Please only ask question here
But id there is a dedicated place on this blog to cater for your question and you don’t use it, then I’ll probably delete your comment/question/post. This blog needs to have some organizational structure about it.
If you upload images do it with the correct orientation. If I have to tilt my head then I’m going to delete the comment/query.
So, fire away….
KYchem 
Why don’t I have 3 questions?
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LOL. Unfortunately that doesn’t count as a question.
I await the three 🙂
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This is a second question?
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Well, if you really have no questions and aren’t just being lazy…. (!) then I guess there’s no point in making you (and others) waste time and synthesising a question. But do some soul searching.
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Sir, how to answer the following question:
14.4 spare questions
q30(b) A proprietary moss killer is a mixture of sand and hydrated iron (II) sulphate. Outline the plan of an experiment to determine the percentage by mass of iron (II) in a sample of this moss killer using a standard solution of potassium dichromate(VI). You must show in your answer how you would calculate the result of the experiment.
(8)
*I don’t quite understand the procedure in the mark scheme
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Hi.
You could do it directly by analysing for Fe+2 (as we have to here) but we could also have done it in other ways, such as gravimetric techniques on the (SO4)2- [e.g. by making a BaSO4 ppt]
Which ever way around you decide to do it, you will need to know a property – either chemical or physical – of the compound concerned.
For Fe 2+(a transition metal ion) we know such ions can be quite easily changed from one oxidation state to another. Analysis of Fe2+ could be done using a redox reaction in which Fe2+ gets changes into another common oxidation state, such as Fe3+
We can use a redox titration using an agent suitable for quantitative analysis that will be able to cause the oxidation of Fe2+ to Fe3+. This specific reaction is mentioned a number of times in A-level and uses standardised potassium permanganate solution, sometimes called the more modern name, potassium manganate (VII). BUT, the question tells us to use potassium dichromate solution :(. Never mind, it doesn’t make much difference.
So we will titrate a sample of the moss killer (conical flask) against K2Cr2O7 solution. This reaction requires a lot of H+ ions. So sometimes these reactions are done with a volume, say 10cm3, of H2SO4 solution in the conical flask before performing the titration.
What concentration of K2Cr2O7 could be used? If too strong, maybe one drop could complete the titration and we’d get a very bad error. If too dilute, we’d have to keep refilling the burette and that would be ‘annoying’. This covered in the PRE-CALC section after the method/procedure. Note: PRE-CALC needs the info in the balanced equation found in the method.
Procedure:
All washing is assumed to involve distilled water.
Wash and dry a weighing container, accurately measuring (3 or 4 s.f) a sample of the moss killer, e.g. 5.00g, and place it in a cleaned, washed, volumetric flask using a cleaned washed funnel. Lets use a voletric flask of capacity = 250 cm3. Carefully wash out the weighing container, transferring the washings into the volumetric flask. Repeat this washing another two times. Rinse the funnel and washings down . Ensure the solid is dissolved and then carefully add distilled water making up to the mark. While securing the stopper with a finger, invert the flask several times to ensure the concentration is homogeneous.
Wash some (3) conical flasks, pipette and a burette. Rinse the burette with the standardised dichromate solution and dispose of the risings. Fill up the burette with fresh dichromate solution.
Rinse the pipette (throw away the rinsings) with the prepared moss killer sample solution, and then pipette a 25cm3 aliquot into each of the conical flasks.
Optional (and better) practice: Rinse a 20ml measuring cylinder with approx 2M H2SO4 solution. Measure out 10 cm3 of the acid and add it to each of the conical flasks, swirling to ensure homogeneity (to make absolutely sure the Fe2+ and dichromate can react in the required acidic medium)
Titrate the sample mixture in the conical flask with the dichromate solution, the end point being the disappearance of the orange colour and the appearance of the blueish-green colour due to the chromium now being present only in the +III state.
Record the volume of chromate solution used. Carry out further titrations until at least two concordant results are obtained (vary by no more than +/- 0.10 cm3).
Calculate the moles of dichromate chromate used, using n = conc x vol in L
Construct a a balanced chem equation for the redox of Fe2+ with (Cr2O7)2-
[it will show 6 Fe2+ to 1 (Cr2O7)2-]
and use the equation to calculate the moles of Fe2+ that must have been present…. IN THE ALIQUOT.
After, calculate (‘scale-up’) the moles of Fe2+ that was present in the 5g sample. by multiplying the moles found in the aliquot by 250/25.
