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

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    • 'A' Level Qn.

       

      What type of reaction is this?

      ROH + RCOOH --> RCOOR + 2HO

       

       

      Solution :

       

      The exam-smart candidate would write all 4 of the following answers, since all of them describe different aspects of the reaction, and are all relevant and correct.

       

      When alcohols react with carboxylic acids with a trace amount of concentrated sulfuric(VI) acid under heat or reflux, this reaction is a

       

      #1 - Esterification reaction, because an ester is generated as the main organic product.

       

      #2 - Condensation reaction (NOT a dehydration reaction) because water is eliminated to join two smaller molecules (the alcohol and carboxylic acid) together into a larger molecule (the ester)

       

      #3 - Nucleophilic acyl substitution (NOT just 'nucleophilic substitution', because those two words alone imply nucleophilic aliphatic substitution, whose mechanism is SN1 or SN2), because nucleophilic substitution is occurring on an acyl group.

       

      #4 - Addition-Elimination reaction, the mechanism by which nucleophilic acyl substitution occurs. The nucleophile is added first (in this case the nucleophile is the alcohol, the nucleophilic atom being the partially negatively charged O atom) as it attacks the electrophile (in this case the electrophile is the carboxylic acid or acyl halide, the electrophilic atom being the partially positively charged C atom of the carboxylic / acyl group), shifting the pi bond upwards to generate a singly bonded O atom with a negative formal charge, which then reforms the carbonyl and the leaving group (in the case of carboxylic acid, it is the protonated OH group, or OH2+; in the case of acyl halide, it is the halogen atom) is eliminated (in this case, as water or as the halide ion).

    • 'A' Level Qn.

       

      Q1) Why are non-metal oxides considered acidic? Why is rainwater slightly acidic, even in non-polluted areas?

       

      Q2) How do you distinguish between a carboxylic acid and an alcohol, using sodium carbonate and limewater? Explain how this test works.

       

       

       

      Solution :

       

      Q1) Non-metal oxides are covalent oxides and hence are acidic oxides.

       

      The reason why the bonds between the oxygen and the heteroelement atoms (eg. C, N, S, P, etc) are covalent rather than ionic, is due to the relatively small electronegativity difference between oxygen and the non-metal heteroelment.

       

      As an example, let's look at carbon dioxide, CO2. Due to the greater electronegativity of oxygen compared to carbon, the carbon is partially positively charged and is hence electrophilic. It consequently invites nucleophilic attack by water molecules, forming carbonic(IV) acid upon hydrolysis :

       

      CO2(g) + H2O(l) ---> H2CO3(aq) ---> H+(aq) + HCO3-(aq)

       

      The protons (H+ ions) will cause the solution to be acidic and the pH to decrease below 7 (at room temperature), which can thusly be detected by the Universal Indicator solution.

       

      Mechanism : Water attacks the C atom of CO2. Pi bond between C and O shifts to form a singly bonded O atom with a negative formal charge. A proton is transferred from the H2O newly bonded to the C atom (notice the O atom has a positive formal charge because it donated a dative bond in its nucleophilic attack, which being electronegative it does not like at all, hence it gladly loses a proton to lose the positive formal charge) to the singly bonded O atom with the negative formal charge (which makes it basic). The product is carbonic(IV) acid, H2CO3(aq).

       

       

      Q2) Add sodium carbonate to the alcohol and separately to the carboxylic acid. No efferverscence is observed with the alcohol, but efferverscene is observed with the carboxylic acid and when the gas evolved is bubbled into limewater, a white precipitate is observed.

       

      The carbonate(IV) ions, CO3 2-, are protonated by the carboxylic acid, forming the conjugate acid carbonic(IV) acid, which exists in equilibirum with, and hence can decompose into CO2(g) and H2O(l). Notice that carbon dioxide, being a gas, leaves the reaction mixture, pulling the position of equilibrium over to the right.

       

      2RCOOH(aq) ---> 2H+(aq) + 2RCOO-(aq)

      2H+(aq) + CO3 2-(aq) ---> H2CO3(aq) ---> CO2(g) + H2O(l)

       

      Notice how this reaction (ie. the 2nd equation) is the exact same reaction as in our discussion on Q1, but in the opposite direction.

        

      The CO2(g) when bubbled into limewater, undergoes hydrolysis to form carbonic(IV) acid, which undergoes acid-base neutralization proton transfer reaction with the calcium hydroxide, to form water and calcium carbonate white precipitate.

       

      If the molarity of the calcium hydroxide limewater is low, then water solvent is present in much higher molarity and thus the carbon dioxide undergoes hydrolysis.

       

      If the molarity of the calcium hydroxide limewater is high, then the hydroxide ion, being a much stronger nucleophile than water, directly attacks the carbon dioxide electrophile.

       

      Either way, you obtain your calcium carbonate white precipitate.

    • 'A' Level Qn - Chemical Equilibria.

       

      Question :

       

      a) the equilibrium constant Kc for the above reaction at 25 deg C is 4.0. Calculate the sample mass of ethanoic acid that must be mixed with 2.0 mol of ethanol to produce 1.5 mol of ethyl ethnoate at equilibrium.

       

      b) Suppose 1 mole of ethanol, 1 mole of ethanoic acid, 3 moles of ethyl ethanoate and 3 moles of water are mixed together in a separate stoppered flask. How many moles of ethyl ethanoate will be present at equilibrium at 25 deg C?

       

       

      Answers :

      157.5g and 2.667mol

       

      (Note : In the exams, always give the 4 or 5 sig fig value first, *then* give the 3 sig fig value. Note that when changing values from 4 or 5 sig fig to 3 sig fig, it's not "approximately equals to", it's "exactly equals to" whatever value, to "3 sig fig".)

       

      Guidance or Notes on Equilibria :

      Whenever you're given non-zero quantities or molarities of both LHS reactants and RHS products, you should determine the Qc value, and compare it to Kc.

