BedokFunland JC's A Level H2 Chemistry Qns (Part 2)

last SG · 2010-07-26T23:40:35+00:00

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UltimaOnline SG

Thanks for ur explanation !!!

I'm not fully understanding the problem yet

Is it :

I want to study the solubility of CaC2O4.

In this case, the solubility of CaC2O4 in high pH(alkaline conditions)

In a way, we want to study the effect of [H+] in the solubility of CaC2O4. (Can be seen from Wilber's equation)

In part iii) , the [H+] plays a role, in protonating anion, in turn having an effect in the solubility of CaC2O4.

However in part iv) , we actually want to study the effect of [H+] in high pH, but however, the NaOH actually causes increased solubility of CaC2O4, as it precipitates out the cation, instead of due to the effect of [H+] , which we want to study.

Is it?

For part iii) , the solution elucidates about the effect of LCP for the explanation. But can i also explain , based on the equation given in part iii, solubility increases, as total [H+] in numerator increases faster than total [H+] in denominator, hence solubility increases

Wrong, in high pH, it's OH- that's relevant, not H+. "We" do *not* want to study the effect of [H+] in high pH, "We" only want to study the effect of pH on solubility, so Wilbur's equation (involving H+) is obsolete and can be retrenched.

Your mathematics-based answer will not be acceptable, because firstly, the H2 Chem syllabus is always testing students about shifting of position of equilibrium (use of acronyms like LCP not allowed in A level exams); and secondly, the actual question (on Wilbur's work, or similar) in the A levels might not give you an mathematical equation or formula at all, so if all you can give are mathematical reasonings (as opposed to chemical reasoning that's required by the mark scheme), you'll be screwed in the A levels.

BedokFunland JC exam tip : If you're exam-smart, you'll give a variety of reasons, whatever you can think of. As long as they don't contradict each other, you'll be given credit for whichever answer Cambridge wanted.

UltimaOnline SG

I understand the discussion about OH- , with the cation, can precipitate out an ionic solid, and , in acidic pH, anion is protonated. But i dont see how these is relevant for the explanation in iv) .

I dont see how "NaOH might precipitate out Ca(OH)2 , which would decrease [Ca2+] , therefore potentially increasing solubility of CaC2O4 at high pH" (given answer solution) addresses the question of iv) , which is to study the effect of (high) pH on solubility.

We want to study the effect of pH on solubility.
For part iv) , understood, Wilber's equation is obsolete for alkaline pH.

For part iv) , Then, why is adding NaOH a problem? it should be Wilber's equation that's the problem.

i dont see how adding NaOH is a problem studying effect of pH on solubility.

Adding NaOH IS A Problem if we want to study effect of pH on solubility, USING Wilber's equation.

I dont see a link here. maybe i dun understand/disagree with the phrasing of the qn, and the ans. might be my command/understanding of english is inadequate.

Ok real sorry, just move on, i dont mean to annoy you, sorry, its ok if u dont reply to this qn
Just move on

source

Just skip the damn question in the A levels if you don't like Cambridge's phrasing. Oh wait, only I can do that and still get an A grade. *You* can't.

Yeah, go chat with (and annoy) your school teacher or private tutor instead.

UltimaOnline SG

Louistanshiehsit asked :

Hi, I would like to ask how do I go about preparing for planning questions? I am doing pretty well in the normal papers but the 5% planning really a big handicap for myself. Im a private candidate and have never been to JC before, so im abit lost at the planning question. Hope to get some advice ! thank youuuuuuuuu :D

Louis, first of all, don't worry too much about the Planning Qn. You'll be better off focusing on polishing your remaining 95%, then the 5% whose preparation is notoriously difficult and not-worth-planning (pun intended) for.

