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

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    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
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    • Originally posted by Flying grenade:

      page 102 organic made easy

      top of page

      what does he mean by conjugated ring exhibits a stabilisation stronger than conjugation alone?

      does he meant that/ he is comparing conjugation between 2 double bond , or 1 double-single-double bond, compared to rings with multiple alternating single and double bonds?

      He means that as compared to conjugation in straight chain, complete conjugation throughout a ring (with same no. of double bonds in the resonance contributors) confers greater stability (this is known as aromaticity). Why? Count the no. of partial double bonds in the resonance hybrid for straight chain vs ring.

    • Originally posted by Flying grenade:

      yet another factor favouring hydrolysis of SiCl4 greater than CCl4  ?

      delta +ve on  Si greater than C. bond str Si-Cl > C-Cl. but SiCl4 much higher propensity to be hydrolysed 

      Minor contributive point. Doesn't make it to the top 3 reasons.

      And instead of overthinking and obsessing over conceptual stuff like these (which the rest of the Singapore JC cohort just memorize and move on, efficiently albeit blindly ; your desire to focus more deeply on understanding is commendable and serves you well up to a certain easy-to-accidentally-pass point, beyond which being anally obsessive is like missing the forest for the trees, allowing you to win the battle but lose the war), and you J2 students really ought to be practicing as many Prelim papers as you can right now. You'll be far better off (as far as your A level results and hence the quality of the rest of your life are concerned) focusing on doing and asking about Pelim paper qns here, instead of OCDing about conceptual ideas too much.

    • Originally posted by Flying grenade:

      2008 p2 qn 1

      noradrenaline qn

      qn : draw the pdt when noradrenaline reacts with bromine dissolved in a suitable solvent 

      super open to interpretation? 

      if Br2 dissolve in water, will have multi sub

      if Br2 dissolve in CCl4, will have mono sub

      pls take a look at pg 233 of organic made easy book, therein writes Br2 in CCl4 will only have mono substitution. Thanks ultima.

      If you're a BedokFunland JC student, you'll be exam-smart enough to write out both answers, with qualifications, ie. "If this solvent is used, this product is obtained ; if that solvent is used, that product is obtained". Cambridge examiner will exclaim, "Excellent student! Must be BedokFunland JC!"

    • Originally posted by gohby:

      Hi UltimaOnline,


      I've a few questions here to consult you.


      Q1: IJC/09/P1/39


      In which sequence are the compounds listed in order of increasing ease of hydrolysis?


      Choice 3 (chlorobenzene, chloromethane, chloroethane) is not a correct choice.


      Remarks: In ascertaining the ease of hydrolysis, I'll first have to see the C-X bond strength. Since it's the same in this choice, I'll have to consider either (I) Resonance within the reactant and (II) stability of the intermediate products.


      In this case, shouldn't 3 also be correct? What's the difference in the electron donating effect between the methyl and ethyl substituent?



      Q2: In relation to VJC/09/P1/29:


      Remarks: How do I establish the relative acidity of an alcohol (CH3CH2OH) to that of an acidic salt (CH3CH2NH3+Cl-)?


      Q3: VJC/09/P1/23


      Remarks: It must have been my blind spot, but why is A not an acceptable answer as well?


      Thank you very much Ultima! 

      Hi Gohby, no prob.

      Q1. The diff btwn Me and Et groups, is the stability (and hence Ea required) of the intermediate (if SN1) or transition state (if SN2). Et group is more strongly electron-donating by induction, which can better stabilize the intermediate or transition state.

      Q2. Acidic salts (notice the adjective?) are always more acidic than alcohol (which is considered neutral, and even less acidic than neutral water).

      Q3. A is the *only* acceptable answer.

    • Originally posted by Flying grenade:

      i thought need to have NaOH (aq) warm, to react with chloroalkane before there will be free Cl- to react with Ag+ in AgCl? is that a printing error, or how can the little amount of OH- present in AgNO3(aq) react with the chloroalkane to produce AgCl ah?

      OH- being a stronger nucleophile than H2O increases the rate of reaction (hydrolysis then ionic precipitation). But the reaction will still proceed (albeit a lot more slowly) at rtp and at neutral pH. The exothermic ionic precipitation shifts the position of equilibrium over to the RHS and makes the reaction thermodynamically feasible. But the kinetics is influenced by temperature and strength of nucleophile.

    • Originally posted by Rapidestlime:

      Hi ultimaonline.

