Basic Chemistry Calculations Index

 Doc Brown's Advanced Level Chemistry  

 Quantitative redox reaction analysis

GCE Advanced Level REDOX Volumetric Analysis Titration Revision Questions

ORIGINAL REDOX TITRATION QUESTIONS

acid–base & other non–redox titration Questions

 Qualitative Analysis * REDOX REACTION THEORY

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 I DO MY BEST TO CHECK MY CALCULATIONS, as you yourself should do,  BUT I AM HUMAN! AND IF YOU THINK THERE IS A 'TYPO' or CALCULATION ERROR PLEASE EMAIL ME ASAP TO SORT IT OUT!

Most of the answers have been rounded up or rounded down to three significant figures (3sf)

Question 1: (a) MnO4–(aq) + 8H+(aq)  + 5Fe2+(aq) ==> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) 

(b) (i) mol MnO4– = 0.0200 x 24.3 / 1000 = 0.000486

therefore mol Fe2+ = 5 x 0.000486 (1 : 5 in equation) = 0.00243 in 20 cm3

so scaling up to 1 dm3, the molarity of  Fe2+ = 0.00243 x 1000 / 20 = 0.122 mol dm–3.

(ii) The end point is the first faint permanent pink due to a trace excess of KMnO4.

(c) mol MnO4– = 0.0200 x 25.45 / 1000 = 0.000509

mol Fe = 5 x 0.000509 = 0.002545,

mass Fe = 0.002545 x 55.9 = 0.1423 g

total Fe in wire = 0.1423 x 10 = 1.423 g (only 1/10th of the made up solution used in titration)

so % Fe = 1.423 x 100 / 1.51 = 94.2 %

(d) The choice of acid is fully discussed in Ex 6.4 of Advanced Redox Chemistry Part 2 but basically you should know that ...

The titrations are done with (i) dilute sulfuric acid present to prevent side reactions e.g. MnO2 formation (brown colouration or black precipitate).

(ii) Nitric acid is an oxidising agent and would oxidise iron(II) ions to iron(III) ions,

and (iii) chloride ions are oxidised to chlorine by manganate(VII) ions (see half-cell potential data in Q3c),

therefore, using either acid (ii) or (iii) would give a false titration result.

(e) (i) 23.87 cm3, i.e. the average titration value, which is statistically more accurate than an individual titration result.

(ii) molarity = mol / volume in dm3

BUT, remember, 1 dm3 = 1000 cm3,

burettes and pipettes work in cm3 and molarity works in dm3,

a bit of pain I know, but live with it and go with the flow!

therefore mol = molarity x vol in cm3 / 1000

mol MnO4– = 0.0200 x 23.87/1000 = 4.774 x 10–4

(iii) From the balanced redox equation, 1 mol MnO4– oxidises 5 mol of iron(II) ions

mol Fe2+ = 5 x 4.774 x 10–4 = 2.387 x 10–3

(iv) You can assume mass of Fe = mass of Fe2+ ion, (2 electrons don't weigh much!)

mol = mass/atomic mass, so mass = mol x atomic mass

mass Fe = 2.387 x 10–3 x 55.8 = 0.1332 g

(v) Total mass of iron in original sample = 10 x 0.1332 = 1.332 g (scaling up by factor of 250/25)

(vi) % iron in the original salt = 1.332 x 100 / 8.25 = 16.1% (1dp, 3sf)


Question 2:(a)

(i)  I2(aq) + 2e– ==> 2I–(aq)   (iodine reduced by thiosulfate ion, reduction is electron gain)
(ii) 2S2O3(aq)  ==>  S4O6(aq) + 2e–  (thiosulfate ion oxidised by iodine, oxidation is electron loss)

(Note that you needed to reverse the half-cell equation as given in question to arrive at the full ionic redox equation)

(i) + (ii) gives:  2S2O3(aq)  +  I2(aq)  ==>  S4O6(aq) + 2I–(aq)

(b) mol S2O3 = 0.0120 x 23.5 / 1000 = 0.000282, mole iodine as I2 = mol  S2O3 / 2 (1 : 2 in equation) = 0.000141,