      Qc, known as Equilibrium Quotient (in contrast to Kc, Equilibrium Constant), has the same formula as Kc (ie. product of molarities of RHS vs molarities of LHS, each raised to the power of their stoichiometric coefficients), except that it can be applied at any time, eg. initial. In contrast, Kc is a value derived from the molarities of reactants and products only at equilibrium point itself.

      Notice in this question, Qc = 9.0 > Kc = 4.0. Imagine you are Qc, standing on a number line with Kc. Do you turn your head to the left or to the right, to look at Kc? Kc represents position of equilibrium. Notice that in this question, since Qc > Kc, you (ie. Qc), have to turn your head to the left to look at Kc (ie. position of equilibrium). Therefore we say "position of equilibrium lies to the left."

      When we say "position of equilibrium lies to the left", it means the rate of the backward reaction will exceed the rate of the forward reaction, until equilibrium is reached, ie. Qc = Kc.

      Hence in your ICE table, your Change will see "+x, +x, -x, -x" instead of the usual "-x, -x, +x, +x".

      In most scenarios, when we add, or take away, some reactants or some products, or increase or decrease total pressure (by changing volume of container), and we say "position of equilibrium has shifted to the...", do you realize it is *not* actually Kc which has changed, but Qc?

      So if Qc shifts to the right (eg. when you add some product), the Kc value or position of equilibrium has appeared to have "shifted to the left", *relative* to Qc (even though the Kc value has not changed at all).

      The only time when Kc value actually changes, rather than the Qc value, is when temperature changes. If the forward reaction is endothermic (ie. heat may be regarded as a reactant), increasing the temperature will increase the Kc value (th Qc value remains unchanged). And therefore, it is indeed the Kc value or "position of equilibrium, has shifted to the right", relative to Qc.

      Basically, Qc is to Kc, as ionic product is to solubility product.

      Edited by UltimaOnline 12 Oct `17, 12:30AM
    • ‘A’ Level Qn : Acid-Base Equilibria

      A sample of sodium hypochlorite is dissolved in 100cm3 of 0.123mol/dm3 of hypochlorous acid solution, generating a buffer solution of pH 6.2. Calculate the number of moles of gaseous hydrogen chloride required to be bubbled into the buffer solution until the pH reaches 6.0. The Kb of the hypochlorite ion at room temperature is 3.162×10-7.

      Answer :
      2.20×10-4 (to 3 sig fig)

      Edited by UltimaOnline 04 Aug `10, 11:31PM
    • 'O' Level Qn - Acids, Bases & Salts; Electrolysis.
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      hmyatthazin<!-- google_ad_section_end -->  posted :
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      Can someone explain me a few chemistry concepts?
      1) What exactly is neutralization reactions? The reactions with ionic equation H+ and OH- that gives H20 or reactions that produce salt and water only? Will the reactions between insoluble base and acid; metal and acid; ammonia and acid; alkali and acidic gas be considered as neutralization? I'm a bit confused!
      2) In a reversible reaction from LHS to RHS gives out 10J of energy, will 10J of energy be absorbed from RHS to LHS?

      3) In a reaction of ammonium chloride (aq) and sodium hydroxide (aq), how do we know if the product sodium chloride is an aq or sol?

      4) Do alcohol and alcoholic acid conduct electricity?why?

      Thanks a lot!!
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      xdolly<!-- google_ad_section_end --> posted :
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      wait for ultimaonline to come in =)
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      UltimaOnline<!-- google_ad_section_end --> posted :
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      Greets xdolly

      Hi hmyatthazin, most of your questions have already been correctly addressed by iBELIEVE, mystes and xdolly; is there any particular question you need more clarification on?

      The term 'neutralization' is actually vague and is used only at 'O' levels, to mean 'acid-base neutralization'. The 'O' level definition of "acid and base", refer to the Arrhenius definition, which is frankly awfully limiting.

      Two more meaningful definitions (which are used at 'A' levels) are the Bronsted-Lowry definition and the Lewis definition.

      While I agree that the Lewis definition may be more appropriate for 'A' levels than 'O' levels, the Bronsted-Lowy definition can certainly be easily understood and appreciated by 'O' level students, and should be taught at 'O' levels.

      Regarding the reaction between metals and acids, it is first and foremost, primarily a redox reaction. But it can also be argued to (secondarily) be a Bronsted-Lowry (and thus also a Lewis) acid-base reaction. Let's have a closer look :

      When a reactive metal, eg. sodium metal, is placed into liquid alcohol / water / carboxylic acid, the following redox reaction occurs :

      Na ---> Na+ + e-

      Sodium is a reactive metal which means it prefers to exist in the oxidized state, ie. Na+. Hence in this reaction, sodium is oxidized from Na to Na+. Who accepts the electrons that it throws off? It is actually the protons (ie. H+ ions), from the alcohol / water / carboxylic acid, that receives the electrons.

      2H+ + 2e- ---> H2

      Hydrogen gas is generated, as a result of protons (ie. H+ ions) being reduced to molecular hydrogen gas.

      The oxidation state (OS) aka oxidation number (ON), of hydrogen is decreased from +1 to 0, while the OS or ON of sodium is increased from 0 to +1.

      So it's clearly a redox reaction. In what way is this also a Bronsted-Lowry (hence Lewis) acid-base reaction?

      Simply because the alcohol / water / carboxylic acid is the proton-donor species (ie. the species that donates the protons; hence the acid here is actually the alcohol / water / carboxylic acid), and upon deprotonation, the alcohol / water / carboxylic acid is converted into its conjugate base (ie. the deprotonated version of the alcohol / water / carboxylic acid), which are the alkoxide (ie. alkyl oxide) anion RO-, the hydroxide anion OH- (more accurately written as HO- or -OH), and the carboxylate anion RCOO-.

      To summarize, reactions between reactive metals and acids (eg. alcohol / water / carboxylic acids) are primarily redox reactions (this answer will suffice at 'O' levels), but are also considered acid-base reactions ('A' level students should give both answers, with explanation).