      I recently did a question on why pka of peroxyethanoic acid(as shown in picture) is more than pka of ethanoic acid. my school answer is :The negative charge on oxygen in CH3CO2-
       is very effectively dispersed through delocalisation within the ‒COO- π system. However, the anion CH3CO3- is not able to delocalise its electrons
      over the C=O group. Hence, CH3CO2‒  is more stable than CH3CO3-.

      May I ask if it is correct to say that since the extra o atom in the peroxyethanoic acid  has p orbital, resonance of pi electron cloud can also form in the ‒COOO- group in the conjugate base of peroxyethanoic acid   , the lone pair of the extra o atom increases electron density of pi electron cloud, hence intensifies the negative charge of conjugate base , making it more unstable, extent of dissociation of peroxyethanoic acid  is lesser, ka is smaller, pka is bigger. correct me if im wrong in my reasoning. thank you.

       Image result for peroxyethanoic acid

      Hi Rapidestlime, to fully understand the answer to this question, why the negative formal charge on the terminal O atom in COOO- cannot be delocalized by resonance, you have to go beyond just simply the idea of sideways overlapping of p orbitals making resonance delocalization of pi electrons and hence delocalization of a formal charge possible.

      You actually have to draw out the resonance contributors in each case, ie. COO- vs COOO-, to explain why resonance delocalization of the negative formal charge is not possible in COOO-. Because unless it involves complete delocalization around a ring, ie. aromaticity, otherwise, lone pairs (in unhybridized p orbitals) cannot delocalize into being lone pairs on other atoms by resonance, lone pairs can only at most delocalize into pi bond pairs.

      In the case of COOO-, the central O atom's lone pair occupying a unhybridized p orbital can indeed delocalize by resonance as a pi bond pair with the acyl sp2 C atom, which can indeed accept the pi bond pair (which happens in the resonance stabilized COO- conjugate base) without having to expand its octet (which it can't do so, being in period 2), because the C=O pi bond can simultaneously delocalize into a lone pair on the O atom.

      However, the lone pair on the terminal O atom bearing the negative formal charge in the COOO- conjugate base, *cannot* at the same time delocalize by resonance to form a pi bond with the central O atom in any resonance contributor (ie. regardless of whether the central O atom has delocalized it's lone pair unto the acyl C atom or not), because the central O atom is unable to accommodate an expanded octet configuration (forcing the central O atom to accept a pi bond from the terminal O- atom will result in the central O atom having either 2 lone pairs + 3 bond pairs in a resonance contributor, or 1 lone pair + 4 bond pairs in another resonance contributor, which cannot be done because period 2 elements only have 1 x 2s orbital and 3 x 2p orbitals, and do not have vacant, energetically accessible 3d orbitals to accommodate an expanded octet).

      If (like all other H2 students) you've no experience with drawing resonance contributors and hybrids, the above will be tough for you to understand. It's best if you get your school teacher or private tutor to draw them out for you.

      Of course, Cambridge will be delighted if you are able to do so (ie. draw out the resonance contributors and hybrids) as part of your explanation, but at the very least, giving your school's (ie. Singapore JCs') simpler answer will be sufficient to obtain the 1 mark for this question. No worries, when the question goes beyond the H2 syllabus and it's not fair to everyone*, it's still fair : it's a bell-curve afterall.

      *Of course, H3 Chem, Olympiad Chem, and BedokFunland JC students will know how to draw out resonance contributors and hybrids. It's still fair because they've put in additional effort to understand Chem deeper. Why else join H3 Chem, Olympiad Chem, and BedokFunland JC?

      Edited by UltimaOnline 22 Aug `16, 4:21PM
    • Originally posted by Rapidestlime:

      Hi ultimaonline.

      Also, is the answer for your cadmium sulfide ppt qns u posted on your website is 5.12x10^(59)mol^(-5)dm^(15)? thank you.

      Very good, Rapidestlime. That is correct.

    • Originally posted by Ephemeral:

      RI 2015 Prelim

      Paper 1

      Q12. How do I infer option B? 

      Q36. Why is option 2 wrong? Cl is reduced frm 0 in cl2 to -1 in cl- and oxidised to +5 in cl03-, so technically it is reduced right? 

      Paper 2

      Q4bi. Why is the rate determining step not the step involving the formation of nitronium ion? Generation of nitronium ion involves 2 neutral molecules while the next step involves a positively charged ion so isn't the reaction faster in second step? 