Ar(I) = 126.9, so Mr(I2) = 2 x 126.9 = 253.8

therefore: mass of iodine = 0.000141 x 126.9 x 2 = 0.0358 g

(c) mol S2O3 = 0.095 x 26.5 / 1000 = 0.002518,

mol of iodine = mole 'thio' / 2 = 0.002518 / 2 = 0.001259 in 25.0 cm3,

scaling up to 1 dm3 gives 0.001259 x 1000 /25 = 0.0504 mol dm–3 of molecular iodine I2.

mass concentration of I2 = 0.0504 x 2 x 126.9 = 12.8 g dm–3 of iodine


Question 3: (a) (i) Sn2+(aq) + 2Fe3+(aq) ==> Sn4+(aq) + 2Fe2+(aq)

(ii) Cr2O7(aq) + 14H+(aq) + 6Fe2+(aq) ==> 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)

(b) for a 25.0 cm3 aliquot titrated

mol Cr2O7 = 0.0200 x 26.4 / 1000 = 0.000528,

mol Fe2+ titrated = 6 x Cr2O7 = 0.003168

(from 1 mol Cr2O7 : 6 mol Fe2+ in the redox equation),

mol Fe2O3 = mol Fe2+ / 2 = mol Fe3+ / 2 = 0.003168 / 2 =  0.001584

(moles Fe2+ or Fe3+ halved, because of 2 moles of Fe3+ in in one mole of Fe2O3)

Mr(Fe2O3) = 159.8 (Fe = 55.9, 0 = 16.0)

so mass of Fe2O3 = 0.001584 x 159.8 = 0.2531 g.

Total mass of Fe2O3 = 0.2531 x 10 = 2.531 g

 Remember only 1/10th titrated, so need to scale up by 10 to account for all the iron oxide in the original sample.

Therefore % Fe2O3 = 2.531 x 100 / 2.83 = 89.4%

Note that that overall the ratio of mol Cr2O7 : mol Fe2O3 is 3 : 1.

Now its ok to solve this problem using the 3 : 1 ratio if you are very confident to use short cuts in this type of calculation. However, I think from a teaching and learning point of view, I prefer to show the full logic of each step in the calculation, particularly when dealing with mole ratios and this is what I've tried to do on this page!

(c) Potassium manganate(VII) isn't used for this titration because it is strong enough to oxidise chloride ions (from the hydrochloric acid) to form chlorine, giving a completely false titration.

(The half-cell potential proves the feasibility of this reaction, with a greater positive potential than chlorine/chloride, the manganate(VII) ion can oxidise the chloride ion to chlorine. Further note, that the dichromate(VI) ion cannot oxides the chloride ion, which is why it would used in this context.)

Note on the question design: There is actually a flaw in this question. In order to ensure all the Fe3+ is reduced, you would need excess Sn2+ solution, BUT, any excess Sn2+ would be oxidised by the Cr2O7 giving a false titration value. Never-the-less, it is a legitimate problem to solve!


Question 4: mol Fe2+ = 0.100 x 25.0 / 1000 = 0.0025,

mol MnO4– = mol Fe2+ / 5 (from equation 1 : 5) = 0.0005 in 24.15 cm3,

scaling up to 1 dm3, molarity of MnO4– = 0.0005 x 1000 / 24.15 = 0.0207 mol dm–3.


Question 5: mol Cr2O7 = 0.0200 x 21.25 / 1000 = 0.000425,

mol of Fe salt = mol Fe2+ titrated = 6 x Cr2O7 = 6 x 0.000425 = 0.00255,

BUT only 1/10th of Fe2+ salt used in titration,

so 1 g of FeSO4.(NH4)2S04 .xH2O is equal to 0.00255 mol.

Scaling up to 1 mol gives a molar mass for the salt in g mol–1 of 1 x 1 /0.00255 = 392.2.

So the formula mass for FeSO4.(NH4)2S04.xH2O is 392.2.

Now the formula mass of FeSO4.(NH4)2S04 = 284.1, this leaves 392.2 – 284.1 = 108.1 mass units.

Mr(H2O) = 18, so 108.1 / 18 = 6.005 mol of water, so x = 6 in the salt formula, FeSO4.(NH4)2S04.6H2O.