      Note to 'O' level students :

      For 'O' levels, alcohols (eg. ethanol) and water are not considered acids, nor are they considered acidic. Please do not be confused by my references in the preceding paragraphs, on alcohols or water being 'acidic'.

      For 'A' levels, you guys of course realize that you were being fed nothing but lies, lies and more lies at 'O' levels. Just like you're still being fed big fat lies (albeit slightly more sophisticated ones) now at 'A' levels. *Evil Laughter*
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      iBELIEVE<!-- google_ad_section_end --> posted :
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      dieeee. how m i going to study uni next year when i don rmb a single stuff UltimaOnline has said. hopefully no chem stuff for me. lol
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      hmyatthazin<!-- google_ad_section_end --> posted :
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      Thanks everyone for explaining, but there are still some more questions that I'm not so sure of.
      First.. alcohol and carboxylic acids are covalent compounds that may be soluble in water, right? But dont carboxylic acid and alcohol dissociate in water to form ions and react with metals..? if so why don't they conduct electricity in water? Quite confusing.
       
      I know the solubility of compounds. But I read in one of the books that NaCl is produced as solid from reacting ammonium chloride and alkali. Quite weird.
      Oh.. another one more qn. What will happen when calcium and magnesium nitrate solution reaction? I thought that the usual decomposition would take place. But in the book says, there is no reaction coz calcium reacts with water to form insoluble calcium hydroxide.
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      UltimaOnline<!-- google_ad_section_end --> posted :
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      Carboxylic acid does dissociate into its protons and its conjugate base, ie. H+ ions and RCOO- ions. So yes, in solution they can conduct electricity.

      Alcohols do not dissociate into its protons and its conjugate base, since alkoxide RO- ions are too unstable to exist in aqueous conditions.

      Regarding your question "calcium and magnesium nitrate solution reaction? I thought that the usual decomposition would take place."

      You should be aware that this is not considered "decomposition". It is better described as "displacement" reaction, or even better, as a "redox displacement" reaction.

      Calcium is indeed significantly more reactive than magnesium (see list of redox potentials, read oxidation potentials as the negative of the reduction potentials listed; the more positive the oxidation potential, the more reactive the metal), however even magnesium is reactive enough such that you cannot easily reduce Mg2+ to Mg in aqueous solution.

      Ca will oxidize itself : Ca ---> Ca2+ + 2e-

      Because this is an aqueous solution, there is plenty of water around. Instead of reducing Mg2+ to Mg (which won't be easy because magnesium is itself a significantly reactive metal; ie. it has a rather positive standard oxidation potential of +2.38V, albeit below that of calcium's +2.87V), the electrons thrown off by the calcium will be used to reduce water :

      H2O + 2e- ---> H2 + 2OH-

      And since calcium hydroxide is only slightly soluble, a precipitate soon coats the calcium metal, slowing down the reaction. Aqueous calcium hydroxide is known as limewater. However, once the molarity of calcium hydroxide gets too high (ie. once ionic product exceeds solubility product), additional calcium hydroxide generated takes the form of a solid precipitate, known as slaked lime.
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      hmyatthazin<!-- google_ad_section_end --> posted :
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      Thanks a lot UltimaOnline!!! Now I've understood. Chemistry is so unpredictable with exceptional cases. how are we to know everything..sigh..
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      darksoul<!-- google_ad_section_end --> posted :
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      (1) hydrochloric acid is about pH 1 or 2
      pure water is pH 7
      is it possible to get like pH 3 or 4 by diluting the acid?
      or is it that the pH is fixed regardless of concentration of the acid?

      (2) under what circumstances will the dissociation of water molecules occur?
      like in electrolysis aqueous sodium chloride has hydrogen ions and hydroxide ions from the water.
      but in a cup of water the water exist as water molecules?
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      UltimaOnline<!-- google_ad_section_end --> replied :
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      Yes, it is possible to get pH 3 or 4 by diluting the hydrochloric acid. the p in pH means "negative log (base 10)". The H in pH means molarity (molar concentration) of H+ ions.

      Hence, if you want pH 4, it means molarity of H+ ions = 10^-4 mol/dm3. Hence dilute the HCl until you get 10^-4 mol/dm3 of HCl, and you have pH4.

      Dissociation of water into protons H+ and hydroxide ions OH- is an endothermic process. The higher the temperature, the greater the percentage of water molecules will dissociate into protons and hydroxide ions. At room temperature, the molarity of H+ ions and OH- ions in 1 dm3 of pure water, is only 1x10^-7 for each. The majority (more than 99%) of water remains as covalently bonded water molecules, not H+ or OH- ions.
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      darksoul<!-- google_ad_section_end -->  posted :
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      i know what if it is like a beaker of distilled water, it exist as H-O-H molecules. but why do they say that in the electrolysis of aqueous sodium chloride as the electrolyte, there are hydrogen ions and hydroxide ions from the dissociation of water molecules (dissociate for no reason? or the sodium chloride cause it to dissociate?).

      So the ions in the sodium chloride (aq) are:
      From sodium chloride: sodium ion and chloride ion
      From water: hydrogen ion and hydroxide ion

      So lets say that 99% of water are water molecules, so since the 1% dissociate (dunno for what), there are 0.5% hydrogen ions and 0.5% hydroxide ions. Due to the selective discharge of ions, the hydroxide ion gets discharged instead of the chloride ion. And 4 moles of hydroxide ions get discharged to form 2 moles of water and 1 mole of oxygen gas. So wont the oxygen produced at the anode be very very little?
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      hmyatthazin<!-- google_ad_section_end --> posted :
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      Yep, something to think about...Then what will happen after this 1% of H+ ions and OH- ions become discharged?
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      UltimaOnline<!-- google_ad_section_end --> posted :
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      The key to solving your riddle "The Mystery of the Limitless H+ & OH- Ions", is to realize that it is the electrons coming into the cathode during electrolysis, that FORCES water to be split into H+ and OH-.

      At the cathode, water molecules split into H+ and OH- ions, because the H+ is needed to accept the electrons. It's like if a gang of robbers whip out parangs and command you, "gimmie your money or I'll kill your girlfriend!". As a responsible boyfriend, naturally you surrender your money.