Use this number of moles, to calculate the mass of FeSO4 present, by multiplying the moles by 152. UPDATE: the question wanted mass of Fe in FeSO4, so, as before, once you know the moles of Fe2+ multiply by the mass of Fe (=55.8), then to get the % of Fe, divide this mass by 5 and multiply by 100. Give the answer to the correct number of s.f.
[OLD: To find the % mass of FeSO4 int he moss killer, divide the mass found for FeSO4 by 5 and then multiply by 100, quoting the answer to the correct number of s.f’s.]
PRE-CALC
As we are told that the moss killer contains FeSO4,
[Mr=56+32+(16×4)=152g]
then at most, a sample of x grams of the moss killer will contain the same mass of FeSO4. i.e. the moss killer will be pure FeSO4 with no impurities.
If the a sample of 5g was taken. So at most, there will be 5/152 = 0.032 moles of FeSO4 present. 1 mole of FeSO4 contains 1 mole of Fe, hence there will be 0.032 moles of Fe2+ present (remember: at most !).
If we dissolved this in 250cm3 but only used a 20cm3 aliquot, then there would be (20/250) x 0.032 moles of Fe2+ present, and according to the balanced equation, that would require 1/6 the number of moles of (Cr2O7)2- to fully react with it.
so 1/6 x (20/250) x 0.032 moles of (Cr2O7)2- would be needed. It would be ‘good’ to able to deliver this number of moles in, say, 35cm3 from the burette, hence the concentration of that solution would be moles / (vol in L) resulting in the kind of concentration we would require is about
[1/6 x (20/250) x 0.032] / 35/1000 = 0.0146 mol dm-3
This kind of ‘PRE-CALC’ often lies behind steps where you are informed that ‘an excess of reagent xyz is added’. It is very rare that it is described how the excess is ensured and not very common for students to have to show how to calculate it.
as for preparation of the ‘acidified’ part of the acidified dichromate solution…
We’ve already calculated that the acidified standardised potassium dichromate solution should contain (Cr2O7)2- at concentration of 0.0146 mol dm-3.
The solution is made by calculating the mass of potassium chromate needed to give _twice_ the concentration of the desired concentration of dichromate solution (because two chromates will combine to produce one dichromate in acidic conditions.
The balanced equation shows 14H+ are necessary for one (Cr2O7)2-. To ensure enough H+ ions are present, we can calculate an amount of conc acid needed to give a 20H+ ratio, and assume this will be be for 100m sm3 of solution, which we can then make up to the mark with distilled water in a 250cm3 volumetric flask to give our acidified potassium dichromate stock solution, for use in the burette.
Extra acid can be added into the conical flask before titration begins to ensure excess acid is present.
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Sir, thank you so much for your detailed explanation. The question is asking about ‘percentage by mass of iron (II)’. Hence, shouldn’t we multiply the number of moles of Fe2+ present in 5g of moss killer by 55.8 to find the mass of Fe2+?
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Oops. 🙂
Yes, you are correct. Q requirement was Fe in FeSO4, not FeSO4 itself (which is what I did). Correction appreciated.
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Hello Sir, I have a question from the recent past year paper that you gave us (MJ 2013/11).
#36 (3) Based on the question, HCl is more thermally stable than HI. What factors contribute to this behaviour.
Would the standard enthalpy of formation of the reactants be a factor here?
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Hi Alyson.
Answer 1: “The strength of the hydrogen-halogen bond”
Strong bonds are more difficult to break, hence compounds with stronger bonds are harder to decompose. That H-Cl did not react in these high thermal conditions but H-I did, reveals that HI and HCl have different bond strengths, and so are a factor here… the answer is valid.
Answer 2: “The size of the halogen atom”
This answer actually is connected to the first factor. The smaller the halogen atom, the stronger the H-X bond strength is. So this answer is correct too and an important part of the reason why the bond strength in H-Cl is greater than that of H-I.
Answer 3 is a trick answer.
(delta)Hf is defined as forming one mole of a compound from its elements. The proposed answer talks about the (delta)Hf of the _products_ of the decomposition. Yet the products of the decomposition are I2 and H2, they are elements themselves,, so the (delta)Hf of those products is actually zero. and actually has no relevance to the reaction that actually happens.
By telling students H2 and I2 are formed, they are trying to trick students into thinking the (delta)Hf mentioned in the proposed answer is the (delta)Hreaction, which it’s not.
If they said (delta)Hreaction (i.e. the decomposition reaction) – which is what you are asking about, then that would be a factor, and be dependent on the difference in bond enthalpy between H2, Cl2 and HCl (and also the bond enthalpy if H2, I2 and HI), which is of course is a function of the bond strengths and hence the size of the halogen atom.