For those of you who are still worried sick about the Planning Qn for H2 Chemistry, there are only 2 reliable sources for this (you can of course, try to buy online from ex-students, various Singapore JC notes on Planning, but note that I said *reliable* sources), namely Chan Kim Seng & Jean Tan's H2 Chemistry Planning Book "Understanding Experimental Planning for Advanced Level Chemistry: The Learner's Approach" (see my website), and Cambridge Mark Schemes on Planning Qns (I've included links to such websites here).

UltimaOnline SG

Several questions!

Inorganic Chem:

For transition metal complex right, how do you determine the coordination number. I understand that most of them are 6 but sometimes it can be 4 or 2, and i get stumped when i see the answer key. so how do i predict the coordination number?

Physical chem:

Do you mind explaining how do you predict solubility? isit as simple as polar solvents dissolve polar reagents and non polar dissolve non polar? refering to TYS 2009 P2 Q2(b) I dont understand why E values are used D:

The topic of complex ions is actually more complex (pun intended) and cannot be fully understood at A levels. At A levels, you only need to know the basic principles and concepts, and you've to memorize the coordination numbers, geometries and colors of the commonly encountered coordination complexes.

For A level purposes, 3 simple concepts are relevant here : size of metal cation (if small, ie. period 2, only 4 ligands are possible, due to lack of vacant, energetically accessible d orbitals to accommodate an expanded octet), size of ligands (eg. water ligands small, so 6 are possible ; chloride ligands large, so only 4 are possible, to avoid excessive steric strain caused by van der Waals repulsions), charge on ligands (eg. water ligands neutral, so 6 are possible ; hydroxide ligands are negatively charged, so only 4 are possible, to avoid excessive steric strain caused by anionic repulsions between ligands).

Beyond these A level concepts, University level chemistry such as the Jahn-Teller effect, will be required to explain deeper questions on coordination complexes, eg. why [Cu(NH3)4]2+ has a square planar geometry instead of tetrahedral geometry.

Solubility is also a complex issue spread over several topics in the H2 syllabus, eg. chemical bonding, thermodynamics, solubility equilibria, Group 2, electrochemistry, organic chemistry, etc. The required answer at A levels depends on the context and H2 Chem syllabus topic that the question falls under. So there's no simple way to predict solubility (simple ways are for O levels, A levels are preparation for Uni level, in which everything becomes increasingly complex*). For the particular question you mentioned, falls under electrochem : only the oxidation potentials of Mg, Zn and Al, but not Cu, are sufficiently positive for their oxidation into aqueous and soluble metal cations when reacted with H+ from HCl (aq), to be thermodynamically feasible.

* In primary school, you thought you already knew everything in the Universe that could be known. In secondary school, you begin to realize you don't everything, but you still feel confident that you know almost everything. In JC and poly, you begin to understand that you don't really know all that much, and that the Universe is more complex than you thought. In University, you're humbled by the understanding that the Universe is so infinitely complex that you, and all sentient beings on Earth, actually know almost nothing relative to the infinitely vast Universe, but we know just enough to begin to appreciate and respect its beauty and complexity.

UltimaOnline SG

I understand this :

By Hess law,

đşHsoln = đşHhyd + đşLEdissociation

Can help me check where i've gone wrong thinking in this way?? :

đşHsoln = energy released on hydration by hydrated ions Minus Energy absorbed by Ionic solid to dissociate

So đşHsoln = (-ve enthalpy change) minus (positive enthalpy change )

If i put modulus on the RHS terms, then it becomes all +ve

Help please thanks!

Your problem is you assume magnitude of hydration enthalpy always outweighs magnitude of endothermic lattice dissociation enthalpy, which is false. Either could outweigh the other.

Hence, without worrying about which magnitude outweighs which, you should simply add up the endothermic lattice dissociation enthalpy change (ie. positive value) with the exothermic hydration enthalpy change (ie. negative value) to obtain solution enthalpy change, which could be either endothermic or exothermic (ie. positive or negative value).

UltimaOnline SG

Chan Kim Seng appears in the video at 3:10 onwards :

http://video.toggle.sg/en/video/series/life-hacks/ep1/384604

UltimaOnline SG

Hi, I need help for this question. Adopted from MJC tutorial.

