Doc Brown's Advanced Level Chemistry Quantitative redox reaction analysis GCE Advanced Level REDOX Volumetric Analysis Titration Revision Questions acidbase & other nonredox titration Questions 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) MnO_{4}^{}_{(aq)} + 8H^{+}_{(aq) }+ 5Fe^{2+}_{(aq)} ==> Mn^{2+}_{(aq)} + 5Fe^{3+}_{(aq)} + 4H_{2}O_{(l)} (b) (i) mol MnO_{4}^{} = 0.0200 x 24.3 / 1000 = 0.000486
(ii) The end point is the first faint permanent pink due to a trace excess of KMnO_{4}. (c) mol MnO_{4}^{} = 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 cm^{3}, i.e. the average titration value, which is statistically more accurate than an individual titration result.
Question 2:(a)
(b) mol S_{2}O_{3}^{2} = 0.0120 x 23.5 / 1000 = 0.000282, mole iodine as I_{2} = mol S_{2}O_{3}^{2} / 2 (1 : 2 in equation) = 0.000141, A_{r}(I) = 126.9, so M_{r}(I_{2}) = 2 x 126.9 = 253.8 therefore: mass of iodine = 0.000141 x 126.9 x 2 = 0.0358 g (c) mol S_{2}O_{3}^{2} = 0.095 x 26.5 / 1000 = 0.002518, mol of iodine = mole 'thio' / 2 = 0.002518 / 2 = 0.001259 in 25.0 cm^{3}, scaling up to 1 dm^{3} gives 0.001259 x 1000 /25 = 0.0504 mol dm^{3} of molecular iodine I_{2}. mass concentration of I_{2} = 0.0504 x 2 x 126.9 = 12.8 g dm^{3} of iodine
Question 3: (a) (i) Sn^{2+}_{(aq)} + 2Fe^{3+}_{(aq)} ==> Sn^{4+}_{(aq)} + 2Fe^{2+}_{(aq)} (ii) Cr_{2}O_{7}^{2}_{(aq)} + 14H^{+}_{(aq)} + 6Fe^{2+}_{(aq)} ==> 2Cr^{3+}_{(aq)} + 6Fe^{3+}_{(aq)} + 7H_{2}O_{(l)} (b) for a 25.0 cm^{3} aliquot titrated
mol Fe^{2+} titrated = 6 x Cr_{2}O_{7}^{2} = 0.003168
mol Fe_{2}O_{3} = mol Fe^{2+} / 2 = mol Fe^{3+} / 2 = 0.003168 / 2 = 0.001584
M_{r}(Fe_{2}O_{3}) = 159.8 (Fe = 55.9, 0 = 16.0)
Total mass of Fe_{2}O_{3} = 0.2531 x 10 = 2.531 g
Therefore % Fe_{2}O_{3} = 2.531 x 100 / 2.83 = 89.4% Note that that overall the ratio of mol Cr_{2}O_{7}^{2} : mol Fe_{2}O_{3} 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 Fe^{3+} is reduced, you would need excess Sn^{2+} solution, BUT, any excess Sn^{2+} would be oxidised by the Cr_{2}O_{7}^{2} giving a false titration value. Nevertheless, it is a legitimate problem to solve!
Question 4: mol Fe^{2+} = 0.100 x 25.0 / 1000 = 0.0025, mol MnO_{4}^{} = mol Fe^{2+} / 5 (from equation 1 : 5) = 0.0005 in 24.15 cm^{3}, scaling up to 1 dm^{3}, molarity of MnO_{4}^{} = 0.0005 x 1000 / 24.15 = 0.0207 mol dm^{3}.
