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Answers to redox volumetric titration questions for Advanced A/AS Level Inorganic Chemistry:
Quantitative redox reaction analysis GCE Advanced Level REDOX Volumetric Analysis Titration Revision Questions ORIGINAL REDOX TITRATION QUESTIONS acidbase & other nonredox titration Questions Qualitative Analysis * REDOX REACTION THEORY If you spot a silly error please EMAIL query?comment?error 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
(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
(d) The choice of acid is fully discussed in Ex 6.4 of Advanced Redox Chemistry Part 2 but basically you should know that ...
(e) (i) 23.87 cm3, i.e. the average titration value, which is statistically more accurate than an individual titration result.
Question 2:(a)
(b) mol S2O32 = 0.0120 x 23.5 / 1000 = 0.000282, mole iodine as I2 = mol S2O32 / 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 S2O32 = 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 dm3 of molecular iodine I2. mass concentration of I2 = 0.0504 x 2 x 126.9 = 12.8 g dm3 of iodine
Question 3: (a) (i) Sn2+(aq) + 2Fe3+(aq) ==> Sn4+(aq) + 2Fe2+(aq) (ii) Cr2O72(aq) + 14H+(aq) + 6Fe2+(aq) ==> 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l) (b) for a 25.0 cm3 aliquot titrated
mol Fe2+ titrated = 6 x Cr2O72 = 0.003168
mol Fe2O3 = mol Fe2+ / 2 = mol Fe3+ / 2 = 0.003168 / 2 = 0.001584
Mr(Fe2O3) = 159.8 (Fe = 55.9, 0 = 16.0)
Total mass of Fe2O3 = 0.2531 x 10 = 2.531 g
Therefore % Fe2O3 = 2.531 x 100 / 2.83 = 89.4% Note that that overall the ratio of mol Cr2O72 : mol Fe2O3 is 3 : 1.
(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.
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 Cr2O72 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 dm3.
Question 5: mol Cr2O72 = 0.0200 x 21.25 / 1000 = 0.000425, mol of Fe salt = mol Fe2+ titrated = 6 x Cr2O72 = 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 mol1 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) + 5C2O42(aq) ==> 2Mn2+(aq) + 8H2O(l) + 10CO2(g)
(b) Mr(H2C2O4.2H2O) = 126.0 total mol H2C2O4.2H2O (or C2O42) = 1.52 / 126 = 0.0120635 but mol of C2O42 in titration = 0.0120635/10 = 0.00120635 (1/10th used, 25 of 250 cm3), mol MnO4 = mol of C2O42 / 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
Mr(KMnO4) = 158.0, so in terms of mass concentration = 0.01965523 x 158 = 3.10555
[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 C2O42 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
[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 dm3.
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 dm3 Mr(Na2C2O4) = 134, so concentration = 0.613 x 134 = 82.1 g dm3
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 dm3 Mr(NaNO2) = 69, so in terms of mass concentration = 0.0646 x 69 = 4.46 g dm3
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 C2O42 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) + 5C2O42(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 S2O32 ('thio') = 0.0500 x 24.1 / 1000 = 0.001205, I2:S2O32 ratio is 1:2 in the titration reaction, so mol I2 = mole S2O32 / 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 dm3, Mr(KIO3) = 214.0, so in terms of mass, concentration = 0.00804 x 214 = 1.72 g dm3. A quicker approach if confident! ratios from all equations involved are: 2S2O32 : 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 starchiodine '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 dm3 (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 dm3.
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+ = EDTA4 = 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) + 2S2O32(aq) ==> S4O62(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 endpoint 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 dm3 for molarity of iodine, formula mass I2 = 2 x 127 = 254, so concentration = 0.0352 x 254 = 8.94 g dm3
Question 18: (a) Cr2O72(aq) + 14H+(aq) + 6I(aq) ==> 2Cr3+(aq) + 3I2(aq) + 7H2O(l) (b) 2S2O32(aq) + I2(aq) ==> S4O62(aq) + 2I(aq) (c) mol 'thio' = 20.0 x 0.100/1000 = 0.002,
(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) Question 19 answers: (a) H2C2O4 + 2NaOH ===> Na2C2O4 + 2H2O
(b) moles used in the NaOH in titration = 0.20 x 8.5/1000 = 0.0017 mol NaOH
(c) (i) involves the gain of 5 electrons, (ii) involves the loss of 2 electrons, so to balance the electron gain and loss you need to add together 2 x (i) plus 5 x (ii)
(d) from part (c): moles H2C2O4 titrated with KMnO4 = 0.0085 in 10 cm3 of the ethanedioic solution.
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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