      2015 RJC P1 Q12. Vol of vessel = 2dm3, hence 0.1mol = 0.05M. Concordantly, at 4 mins, molarity increases from 0.05 to 0.10.

      2015 RJC P1 Q36. Yes, but "reduced" implies "reduced only" while here Cl is "disproportionated", ie. "both reduced and oxidized". Blame RJC teacher if you disagree, but even Cambridge often 'chut' ambiguous stunts like these, and Singapore JC teachers often disagree on the best answer.

      2015 RJC P2 Q4bi. Because even temporarily losing aromaticity is a hugely destabilizing, reluctant step.

      2015 HCI P1 Q25. The MCQ asks for "best describes". NO2 isn't itself a greenhouse gas (that would be N2O, not NO2), it only contributes to the formation of tropospheric ozone which is a greenhouse gas (in contrast, the beneficial ozone layer refers to stratospheric ozone). Instead, NO2 pollutant is far more damaging to humans and the environment as a major contributor to acid rain.

    • Originally posted by ACA-Milkshake:

      2012 TYS Paper 2 

      Qns 5b(i): The answer that my school gave was that "All molecules present in the same chirality, that us, only one enantiomer is present." But can I just say that the lactic acid contains chiral carbons that's why it causes the rotation of th eplane polarised light?

      Your school teacher's answer and your own answer are both incomplete. You also need to define what a chiral carbon is, and/or what an optically active molecule entails, as part of your answer.

    • Originally posted by Fxwhy:

      MJC 2015 Prelims Paper 2 

      1e) For the safety hazard, can I write chemical reagents like ethanal are corrosive when it is in contact with the skin and eyes? 

      3a) Where did the ans get 2/4 from? I am also quite weak at approaching this type of qns where do you start?

      3d) Why doesnt the ans include bond length? How do you know when the ans requires you the write bond length or bond energy? 

      Will I be penalised if I write PCl5 at rtp instead of anhydrous PCl5 at rtp?

      Q1e) Ethanal is toxic and carcinogenic, not corrosive.

      Q3a) Instead of blindly memorizing and applying a mathematical formula, use common sense. For a uninegatively charged species : when mass is 37g, deflection is merely -1 degrees. Hence when mass is 1g (ie. much lighter), deflection should concordantly be -37 degrees (ie. larger magnitude deflected). Since mass of He is 4g, hence deflection should be -9.25 degrees (if uninegatively charged). Since He nuclei is dipositively charged, hence deflection should be +18.5 degrees.

      Q3d) Bond length here is only indirectly relevant (since bond length is inversely proportional to bond strength, ceteris paribus). It is bond strength which is directly relevant. Hence, there is no need to (ie. no marks awarded for) mentioning bond length.

      Q4ai) It's ok if you don't specify "anhydrous", as long as you do *NOT* specify aqueous (aq) (which would be marked wrong).

      Same for all reagents which need to be anhydrous, eg. LiAlH4 reducing agent, acyl halides for nucleophilic acyl substitution, Grignard nucleophilic & reducing reagents, AlCl3 or FeBr3 Lewis acid catalysts for halogenation via electrophilic aromatic substitution and for Friedel-Crafts alkylation & acylation, etc.

      Edited by UltimaOnline 25 Aug `16, 12:19AM
    • _

      Edited by UltimaOnline 13 Oct `17, 2:14AM
    • Originally posted by Flying grenade:

      pg 57 and 58 made easy inorganic book

      hydrolysis of both Mg2+ and Al3+ produces 1 mol of HCl.

      However, pH of soln containing Al3+ is lower than Mg2+. means concentration of H+ is higher for the hydrolysis of Al3+ compared to Mg2+.

      so what does 1 mol entails? it doesn't really mean equal amounts of a particular substance, even if two seperate solns have 1 mol of that particular substance? always must look at position of equilibrium, which one lies further to the right and, which equilibrium has its forward reaction of a greater extent to the right? 


      one way is to use ICE table?


      You cockanaden! If hydrolysis of both Mg2+ and Al3+ produces 1 mol of HCl, wouldn't the pH be the same?!? Obviously the extent of hydrolysis is significantly greater for AlCl3, and lesser for MgCl2, hence the different pH values. Cambridge will require you to explain why, and the balanced equations will (of course) be required.