"Graphite is one of the allotropes of carbon. The graphite structure consists of hexagonal layers. Each carbon atom has one electron that can be delocalized parallel to the layers to make graphite a good electrical conductor.

Different atoms can be fitted in between the layers of carbon atoms to produce an usual set of âgraphite compoundsâ. When the graphite lattice contains atoms of alkali metal such as caesium, then a bronze-coloured solid is formed which has a greater conductivity than pure graphite.

In hexagonal boron nitride , the structure is composed of layers of hexagonal B3N3 ring. The boron atom utilizes all the three valence electrons to form covalent bonds with three neighbouring nitrogen atoms. Each nitrogen atom still has a lone pair of electrons which it utilizes to form a dative bond with an adjacent boron atom. Although graphite and hexagonal boron nitride have similar structures, they differ in electrical conductivity. Unlike graphite, boron nitride is an insulator.

When heated under pressure, this form of boron nitride is converted into another form which is an extremely hard solid .

(g)(i) Suggest and describe the structure adopted by the new material formed"

Giant covalent tetrahedral lattice structure similar or analogous to diamond, in which each N atom has a unipositive formal charge, and each B atom has a uninegative formal charge, because all N and B atoms have a stable octet of 4 bond pairs and 0 lone pairs (just like the C atoms in diamond).

UltimaOnline SG

Hi Ultima, i have some queries

Page 97 cs toh advanced guide

Btm of the page "C=O bond energy in CO2 is significantly stronger than average bond energies given in data booklet"

Is the reason that why not all C-H or C=O bonds are the same, because the C or O atom might be attached to another (different) atoms in different compounds?

Thanks for helping

Yes, that's right.

UltimaOnline SG

https://www.dropbox.com/s/81xap6m34negx3m/20160508_154845.jpg?dl=0

Pretentious packaging? Is the common person suppose to infer, an amino acid that is branched, is very good?

Anyway, what's so good about branched chain amino acid (BCAA)?

Yes, BCAAs are useful specifically for sports and bodybuilding purposes, to prolong stamina and reduce DOMS after workouts.

https://en.wikipedia.org/wiki/Branched-chain_amino_acid#Claims_in_Bodybuilding

http://reviews.bodybuilding.com/SciVation/Xtend/

UltimaOnline SG

May I ask a general qns? My school tutor claimed that for energy lvl diagram when the 3d and 4s orbital is filled, 3d orbital will have a higher energy than 4s orbital but when unfilled, the 4s orbital will have a higher energy than 3d orbital. Is that true?

Yes, this is true for the transition metals, as the electrons in the inner 3d orbitals will repel the electrons in the outer 4s orbitals, raising the 4s orbital to a higher energy level than 3d. Consequently, during ionization process, electrons are removed from the outer 4s orbital before electrons are removed from the inner 3d orbitals.

UltimaOnline SG

Hi ultima,

I read cs toh book page 108 and understood *why* O2(g) is a *better choice* than F2(g) as an oxidiser.

But i thought of comparing oxidising strength between F2(g) and O2(g)

I know F2(g) is the strongest oxidising agent of all known chemical compounds (or is it just the strongest, common oxidising agent?)

Question : how to determine oxidising power of O2? In the data booklet, there isn't any equation for O atom , e.g. O2 + 4e- -> 2O2- , unlike F, which has a eqn F2 + 2e- -> 2F- , with a given E°value.

Is oxidising power related/affected to electronegativity? Are there other factors?

2. Came across this "Why is F2 a stronger oxidising agent than Cl2 even though F2 has a less exothermic E.A than Cl2? "

https://www.quora.com/Why-is-F2-the-most-oxidising-agent-instead-of-Cl2-even-though-F2-has-less-electron-affinity-than-Cl2

But don't really understand fully

Why is F2 a stronger oxidizing agent compared to Cl2 even though F2 has a less exothermic EA compared to Cl2?