Question 5: mol Cr_{2}O_{7}^{2} = 0.0200 x 21.25 / 1000 = 0.000425, mol of Fe salt = mol Fe^{2+} titrated = 6 x Cr_{2}O_{7}^{2} = 6 x 0.000425 = 0.00255, BUT only ^{1}/_{10}th of Fe^{2+} salt used in titration, so 1 g of FeSO_{4}.(NH_{4})_{2}S0_{4} .xH_{2}O 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 FeSO_{4}.(NH_{4})_{2}S0_{4}.xH_{2}O is 392.2. Now the formula mass of FeSO_{4}.(NH_{4})_{2}S0_{4} = 284.1, this leaves 392.2 284.1 = 108.1 mass units. M_{r}(H_{2}O) = 18, so 108.1 / 18 = 6.005 mol of water, so x = 6 in the salt formula, FeSO_{4}.(NH_{4})_{2}S0_{4}.6H_{2}O.
Question 6: (a) 2MnO_{4}^{}_{(aq)} + 16H^{+}_{(aq) }+ 5C_{2}O_{4}^{2}_{(aq)} ==> 2Mn^{2+}_{(aq)} + 8H_{2}O_{(l)} + 10CO_{2(g)}
(b) M_{r}(H_{2}C_{2}O_{4}.2H_{2}O) = 126.0 total mol H_{2}C_{2}O_{4}.2H_{2}O (or C_{2}O_{4}^{2}) = 1.52 / 126 = 0.0120635 but mol of C_{2}O_{4}^{2} in titration = 0.0120635/10 = 0.00120635 (^{1}/_{10}th used, 25 of 250 cm^{3}), mol MnO_{4}^{} = mol of C_{2}O_{4}^{2} / 2.5 (2:5 or 1:2.5 ratio), mol MnO_{4}^{} = 0.00120634 / 2.5 = 0.00048254 (in 24.55 cm^{3}), scaling up to 1 dm^{3} the molarity of MnO_{4}^{} = 0.00048254 x 1000 / 24.55 = 0.0196554
M_{r}(KMnO_{4}) = 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 KHC_{2}O_{4}.H_{2}C_{2}O_{4}.2H_{2}O (M_{r }= 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 C_{2}O_{4}^{2} in salt), so mol MnO_{4}^{} in titration = 0.000590087 / 1.25 = 0.000472069 in 23.2 cm^{3}, scaling up to 1 dm^{3} gives for [MnO_{4}^{}] = 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 cm^{3}]
Question 8: (a) 2MnO_{4}^{}_{(aq)} + 6H^{+}_{(aq) }+ 5H_{2}O_{2(aq) }==> 2Mn^{2+}_{(aq)} + 8H_{2}O_{(l)} + 5O_{2(g)} (b) in titration mol MnO_{4}^{} = 0.0200 x 20.25 / 1000 = 0.000405, MnO_{4}^{}:H_{2}O_{2} ratio is 2:5 or 1:2.5, so mol H_{2}O_{2} in titration = 0.000405 x 2.5 = 0.0010125, scaling up for total mol H_{2}O_{2} in diluted solution (of 1 dm^{3} or 1000 cm^{3}) = 0.0010125 x 1000 / 25.0 = 0.0405 mol, but in the original 50 cm^{3} solution, therefore scaling up to 1 dm^{3}, the original molarity of H_{2}O_{2} is 0.0405 x 1000 / 50 = 0.810 mol dm^{3}.
Question 9: (a) Zn_{(s)} + 2Fe^{3+}_{(aq)} ==> Zn^{2+}_{(aq)} + 2Fe^{2+}_{(aq)} (b) mol MnO_{4}^{ }in titration = 0.0100 x 26.5 / 1000 = 0.000265, mol Fe (Fe^{2+}) = mol MnO_{4}^{ }x 5 = 0.001325 in 20.o cm^{3} 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 MnO_{4}^{ } in titration = 0.05 x 24.5 / 1000 = 0.001225, ratio MnO_{4}^{}:Na_{2}C_{2}O_{4} is 2:5 or 1:2.5, so mol Na_{2}C_{2}O_{4} titrated = 0.001225 x 2.5 = 0.003063 in 5 cm^{3}, scaling up to 1 dm^{3}, molarity Na_{2}C_{2}O_{4} = 0.003063 x 1000 / 5 = 0.613 mol dm^{3} M_{r}(Na_{2}C_{2}O_{4}) = 134, so concentration = 0.613 x 134 = 82.1 g dm^{3}
Question 11: mol KMnO_{4} = 0.0100 x 43.85 / 1000 = 0.0004385, mol Fe (Fe^{2+}) = mol KMnO_{4} x 5, mol Fe = 0.0004385 x 5 = 0.0021925, so mol FeSO_{4}.xH_{2}O is also 0.0021925, in the titration ^{1}/_{20}th of the salt was used (^{25}/_{500}), so ^{1}/_{20}th 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 FeSO_{4}.xH_{2}O is 277.8, now the formula mass of FeSO_{4} is 152.0, so the formula mass of xH_{2}O = 277.8 152.0 = 125.8, M_{r}(H_{2}O) = 18, so x = 125.8 / 18 = 6.989, so x = 7 and the formula of the salt is FeSO_{4}.7H_{2}O, i.e. seven molecules of water of crystallisation.
