1. Section 4
Volumetric Analysis [solutions]
2002
(a)
(b)
(c)
(d)
Question 1
COMPOUND: ethanol / C2H5OH
PROCESS: oxidation
DESCRIBE: pipette (burette) vinegar[ can be shown on diagram]
into volumetric flask [can be shown with diagram provided line on neck present]
add deionised water
when near mark, add dropwise (using dropper/read bottom of meniscus)
NAME: burette
WASHING & FILLING: with deionised water / then solution (ethanoic acid) / use of funnel / ensure
that the area below the tap is filled
INDICATOR: phenolphthalein Cancelling applies to INDICATOR
CONC DIL VIN: 0.15
[ 20.5 x M = 25 x 0.12 (3) M = 0.15 (3) ]
ORIG VIN: 1.5
[ 0.15 x 10 (3)]
PERCENTAGE: 9
[ 1.5 x 60 = 90 g l–1 (3) 90 ÷ 10 = 9 % (3) ]
2003
(3 x 3)
(3)
(6)
(3)
(6)
Question 1
prevents oxidation by air of iron(II) {Fe2+}
Allow (3) if ‘air’ is not mentioned.
tablets crushed with mortar and pestle //
washed into beaker // stirred to dissolve //transferred into flask using funnel //
rinsings of beaker added to flask // mark at eye-level // add drop-by-drop // until bottom of
meniscus level with mark // invert
ANY SIX
(6 × 3)
to ensure complete conversion of MnO4 – to Mn2+ / to fully reduce MnO4 – / to prevent formation of
manganese (IV) { MnO2} /
Allow (3) for “to prevent the formation of a brown pptte”
first permanent pink (purple) / pink (purple) remains /
0.0278 / 0.028 M
25 × X = 13.9 × 0.01(0)
5
1
X = 0.0278 / 0.028 (3)
Second (3) not available if (0) obtained for first (3)
(a)
(b)
(c)
(d)
(e)
(4)
(4)
(3)
(3)
(3)
(3)
(3)
(i)
(5)
(18)
(6)
(3)
(6)
(ii)
(9)
(iii)
2004
0.0973 – 0.0982 g
0.0278 × 56* = 1.5568 / 1.56 / 1.568 / 1.57 g l–1 (3) [*56 essential]
Accept to two significant figures. {55, 57, 65 (– 1)}
1.5568 ÷ 4 = 0.3892 – 0.3925 g in 250 cm3 (in four tablets) (3)
0.3892 ÷ 4 = 0.0973 – 0.0982 g in one tablet (3)
27 %
0.0973* × 100 = 27
0.36
(answer that would give 27 if given to two significant figures) (3)
(3)
Question 1
(a)
(b)
(c)
(d)
(i)
(ii)
(iii)
NAME: eriochrome black
COL. CHANGE: wine red (wine) to deep blue
DESCRIBE: rinse with deionised water // rinse with solution //
clamp vertically // use funnel / remove funnel after filling //
open tap to fill below tap / remove air bubbles [Allow ‘tap is full’] //
set bottom of meniscus on mark / read bottom of meniscus
ANY FOUR (6 + 3 × 3)
SOLUTION: buffer
to keep pH above 9 / [Accept ‘to keep pH at a certain value (from changing)’]
CALCULATE: 0.00081 mol l–1
100 × X = 8.1 × 0.01 / 100 × X = 8.1 × 0.01 (6)
X = 0.00081 (3)
1
1
0.081 g l–1
0.00081 × 100* = 0.081 (3)
[*100 essential unless calculation shown e.g. 40 + 12 + 3 x 16 = 98 (slip)]
81 p.p.m.