Oxidizing strength is primarily related to electronegativity, but secondarily to hydration enthalpy, because in the natural world, redox reactions occur mostly in the aqueous environments (ie. water being the solvent that brings the reactants together and allowing chemical reactions to occur), hence the redox potentials that are relevant are in the aqueous state.

Yes, F2 a stronger oxidizing agent than Cl2 even though F2 has a less exothermic EA compared to Cl2. First, understand why Cl2 has a more exothermic EA compared to F2 : it's because even though F is more electronegative than Cl, but the charge density of F- is too high due to small ionic radius, concordantly the inter-electron repulsions within the small F- ion are highly destabilizing. Consequently, the EA for F isnt as exothermic, as F- isn't as stable as might be imagined solely due to electronegativity (as would be the case with Cl to Cl-), as there is also a destabilizing (ie. endothermic) component to the EA process for F (but not for Cl), ie. the inter-electron repulsions of the high anionic charge density F- ion.

Next, if the EA of Cl is more exothermic compared to that of F, shouldn't Cl2 be a stronger oxidizing agent (ie. more positive reduction potential) compared to F2? Yet the Data Booklet clearly shows F2 having a more positive reduction potential compared to Cl2, and F2 is thus the stronger oxidizing agent. This is because as I mentioned earlier, we're interested in redox reactions occurring in the aqueous solvent or phase, and thus concordantly by Hess Law, reduction enthalpy = electron affinity enthalpy + hydration enthalpy ; and likewise oxidation enthalpy = ionization enthalpy + hydration enthalpy.

Because F- has a much higher anionic charge density compared to Cl-, it's hydration enthalpy is also that much more exothermic (ie. much stronger ion - permanent dipole interactions formed). And after Hess Law is applied, the overall reduction potential of F2 to F-, is still more positive compared to the reduction potential of Cl2 to Cl-, and thus in aqueous environments, F2 is still a much stronger oxidizing agent compared to Cl2.

UltimaOnline SG

Question : how to determine oxidising power of O2? In the data booklet, there isn't any equation for O atom , e.g. O2 + 4e- -> 2O2- , unlike F, which has a eqn F2 + 2e- -> 2F- , with a given E°value.

I told you before : in the natural world, these redox reactions occur mostly in aqueous environments / solvents / phases. Do you think the highly unstable O2- ion is able to exist in aqueous environment? Of course not! It will kena immediately hydrolyzed to OH-, and (depending on pH) can be further protonated to H2O.

So for the oxidizing power of O2, look in the Data Booklet for the reduction potential of O2 to either OH- or H2O (depending on pH). (Ignore obscure & rare reduction products like peroxides, super oxides, ozonides, etc, unless otherwise specified by the question).

UltimaOnline SG

Hi, is being able to solve A level questions sufficient? I am doing well on the TYS but prelim papers are ridiculously tough, isit unneccessarily tough?

Being able to solve A level questions is necessary but not necessarily sufficient. If you stick to doing A level papers only, without Prelim papers, you may or may not get A grade. Of course, just by doing Prelim papers (with or without A level papers) is by itself no guarantee that you'll get A grade either. Every year, there are dozens (sometimes over a hundred) of unfortunate students who do worse (eg. B grade) for their A levels compared to their Prelim exams (eg. A grade).

My advice would be to focus on the style of A level papers, but at the same time broaden your understanding by learning from the "ridiculously" difficult questions and mark schemes from the JCs. Unless of course, you already find A level papers difficult, in which case confusing yourself with the even tougher Prelim papers would be counter-productive.

UltimaOnline SG

A BedokFunland JC original H2 Chemistry question

Source : https://pixabay.com/en/photos/rain%20forest/

A drop of water hangs from a leaf. Over the next few hours, depending on environmental conditions, the drop of water either increases or decreases in size (ie. mass and volume). Identify, describe and explain 2 of these environmental conditions (excluding wind and rain). Your answer must include all of the following terms : "Le Chatelier's principle", "enthalpy", "entropy" and "Gibbs free energy".