      If a single source or book confuses you, cross-reference with other sources and books, to come to a better understanding of the subject matter. If you did so in this case, and you *still* came to the cockanaden conclusion that "hydrolysis of both Mg2+ and Al3+ produces 1 mol of HCl", then you might as well voluntarily withdraw from the 2016 A level exam, save yourself the pain of waiting for your U grade results in March 2017.

    • Originally posted by supercat:

      Hi, sorry to disturb again. For A level 09 P1 Q1, may I clarify if a cyclic structure is part of an "unbranched chain"? I drew a cyclohexane with -OH attached to all carbon atoms, which also fulfils the C6H12O6 formula. 

      Cambridge was unfortunately somewhat ambiguous in this MCQ. Instead of "unbranched", Cambridge meant (and should have used the more correct term) "acyclic". As such, this TYS question required you to draw the acyclic isomeric form (below left). Being acyclic with 1 degree of unsaturation, you can deduce the double bond is C=O, since the question specified no C=C present. Hence, the answer to this MCQ is 5 mol of Na, 2.5 mol of H2.


      Source : http://2012books.lardbucket.org/books/introduction-to-chemistry-general-organic-and-biological/s19-04-cyclic-structures-of-monosacch.html

    • Originally posted by Flying grenade:


      can this molecule exist?

      is it unstable, and may decompose? what will it decompose into?



      other miscellaneous thoughts 

      not industrially useful hence dont have molecule's name or name of the class of cpds

      difficult to synthesise

      molecule not stable 


      Enols exist in equilibrium with their prototropic tautomer, ketones / aldehydes. Draw the mechanism (either neutral pH or acid catalysed, up to you) to prototropically tautomerize ethenol to its carbonyl tautomer, ethanal.

      As a challenging A grade question (which I will not provide the answer here, go ask your school teacher or private tutor), Cambridge may ask you to explain where the position of equilibrium lies, and (more challengingly) exactly why.

    • A BedokFunland JC H2 Chemistry Challenge Question

      Q1. Why does the position of equilibrium lie on the side indicated?
      Q2. In organic solvents, the cis-isomer is favored, whereas in water the trans-isomer predominates. Explain.
      (I won't reveal the answers here and spoil the fun, you can go ask your school teacher or private tutor.)



      Edited by UltimaOnline 30 Aug `16, 7:32PM
    • Originally posted by Nikkilyx:


      Hi I need your help again.




      For f, since deltaH = deltaH2 - deltaH3 + deltaH(H2O), am i correct to say that residual heat from experiment 1 (hcl+naoh) causes the temperature change for experiment 2 (mg+hcl) to be higher than expected, making the value more exothermic. Furthermore, heat lost to the surroundings (mgo+hcl) causes the temperature change for experiment 3 to be lower than expected, making the value less exothermic. I feel that my answers are contradicting themselves as there should be residual heat from both experiment 1 and 2 causing the enthalpy change for 3 to be incorrect.


      For g, i have absolutely no clue on how to approach the question.

      f) The discrepancy is due to the differing state symbol of water. In the experiment, it's H2O(g). In the calculations, it's H2O(l).

      g) Since the enthalpy change of hydration applies equally for both alternative intermediate steps in our enthalpy cycle generating [Mg(H2O)n]2+(aq), they in effect cancel out (when applying Hess' Law) and hence does not affect our enthalpy change calculations of Mg to MgO.

    • Originally posted by Flying grenade:


      i worked out d.o.u. of benzene =4 .

      D.O.U.(benzene) = (14-6)/2 =4

      Look at the *structure* (not the Singapore JC guniang-sissy version with the circle in the middle, ie. resonance hybrid ; but use the Uni level garang-gungho version with alternating double and single bonds, ie. resonance contributor) of benzene, and explain exactly why the d.o.u. for each benzene ring is 4.

      Originally posted by Flying grenade:

      anyway im more stress abt this qn now : for the A level 2015 p3 qn 1 question, we calculated dou is 3. but the molecule only gt 2 double bonds. one double bond = 1 dou

      what shld i do?? what shld i learn from 3 dou? teach me ultima T-T

      Figure this out yourself. What's the definition of D.O.U.???

      Super Coconaden Hint for Super Coconaden pple : obviously it can't be just double bonds which affect degree of unsaturation!!! *Sheeeesh*


      Originally posted by Flying grenade:

      can't find definition of d.o.u. online :'(

      Wow, Wikipedia must have moved to the dark deep web, no wonder you can't find it.