      When water molecules at the cathode see a rush of electrons zooming towards them, they realize they have no choice but to surrender a part of themselves, the H+ ions.

      Hence, H2O ---> H+ + OH-

      The H+ accept electrons : 2H+ + 2e- ---> H2

      What's left over, is the hydroxide ions OH-.

      Similarly, at the anode where oxidation occurs, it's like the mafia collecting protection money. You know if you don't pay up, your friends and family members will suffer. So you dutifully pay up.

      At the anode, water molecules realize they have to give up electrons. Only negatively charged OH- ions have excess electrons to give. Thus water again carries out the noble sacrifice : H2O ---> H+ + OH-

      The OH- ions are the ones who allow themselves to get oxidized to O2 gas and electrons (which the mafia is waiting impatiently for).

      4OH- ---> O2 + 2H2O + 4e-

      Capisce? (pronounced "ka-peesh". Italian for "understand?")

      P.S.
      hmyatthazin, you're totally welcome. Keep enjoying Chemistry! <!-- google_ad_section_end -->
      Edited by UltimaOnline 08 Aug `10, 11:35PM
    • 'O' Levels / 'A' Levels Qn : Qualitative Analysis.

       

      Why is it necessary to acidify our test solution before testing for the presence of Cl- ions?

       

      and

       

      How does the test for nitrate(V) NO3- ions work? Why does adding aluminium foil and sodium hydroxide to a solution of nitrate(V) ions, result in ammonia gas being released?

       

       

      Here's the Solution (pun!)

       

      --------------------------

       

      On why test solutions are acidified before testing for Cl- ions :

       

      Acidifying the test solution removes any carbonate ions or hydrogencarbonate ions present, which might give a false precipitate result.

       

      CO3 2- + 2H+ ---> H2CO3 ---> CO2 + H2O

      or

      HCO3- + H+ ---> H2CO3 ---> CO2 + H2O

       

      Carbon dioxide, being a gas, leaves the reaction mixture, pulling the position of equilibrium over to the right.

       

      But why would any carbonate ions be present in the first place?

       

      You see, the earth's atmosphere (that includes the laboratory where this experiment is being carried out) contains carbon dioxide. Inevitably, a couple of carbon dioxide molecules will find their way into our solution, whereby hydrolysis of carbon dioxide occurs.

       

      Carbon, being a non-metal oxide, is a covalent oxide (due to the relatively small magnitude of electronegativity difference between oxygen and the non-metal, eg. carbon), and hence is an acidic oxide.

       

      The reason for this is that as electron density is withdrawn from the less electronegative C atoms to the more electronegative O atoms, in carbon dioxide. This renders the C atom partially positively charged, and hence electrophilic.

       

      Water molecules are nucleophilic (because the electron density from the two H atoms are inductively drawn towards the more electronegative O atom, making it partially negatively charged, and its lone pairs available for nucleophilic attack), and hence attack the carbon dioxide electrophile.

       

      The result is carbonic(IV) acid, H2CO3, which naturally dissociates a proton to form the hydrogen carbonate ion, HCO3-. The dissociated protons, explains why rain water, even in unpolluted areas, are naturally slightly acidic.


      And if any hydroxide ions are added, any hydrogen carbonate ion present will be deprotonated, and there we have it - carbonate ions present in solution. 

       

      --------------------------

       

      On the QA test for nitrate(V) ions :

       

      Al is a reactive metal and hence has a significantly positive oxidation potential. Hence Al oxidizes itself to Al3+.

       

      Al ---> Al3+ + 3e-

       

      The electrons thrown off, are used to reduce NO3- to NH4+.

       

      NO3- + 10H+ + 8e- ---> NH4+ + 3H2O

       

      Thereafter, the ammonium ions are deprotonated by the hydroxide ions, generating ammonia :

       

      NH4 + OH- ---> NH3 + H2O

       

      A little heat is applied to overcome the hydrogen bonding between ammonia and water, to vaporize the ammonia, whereupon it can be detected by moist red litmus paper :

       

      NH3 + H2O ---> NH4 + OH-

       

      The hydroxide ions turn red litmus paper blue.

    • 'A' Level Qn.

       

      Suggest what interaction occurs when the NH2- ion is placed into water.

      A) Hydrogen bonding

      B) Permanent dipole - permanent dipole attraction

      C) Ion - permanent dipole attraction

      D) None of the above

       

      Answer :

       

      D) None of the above.

       

      Options A, B and C refer to hydration interactions. NH2-, being unstable in water and hence being a strong base (bear in mind that Bronsted-Lowry bases are proton acceptors; their mission in life is to seek out protons (H+ ions) in order to datively bond with them, in a desperate bid to stabilize themselves (since bases will always have either a negative partial charge or a negative formal charge; and the central dogma of Organic Chemistry is "Charge is Destabilizing"). Hence the more unstable the base, the more desperately it will seek out protons, and hence the stronger it is as a base), NH2- will undergo HYDROLYSIS (not hydration) in aqueous solution, specifically a Bronsted-Lowry acid-base proton transfer reaction, as follows :

       

      NH2- + H2O ---> NH3 + OH-

       

    • 'A' Level Qn.

       

      Explain why, when heated, NH4Cl will decompose readily into NH3 and HCl gases; unlike most other ionic solids which are thermally stable.

       

      Solution :

       

      It is true that most solid ionic compounds are somewhat stable, due to the (usually strongly) exothermic lattice enthalpy / energy which is the result of ionic bond formation, ie. the strong electrostatic forces of attraction between cations and anions.

      However, NH4+ is acidic, and Cl- is basic/nucleophilic. When sufficient activation energy is applied in the form of the heating process, it is an easy matter for the Cl- base to abstract away a proton from its neighbour in close proximity - the acidic NH4+ ion.