UltimaOnline SG

A classmate of mine asked a qn in class in which the tcher was stumped : why electron shell closer to nucleus, has lower energy level?

For a simplified version (which will suffice for A level purposes), you can think of it as : to maintain your orbit further from the nucleus, you (as an electron in the outer electron shells) naturally require more energy to sustain your further orbit and resist the electrostatic pull of the positively charged nucleus.

Going any deeper like the classic scientific enigmas of "Why don't electrons fall into the nucleus?", and "Why don't the protons in the nucleus repel each other and cause the nucleus to fall apart?" (also see Cordus physics) that goes significantly beyond A levels (so we we're *not* going discuss it here, and it's a safe bet that neither your school teacher nor your private tutor will entertain you on this), and at University levels, physicists and chemists have come up with a few different theories or explanations or ways of looking at these enigmas. You may come across some of these ideas in the future if you choose to study Physics or Chemistry in the University. For now, if you're still curious (though you should of course focus on your A levels), you can google out some preliminary reading for your entertainment and satisfaction.

UltimaOnline SG

Why is 4-aminobenzenesulfonic acid a much weaker acid than benzenesulfonic acid?

2 separate points :

The amino group is electron-donating by resonance, hence destabilizing the negatively charged sulfonate group in the conjugate base. In addition, 4-aminobenzenesulfonic acid exists as a zwitterion in aqueous and solid states, and because ammonium groups (ie. conjugate acid of amines) are very weak acids with small Ka values, hence zwitterionic 4-aminobenzenesulfonic acid is a much weaker acid than non-zwitterionic benzenesulfonic acid.

UltimaOnline SG

Hi Nelexus, glad to hear you chose to read Chemistry in the University. Btw, while both NUS and NTU Chemistry are (approximately) equally good, the Science bookstore at NUS is better stocked compared to the Science bookstore at NTU. Fyi.

Generally (for everyone reading this), please do not PM me Chemistry questions (PM me only if you're interested in joining my BedokFunland JC tuition), but post Chemistry questions on the forum for discussion so that everyone can participate, learn and benefit together.

Next, for Uni level Chemistry questions, please do not post them here, as my professional interest (ie. my job) is specifically A level Chemistry.

But no worries (this suggestion is for all JC graduate students taking Chemistry or Chemistry related courses in the University), I'll recommend you an excellent Chemistry forum (probably the best on the internet) where you can get help for all your Chemistry questions, all the way to Masters, PhDs and professional Chemical industry questions.

Presenting.... the Chemical Forums (yeah that's the name, strictly functional and down-to-business, nothing flowery or fanciful).

http://www.chemicalforums.com/

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Nelexus PMed me :

Hi Ultima, I have a few Chemistry questions that I would like to clarify with you. Am a graduated JC student but intending to read chemistry in uni. So I have been reading up on Chemistry before Uni starts. Some of the questions I am asking will be covering advance concepts far beyond the h2 chem syllabus.

1. Resonance Vs Conjugation

What is the difference between resonance and conjugation? From what I've read online, a continuous overlapping of adjacent p orbitals forms a conjugated system. The overlapping of p orbitals allows electrons to "move" in the conjugated system, resulting in delocalisation of electrons in that conjugated system. This is illustrated by the diagram below I've picked up from the internet.

Therefore, from this scenario above, conjugation allows resonance to occur.