Question 12: (a) 2MnO_{4}^{}_{(aq)} + 6H^{+}_{(aq) }+ 5NO_{2}^{}_{(aq)} ==> Mn^{2+}_{(aq)} + 5NO_{3}^{}_{(aq)} + 3H_{2}O_{(l)} (b) mol KMnO_{4} in titration = 0.0250 x 25 / 1000 = 0.000625, mol ratio MnO_{4}^{}:NO_{2}^{} is 2:5 or 1:2.5, so mol NO_{2}^{} in titration = 0.000625 x 2.5 = 0.0015625 in 24.2 cm^{3}, scaling up to 1 dm^{3} gives a molar concentration of NaNO_{2} of 0.0015625 x 1000 / 24.2 = 0.0646 mol dm^{3} M_{r}(NaNO_{2}) = 69, so in terms of mass concentration = 0.0646 x 69 = 4.46 g dm^{3}
Question 13: M_{r}(FeC_{2}O_{4}) = 143.9, mol FeC_{2}O_{4} in original solution = 2.68 / 143.9 = 0.01862, scaling down the mol FeC_{2}O_{4} in the titration = 0.01862 x 25 / 500 = 0.000931, mol KMnO_{4} in titration = 0.0200 x 28.0 / 1000 = 0.00056, so ratio KMnO_{4}:FeC_{2}O_{4} is 0.00056:0.000931 = giving the 'not so easy to spot' 3:5 the reacting mole ratio. FeC_{2}O_{4} is made up of a Fe^{2+} ion and a C_{2}O_{4}^{2} ion, and the full redox equation is: 3MnO_{4}^{}_{(aq)}+ 5FeC_{2}O_{4(aq)} + 24H^{+}_{(aq) }==> 3Mn^{2+}_{(aq)} + 5Fe^{3+}_{(aq)}+ 12H_{2}O_{(l)} + 10CO_{2(g)} or 3MnO_{4}^{}_{(aq)}+ 5Fe^{2+}_{(aq)} + 5C_{2}O_{4}^{2}_{(aq)} + 24H^{+}_{(aq) }==> 3Mn^{2+}_{(aq)} + 5Fe^{3+}_{(aq)}+ 12H_{2}O_{(l)} + 10CO_{2(g)}
Question 14: (a) IO_{3}^{}_{(aq)} + 5I^{}_{(aq)} + 6H^{+}_{(aq)} ==> 3I_{2(aq)} + 3H_{2}O_{(l)} (b) mol I^{} titrated = 0.100 x 20.0 / 1000 = 0.002, mole ratio IO_{3}^{}:I^{} is 1:5, so mole IO_{3}^{} reacted = 0.002 / 5 = 0.0004, so 0.0004 = 0.012 x (volume IO_{3}^{} required) / 1000, volume IO_{3}^{} required = 0.0004 x 1000 / 0.012 = 33.3 cm^{3} (c)(i) mole S_{2}O_{3}^{2} ('thio') = 0.0500 x 24.1 / 1000 = 0.001205, I_{2}:S_{2}O_{3}^{2} ratio is 1:2 in the titration reaction, so mol I_{2} = mole S_{2}O_{3}^{2} / 2 = 0.001205 / 2 = 0.0006025, now the IO_{3}^{}:I_{2} reaction ratio is 1:3, so mol IO_{3}^{} reacting to give iodine = mole I_{2} formed / 3 = 0.0006025 / 3 = 0.000201 in 25 cm^{3}, so scaling up to 1 dm^{3} the molarity of the KIO_{3} (IO_{3}^{}) = 0.000201 x 1000 / 25 = 0.00804 mol dm^{3}, M_{r}(KIO_{3}) = 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: 2S_{2}O_{3}^{2} : I_{2} : ^{1}/_{3}IO_{3}^{}, 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 KMnO_{4} = 0.0200 x 22.5 / 1000 = 0.00045, mol Fe^{2+} = mol KMnO_{4} x 5 = 0.00225 