      Furthermore, the activation energy supplied in the form of the heat energy provided, will allow the resulting covalent molecular products (of the reaction which may be thought of as a Bronsted-Lowy acid-base proton transfer reaction, and/or a thermal decomposition reaction) of NH3 gas and HCl gas, to gain sufficient kinetic energy to escape from the electrostatic confines of the ionic lattice structure.

      This explains why, unlike most other ionic solids, NH4Cl can readily decompose into NH3 gas and HCl gas, when heated.

    • 'A' Level Qn : Acid-base Equilibria.

       

      Hydrazine, N2H4, is a weak base with the following Kb values : Kb1 = 8.5 x 10-7 and Kb2 = 8.9 x 10-16). Solution A is 0.10M solution of hydrazine. Solution B is 0.20 mol/dm3 of hydrochloric acid. 25cm3 of solution A was titrated with solution B until the 1st equivalence point was reached.
      (a) Calculate the initial pH of solution A.

      (b) Calculate the pH of the solution when 6.25 cm3 of solution B was added to solution A.
      (c) Calculate the pH of the solution at 1st equivalence point.

      (d) Sketch the titration curve from initial to 1st maximum buffer capacity to 1st equivalence point, labeling all relevant values on both axes.
      (e) Calculate the new pH when 2.5 cm3 of 0.2 mol/dm3 sodium hydroxide solution is added to the mixture in (b).
      (
      f) Draw the Kekule structures of hydrazine, its conjugate acid and its conjugate base. Indicate formal charges.

      (g) Draw the dot-&-cross structures of hydrazine, its conjugate acid and its conjugate base. Let dots represent electrons of nitrogen atoms, and crosses represent electrons of hydrogen atoms.
      (h) Hydrazine can disproportionately decompose into two common nitrogen-containing products (no other products are formed). By stating clearly the oxidation states of nitrogen in hydrazine and in these 2 products, explain why this is a disproportionation reaction. Write the balanced redox equation for this reaction.
      (i) By using Data Booklet bond energies, calculate the enthalpy of disproportionation per mole of hydrazine.

       

       
      Answers :
      a) 10.5

      b) 7.93

      c) 4.55

      d) (graph)

      e) 8.30

      f & g) Kekule structures and dot-&-cross structures for the conjugate base must have a negative formal charge and a dot-dot lone pair on the N atom, respectively; Kekule structures and dot-&-cross structures for conjugate acid must have a positive formal charge and a dot-dot bond pair with a proton on the N atom, respectively.

      h) 3N2H4 --> 4NH3 + N2

      i) -171.3 kJ/mol

      Edited by UltimaOnline 25 Aug `10, 12:30AM
    • 'A' Level Qn : Hess Law.

      2)         Determine the formation enthalpy of NO(g), given :

       

      N2(g) + 3 H2(g) à 2 NH3(g)                               ΔH = -91.8 kJ / mol

      4 NH3(g) + 5 O2(g) à 4 NO(g) + 6 H2O(l)        ΔH = -906.2 kJ / mol

      H2(g) + ½ O2(g) à H2O(l)                                 ΔH = -241.8 kJ / mol

       

       

      3)         Determine the enthalpy change of reaction for :  C(s) + CO2(g) à 2CO(g)

      Given :

      Fe2O3(s) +  3/2 C(s) à 2 Fe(s) + 3/2 CO2(g)          ΔH = +234 kJ / mol

      Fe2O3(s) + 3 CO(g) à 2 Fe(s) + 3 CO2(g)            ΔH = -24.8 kJ / mol

       

       

      Answers :
      Qn2) +90.25 kJ/mol

      Qn3) +172.5 kJ/mol

      Edited by UltimaOnline 25 Aug `10, 12:32AM
    • 'A' Level Qn : Redox & Structures

       

      Draw the uninegative iodine dichloride ion, indicating all formal charges and oxidation states of all atoms within the ion. Prove that ionic charge = sum of oxidation states. State its electron and ionic geometry. Write the relevant half-equations and balanced redox equation to represent the redox reaction that occurs (under acidic conditions) between the iodate(V) ion, the iodide ion, and the chloride ion, to generate the uninegative iodine dichloride ion.

      Answers :

      6Cl- + 6H+ + 2I- + IO3- ---> 3[Cl-I-Cl]- + 3H2O

      Structure : Cl-I-Cl; uninegative formal charge on I, no formal charges on Cl atoms; oxidation states of atoms Cl (-1), I (+1), Cl (-1).

      Edited by UltimaOnline 25 Aug `10, 12:38AM
    • 'A' Level H2 Notes on "Chemical Bonding" and "Solubility of Grp II compounds"

       

      Originally posted by ohnoez!:

      does hydration energy become less exothermic down group 2? is it because of the weaker ion-dipole attractions due to lower charge densities?

      will compounds with hydrogen bonding have pd pd also?

       

      Both the endothermic lattice dissociation enthalpies and the exothermic hydration enthalpies are decreased in magnitude as you go down Group II, for the same reason : decrease in charge density results in weaker ionic bonds and weaker ion-dipole interactions.

       

      If the anion is small (eg. OH-), the rate of decrease in endothermic lattice dissociation enthalpy outweighs the rate of decrease of hydration enthalpy, as you go down Group II; consequently, solution enthalpy becomes more exothermic (or less endothermic) as you go down Group II. Therefore solubility of Group II hydroxides increase down the group.

       

      If the anion is large (eg. CO3 2- or SO4 2-), the rate of decrease of exothermic hydration enthalpy outweighs the rate of decrease of endothermic lattice dissociation enthalpy, as you go down Group II; consequently, solution enthalpy becomes more endothermic (or less exothermic) as you go down Group II. Therefore solubility of Group II carbonates & sulfates decrease down the group.

       

      However, this is only half of the full picture. The other half is entropy. Gibbs free energy completes the picture. Near the top of the group (ie. high charge density), the decrease in entropy during the hydration process outweighs the increase in entropy during the lattice dissociation process. Near the bottom of the group the opposite occurs.