Now consider the conjugate base of benzoic acid and pyruvic acid. I believe they are all conjugated system. The thing is, for the benzoate ion, the electrons on oxygen in the carboxyl group cannot be delocalised into the benzene ring no matter how I draw, thus no resonance structure forms arises for the benzoate ion. Likewise for the pyruvic acid, the electrons cannot be delocalised into the ketone group. So while they are conjugated, there are no resonance structure for . This sounds abit strange, considering that for a conjugated system with p orbitals overlapping with each other, electrons can "move" around the system, yet this is not possible for the electrons on the oxygen atoms of benzoate and pyruvate ion. My point is ilustrated below (pardon the big image)

For this amine below that I've drawn, there's a conjugated system and the delocalisation of electrons can happen. My confusion is tha tlogically, conjugation which is the overlapping of p orbitals, should allow electrons to delocalise, and thus entails resonance. Yet this is not the case for some system. Why?

2. pH of equivalenve points of polyprotic acids

The diagram below shows the titration curve of sulfurous acid with sodium hyroxide.

For he second equivalence point where only SO3 2- is present, the equivalence point can simply be obtained by obtaining the Kb of SO3 2-, and then find [OH-] using Kb = [OH-]^2/ [SO3 2-] am I right?

Now the tricky part is the first equivalence point. The species present in the first equivalence point is HSO3- which is an amphoteric species. Meaning HSO3- can act both as a base and an acid as follow

HSO3 - + H2O --> SO3 2- + H3O+ Ka
HSO3 - + H2O --> H2SO3 + OH- Kb

Is it possible to calculate the pH of the first equivalence point?

I understand that for amino acids, theres this formula pH = 1/2 (pka1+pka2) which is used for calculating the pI and the so called equivalence point as shown below.

Logically amino acids are polyprotic acids as well, so would the same formula work? Or do I have to physically inspect whether Ka or Kb is larger, and then proceed to using the method I used for finding the second equiv point?

3. Amino Acids

For amino acids, why is it that amino acids always exist in its protonated form? For example, in the titration of amino acids, the major species in the beginning is always NH3+ -R - COOH at a low pH. Should it be the case that say in the laboratory, the amino acids you work with are powders in the form of zwitterions NH3+ - COO- and then when you add it into water, the carboxylate group gets protonated and a basic solution is formed. Yet this is not the case for the examples I've seen on the internet, where the amino acids exist as NH3+ - R - COOH at pH < 7?

Secondly, to say the isoelectronic point of an amino acids, does it mean to have equal concentrations of both the cation and anion, or the maximum concentration of the zwitterions? Some textbooks obtain the formula of pH = 1/2 (pka1+pka2) using the assumption that [cation] = [anion], which I think does not make sense.

I quote from Essentials of Organic Chemistry by Dewick

"The pH at which the concentration of the zwitterion is a maximum is equal to the isoelectric point pI, strictly that pH at which the concentrations of cationic and anionic forms of the amino acid are equal. With a simple amino acid, this is the mean of the two pKa.

in the derivation, how is the statement " cation = anion " true? Doesn't the dissociation of amino acids occur stepwise, meaning cation --> pka1 where cation = zwitterion --> equivalence where only zwitterion --> pka2 where zwitterion = anion --> 2nd equivalence where there is only anion.

I understand that these questions are far beyond the h2 chem syallbus which you may not be obliged to answer, but I really hope you do. I've scoured countless materials and resources on the internet to find no answer. Thank you in advance!

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UltimaOnline here again. I will give brief guidance & comments regarding Nelexus' questions. For deeper discussion, you're advised to discuss on the Chemical Forums.

Conjugation (not to be confused with hyperconjugation) *allows* for delocalization of electrons to occur via resonance.

In the benzoate ion, the negative formal charge on the O atom, cannot be delocalized by resonance into the benzene ring, as evidenced from the impossibility of drawing the required mechanism to delocalize the negative formal charge into the benzene ring (since resonance cannot involve the breaking or forming of sigma bonds).

However, note that in the conjugate acid form, the pi electrons in the benzene ring can indeed be delocalized by resonance to the COOH group, ie. the COOH group is electron-withdrawing from the benzene ring by resonance, and is hence deactivating and meta directing. When deprotonated however, the COO- group, being electron-rich, ceases to be electron-withdrawing by resonance from the benzene ring, as the high electron charge density of the dinegatively charged COO2- resonance contributor is excessively destabilizing.