Question 6: (a) 2MnO4–(aq) + 16H+(aq)  + 5C2O4(aq) ==> 2Mn2+(aq)  + 8H2O(l) + 10CO2(g)

or  2MnO4–(aq) + 6H+(aq)  + 5H2C2O4(aq) ==> 2Mn2+(aq)  + 8H2O(l) + 10CO2(g)

(b)  Mr(H2C2O4.2H2O) = 126.0

total mol H2C2O4.2H2O (or C2O4)  = 1.52 / 126 = 0.0120635

but mol of C2O4 in titration = 0.0120635/10 = 0.00120635 (1/10th used, 25 of 250 cm3),

mol MnO4– = mol of C2O4 / 2.5 (2:5 or 1:2.5 ratio),

mol MnO4– = 0.00120634 / 2.5 = 0.00048254 (in 24.55 cm3),

scaling up to 1 dm3 the molarity of MnO4– = 0.00048254 x 1000 / 24.55 = 0.0196554

= 0.01966 mol dm-3 (5dp, 4sf) OR 0.0197 mol dm–3 (4dp, 3sf)

Mr(KMnO4) = 158.0, so in terms of mass concentration = 0.01965523 x 158 = 3.10555

= 3.106 g dm-3 (3dp, 4sf) OR 3.11 g dm–3 (2dp, 3sf)

[NOTE: The titration volume, mass and formula masses are quoted to four significant figures (4sf), so it might be considered legitimate to quote the answer to four significant figures]


Question 7: mol KHC2O4.H2C2O4.2H2O (Mr = 254.2) = 0.150 / 254.2 = 0.000590087

ratio of tetroxalate to manganate(VII) is 2:2.5 or 1:1.25 (note equiv of 2 C2O4 in salt),

so mol MnO4– in titration = 0.000590087 / 1.25 = 0.000472069 in 23.2 cm3,

scaling up to 1 dm3 gives for [MnO4–] = 0.000472069 x 1000 / 23.2 = 0.02034781

= 0.02034 mol dm-3 (5dp, 4sf) OR 0.0203 mol dm–3 (4dp, 3sf)

[NOTE: Quoting the concentration to 4dp, 3sf is more appropriate here because the mass is only quoted to 3sf and the titration is only likely to be accurate to the nearest 0.05 cm3]


Question 8: (a)  2MnO4–(aq) + 6H+(aq)  + 5H2O2(aq)  ==> 2Mn2+(aq) + 8H2O(l) + 5O2(g)  

(b)  in titration mol MnO4– =  0.0200 x 20.25 / 1000 = 0.000405,

MnO4–:H2O2 ratio is 2:5 or 1:2.5, so mol H2O2 in titration = 0.000405 x 2.5 = 0.0010125,

scaling up for total mol H2O2 in diluted solution (of 1 dm3 or 1000 cm3) = 0.0010125 x 1000 / 25.0 = 0.0405 mol,

but in the original 50 cm3 solution,

therefore scaling up to 1 dm3, the original molarity of H2O2 is 0.0405 x 1000 / 50 = 0.810 mol dm–3.


Question 9: (a) Zn(s) + 2Fe3+(aq) ==> Zn2+(aq) +  2Fe2+(aq)

(b) mol MnO4– in  titration = 0.0100 x 26.5 / 1000 =  0.000265,

mol Fe (Fe2+) = mol MnO4– x 5 = 0.001325 in 20.o cm3 of the alum solution,

scaling up gives total mol Fe = 0.001325 x 500 / 20 = 0.033125,

total mass Fe in the 13.2 g of alum = 0.033125 x 55.9 = 1.852,

so % Fe = 1.852 x 100 / 13.2 = 14.0%


Question 10: mol MnO4– in titration = 0.05 x 24.5 / 1000 = 0.001225,

ratio MnO4–:Na2C2O4 is 2:5 or 1:2.5, so mol Na2C2O4 titrated = 0.001225 x 2.5 = 0.003063 in 5 cm3,

scaling up to 1 dm3, molarity Na2C2O4 = 0.003063 x 1000 / 5 = 0.613  mol dm–3

Mr(Na2C2O4) = 134, so concentration = 0.613 x 134 = 82.1 g dm–3


Question 11: mol KMnO4 = 0.0100 x 43.85 / 1000 = 0.0004385, mol Fe (Fe2+) = mol KMnO4 x 5,