       

      The result, is a complex balance between enthalpy versus entropy, lattice dissociation versus hydration. Group II carbonates decrease in solubility from Be to Sr (because thus far, enthalpy effect outweighs entropy effect), thereafter solubility increases from Sr to Ba (because from here on, entropy effect outweighs enthalpy effect).

       

      For 'A' level H2 Chemistry, usually discussing enthalpy effect will suffice. Only in cases where enthalpy effect does not adequately explain the solubility trend, then bring in entropy, effect. As a further example, the solution or dissolving process of ammonium salts are endothermic (ie. endothermic lattice dissociation enthalpy outweighs exothermic hydration enthalpy), yet all ammonium salts are soluble. This is due to the favourable (ie. positive) entropy effect outweighing the unfavourable (ie. endothermic) enthalpy effect.

       

      -----------------------

       

      "Permanent Dipole - Permanent Dipole" Van der Waals interactions (or more specifically, forces of attraction) are between polar molecules, which are not capable of hydrogen bonding.

       

      "Permanent Dipole - Induced Dipole" Van der Waals interactions (or more specifically, forces of attraction) are between polar molecules and non-polar molecules.

       

      "Instantaneous Dipole - Induced Dipole" Van der Waals interactions (or more specifically, forces of attraction) are between non-polar molecules. (Equivalently, terms such as "London forces" or "Dispersion forces" or "temporary dipoles" or "induced dipoles" or "induced dipole-dipoles" or "induced dipole - induced dipole forces", or simply "Van der Waals forces" may be used to refer to such interactions between non-polar molecules).

       

      Hydrogen bonding, refers to the particularly strong permanent dipole - permanent dipole interactions that exist between molecules capable of hydrogen bonding. These are electrostatic attractions (with some degree of covalent nature) between a partially positively charged H atom (for the partial positive charge to be of sufficient magnitude for H-bonding, the H atom must be covalently bonded to an electronegative N, O or F atom) and a partially or formally negatively charged N, O or F atom (if the N, O or F atom has a partial or formal positive charge, no hydrogen bonding is possible with a partially positively charged H atom, because positive-positive like-charges repel) with at least one lone pair available.

       

      Ion-dipole interactions are similar in strength to hydrogen bonding; usually slightly stronger, though this varies depending on individual charge densities of the ions involved. There is some overlap between some ion-dipole interactions and hydrogen bonding.

       

      For instance, when aqueous sodium hydroxide is added to phenol, the phenol ppt dissolves. This is because when phenol is deprotonated, the resulting solute-solvent interactions between the phenoxide ion and water solvent, are stronger than the limited hydrogen bonding between phenol and water. These stronger interactions may be thought of as ion-dipole interactions, or as particularly strong hydrogen bondings (there is an upgrade in strength of hydrogen bonding because the partially negatively charged O atom in phenol has been upgraded to a formally negatively charged O atom in phenoxide ion. The O atom in phenoxide ion has one bond pair and 3 lone pairs, conferring on the O atom a negative formal charge. Yes, there is delocalization by resonance of the negative charge into the benzene ring, specifically the ortho and para carbon atoms; but because carbon is not as electronegative as oxygen, most of the negative charge in the resonance hybrid is still borne by the O atom).

       

      Similarly do hydroxide ions have ion-dipole or hydrogen bonding with water? The interaction between the partially positively charged H atom of OH-, and water, is obviously hydrogen bonding. But the interaction between the negatively formally charged O atom in OH-, and water, may be thought of, or argued to be, either ion-dipole or hydrogen bonding. Both ways of looking at this, are acceptable by Cambridge at 'A' levels.

       

      The interaction between NH4+ and H2O though, should be more accurately regarded as hydrogen bonding rather than ion-dipole. This is because the positively formal charged N atom does not have the opportunity for ion-dipole interaction with water. Why? Because of the 4 partially positively charged H atoms arranged tetrahedrally about the N atom. Before any water molecule can get close enough to the central positively charged N atom, these tetrahedrally arranged H atoms will seize the opportunity to hydrogen bond with the water molecule, denying the N atom the chance for any such interaction.

       

       

    • 'A' Level Qn : Acid-Base Equilibria.

       

      The Ka of hypochlorous acid {latin name}, a.k.a. chloric(I) acid {stock name} is 10-7.5. A sample of solid sodium hypochlorite {latin name}, a.k.a. sodium chlorate(I) {stock name}, was dissolved in 100cm3 of 0.123mol/dm3 of hypochlorous acid {latin name}, a.k.a. chloric(I) acid {stock name}, generating a buffer solution with a pH of 6.20. Some hydrogen chloride gas was then passed into the buffer solution. Calculate the number of moles of hydrogen chloride gas that was required to reduce the pH of the buffer solution to 6.0.

      Answer :

      2.20 x 10^-4 moles

    • 'A' Level Qn : Acid-Base Equilibria.

       

      Q1a)    The conjugate acid of a weak base B, has a pKa value of 9.2014 at room temperature. What volumes of 0.1 mol/dm3 of HCl(aq) and 0.05 mol/dm3 of a solution of B, may be mixed to generate 500cm3 of a buffer solution with a pH of 8.90?

       

      b)          A chemical engineer suggests that, one method of distinguishing between two beakers of solutions with the same pH (where one is a dilute solution of a weak base, while the other is a buffer solution consisting of this base and its conjugate acid), is to add a large volume of water to both beakers, and observe any changes in pH using electronic data loggers. Determine if the chemical engineer’s method works, by calculating the initial and final pH values of 500cm3 of a solution of B, after the addition of 5dm3 of water; as well as the initial and final pH values of 500cm3 of the buffer solution generated in part (a), after the addition of 5dm3 of water.

       

      Answers :

      a) 125cm3 of 0.1 mol/dm3 of HCl, and 375cm3 of 0.05 mol/dm3 of B, are required to generate 500cm3 of buffer solution with pH 8.9 pH.

      b) Buffer : initial 8.9, final 8.9

      Base : initial 10.95, final 10.43

      The pH of buffer solutions do not change upon addition of water. The pH of acidic and basic solutions increase and decrease respectively, upon addition of water.