The real reason why the benzene ring can help to stabilize the benzoate ion conjugate base (thus making benzoic acid a stronger acid than say, ethanoic or propanoic acid), is because the sp2 C atoms of the benzene ring have a higher % s orbital character, and is thus electron-withdrawing by induction (but not by resonance).

Yes, the formula pH = 1/2 (pKa1 + pKa2) or pH = 1/2 (pKa2 + pKa3), is applicable for all solutions containing only an amphiprotic species, eg. at equivalence point for polyprotic or multiprotic acids, such as zwitterionic amino acids.

A solution at its isoelectric point contains the highest possible molarity of its zwitterionic form (ie. no net ionic charge), and very low and equal molarities of both the cationic (ie. conjugate acid) and anionic forms (ie. conjugate base) forms. The molarities of the ionic forms at pI can be calculated using the relevant Ka values.

Amino acids can (depending on the exact amino acid's R group) exist as dinegative anionic, uninegative anionic, neutral zwitterionic, unipositive cationic, and dipositive cationic. So if you were carry out a titration of an amino acid, you would usually start with either the fully deprotonated anionic form (if you're adding acid from burette), or the fully protonated cationic form (if you're adding alkali from burette).

UltimaOnline SG

I don't have the photo of the qn, but i wrote it down, so I'll type it out

Qn : write the reaction for titration of 100ml of 0.100M anilinium bromide (aminobenzene • HBr) with 0.100M NaOH. Sketch the titration curve for points Vb=0, 0.1Ve, 0.5Ve, 0.9Ve, Ve, 1.2Ve

Species present at equivalence pt is phenylamine.

Oh use equation of dissociation of phenylamine ?

Why cannot use amphiprotic formula in this case?since phenylamine is the cj base of phenylammonium cation

So not every time at equivalence point use amphiprotic formula, is it?

Thanks for those guidance in thought process that you've written down !!

Use amphiprotic formula only when the species present is amphiprotic. At the various equivalence points for acid-base titrations, the species present could be either only acidic (eg. H3PO4), only basic (eg. PO4 3-), or amphiprotic (eg. H2PO4- or HPO4 2-).

Instead of blindly memorizing formulae without understanding (like the majority of pitiful Singapore JC students), you should intelligently understand what's going on, intelligently identify on a case-by-case basis what's the nature of the species present, and hence intelligently apply the relevant formula.

And btw, phenylamine is a base, bases do not undergo dissociation (as erroneously taught in Singapore JCs), bases undergo hydrolysis.

UltimaOnline SG

Hi ultima , i would like to ask,

I have written down the kb value from the soln book,

I am confused by this part

In part i), the ka value is for phenylammonium cation, correct?

For the part of vol base added = vol equivalence, i used kw/ka(phenylammonium) to get kb(phenylamine)

So, when i use kw/ka is get kb of the cj base ah??

Damnit, all the while i thought its ka and kb of the same molecule !

When i apply kw/ka to find kb, is the kb the kb of same molecule or kb of its cj base!

Ka of HA x Kb of A- = Kw

Ka of BH+ x Kb of B = Kw

Ka of conjugate acid x Kb of conjugate base = Kw

UltimaOnline SG

Why is Mno4- a stronger oxidising agent compared to Cr2O72-?

Because of a complex combination of reasons, which you needn't worry about for A levels. Most relevantly, the oxidation state of the heteroatom in the oxidizing agent (ie. +7 versus +6), but also the electronegativity, the charge density, the electron affinity enthalpy, hydration enthalpy, electron configuration, reaction mechanism, etc.

So when you explore these issues at Uni level, you'll realize there are no longer any short, simple answers (those are the O level days of naivety), but a beautifully complex balance of separate (and oftentimes opposing factors). A levels is a transition state or intermediate between naively simplistic O levels, and beautifully complex Uni levels.

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