BedokFunland JC's A Level H2 Chemistry Qns (Part 2)
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Originally posted by Kahynickel:
There are two possible shapes for PCl5 i.e tripgonal bipyramidal and square pyramidal. The first one is non-polar overall despite the presence of polar P-Cl bonds.
But what about square pyramidal. The dipoles do not cancel out each other so it overall polar shape. Can you through some light on this issue.
If PCl5 were square pyramidal (which it is not), then yes it would be polar. But the square pyramidal molecular/ionic geometry is based on the octahedral electron geometry, which requires 6 electron charge clouds (eg. 5 bond pairs and 1 lone pair).In PCl5 (liquid and gaseous states), the P atom has 5 bond pairs and 0 lone pairs (since PCl5 is a neutral molecule with no net ionic charge which implies all atoms have no formal charges which accordingly implies that the P atom has 5 electrons from 5 bond pairs, as P is a Grp V element). Hence the PCl5 molecule has the trigonal bipyramidal (electron and molecular) geometry.
In the mechanism for forming the solid state PCl5 (from the liquid state), a Cl- ion is eliminated from one PCl5 molecule, to nucleophilically add onto the electrophilic P atom of a neighbouring PCl5 molecule, generating the PCl4+ cation (with tetrahedral ionic geometry) and the PCl6- anion (with octahedral ionic geometry).
Therefore in accordance to their geometries, all 3 species, ie. the trigonal bipyramidal PCl5 molecule, the tetrahedral PCl4+ cation, and the octahedral PCl6- anion, are all non-polar species.
image credits : http://en.wikipedia.org/wiki/Phosphorus_pentachloride
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[True Story] - The story of Anna's JC life
http://www.domainofexperts.com/2014/01/the-ugly-truth-about-life-as-jc-student.htmlI recall quite a number of other similar cases, such as Kevin Wee (the RJC guy who had a total breakdown and YouTubed himself BBQing and burning away his A level cert in his room) and XinLin (the HCJC girl who shared with the world how her JC life were the worst 2 years of her life).
For Kevin, XinLin and Anna, and countless others like them, our sympathies and empathies, and wish all of them the best for their future endeavours.
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The paramount importance of a science education (including the very useful H2 Chemistry) with an intelligent (not just highly educated or qualified) open-minded approach in life :
"It's 3 a.m., Is that a Parasitic Worm in Your Cheek?"
http://sg.news.yahoo.com/3-m-parasitic-worm-cheek-op-ed-172448068.html -
H2 Chemistry Qn... (well, kinda)
An amphiprotic species, eg. the HS- ion, is a species that can accept a proton, or donate a proton, depending on its environment.
An ambidentate ligand, eg. SCN-, is a ligand species that can donate a dative bond from either of two different donor atoms (in this case the S atom and the N atom, due to the uninegative formal charge delocalized by resonance over the S and N atoms), but not from both atoms simultaneously (assuming the ligand is monodentate ambidendate).
Question : Since both prefixes have the same meaning, then why is the prefix "amphi" in one case, and "ambi" in the other case?
Answer : Yes, both prefixes mean the same thing, it's just that "ambi" is a Latin word, while "amphi" is a Greek word. Due to the historical evolution of the English language, Chemists (and everyone else, for that matter) continue to borrow upon both the Latin and Greek languages. -
The correct way of determining orbital hybridization, is by looking at the electron geometry (and has nothing to do with sigma vs pi, single vs double vs triple bonds, as is often erroneously taught in schools).
For instance, the N atom is NH3 has 1 lone pair and 3 bond pairs, hence electron geometry is tetrahedral, while molecular geometry is trigonal pyramidal.
In determining orbital hybridization, we focus on the electron geometry.
Since the electron geometry about the N atom in NH3 is tetrahedral, the orbital hybridization is sp3.
sp = s + p = 2 electron charge clouds = linear.
sp2 = s + p + p = 3 electron charge clouds = trigonal planar.
sp3 = s + p + p + p = 4 electron charge clouds = tetrahedral.
sp3d = s + p + p + p + d = 5 electron charge clouds = trigonal bipyramidal.
sp3d2 = s + p + p + p + d + d = 6 electron charge clouds = octahedral.
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Original Qn :
When 11.7 g of sodium chloride (Mr 58.5) dissolves in 100 g of water, the temperature of the solution falls by 2.4 deg C. Given the specific heat capacity of the solution to be 4.2 J g-1 K-1, what is the enthalpy change of solution of sodium chloride?