       

       

       

    • 'A' Level Qn : Acid-Base Equilibria

       

      If the Ksp of A3B2(s) = Kw of water at 298 K, what is the molarity of A2+(aq)
      in a saturated solution at 298 K?           A3B2(s) ßà 3A2+(aq) + 2B3-(aq)

      Answer : 1.864 x 10^-3 mol/dm3

    • 'A' Level Qn : Acid-Base Equilibria

       

      Calculate the pKa of dihydrogen monoxide at 298K.

       

      Answer :

      15.7

       

    • 'A' Level Qn : Acid-Base Equilibria

       

      Calculate the molarity of aqueous ammonia required to initiate the precipitation of iron(II) hydroxide from a 0.003 mol/dm3 aqueous iron(II) chloride. pKa of NH4+ = 9.2553, solubility of Fe(OH)2 = 1.4255 x 10-3 g/dm3.

       

      Answer :

      molarity of aqueous ammonia needs to be greater than 2.60 x 10^-6 mol/dm3 

    • 'A' Level Qn : Acid-Base Equilibria

       

      Q6.      Which of the following is true of pure water at               (i) 274 K?                    (iii) 299 K?

      a)      pH = 7               b)   pH < 7                   c)  pH > 7

      d)   pOH = 7            e)   pOH < 7                f)  pOH > 7

       

      Answer :

      (i) c & f

      (ii) b & e

    • 'A' Level Qn : Organic Chemistry

       

      A molecule contains a primary alcohol group, an aldehyde group, an alkene group, a carboxylic acid group, and an ester group. How would you carry out the following reactions?

      a) Oxidize the primary alcohol group without oxidizing the aldehyde group?

      b) Oxidize the aldehyde group without oxidizing the primary alcohol group?

      c) Simultaneously reduce the aldehyde group and the ester group without reducing the alkene group?

      d) Reduce the aldehyde group without reducing the ester group?

      e) Reduce the alkene group and the aldehyde group without reducing the carboxylic acid and ester group?

       

      Answers :

      a) Pyridinium chlorochromate

      b) Fehling's / Benedict's / Tollen's

      c) LiAlH4 (in dry ether), followed by protonation or hydrolysis

      d) NaBH4 (in dry ether), followed by protonation or hydrolysis

      e) Hydrogen gas and nickel / platinum / palladium catalyst

       

       

    • 'A' Level Qn : Organic Chemistry

       

      With relevant diagrams and/or equations, explain the chemistry behind “rebonding” and “perming” hair.

      Answers :

      http://www.humantouchofchemistry.net/hairstyling

    • 'A' Level Qn : Chemical Structure & Bonding

       

      Cubane is a hydrocarbon with the following cube-like structure :

      http://upload.wikimedia.org/wikipedia/commons/7/75/Cuban.svg

       

      Explain why cubane is unstable and hence reactive.

       

       

      Answer :

      Each C atom in cubane has 4 bond pairs and 0 lone pairs, which ideally means it would be sp3 hybridized and have a tetrahedral geometry with bond angles of 109.5 degrees, to minimize inter-electron bond pair repulsions. But the C-C-C bond angles in cubane are forced to be orthogonal (ie.of 90 degrees), and consequently there is considerable angle strain due to inter-electron bond pair repulsions.

       

    • H2 Chem Equilibria Qn.

      0.2 mols of SbCl5 is placed in a beaker fixed with a piston, pressure maintained at 0.2bar, temp maintained at 448 degrees. Calculate the volume of the equilibrium mixture. Assume all gases are ideal.

      SbCl5 <==> SbCl3 + Cl2

      Kp = 1.48 at 448 degrees<!-- google_ad_section_end -->

       

      Solution :

      Our ICE will be in moles.

      SbCl5 <==> SbCl3 + Cl2
      I | 0.2 | 0 | 0
      C | -x | +x | +x
      E | 0.2 - x | x | x

      Total no of moles of gas = (0.2 + x)

      P V = n R T
      (0.2 x 10^5 Pa) (Vm3) = (0.2 + x) (8.31) (746)
      Make V the subject, in terms of x.

      P(SbCl3) = mole fraction x total pressure = (x / (0.2 + x)) x (0.2 x 10^5 Pa)
      P(Cl2) = mole fraction x total pressure = (x / (0.2 + x)) x (0.2 x 10^5 Pa)
      P(SbCl5) = mole fraction x total pressure = ((0.2 - x) / (0.2 + x)) x (0.2 x 10^5 Pa)


      Kp = ( P(SbCl3) x P(Cl2) ) / P(SbCl5) = 1.48
      ==> ( (x / (0.2 + x)) x (0.2 x 10^5 Pa) x (x / (0.2 + x)) x (0.2 x 10^5 Pa) ) / ( ((0.2 - x) / (0.2 + x)) x (0.2 x 10^5 Pa) ) = 1.48

      Solve for x.

      Substitute into expression for V, and solve for V.<!-- google_ad_section_end -->

    • hesaniga posted :

       

      "Is the lone pair of electrons on a double bonded O atom ever available? As in can it act as nucleophile or as ligand?"

       

       

      ThunderFbolt replied :

       

      "yeap, if not oxygen won't be able to bind to hemoglobin."

       

       

      Ultima Online added :

       

      True, that.

      Although there's quite a bit more to oxygen.

      For instance, bear in mind that for some ligands/nucleophiles, depending on the structure, there may be resonance effect to be considered. Deduce the resonance contributors and elucidate the resonance hybride. The greater the electron density of the O atom in the resonance hybrid, the stronger it's nucleophilic or ligand capacity.

      Additionally, in the context of molecular orbital theory, the ground state of molecular oxygen is triplet rather than singlet. The triplet state of oxygen is diradical, meaning each oxygen atom (in O2) has an unpaired electron. So the simplified O=O dot-&-cross, lewis or kekule structures taught at 'O' and 'A' levels give only a partial picture of the real properties of oxygen.