      Even if Wikipedia or other sites don't directly spoonfeed you the definition, Cambridge can (in a challenging A grade H2 Chem A level exam question) give you formula for degree of unsaturation, and ask you to concordantly suggest a definition for the term.

      Edited by UltimaOnline 31 Aug `16, 8:39PM
    • Originally posted by gohby:

      Hello UltimaOnline,

      Regarding Q3 (VJC/09/P1/23), What's wrong with choice B in fulfilling all the conditions? It reacts with bromine because of the C=C bond, it has 2 Cl substituents which will become 2 alcohol groups in the presence of hot alkaline and it reacts with PCl5 because of the carboxylic acid.


      Which site gets deprotonated first? I'd have thought that in a neutralisation reaction with a base the most acidic component will be deprotonated first (i.e. the NH3+), followed by the COOH not connected to the CH2, followed by the last COOH. Is my understanding correct? (the answer scheme suggests otherwise)

      Thank you :)

      Hi Gohby,

      VJC qn : geminol diols spontaneously dehydrate into into carbonyl compounds with thermodynamically favourable positive entropy change. H3 syllabus stuff, but has been asked many times in H2 Prelim papers.

      HCJC qn : yes you're right that the most acidic proton is abstracted away first. Most acidic to least : alpha COOH, R group COOH, alpha NH3+, R group NH3+.

      No prob :)

      Edited by UltimaOnline 03 Sep `16, 8:03PM
    • Originally posted by supercat:

      Hi, I have a question regarding NJC 2013 Prelim P1 Q14.

      How to derive half life? Since it's a first order reaction, I know k = (ln2)/half-life.

      I know that 10 mins into reaction, Pa + Pb = 600Pa,

      And at the end of reaction, Pa + Pb = 400Pa.

      What should I do with the data given?

      At end of reaction, partial pressure of B is 400 Pa. Hence initial partial pressure of A is 800 Pa. Hence, 1 half-life into the reaction, partial pressure of A is 400 Pa, and B is 200 Pa. That's a total pressure of 600 Pa. Hence, each half-life must be 10 min. Since half-life = ln 2 / k, hence k = ln 2 / half-life, which is calculated out to be 0.0693 min-1.

      For A level purposes, such questions will be kept mathematically simple (such as in this case), to avoid necessitating the use of calculus and integrated rate laws.

      A possible (but still kept mathematically simple) extension of the above question would be : modifying the total pressure after 10 min to 500 Pa (instead of 600 Pa), then what would the half-life be?

      Edited by UltimaOnline 04 Sep `16, 9:51PM
    • Originally posted by Flying grenade:

      for 2015 chem olympiad qn 3 on Ritalin, 

      for the structure of compounds,

      for cpd F to ritalin, why isn't benzene ring reduced? and the imine reduce to secondary amine? 

      Aromaticity resists hydrogenation. For A level purposes, unless hinted by the question with extraordinarily high temperatures and pressures, you are to take it that the benzene ring does not get hydrogenated under normal reaction conditions required to hydrogenate or reduce other organic functional groups.

      Originally posted by Flying grenade:

      from Internet, benzene ring gets hydrogenated to form cyclohexane, under similar conditions 

      but you're the most trustworthy source.

      Okay, thank you Ultima for your kind replies !

      Read your CS Toh Adv Study Guide, compare the exact temperature and pressure required to hydrogenate benzene vs alkene vs alkyne vs ketone vs aldehyde vs nitrile vs amide etc. Don't need to post here, just read silently for yourself.

      Edited by UltimaOnline 05 Sep `16, 12:45AM
    • Originally posted by Flying grenade:

      what's the geometry of methyl radical?

      trigonal pyramidal or quasi-trigonal planar? 

      for A levels, we draw trigonal pyramidal and write bond angle more than 109.5° or between 109.5 and 107.5° ?


      If CF3 radical, then trigonal pyramidal. If CH3 radical, then trigonal planar. Reasons for the difference are beyond the H2 syllabus (a hint for A grade students : ask yourself what possible thermodynamic advantage could a CH3 radical gain by adopting sp2 hybridization, versus what possible thermodynamic advantage could a CF3 radical gain by adopting sp3 hybridization, taking into consideration the contrasting natures of s and p orbitals), go ask your school teacher or private tutor if you're interested.