0.081 × 1000 = 81 (3)
(4)
(4)
(15)
(3)
(3)
(9)
(3)
(3)

2. EQUATION: CaCO3 + 2HCl = CaCl2 + H2O + CO2 /
CaCO3 + 2HCl = CaCl2 + H2CO3
2005
Question 1
WHY: so that oxygen content doesn’t increase due to photosynthesis /
Note: do not award marks for attempts like “may absorb oxygen from the air”
WHY CONC.: to minimise the amount of the water sample that is displaced
DESCRIBE: remove a few cm3 of river water from the bottle / addition made so that water overflows
from the bottle // make additions under the level of the water // using a dropper (pipette, syringe) //
do not bubble air (oxygen) into the water in the process //
ANY TWO: (2 × 3)
(a)
(b)
(c)
WHAT: do not trap air (oxygen) bubbles
DESCRIBE: rinse with water followed by iodine // fill using a pipette filler // bottom of meniscus
on mark / read at eye level // remove droplets adhering to outside // drain into titration flask //
touch tip of pipette against side of flask // do not blow out drop inside pipette
ANY THREE (3 × 3)
INDICATOR: starch
WHEN: when the solution is straw-coloured)
COLOURS: blue-black to colourless (do not accept “clear”)
(d)
(e)
–4
(f)
CALCULATE: 0.0006 / 6 x 10 mol l
50 × X = 6.0 × 0.01 (3)
1
2
CALCULATE: 9.6 p.p.m
0.0006 ÷ 2 = 0.0003 (3)
(g)
2006
(c)
(i)
(ii)
(d)
(i)
(ii)
(e)
(i)
(ii)
(a)
(b)
(c)
(d)
–1
(5)
(6)
(6)
(3)
(9)
(3)
(3)
(3)
(6)
–4
X = 0.0006 / 6 x 10 (3)
.(6)
0.0003 × 32 × 1000 = 9.6 (3)
Question 1
(a)
(b)
2007
(6)
Formulas (3) Balancing (3)
IDENTIFY: anhydrous sodium carbonate (Na2CO3) [Allow (3) for sodium carbonate.]
NAME: indicator methyl orange
colour change
orange (yellow) // to red (pink) (2x3)
EXPLAIN: indicator is a weak acid / indicator is a weak base
DESCRIBE: rinse with deionised (distilled) water // rinse with solution it will contain
WHY: air will be displaced by the solution (reagent) as tip fills
[Accept ‘air bubbles’ for ‘air’]
MOL/LITRE: 0.05731 / 0.0573 / 0.057 M [0.06 (-1)*] *Not deducted if more accurate value also given.
However, lost later if 0.06 used in later calculations
25 × X = 26.05 × 0.11 (3)
X = 0.05731 / 0.0573 / 0.057 M (3)
1
2
g/LITRE: 6.042 to 6.075 g l-1
0.0573 × 106* = 6.075 (3) [* Addition must be shown for error to be treated as a slip.]
CALCULATE
62.9 to 63.2% (3)
10.325 × 100 = 62.9 (3)
16.4
10 [Accept answers giving 10 when rounded off to nearest integer]
Hydrated form= 8.02 X 2 = 16.4 gl-1
Water content = 16.4 – 6.075 = 10.325
6.075/106 : 10.325/18
0.0573 : 0.573
1 : 10
[Note: If no marks have been got in (e) (ii), 3 marks to be awarded if Mr of Na2CO3 (106) appears in
the candidate’s calculations.]
(5)
(3)
(6)
(3)
(2x3)
(6)
(6)
(3)
(3)
(9)
Question 1
PRIMARY: pure with / high molecular (molar) mass (Mr) //
from which solutions of known concentration (molarity) can be made
DESCRIBE: rinse from clock glass into beaker containing deionised water // stir // dissolve // pour
through funnel into volumetric flask // add rinsings of beaker // add deionised water until bottom of
meniscus on (level with) mark / read at eye level // stopper and invert* several times [* Do not allow
“shake” for “invert”] ANY FIVE: (5×3)
WHAT: source of iodide (I-) ions (potassium iodide, KI)
COLOURS: red / brown
straw coloured
blue-black / blue
colourless [Do not accept ‘clear’]
(5+3)
(15)
(3)
(3)
(3)
(3)
(3)

3. MOLARITY: 0.125 mol l–1
20 x M = 25 x 0.05 (3)
M = 0.125 (3)
2
1
GRAMS l–1: 31 g l–1
0.125 x 248* (3) = 31 g l–1 (3)
*addition must be shown for error to be treated as slip.