Qn on the Qn : Should the mass used in the formula Heat = MC delta T, be (100 + 11.7)g or 100 g?
Ans for the Qn on the Qn :
While the overall mass of the solution would indeed be (100 + 11.7) g, however the mass of the water which absorbs or releases the heat (and which therefore must be used in our calculations) remains as 100g.
Students are expected (though not many JCs teach this well) to be able to discern that out of the (100 + 11.7) g of the solution that absorbed or released the heat, it is to be regarded that only the 100g of water that registers this thermal transfer process (ie. that allows us to determine via calculations the magnitude of heat transfer), and not the 11.7g of the salt (which for all relevant A level purposes and intents, is identical to an external heat source, say combustion of a fuel). -
4f orbitals have a poorer shielding/screening effect compared to 3d orbitals which have a poorer screening/shielding effect compared to 2p orbitals which have a poorer shielding/screening effect compared to1s orbitals, because of two reasons :
The further away the (principal quantum) electron shell the orbital belongs to, the larger and hence more diffused the orbital and therefore the poorer the shielding/screening effect.
The shape of the orbitals also determine their effectiveness of shielding/screening : the shape of the orbitals are such that f orbitals are poorer shielders/screeners compared to d orbitals which are poorer shielders/screeners compared to p orbitals which are poorer shielders/screeners compared to s orbitals, even when in the same (principal quantum) electron shell.
In the context of A level Chemistry (instead of say, physics or engineering or Marvel's Agents of S.h.i.e.l.d), there is no difference between shielding effect vs screening effect. Cambridge will accept both terms eagerly with open arms.
Originally posted by bluepie:The atomic radius of aluminium is bigger than gallium due to the poor shielding effect of the 3d orbitals.
Why does the electron in 3d orbital have a poor shielding effect than say 1s or 2s or 2p?
on an interesting note, atomic radius of lanthanide series decreases with increasing proton number, also due to the poor shielding effect by 4f electrons.
again, why does 4f electron have a poorer shielding effect? lastly, what is the difference between shielding effect and screening effect?
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Originally posted by bluepie:
Hi i have another question relating to the atomic radius of transition metals.
for element 1-20, going across a group, screening effect remains constant and the only factor affecting atomic radius is due to the increasing nuclear charge and thus increasing attraction between valence electrons and positive nucleus.
However, such a trend is not observed for transition metals; their atomic radius stays relatively unchanged even with increasing proton no. This is because electrons are added into the inner 3d subshells and not 4s subshell, and hence the screening effect increases. (is it correct to say this?) The increasing screening effect offsets the the increase in nuclear charge and thus attraction between 4s electrons and nucleus is relatively same. Hence, atomic radius remains relatively unchanged.
Are my explanations correct or is there a better way of phrasing it?
Your answer is correct.
JC students commonly have difficulty with this topic, because it's just like the classic accountant joke (the boss is hiring and interviewing a physicist, a mathematician, a IT guy and an accountant; the boss asks, "what's 2 + 2?" The physicist, a mathematician, and IT guy each give a similar expected answer, with technical variants depending on their different backgrounds. When it comes to the accountant, the accountant replies, "What do you want it to be, boss?". The accountant got the job).
For the qn, "Why do d block elements have smaller atomic radii compared to s block elements?", students are expected to state, "3d orbitals are poor shielders!"
In contrast, for the qn, "Why do the atomic radii and 1st ionization energy of d block elements remain relatively invariant across the period?", students are expected to state, "3d orbitals are effective shielders!"
The key to understanding this apparent contradiction, is of course (Einstein would be proud) : Relativity.
For the same electron shell, when compared to the 3s and 3p orbitals, 3d orbitals are poor shielders because of their shape. Hence d block elements have smaller atomic radii compared to s block elements.
At the same time, 3d orbitals (being part of the less diffused 3rd electron shell) provide more effective shielding compared to 4s orbitals (being part of the more diffused 4th electron shell). Accordingly, as we go across the period, because the additional electron is added into the 3d orbitals, which provide relatively more effective shielding compared to the 4s orbitals, and consequently the slight increase in relatively effective shielding by the additional 3d electron, partially offsets or counters the increase in nuclear charge, resulting in the relative invariance of the atomic radii and 1st ionization energy of the d block elements across the period.
Anyway, that's just my own phrasing. Your phrasing is already perfectly fine, and will gain you full credit by Cambridge in the A levels.