      Cambridge is reasonable. As long as you give chemically reasonable justifications for your answer, Cambridge (in most cases) will accept a variety of reasonable answers (eg. a range for bond angles, explanation & justification for either geometry for either radical), especially for beyond-syllabus questions such as these.

    • Originally posted by Flying grenade:



      why option B cannot? carboxylate salt not basic enough, cannot abstract H+ from water?

      Most carboxylate salts are basic enough, but here the 3 Cl atoms are collectively strongly electron-withdrawing by induction, resulting in the CCl3COO- ion being too weak a base to undergo significant hydrolysis.

    • Originally posted by Flying grenade:


      can resonance occur in the straight chain of the molecule(and therefore throughout the whole molecule including the benzene ring) ? the region indicated by "z" in the qn

      if yes, then that molecule shown in the qn, is one of the many resonance contributors? 

      Correct. This is a really lousy question set by Cambridge. The question should specify "sigma bonds", because all 4 bonds specified actually have partial double bond character in the molecule's resonance hybrid, ie. sp2-sp2 sigma bonding, with partial p-p pi bonding for all 4 bonds.

    • Originally posted by Flying grenade:

      2013 paper 2 qn 5bi

      i dont understand how to get the primary structure of cys-tyr-ile-gln-asn-cys-pro-leu-gly

      why cannot(or can?) the other way round , gly-leu-pro-cys-asn-gln-ile-tyr-cys ?

      i understand by convention, N terminal is on the left, C terminal is on the right, but i can't tell/dk how use this convention here


      part bii)

      both phenol(at tyr there) and amine (at cys there) can be deprotonated and protonated at pH 7?

      The trickiness of this question (other than the disulfide bond / linkage / bridge), is that the last amino acid residue gly's COOH terminus has been acylated (ie. biochemically catalyzed nucleophilic acyl substitution, addition-elimination, condensation) with NH3 to generate a terminal amide, which Cambridge deliberately employed to confuse students as to whether or not to begin with Gly or Cys as the N terminus.

      This is known as post-translational modification of proteins (H2 Biology). Since NH3 isn't an amino acid, the C terminus remains as Gly, and hence the only correct acceptable answer is to begin with the correct N terminus, ie. Cys.

      Yes, Cambridge will accept either phenol or amine as the answer. Don't be confused : phenol and amine exist in equilibrium with phenoxide and ammonium respectively, it doesn't mean at pH 7, phenol is completely deprotonated (it's not), or that amine is completely protonated (it's not). That's why Ka and Kb values exist, to describe the position of equilibrium and extent of hydrolysis for acidic and basic groups.

      Edited by UltimaOnline 07 Sep `16, 5:11PM
    • Originally posted by Flying grenade:

      for the titration using KMno4 , for example, KMno4 with SO2 soln in 2013/p3/2cii , for detection of end point, is the observation from purple to colorless after adding KMno4 dropwise, better than, observing the first permanent pink color after adding one excess drop of KMno4 to a colorless soln? 

      [1 mark] ~ Titrate with KMnO4 from the burette

      [1 mark] ~
      Until the first permanent pink colour
      Until colour change from colourless to pink

      [1 mark] ~ Repeat until two titres are concordant within 0.1 cm3

      Originally posted by Flying grenade:

      yes i know that..

      but my doubt is why must drop 1 more kmno4 to turn it from colorless to pink

      wouldn't when kmno4 decolorises , i.e. purple to colorless, suffice, for detection of end pt

      Sekali it's not totally colorless, but only looks colorless coz it's a dilute solution, then how?

      Originally posted by Flying grenade:

      err.. so do we take Volume of KMnO4 to turn soln into a permanent pink color, minus 1cm3 ? since that 1 extra drop is in excess

      No. Becoz in titration, no matter how many drops (and each drop isn't exactly 1cm3 in the first place) you use to reach endpoint, the moles of titrant used will always technically be either slightly before, or slightly after, the equivalence point anyway.

      So if you minus 1 drop, your moles of titrant used will be slightly before equivalence point. If you don't minus 1 drop, your moles of titrant used will be slightly after equivalence point. Since either way it'll be slightly inaccurate (can't be helped), so it makes more sense to just use the titre volume to reach endpoint, without further complicating errors by minusing away 1 drop.

      Now you understand why we call this 'endpoint' instead of 'equivalence point'? Endpoint merely approximates equivalence point.

      Edited by UltimaOnline 07 Sep `16, 8:07PM
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