(e)
2008
(c)
(i
)
(ii
)
(e)
2009
(a)
(b)
(c)
(d)
(6)
Question 1
(a)
(b)
(d)
(6)
WHY: vinegar too concentrated
DESCRIBE: rinse pipette with water // then with vinegar // fill with pipette filler / bottom of
meniscus on mark / read pipette at eye level // add 25 cm3 to 250 cm3 volumetric flask //
add deionised water until level of water near mark // add dropwise // bring bottom of
meniscus to mark / vol. flask at eye-level // stopper and invert several times
ANY FIVE: (5 × 3)
NAME: phenolphthalein
JUSTIFY: titration of weak acid-strong base
STATE: colour before (in base // colour after i.e. phenolphthalein // pink // colourless [or
other correct]
0.11 mol l–1 [Multiplied (or divided) by 4:loses 3 marks.] (6)
Mean titre = (22.6 + 22.7)/2 = 22.65 [Loses 3 if incorrect]
22.65 × M = 25.0 × 0.10 (3) M = 0.11 (3).
6.6 g l–1 (3)
0.11 × 60* = 6.6 (3)
* Addition must be shown for error to be treated as a slip
STATE: 66 g l–1
6.6 × 10 = 66 (3)
EXPRESS: 6.6 % (w/v)
66 ÷ 10 = 6.6 (3)
IDENTIFY: methanoic (formic) acid / HCOOH (3)
[If name & formula are given and one is incorrect, award marks on basis of first answer given.]
(5)
(15)
(3)
(3)
(2 × 3)
(15)
(3)
Question 1
WHY: to prevent anaemia / for haemoglobin / for red corpuscles / oxygen transport / specific
reference to blood disorder / needed for the blood (5) [No marks for the word ‘anaemia’ on its
own. Must have more e.g. ‘prevent anaemia’, ‘for anaemia’, ‘treat anaemia’. Allow 3 marks for ‘ iron
deficiency’ or ‘essential element’]
WHY MUST: not primary standard / to find concentration / reagent (KMnO4) impure (3)
WHY WAS: unstable / decomposed / reacts with water (3)
WHAT: ammonium iron(II) sulfate / sodium ethanedioate (oxalate) / ethanedioic (oxalic) acid /
excess iodide => iodine titrated with thiosulfate [Name or formula] (3)
[If formulas of ions, e.g. of thiosulfate, are used, the correct charges must be shown]
DESCRIBE: tablets crushed and dissolved (3)
transferred (added, poured, etc.) with rinsings to 250 cm3 volumetric flask (3)
bottom of meniscus on mark / brought to mark (3)
[If none of these (3)s has been awarded, allow (3) for any one of the following: “use of mortar and
pestle”, “rinse (wash) into beaker”, “stir to dissolve”, “transfer to flask using funnel (glass rod)”,
“flask on level surface (mark at eye-level)”. “add to flask drop-by-drop {using dropper (pipette, wash
bottle), top up carefully}”, “invert volumetric
flask (mix, shake – but not swirl)”.]
WHY: prevents oxidation by air (atmosphere, oxygen) of iron(II) (3)
EXPLAIN: to ensure complete conversion (reduction) of MnO4– // to prevent formation of
manganese(IV) (MnO2) / to prevent formation of a brown pptte (2 x 3) If neither of the above
(3)s has been obtained, give (3) marks for “to allow complete reduction (not oxidation or reaction)”
(i)
0.0375 M (6)
(25 x M)/5 = (18.75 x 0.01) / 1 (3)
M = 0.0375 (3)
(ii) 0.525 g [Allow Ar of iron from Maths Tables] (6)
0.0375 x 56 = 2.1 (3)
2.1 ÷ 4 = 0.525 (3)
(iii) 32.3 % [Apply –1 for rounding off in (i), (ii), (iii). Apply once only.] (6)
0.525/(0.325 x 5) = 0.323 or / (0.525 ÷ 5)/ 0.325 = 0.323 (3)
0.323 x 100 = 32.3 (3)
(5)
(9)
(12)
(24)