Edited : Oh, in case some students are still wondering, "If for the 2nd period, the shielding effect remains constant, then why doesn't the shielding effect also remain constant for the d block elements?"
The reason is : even for period 2, the shielding effect isn't absolutely constant. It does increase slightly across the period, when more electrons are added. But because electrons are added to the 2nd electron shell (which is the outermost valence shell), hence shielding effect from the 2nd electron shell is relatively negligible (shielding effect is mostly effective from inner shell electrons). In contrast, for the d block elements, the additional electron is added into the 3d orbital (which is part of an inner electron shell), which does provide significantly effective additional shielding effect for the outer 4s electrons, which are the electrons involved in determining atomic radii and 1st ionization energy. -
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Originally posted by bluepie:
The hybridisation of sulfur in anion SO42- is given to be sp3. Why is it so? Sulfur has two double bond and two single bond with four oxygen atoms, meaning there are a total of 12 electrons surrounding sulfur. sp3 only accomodates up to 8 electrons. besides, sp3 means that we are assuming there are only four single S-O bonds, which is not the case since resonance structure for it exists.
Also given that we know resonance structure exists for the sulfate ion, it must mean there is an unhypridised p orbital on sulfur. if sulfur were to be sp3 hybridised, there will not be any unhybridised p orbital to overlap with other p orbitals of oxygen. shouldn't the hybridisation of sulfur have beeen spd2? two unhybridised p orbitals, as well as four spd2 orbitals, so there will be 4 hybridised orbital to overlap with the ones in oxygen, and 2 overlap of p orbitals to form two double bonds?
Edit: in order for resonance structure for sulfate ion exists, does it mean that one sulfur atom would need to have 4 unhybridised p orbitals to overlap with each p orbitals from 4 oxygen atom? or would 1 unhybridised p orbital for sulfur will do? [likewise for CO32-, NO3-]
As the electron geometry about the S atom in SO4 2- is tetrahedral (in order to maximize stabilities by minimizing electron pair repulsions), hence sp3 hybridized orbitals are required for the sigma bonding about the S atom.
At A levels, the usual major resonance contributor taught to students, is that of an expanded octet with no separation of formal charges (at least in part because formal charges are not usually taught to most H2 students in the first place), ie. Fig 1 below.
But experimental evidence has indicated that such a resonance contributor, is actually only a minor resonance contributor. The major resonance contributor, is actually that of a dipositively formal charged S atom, singly bonded to 4 O atoms with uninegative formal charges, ie. Fig 2 below.
The (almost negligible) pi bonding that exists (for the minor resonance contributors), utilize the d orbitals of the S atom to overlap (poorly) with the p orbitals of the O atoms.
In other words, the bonding in the SO4 2- ion, is actually covalent with significant ionic character (due to the formal charges present in the major resonance contributor).
Experimental evidence has shown that the actual charge on the S atom is only slightly less than its formal dipositive charge, as such an S 2+ atom would accordingly be strongly electron-withdrawing by induction, withdrawing electron-density from the singly bonded O- atoms, towards the central S 2+ atom.
Concordantly, consequently and most interestingly, you could say that the (slight) partial double bond character (ie. in terms of bond length and bond strength) of the S=O bonds in the SO4 2- ion, is partly (only slightly) due to resonance, and partly (somewhat moreso) due to induction.
Originally posted by bluepie:Edit: in order for resonance structure for sulfate ion exists, does it mean that one sulfur atom would need to have 4 unhybridised p orbitals to overlap with each p orbitals from 4 oxygen atom? or would 1 unhybridised p orbital for sulfur will do? [likewise for CO32-, NO3-]
1 unhybridized p orbital would be sufficient, but because only one unhybridized p orbital is available, the sideways overlap with the p orbitals of the surrounding atoms to form pi bonds is only *partial*. This is concordant with the observation that in the resonance hybrid, each C=O bond in CO3 2-, and each N=O bond in NO3 -, only has *partial* double bond character.
One last point : for elements beyond period 2, while their atoms often appear to achieve an expanded octet (using vacant, energetically accessible 3d orbitals), which is what is taught at A levels, however experimental evidence has shown this to be inaccurate. In point of fact, all molecules/ions with central atoms that appear to have an expanded octet, actually have as their major resonance contributor, the central atom with only a stable octet, and (accordingly) with separation of formal charges. In other words, in the actual structure or resonance hybrid of the molecule/ion, the bonding is always covalent with significant ionic character.