      But if the TS/OP is asking at H2 / A levels, then :

      For nucleophiles, O atoms with a negative formal charge are of course stronger nucleophiles (eg. hydroxide ion), than O atoms with only a negative partial charge (eg. water molecule). This is the reason why nucleophilic substitution of alkyl halides into alcohols are called "hydrolysis" (meaning "chemical reaction with water"), even though the hydroxide ion is often used (since the hydroxide ion is a stronger nucleophile, and can greatly speed up the rate of reaction).

      In fact, in the acid-catalyzed hydrolysis of esters & amides, water (with only a partial, not formal, negative charge) is the nucleophile. In the base-promoted hydrolysis of esters & amides, the hydroxide ion is the nucleophile. (One may errorneously conclude that this means alkaline hydrolysis might be preferred over acidic hydrolysis; not necessarily so, because in the elimination step (of nucleophilic acyl substitution of hydrolysis of esters & amides) of the leaving group has a significantly lower activation energy in the case of the protonated OH2+ or NH2+ group during acid-catalyzed hydrolysis, compared to base-promoted hydrolysis, of esters and amides).

      For ligands, if such are present, you should preferentially use O atoms with a negative formal charge (since these are more electron rich) to donate coordinate dative bonds. (Of course, bear in mind there's resonance delocalizaton of the -ve charge over both O atoms in each carboxylate group). For instance, think EDTA and oxalate (ethandioate) ligands. In these ligands, it is the O atoms with the -ve formal charge donate dative coordinate bonds to the metal ion (since these are more electron rich than the doubly bonded O atoms with a -ve partial charge). Of course, if no such formally negatively charged O atoms are present, a partially negatively charged atom such as in water, will have to do. Water molecules are the most common, but naturally one of the weakest, ligands around.

      However, when an oxygen based nucleophile or ligand is preferred, we should deprotonate any OH group whenever possible. Eg. phenol is a weak nucleophile or ligand, so we could deprotonate it (by use of an alkali such as aqueous sodium hydroxide) to generate it's conjugate base, phenoxide ion, which is a significantly stronger nucleophile and ligand than phenol. The formation of the hexaphenoxoferrate(III) complex ion results in the purple colouration that indicates the presence of phenol with neutral FeCl3(aq).

      To generate the very strong alkoxide ion nucleophile or base (the negative formal charge on the O atom cannot be delocalized by resonance without violating C's octet; furthermore the electron donating by induction alkyl group further intensifies and destabilizes the -vely charged O atom), we cannot use the hydroxide ion to deprotonate alcohol (since position of equilibrium lies to the side of alcohol and hydroxide, which are more stable species than alkoxide and water), and instead we have to use sodium metal (the standard oxidation potential of Na to Na+ is +2.71V, hence Na is a powerful reducing agent) which oxidizes itself, throws away its valence electron towards the alcohol, which is forced to deprotonate itself and to use its proton to accept the incoming electron, hence becoming its conjugate base the alkoxide ion, and hydrogen gas.

      Na ---> Na+ + e-
      ROH ---> RO- + H+
      H+ + e- ---> 0.5 H2

      <!-- / message --><!-- sig -->

    • H2 Chem

      Inorganic Chem Trends Qn / Organic Chem Mechanisms Qn

       

       

      Longchamp3 asked :

      "The trichlorides of nitrogen and phosphorus each react with water to give two products but they react in different ways. Phosphorus trichloride, PCl3 reacts to give HCl as the only chlorine containing product. NCl3, produces HOCl as the only chlorine containing product. In each case, predict the other product and write an equation for its production. Suggest an explanation for this difference in behaviour. I know that the other products are H3PO3 and NH3, but im unable to explain the difference in behaviour.
      Once again, help is greatly appreciated!"

       

       

      UltimaOnline replied :

       

      The different behaviours of the two Grp V chlorides are the consequence of the different electronegativities of N and P versus Cl, resulting in NCl3 being nucleophilic/basic, and PCl3 being electrophilic/acidic.

      One other factor would be that P has vacant, low lying, energetically accessible 3d orbitals to expand its octet. Notice that P has an expanded octet in H3PO3. This is not possible for N.

      The best way to answer such a question would be to draw the mechanisms for the hydrolysis in each case, which will clearly explain and account for their different behaviours.

      Water nucleophile attacks P of PCl3, Cl atom leaves, abstracts a proton from +ve formal charged O atom of H2O just added, repeat, repeat, one of the O atoms forms pi bond with P, +ve formal charged O atom loses proton, -ve formal charged P atom acceps proton; we have H3PO3 and 3 HCl.

      N atom of NCl3 abstracts a proton from H2O, OH- nucleophile thusly generated attacks Cl atom which cleaves away from +ve formal charged N atom, repeat, repeat; we have NH3 and 3 HOCl (hypocholorous acid aka chloric(I) acid).

       

       

    • H2/H1 Chem

      Acid-Base Equilibria Qn.

       

       

      Animizer asked :

      "10mL of 0.10 H2SO3 is titrated with 0.10M KOH. Determine the pH at the 1st equivalence point. Ka1 = 1.2 x 10^-2"

       

       

      UltimaOnline replied :

       

      Moles of HSO3- at equivalence point = 1 x 10^-3
      Volume of solution at equivalence point = 20cm3
      [HSO3-] at equivalence point = 0.05

      Since Ka2 not given, we assume Kb1 > Ka2. In other words, we assume salt is basic rather than acidic.

      (Note : If Ka2 is given, we have to compare Kb1 vs Ka2, to determine if HSO3- is more acidic or basic, and calculate pH accordingly.)

      On the assumption that HSO3- is basic rather than acidic (ie. Kb1 >  Ka2), HSO3- will undergo hydrolysis and generate OH-

      After doing the ICE table at equivalence point, for the hydrolysis of HSO3- acting as a base, we have :

      Kb = 8.333 x 10^-13 = [OH-]^2 / [HSO3-]
      [OH-]^2 = 4.167 x 10^-14
      [OH-] = 2.041 x 10^-7

      Since pOH = 6.69, hence pH = 7.31

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