in 25 cm^{3}, scaling up to 1 dm^{3}, molarity of the original Fe^{2+} = 0.00225 x 1000 / 25.0 = 0.090 mol dm^{3} (ii) the 2nd titration gives the total concentration of Fe^{2+} + Fe^{3+} because any Fe^{3+} has been reduced to Fe^{2+}, mol KMnO_{4} = 0.0200 x 37.6 / 1000 = 0.000752, total mol Fe^{2+} titrated = mol KMnO_{4} x 5 = 0.00376 in 25 cm^{3}, scaling up to 1 dm^{3}, total molarity of Fe^{2+} + Fe^{3+} in original solution = 0.00376 x 1000 / 25.0 = 0.150 mol dm^{ 3}, so using the result from (a) the Fe^{3+} concentration = 'Fe' total Fe^{2+} = 0.150 0.090 = 0.060 mol dm^{3}.
Question 16: you can ignore the 25 cm^{3} 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 Fe^{2+} = 5 x MnO_{4}^{} = 5 x 0.0200 x 16.9 / 1000 = 0.00169 mol = unreacted iron (which dissolved in the acid to form Fe^{2+}). (b) mol Fe^{3+} = EDTA^{4} = 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 Fe^{2+}. Calculation (b) gives the total of unreacted Fe + the rust i.e. Fe^{3+}, because any Fe^{2+} formed from Fe has been oxidised to Fe^{3+}. So from the original mixture (in terms of the 25 cm^{3} 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) I_{2(aq)} + 2S_{2}O_{3}^{2}_{(aq)} ==> S_{4}O_{6}^{2}_{(aq)} + 2I^{}_{(aq)} or I_{2(aq)} + 2Na_{2}S_{2}O_{3(aq)} ==> Na_{2}S_{4}O_{6(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 I_{2} = 0.00176 χ 2 = 0.00088 in 25 cm^{3}, scaling up gives 0.00088 x 1000 χ 25 = 0.0352 mol dm^{3} for molarity of iodine, formula mass I_{2} = 2 x 127 = 254, so concentration = 0.0352 x 254 = 8.94 g dm^{3}
Question 18: (a) Cr_{2}O_{7}^{2}_{(aq)} + 14H^{+}_{(aq)} + 6I^{}_{(aq)} ==> 2Cr^{3+}_{(aq)} + 3I_{2(aq)} + 7H_{2}O_{(l)} (b) 2S_{2}O_{3}^{2}_{(aq)} + I_{2(aq)} ==> _{ }S_{4}O_{6}^{2}_{(aq)} + 2I^{}_{(aq)} (c) mol 'thio' = 20.0 x 0.100/1000 = 0.002, therefore 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) M_{r}(K_{2}Cr_{2}O_{7}) = 294.2 mass K_{2}Cr_{2}O_{7} titrated = 0.000333 x 294.2 = 0.0980 g (3 sf) (f) Since the aliquot of 25.0 cm^{3} is 1/10th of the total solution in the flask, the total mass of the K_{2}Cr_{2}O_{7} in original sample dissolved in the flask solution = 10 x 0.0980g = 0.98g and the % purity of the K_{2}Cr_{2}O_{7} = 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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