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Originally posted by Kahynickel:
The electronegativity of Nitrogen is 3.0 and of chlorine is 3.2. I am not sure about the exact values. anybody knows the actual values.
There are multiple systems of electronegativity ratings using different experimental systems and calculation methods. As such, you will come across different systems (eg. Pauling, Mulliken, Allred–Rochow, Sanderson, Allen, etc) giving you different electronegativity values for each element.
For A level purposes, N is certainly regarded as more electronegative than Cl, and as such, only NH groups (in addition to HF and OH groups), and not HCl, are considered capable of donating H bonds.
A level students are not expected to memorize electronegativity values from any particular system. Cambridge may give electronegativity values (from any of the different electronegativity systems) in the A level question itself, to examine if the A level candidate is able to interpret the relative values of the various elements. For instance :
Carbonyl sulfide has the formula O=C=S. If the electronegativity of O, C and S are 3.5, 2.5 and 2.4 respectively, which bond is more polar? The O=C bond or the C=S bond? Indicate the partial charges and dipoles on the O=C=S molecule. -
UltimaOnline posted :
Every year, many thousands of JC students with all straight As, and hundreds of Poly students with perfect GPAs, have to compete tooth and nail for just 200 places in NUS+NTU Medical School. Every year (for last few decades), hundreds of these straight As students are forced to study Medicine abroad (and these are the lucky ones whose family can afford the $1 million fees + expenses to study Medicine overseas), and the fact that recognized Medical schools overseas (including many even better than NUS) accept them, proves they're actually good enough to be medical doctors. And yet, every year PAP says "Jialat lah! we not enuff medical doctors in our hospitals leh! must import a lot more FT doctors!" and keep bringing in thousands of FT doctors every year. When our own Singaporean students who are good enough to study Medicine, are deprived of the opportunity to study Medicine locally.
mojito posted :
A poor sentiment, my friend. You see, if the PAP did not import more doctors, we would have an acute shortage of doctors as the number of patients spiked in the recent years. Medicine is an important vocation, too important in fact, to allow local students without the necessary years of housemanship to take on key roles as specialists. As you must now realize, the PAP had to make difficult trade offs. Without them we would have let our emotions get in the way of doing what is best for the people. For that, you have to thank the PAP.
halsey02 posted :
You weren't a medical student back, in the 70's & 80's...if not, you do not know what you are talking about....the PAP f$#$$^$ all those wanting to study for medicine here...many have to go abroad to study that...I know a couple classmates...they never returned... So, try not to bluff too much huh!.... they are people like me of that generation who went through the system....& still around & many still around scattered all over the world...
freedalas posted :
I have 2 nephews who were scored straight As for the A levels and yet were denied admission to our medical school here. Both had to study abroad and their respective university assessed them to so very good after their first and second year that they were both granted scholarships for the remaining years of the course. Now both of them are practicing overseas and really doing very well. Asked them if they ever want to come back to practice in Spore, they both said "tell the PAP government to go fuck itself". They very much prefer to contribute to the overseas country that treated them so much better than Singapore. My 2 nephews are just tip of the iceberg. I am sure many many more Singapore students who even better than them have been denied entry by PAP to medicine school here. The PAP would instead open their assholes to freely let Pinoys, Ah Nehs and Burmese with fake degrees in. And as a side point, if one ever get attended by a Pinoy doctor, be very very careful as almost 100% he or she is a quack. Of all foreign trash here, the Pinoys are the biggest bull shitter, even worse than Ah Nehs. -
BedokFunland JC H2 Chemistry Qn
State the formula and colours of the manganate(VI) and manganate(VII) ions, and suggest with explanation, whether the Mn-O bond length in manganate(VI) ion is expected (ie. don't Wikipedia the answer immediately, but first think over it critically based on what you've learnt in the H2 syllabus) to be longer or shorter, compared to the manganate(VII) ion.
