Sodium hydrogencarbonate, NaHCO3, is used as baking soda in cooking. Baking soda may also contain small amounts of other chemicals. In this experiment, you will determine the percentage purity by mass of an impure sample of NaHCO3 by titration with sulfuric acid. FA 1 is 0.0500 mol dm⁻³ sulfuric acid, H2SO4. FA 2 is impure NaHCO3. methyl orange

Mark scheme

Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.

1. [1] I Correct headings The following data are recorded in the space provided ∞ mass of container with FA 2 ∞ mass of (empty) container ∞ mass of FA 2 'Mass' must be stated for each piece of data. Unit/g (etc.) must be given for each piece of data. Subtraction for mass of FA 2 used must be correct.
2. [1] II All the following data are recorded ∞ two burette readings and titre for the rough titration ∞ initial and final burette readings for two (or more) accurate titrations
3. [1] III Titre values recorded for accurate titrations, and Appropriate headings and units in the accurate titration table ∞ initial / start (burette) reading / volume ∞ final/end (burette) reading / volume ∞ titre or volume / FA 1 and used/added ∞ unit: / cm³ or (cm³) or in cm³ (for each heading) or cm³ unit given for each volume recorded
4. [1] IV All accurate burette readings are recorded to the nearest 0.05 cm³. The requirement to record to 0.05 applies to burette readings, including 0.00 cm³ (if this was the initial reading), but it does not apply to the titre. This mark is not awarded if: ∞ 50(.00) is used as an initial burette reading ∞ more than one final burette reading is 50.(00) α any burette reading is greater than 50.(00)
5. [1] V The final accurate titre recorded is within 0.10 cm³ of any other accurate titre. ∞ Do not include a reading if it is labelled “rough”. α Do not award the mark if any ‘accurate' burette readings (apart from initial 0 cm³) are given to zero dp.
6. [1] Award VI, VII and VIII if δ ≤ 0.2 (cm³ g⁻¹)
7. [1] Award VI and VII if 0.2 < δ < 0.4
8. [1] Award VI, only, if 0.4 < δ < 0.6
9. ∞ Spread penalty: if the two “best” (corrected) titres used by the Examiner were ≥ 0.50 cm³ apart, maximum 2 accuracy marks.
10. α If only a rough titration is shown, award Q marks based on that, maximum 2 accuracy marks.
11. [1] Candidate calculates the mean correctly. ∞ Candidate must take the average of two (or more) titres that are within a total spread of not more than 0.20 cm³. ∞ Working/explanation must be shown or ticks must be put next to the two (or more) accurate readings selected. ∞ The mean should be quoted to 2 dp, and be rounded to nearest 0.01 cm³. (e.g. 26.665 cm³ must be rounded to 26.67 cm³) Two special cases, where the mean need not be to 2 dp: ∞ Allow mean expressed to 3 dp only for 0.025 or 0.075 (e.g. 26.325 cm³) ∞ Allow mean if expressed to 1 dp, if all accurate burette readings (apart from initial 0) were given to 1 dp and the mean is exactly correct. (e.g. 26.0 and 26.2 = 26.1 is allowed) (e.g. 26.0 and 26.1 = 26.1 is wrong – should be 26.05) This mark is not awarded if: ∞ The rough titre was used to calculate the mean. ∞ The candidate did only one accurate titration. ∞ Burette readings were incorrectly subtracted to obtain any of the accurate titre values. ∞ All burette readings used to calculate the mean were recorded as integers Note: the candidate's mean will sometimes be marked correct even if it was different from the mean calculated by the Examiner for the purpose of assessing accuracy.
12. [1] No of moles of H2SO4 used = 0.05(0) × (b) / 1000 to minimum 2 sf
13. [1] 2NaHCO3 + H2SO4 → Na2SO4 + 2CO2 + 2H2O and No of moles of NaHCO3 = 2 × answer (i)
14. [1] Mass of NaHCO3 = answer (iii) × 10 × 84
15. [1] % = answer (iv) / mass of FA 2 used × 100 All answers attempted in (i), (iii), (iv) & (v) are shown to 3 or 4 sf Minimum 3 answers attempted to gain the mark
16. [1] Any one of the following answers. ∞ the impurity does not react with (sulfuric) acid / FA 1 / NaHCO3 ∞ the impurity is not alkaline / acidic ∞ the impurity is neutral
17. [1] % error (= 0.1 / 250 × 100) = 0.04%