mol Fe = 0.0004385 x 5 = 0.0021925, so mol  FeSO4.xH2O is also 0.0021925,

in the titration 1/20th of the salt was used (25/500), so 1/20th of 12.18 g = 0.0021925 mol of the salt = 0.609 g,

scaling up the mass of 1 mole of the salt is 0.609 x 1 / 0.0021925 = 277.8,

so formula mass of FeSO4.xH2O is 277.8, now the formula mass of FeSO4 is 152.0,

so the formula mass of xH2O = 277.8 – 152.0 = 125.8,

Mr(H2O) = 18, so x = 125.8 / 18 = 6.989,

so x = 7 and the formula of the salt is  FeSO4.7H2O, i.e. seven molecules of water of crystallisation. 


Question 12:  (a) 2MnO4–(aq) + 6H+(aq)  + 5NO2–(aq) ==> Mn2+(aq) + 5NO3–(aq) +  3H2O(l)

(b) mol KMnO4 in titration = 0.0250 x 25 / 1000 = 0.000625,

mol ratio MnO4–:NO2– is 2:5 or 1:2.5, so mol NO2– in titration = 0.000625 x 2.5 = 0.0015625 in 24.2 cm3,

scaling up to 1 dm3 gives a molar concentration of NaNO2 of  0.0015625 x 1000 / 24.2 = 0.0646 mol dm–3

Mr(NaNO2) = 69, so in terms of mass concentration = 0.0646 x 69 = 4.46 g dm–3


Question 13: Mr(FeC2O4) = 143.9, mol FeC2O4 in original solution = 2.68 / 143.9 = 0.01862,

scaling down the mol FeC2O4 in the titration = 0.01862 x 25 / 500 = 0.000931,

mol KMnO4 in titration = 0.0200 x 28.0 / 1000 = 0.00056,

so ratio KMnO4:FeC2O4 is 0.00056:0.000931 = giving the 'not so easy to spot' 3:5 the reacting mole ratio.

FeC2O4 is made up of a  Fe2+ ion and a  C2O4 ion, and the full redox equation is:

3MnO4–(aq)+ 5FeC2O4(aq) + 24H+(aq)  ==> 3Mn2+(aq) + 5Fe3+(aq)+ 12H2O(l) + 10CO2(g)

or 3MnO4–(aq)+ 5Fe2+(aq) + 5C2O4(aq) + 24H+(aq)  ==> 3Mn2+(aq) + 5Fe3+(aq)+ 12H2O(l) + 10CO2(g)


Question 14: (a) IO3–(aq) + 5I–(aq) + 6H+(aq) ==> 3I2(aq) + 3H2O(l)

(b) mol I– titrated = 0.100 x 20.0 / 1000 = 0.002, mole ratio IO3–:I– is 1:5,

so mole IO3– reacted = 0.002 / 5 = 0.0004,

so 0.0004 = 0.012 x (volume IO3– required) / 1000,

volume IO3– required = 0.0004 x 1000 / 0.012 = 33.3 cm3

(c)(i) mole S2O3 ('thio') = 0.0500 x 24.1 / 1000 = 0.001205,

I2:S2O3 ratio is 1:2 in the titration reaction, so mol I2 = mole S2O3 / 2 = 0.001205 / 2 = 0.0006025,

now the  IO3–:I2 reaction ratio is 1:3,

so mol IO3– reacting to give iodine = mole I2 formed / 3 = 0.0006025 / 3 = 0.000201 in 25 cm3,

so scaling up to 1 dm3 the molarity of the KIO3 (IO3–) = 0.000201 x 1000 / 25 = 0.00804 mol dm–3,

 Mr(KIO3) = 214.0, so in terms of mass, concentration = 0.00804 x 214 = 1.72 g dm–3.

A quicker approach if confident! – ratios from all equations involved are: 2S2O3 : I2 : 1/3IO3–, means that the overall mole iodate(V) = mole thiosulphate / 6, so you can 'jump' from line '1' to the last 'few' lines. However in exams these days all the stages (i.e. , to , !) are often 'broken down' for you and it might be best you work through the problem thoroughly and methodically.

(ii) Starch indicator is used for the titration, when the last of the iodine reacts with the thiosulphate, the blue colour from the starch–iodine 'complex' is discharged and the solution becomes colourless.