Hope that helps.
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Hello sir, I have a question from May June 2011 paper 43
3(c) the question requires us to determine the products of electrolysis on cathode and anode based on information from data booklet for AgF, FeSO4, MgBr2
I tried using the standard electrode potential but could not obtain the answer given.
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Update #2 Monday 09:01 30-March-2015
I’m redoing my answer. I see my error now.
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Ermmm. I corrected my ‘error’ (which wasn’t that much of an error) and I was unable to get the right answer. I changed the method (ignoring auto-ionisation) and then the Br2 discharge worked. but then The FeSO4 didn’t work.
So I’m currently stuck….. I’ve began a discussion of it on the CIE teachers support forum and am awaiting a reply and further discussion. I’ve asked others about it and they too are at odds with it. So all i can do is keep working on it and let you know later.
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I’ve been through all possible options that I can think of, and I simply cannot figure out how the data book (as instructed by the question) can generate the right answer. I calculate O2 to be discharged at the anode and not Br2, but the mark scheme gives Br2. The mark scheme and the examiners report are virtually useless in shedding light on the matter. I suspect CIE put down the wrong answer to the question given, but can’t admit it.
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Well, here it is…. The extremely unsatisfying answer:
Apply the normal method by considering all the cations which migrate to the cathode to undergo reduction and remember water can also be reduced too!. Note all of their reduction potentials.
The species with the most positive potential will be discharged (produced) at the cathode.
Do the same for the anode. List down all anions that will migrate there and remember water can also be oxidised too. List down all their reduction potentials. SWAP the equations around so they now read as oxidations. The most positive of those will be the one that’s discharged. That’s what you would do with the data book values.
It generates the right answers apart from the MgBr2.
Now here’s where you have to DEFY the question and use MEMORY despite it not being in the syllabus.
…..IF…… there are anions which appear on the series that appears on page 291 of the orange CIE book, then that anion may supersede any other ion that you’ve got. In the case of MgBr2 you have Br- and OH- (from the auto ionisation of water) and water itself that may be reduced. Br- and OH- appear on the series, which states that Br is oxidised in preference to OH-
I know it’s a stupid unscientific answer but that’s the best that any of us can to to read the mind of the CIE examiners preferring to ask silly questions rather than asking clear questions about what they state in the syllabus you are supposed to learn..
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Thank you for your replies, sir.
But the question asked us to use relevant information from the data booklet which does not include the series in the orange coursebook.
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I totally agree with you. The question is sub-standard. No excuses.
I suspect they noticed this when the examiners were marking the papers and they contacted CIE. I also suspect CIE would have either told the examiner to accept O2 at the anode as a correct answer OR adjusted the grade boundaries appropriately.
The thing is, they may ask something similar again but not directly specify use use data book, thereby drawing upon the series stated.
CIE are humans too.
(so the rumor goes anyway)
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OK. After consultation, this is how you should answer questions in the future:
For the possible reduction of water at the cathode in an electrolysis cell, use this equation:
2H2O + 2e– ⇌ H2 + 2OH–, reduction potential = –0.83 V
for the possible oxidation of water at the anode in an electrolysis cell, use this reduction potential equation: O2 + 4H+ + 4e– ⇌ 2H2O, reduction potential = +1.23 V
(of course if you want to use this to oxidise water, you will have to swap the equation around, having H2O as the reactant and visibly producing ‘free’ e-).
i would say that only in “special circumstances” would you use this: H2O2 + 2H+ + 2e– ⇌ 2H2O, reduction potential = +1.77 for the possible odidation of water (again you’d have to swap the equation around for oxidation). You’d probably be given specific directions in the question if this equation had to be used OR, more obviously, if the question involved the reduction of hydrogen peroxide.
ONLY use this equation: O2 + 2H2O + 4e– ⇌ 4OH–, reduction potential +0.40 V, for “obviously” alkaline solutions
I used it previously as a possible candidate for oxidation at the anode, where I considered the presence of OH- from the auto-ionisation of water. do NOT do this. Ignore OH- from auto-ionisaation of water.
[But you do consider H+ from the autoionidation of water – which allows H2(g) to be formed at the cathode] and this was done in the FeSO4 part.
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Sir, according to the CIE revision guide by David Bevan (the one that you gave us recently) page 66, it stated that the two half equation used to decide which ions are discharged in aq solution are:
4H+ +4e 2H2 +0.00V
O2 + 2H2O +4e 4OH- -0.40V
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Hummmm….