Hints :
Draw the first couple of resonance contributors, then the resonance hybrid, for the 2 ions. Hence state the bond order (eg. 1.5 bonds, or 1.75 bonds, etc) for each individual Mn-O bond in the resonance hybrid for the 2 ions. -
I've always been supportive of the private candidacy route, as I've always been of the opinion that the majority of time and energy spent in JC life is wasted (eg. how many students really learn and benefit from JC lectures? Lectures should be replaced completely by tutorials), majority of JC students suffer from perpetual chronic sleep deprivation (it's a medically proven fact that everyone on average requires a minimum of 8 hours of uninterrupted sleep, and daily sleep deprivation of less than 8 hours of sleep results in accumulative damage to your brain and body resulting in damaged intellectual faculties and semi-permanently lower IQ (damaged can be partially reversed if you successfully manage 8 hours of sleep daily from now on) and consequently significantly poorer A level results, and it's also therefore no wonder that Singaporeans have the highest incidence of diabetes, heart attacks and strokes, the risk of these diseases greatly exacerbated by chronic lack of sleep) and the combined (mental, emotional, psychological) stress from all the different JC subjects, CCAs, projects and other committments, can actually do more harm than good (and in extreme cases can almost ruin one's entire life, even for students in or perhaps especially students from top JCs, such as Kevin Wee of RJC, and XinLin of HCJC, etc. Hope Kevin and XinLin are both doing well in their lives now).
Better to leave now and go the private route and take back control of your life, than risk a severe psychological breakdown that will not only ruin your A level results anyway, but may even cost you your phsyical, emotional, mental and psychological health, and in tragically worst case scenarios (there have been several cases in the last few years), it may even cost you your life.
If you can afford it, go for tuition (group tuition is always a lot cheaper than solo tuition) for your H2 subjects. But even if you can't afford any tuition at all, even solely self-study will still (in all likelihood) gain you a better performance at the A levels, compared to if you were to suffer a psychological breakdown in the unhealthy JC environment.
Final disclaimer : obviously this is just my opinion. I'm *not* telling you to quit JC and take the private route (coz this is your own life and therefore you must make your own decisions for yourself). I'm just saying that if that's what you decide for yourself that it's you want to do, then as far as I'm concerned, there's certainly a lot of sense in doing so. Obviously there will be some JC and ex-JC students will say, "What a load of rubbish! I'm totally enjoying JC life! No stress at all for me! No sleep deprivation at all for me! I love JC life, which has been socially, mentally, emotionally and personally enriching and fulfilling for me in every way! I just wish it was at least 4 to 6 years instead of just 2 short years!", but just as there are such students on one end of the spectrum, there are bound to be students on the other end of the spectrum, for which private candidacy would make a lot of sense.
Decide for yourself, and no matter the outcome, you can respect yourself for taking back control and responsibility for your own life. -
Sure no prob.
Q1. Yes, that's correct.
Q2. No, that's not quite correct (you stated two facts correctly, but there's no direct link between them, or the link is indirect and tenuous at best). Different isomers (eg. structural, functional group, geometric, etc) have different thermodynamic stabilities and hence possess different potential enthalpies, and therefore their combustion enthalpies (ie. magnitude of enthalpy difference between the isomers, and the combustion products) are different. A good example at A levels would be terminal alkenes vs internal alkenes (internal alkenes have greater number of stronger C-C sigma bonds with the higher % s orbital character), or cis alkenes vs trans alkenes (cis alkenes have greater steric van der Waals strain or repulsion between bulky alkyl groups on same side of the double bond).
Basic idea is to draw an energy profile diagram with y-axis as energy/enthalpy level. Near the top would be the elements, near the bottom would be the combustion products, and in the middle (with varying energy/enthalpy levels and thermodynamic stabilities) would be the different isomers. By Hess Law, overall (formation + combustion) enthalpy would be the same. But if the isomer is less stable = more energy, its formation enthalpy would be less exothermic but its combustion enthalpy would be more exothermic; if the isomer is more stable = less energy, its formation enthalpy would be more exothermic but its combustion enthalpy would be less exothermic. Examples would be (as I mentioned earlier) internal vs terminal alkenes, and cis vs trans alkenes.
At O levels this question shouldn't, and wouldn't, be asked by Cambridge at all. Whoever asks this question at O levels (eg. school prelim paper, Cambridge, etc) should be shot, as the correct answer requires a minimum of A level Chemistry.
Q3. Correct, per unit volume of the fuel, there is a greater number of C and H atoms, and hence greater number of C and H atoms oxidized (exothermically via combustion) to CO2 and H2O, and concordantly, larger amount of heat released during combustion. An additional secondary effect, results from having stronger van der Waals attractions between molecules with larger number of atoms, allowing for closer packing of the molecules, ie. increasing density of fuel, hence resulting in greater number of fuel molecules per unit volume.