Question 15: (i) mol KMnO4 = 0.0200 x 22.5 / 1000 = 0.00045,

mol Fe2+ = mol KMnO4 x 5 = 0.00225 in 25 cm3,

scaling up to 1 dm3, molarity of the original Fe2+ = 0.00225 x 1000 / 25.0 = 0.090 mol dm–3

(ii) the 2nd titration gives the total concentration of Fe2+ + Fe3+ because any Fe3+ has been reduced to Fe2+,

mol KMnO4 = 0.0200 x 37.6 / 1000 = 0.000752, total mol Fe2+ titrated = mol KMnO4 x 5 = 0.00376 in 25 cm3,

scaling up to 1 dm3, total molarity of  Fe2+ + Fe3+ in original solution = 0.00376 x 1000 / 25.0 = 0.150 mol dm –3,

so using the result from (a) the  Fe3+ concentration = 'Fe' total – Fe2+ = 0.150 – 0.090 = 0.060 mol dm–3.


Question 16: you can ignore the 25 cm3 of the solution because you use the same volume in each titration and you can work on the ratio of the moles of 'Fe' out of the (a) and (b) titration calculations.

(a) mol Fe2+ = 5 x MnO4– = 5 x 0.0200 x 16.9 / 1000 = 0.00169 mol = unreacted iron (which dissolved in the acid to form Fe2+).

(b) mol Fe3+ = EDTA = 0.100 x 17.6 / 1000 = 0.00176 mol = total mol iron in the sample titrated.

(c) calculation (a) gives the relative moles of unreacted iron Fe, as it dissolved to form the titratable Fe2+.

Calculation (b) gives the total of unreacted Fe + the rust i.e. Fe3+, because any Fe2+ formed from Fe has been oxidised to Fe3+.

So from the original mixture (in terms of the 25 cm3 sample), mol unreacted Fe = 0.00169,

mol of reacted iron = 0.00176 – 0.00169 = 0.00007.

Therefore the % rusted iron = 0.00007 x 100 / 0.00176 = 3.98 % rusted iron.


Question 17: (a) I2(aq) + 2S2O3(aq) ==> S4O6(aq) + 2I–(aq)

 or I2(aq) + 2Na2S2O3(aq) ==> Na2S4O6(aq) + 2NaI(aq)

(b) Starch indicator is used, starch gives a blue/black colour with iodine, this colour disappears when the last of the iodine is titrated, so a blue to colourless sharp end–point is observed.

(c) mole 'thio' = 0.100 x 17.6/1000 = 0.00176,

mol I2 = 0.00176 χ 2 = 0.00088 in 25 cm3,

scaling up gives 0.00088 x 1000 χ 25 = 0.0352 mol dm–3 for molarity of iodine,

formula mass I2 = 2 x 127 = 254, so concentration = 0.0352 x 254 = 8.94 g dm–3


Question 18: (a) Cr2O7(aq) + 14H+(aq) + 6I–(aq) ==> 2Cr3+(aq) + 3I2(aq) + 7H2O(l)

(b) 2S2O3(aq)  +  I2(aq)  ==>  S4O6(aq) + 2I–(aq)

(c) mol 'thio' = 20.0 x 0.100/1000 = 0.002,

Basic Chemistry Calculations Indextherefore from equation (b), mol iodine = mol 'thio'/2 = 0.001

(d) From equation (a) mol dichromate(VI) reacting = mol iodine liberated/3 = 0.000333 (3sf)

(e) Mr(K2Cr2O7) = 294.2

mass K2Cr2O7 titrated = 0.000333 x 294.2 =  0.0980 g (3 sf)

(f) Since the aliquot of 25.0 cm3 is 1/10th of the total solution in the flask, the total mass of the K2Cr2O7 in original sample dissolved in the flask solution = 10 x 0.0980g = 0.98g

and the % purity of the K2Cr2O7 = 0.98 x 100/1.01 = 97.0 % (3 sf)


 I DO MY BEST TO CHECK MY CALCULATIONS, as you yourself should do,  BUT I AM HUMAN! AND IF YOU THINK THERE IS A 'TYPO' or CALCULATION ERROR PLEASE EMAIL ME ASAP TO SORT IT OUT!


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