It conflicts with what the CIE teachers support forum stated.
Which one is correct?? Let me re-assess ALL the info (again). I’ll mention “the CIE endorsed revision book states the use of the 4 OH- equation” on the forum and see what they say…
So sorry about this. I’m could give my opinion, but I want to give you CIE’s opinion (if they were different), so once again… hold tight…
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Why OH- ion does not react with chlorobenzene?
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As you know Nicholas, I already steered you towards the answers in class. But for the benefit of others…
OH- is a nucleophile. It needs an electron deficient centre (i.e. a sufficiently electron deficient centre) in order to react. Such a condition is present in the normal haloalkanes, but in chlorobenzene, the ring contains 6 delocalised (hence ‘mobile’) electrons and is therefore electron rich. In chlorobenzene therefore, there is no OH- ion to attack hence no reaction. This is one reason who arylhalides maintain/persist in their toxicity in the environment – because of their lack of reactivity.
It’s sufficiently power electrophiles that react with aryl groups and not nucleophiles.
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Hello sir. On paper 41/mj/10, question 3(c) talks about reaction between CCl4 and water. Does CCl4 react with water? Isn’t the bulky chlorine atom shielding the small carbon atom from attack by water molecules?
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Sorry sir, read the question wrongly. It’s CCl2 not CCl 4
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🙂
Glad you came here anyway and asked.
Now Malaysia is not falling so far behind Zanzibar in location of visitors. 🙂
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Aluminium carbide, Al 4C3, reacts readily with aqueous sodium hydroxide. The two products of the
reaction are NaAl O2 and a hydrocarbon. Water molecules are also involved as reactants.
What is the formula of the hydrocarbon?
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Is this a CIE question? I would presume methane is formed so the answer is CH4?
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Sir, it is a CIE paper 1 question. How to get CH4?
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Hummm… How to answer this from the perspective of a student…
The ionic view: (Al is a metal hence forms cations, carbide suggests an anion or ‘negative carbon’ so an assumption that it’s ionic isn’t too bad)
and yes, I know aluminium chloride is covalent yet the chlorine also carries an ‘ide’ ending, and anions of C are harder to make than anions of Cl…. but other than having learned about this compound, I can’t see any other way in which a student could attempt to answer it.
4 Al’s therefore +12 charge. 3C’s must have an average charge of -4, that is a C atom with 4 lone pairs in the n=2 shell, obviously highly unstable in the presence of a possible/prospective reactant.
C will want to share those one pairs with something capable of accepting an e-pair into its valence shell. In NaOH (and H2O which was mentioned as also being a reactant in your info), H is the most likely atom to do this e- pai accepting, so the C will likely pick up /attack the H. We said earlier the average charge of the C’s was -4; Given we can’t get a C with a -5 charge, this must mean that actually ALL the C’s must be -4. (remember we are assuming ionic species here) so one (C)4- “ion” must attack 4 H’s and so methane would be the product.
Which paper was that question from Nick?
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I suspect the question comes from a different syllabus where carbides were mentioned in the learning requirements…
although with Cambridge’s tendency to ask questions like “How many moles of carbon are consumed by the aliens of Mars each day assuming they received a 3 tonne batch from their cousins from Pluto three weeks ago” I guess I can’t rule out the possibility it comes from the 2015 syllabus exams.
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Thank you sir. WInter 14 11 question 6
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Thanks Nick. Appreciate it.
If anyone can give any justification as to how such a question possibly complies to the learning of the 2015 syllabus, then I’ll give you a monetary reward.
The answer to the question above was zero… because the evil Zogonians from Saturn secretly broke into the shipment container and consumed it all in transit of course!.
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Ques: Describe and explain how the solubilities of the sulfate of Group ll elements vary down the group.
Ans: “As you go down the group, the solubility of Group ll elements decreases. The larger the cation, the less it will polarize the anion. The e- of the compound is widely distributed, which decreases its e- density. The e- has more space to travel which causes water molecules to find difficulty in forming ion-dipole bonds as e- are not condensed in one place. The energy produced by formation of ion dipole bond is not strong enough to break the ion apart.”
Why isn’t this answer acceptable in marking scheme?
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Hi Silver Bunny 🙂
The question features ionic salts that show a range of volubilities within a periodic group.
Solubility depends upon the balance of Lattice Enthalpy (LE) and Hydration Enthalpy (HE) of the constituent ions. You need to mention these two factors. Also how each of them varies when you have a very large anion and an increasingly large cation. You should also specify DELTA(Enthalpy of Solution)= DELAT(HE) – DELTA(LE)
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