying the volume of Q from the pipette.

Things to Be Careful About

Read the burette at eye level against the bottom of the meniscus. Record readings to one decimal place, including the final zero (for example $0.0$ cm³, $20.2$ cm³). Use a white tile behind the flask to see the end point colour change clearly. Be consistent: the colour change should come in a single drop, not in pieces. This precision is essential for accuracy marks.

Walk

Concentration is in mol/dm³, so a volume written in cm³ must be divided by 1000 to become dm³.

$25.0\ \text{cm}^3 = 25.0/1000\ \text{dm}^3 = 0.0250\ \text{dm}^3$

Then use the triangle: moles = concentration × volume.

$n = cV = 0.527 \times 0.0250 = 0.0132\ \text{mol}$

The titration uses this amount of sodium hydroxide in every portion of Q.

Key Takeaways

• Always convert cm³ to dm³ when using mol/dm³.
• Concentration is a ratio: amount of solute in 1 dm³ of solution.

Common Mistakes

• Forgetting to divide the volume in cm³ by 1000.
• Holding the conversion until after the multiplication.
• Reading the concentration as 0.5270 mol/dm³ instead of the value in the paper.

Things to Be Careful About

For the mark, show the volume divided by 1000 explicitly. If the candidate actually pipette 20.0 cm³ of Q, the answer would be $20.0/1000 \times 0.527 = 0.0105$ mol; the working here follows the 25.0 cm³ used by the marking scheme.

Walk

The neutralisation equation shows one mole of NaOH requires one mole of HCl. At the end point of the titration, an equal number of moles of acid and base have been used. Since part (b) gave 0.0132 mol of NaOH, the number of moles of HCl reacting with it is also 0.0132 mol.

Key Takeaways

• The balanced equation gives the mole ratio for the titration.
• Use of a titration allows you to find an unknown amount of acid from a known amount of alkali.

Common Mistakes

• Applying a 2:1 ratio from Mg(OH)2 equation here (incorrect for NaOH + HCl).
• Correctly writing the equations but using the ratio only after looking at 1:2 from the other reaction.

Things to Be Careful About

The neutralisation equation is not the same as the magnesium hydroxide reaction; this is just NaOH and HCl, in a 1:1 ratio. The number must match the answer in part (b).

Bout

The titre of P contained 0.0132 mol HCl, but that was only the small volume that reacted with the NaOH. The whole solution P is 110 cm³. The acid is evenly mixed, so the number of moles is simply proportional to volume.

Scale factor: multiply the moles in the titre by the total volume of P divided by the titre volume.

$n_{110} = 0.0132 \times \frac{110}{V}$

If your sample titre is different from 20.2 cm³, the number will change accordingly. The important idea is that the answer is bigger than 0.0132 because 110 cm³ is more than the volume used in the titration.

Key Takeaways

• This is a ratio/proportion calculation, not a mole-ratio from a new equation.
• Use the same unit (cm³) on top and bottom.

Common Mistakes

• Scaled in the wrong direction (dividing by 110 instead of multiplying).
• Using the 100 cm³ of original acid instead of the total 110 cm³ of P.
• Forgetting the unit while writing.

Things to Be Careful About

The total solution P is 110 cm³, not the original 100 cm³ of acid. The acid remaining is diluted because the milk of magnesia added 10 cm³. Also, keep the volume in cm³ if you have already used a factor of 1000 in the earlier part.

Walkthrough

Original acid was $100$ cm³ of $1.00$ mol/dm³ HCl. Convert to dm³: $100/1000=0.100$ dm³. Then moles = concentration × volume = $1.00 \times 0.100 = 0.100$

This is the total amount of HCl that was present before any Mg(OH)₂ reacted. Later it is compared with amount left after the milk of magnesia had neutralised some acid.

Key Takeaways

• The same $n=cV$ formula is used again.
• This value is the starting acid before any reaction.

Common Mistakes

• Writing volume as 100 dm³ instead of 0.100 dm³.
• Forgetting the factor 1000.
• Confusing this with physical acid left after the reaction.

Things to Be Careful About

The acid solution was 1.00 mol/dm³, but the volume is 100 cm³, not 1 dm³. The answer is 0.100 mol, not 100 mol.

Walk

Before any reaction, $100$ cm³ of acid contained $0.100$ mol HCl. After Mg(OH)₂ had reacted, some acid had been consumed and the remaining acid was titrated, found to be the amount in (d). So

reacted acid = initial acid − remaining acid.

The acid consumed is only that which reacted with the magnesium hydroxide; the rest remained free.

Key Takeaways

• Titration of the mixture gives the leftover acid.
• The amount of acid consumed is the difference between the initial and leftover amounts.

Common Mistakes

• Adding the two quantities instead of subtracting.
• Trying to use the 110 cm³ again as if the acid remaining was potentially from the initial acid.
• Not using the value from part (d) because it is not the volume unit.

Things to Be Careful About

Use exactly the same units: both are moles. Do not subtract concentrations or volumes; subtract the numbers of moles. The percentage for sample titre is $0.100-0.0719=0.0281$ mol. If the titration result was different, this difference should change, not the mole ratio.

Walkthrough

The acid that reacted was used to neutralise the Mg(OH)₂. The balanced equation tells us two moles of HCl react with one mole of magnesium hydroxide. Therefore the number of moles of Mg(OH₂ that reacted is half of the HCl moles. This Mg(OH₂ came from the 10 cm³ of milk of magnesia initially.

To convert this to mass, multiply by the relative formula mass, 58. That gives the mass in grams of Mg(OH₂ in the original 10 cm³ sample. Concentration in g/dm³ is grams of solute in 1 dm³. Since 1 dm³ = 1000 cm³, multiply the 10 cm³ mass by 1000/10.

The worked formula in the marking scheme allows one mark for each basic conversion: halving the moles of HCl, converting it to grams with $M_r$, and scaling the volume to a dm³.

Key Takeaways

• The 2:1 equation ratio halves the acid moles to get the hydroxide moles.
• Concentration in g/dm³ = mass of solid in 1 dm³.
• Use of $M_r$ changes moles to mass.

Common Mistakes

• Forgetting to halve the moles of HCl before applying the Mr.
• Multiplying by $\frac{10}{1000}$ instead of $\frac{1000}{10}$.
• Confusing g/dm³ with mol/dm³.

Things to Be Careful About

The milk of magnesia sample was only 10 cm³, but the concentration is quoted per dm³, so the factor is 100, because $10/1000 = 0.01$; i.e. 1 dm³ is 100 times bigger than 10 cm³. The $M_r$ given (58) is already in g / mol, so no extra unit is needed until the final answer.