Originally posted by ArJoe:Hi ultimaonline, thank you for taking your time to reply, really appreciate it. I have a few o level questions to ask. 1) what are the difference between addition and condensation polymerisation? Can i state addition produces polymers that have the same empirical formula as the monomer itself whereas condensation does not? I am very certain about the other 2 differences.,
2) why do isomers(with the same molecular formula) have different enthalpies change of combustion(heat energy generated from combustion)? can i state because they have different physical properties(melting and boiling point), therefore different amount of heat energy is absorbed and produced during bond breaking of reactants and bond making of products, therefore enthalpy change is different?3) why do alcohols with increasing carbon atoms produce more heat energy during combustion? Because larger volume of co2 and h2o are produced, therefore more heat energy is released during bond making....?
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Rachel Lee JiaLi asked BedokFunland JC this question 2 days before her 2014 A levels Paper 2 exam : "If you're given 25.0 cm3 of 0.040 M of CH3COOH with pKa = 4.76, and 50.0 cm3 of 0.050 M Ba(OH)2, and you carry out titration (ie. limiting reactant in conical flask, excess reactant drip down from burette), what's the pH at equivalence point?"
Observe that CH3COOH acid is monoprotic and Ba(OH)2 base is diprotic.
Moles of H+ available = 25/1000 x 0.04 = 1 x 10^-3 mol
Moles of OH- available = 2 x (50/1000 x 0.05) = 5 x 10^-3 mol
Hence limiting reactant is CH3COOH = 1 x 10^-3 mol
Hence moles of OH- required to reach equivalence = 1 x 10^-3 mol
Hence moles of diprotic Ba(OH)2 required to reach equivalence = 5 x 10^-4 mol
Hence to find volume of diprotic Ba(OH)2 required to reach equivalence :
Molarity = Moles / Volume
0.05M = 5 x 10^-4 mol / Volume dm3
Hence volume of diprotic Ba(OH)2 required to reach equivalence = (5 x 10^-4) / (0.05) = 1 x 10^-2 dm3
Hence total volume in conical flask at equivalence point
= volume of CH3COOH (originally in conical flask) + volume of Ba(OH)2 (added in from burette)
= (25/1000) dm3 of CH3COOH + (1 x 10^-2) dm3 of Ba(OH)2
= 3.5 x 10^-2 dm3
Moles of CH3COO- ions from the soluble salt Ba(CH3COO)2 at equivalence point = 1 x 10^-3 mol
(explanation : since that's how much CH3COOH acid we used ; don't let the stoichiometry of the formula of the barium ethanoate salt Ba(CH3COO)2 confuse you, focus on the moles and molarity of the active species which determines the pH, ie. the CH3COO- ion)
Hence molarity of CH3COO- ions at equivalence point
= moles of CH3COO- / total volume at equivalence point
= (1 x 10^-3) / (3.5 x 10^-2)
= 2.8571 x 10^-2 mol/dm3
From pKa of 4.76 for CH3COOH, we find pKb for CH3COO-
= 14 - 4.76
= 9.25
Hence Kb for CH3COO- = 10^-(pKb) = 5.7544 x 10^-10
Approximating equilibrium molarity of CH3COO- back to it's initial molarity (in the ICE table) since Kb value for the weak base is much smaller than initial molarity of the base CH3COO-, that is to say, X or [OH-] is negligible compared to 2.8571 x 10^-2 mol/dm3,
Kb = [CH3COOH] x [OH-] / [CH3COO-]
5.7544 x 10^-10 = X^2 / 2.8571 x 10^-2
X or [OH-] = square root of (1.6441 x 10^-11)
[OH-] = 4.0547 x 10^-6
Hence pOH = 5.392
Hence pH = 14 - 5.392
= 8.608 (Note : 8.608 is actually to 3 significant figures as following the data given in the question ; for pH, significant figures are counted only after the decimal point of the pH value). -
A few hours before 2014 A Level Paper 3, LeeJunEr asked BedokFunland JC, "If we need to convert para-methylnitrobenzene into para-aminobenzoic acid, does the order in which we carry out oxidation of the methyl group, versus the reduction of the nitro group, matter?"
BedokFunland JC to the rescue :
Yes, it does matter.
Although not taught in the H2 syllabus, KMnO4 can indeed oxidize phenylamine into nitrobenzene (at neutral pH), or azobenzene (at alkaline pH), or polyphenylamine (at acidic pH).
Therefore, the methyl group needs to be oxidized using acidified KMnO4 first (which doesn't affect the NO2 group), and thereafter the nitro group needs to be reduced using Sn and excess concentrated HCl (which doesn't affect the COOH group, since carboxylic acids can only be reduced by LiAlH4, and not by H2 gas generated by acid and metal reaction). -
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