'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Sun, 15 Nov 2009 13:27:26 +0800

UltimaOnline — Sun, 15 Nov 2009 13:27:26 +0800

2009 ‘A’ Level Exam Qn.

Methylbenzene requires concentrated sulfuric(VI) and nitric(V) acids and 30 degrees celsius to undergo nitration whereas phenol only requires dilute nitric acid. Give a reason for the difference in conditions and reagents.

Solution :

Phenolic hydroxy group is electron donating by resonance and is hence a strong activator, enriching the pi electron density of the benzene ring, making it a much stronger nucleophile (than ubsubstituted benzene). Hence only dilute nitric(V) acid is required for electrophilic aromatic substitution and nitration to occur.

Methyl group is electron donating by induction (and hyperconjugation) and is hence a weak activator, only slightly enriching the pi electron density of the benzene ring, making it only a slightly stronger nucleophile (than ubsubstituted benzene). Hence a nitrating mixture of concentrated nitric(V) and sulfuric(VI) acids are required to generate the nitronium cation (NO2+) for electrophilic aromatic substitution and nitration to occur.

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Sun, 15 Nov 2009 12:46:50 +0800

UltimaOnline — Sun, 15 Nov 2009 12:46:50 +0800

2009 ‘A’ Level Exam Qn.

When methylbenzene is reacted with Cl2 and AlCl3, a monochloro compound K is formed. Treatment of K with more Cl2 in the presence of light produces compound L. When L is heated with NaCN in ethanol, compound M, C8H6ClN is formed. M can be converted into an acidic compound N by heating under reflux with dilute H2SO4. Heating L with NaOH(aq) produces compound P, C7H7ClO. When a mixture of N and P is heated with a small amount of concentrated H2SO4, compound Q, C15H12Cl2O2, is produced.

Elucidate K to Q. State the type of each reaction described above.

Solution :
For questions such as these, where it is not clearly specified what they mean by “type of reaction”, students are advised to give both the type of reaction in terms of the mechanism, and in terms of the end product. You will never be penalized for giving both, unless one of them is wrong and/or contradicts the other.

To produce K, the reaction is : electrophilic aromatic substitution resulting in chlorination of benzene.
To produce L, the reaction is : free radical substitution resulting in chlorination of alkyl side chain.
To produce M, the reaction is : SN2 nucleophilic substitution (cyanide ion as nucleophile) resulting in the formation of a cyano compound.
To produce N, the reaction is : repeated nucleophilic substitions during hydrolysis (water as nucleophile) to generate a carboxylic acid.
To produce P, the reaction is : SN2 nucleophilic substitution (hydroxide ion as nucleophile) resulting in the formation of a primary alcohol.
To produce Q, the reaction is : addition-elimination (not nucleophilic substitution) resulting in the removal of water (ie. condensation reaction) and the formation of an ester (ie. esterification reaction).

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Sun, 15 Nov 2009 12:28:03 +0800

UltimaOnline — Sun, 15 Nov 2009 12:28:03 +0800

‘A’ & ‘O’ Level Qn.

In the industrial electrolysis of brine (concentrated sodium chloride solution), a diaphragm cell is used. Write equations that occur at the cathode and anode. What are the ratio of useful products generated in this process? What is the purpose of the diaphram? Write equations for the undesirable reactions that would occur if the diaphram was absent.

Solutions :
Cl2, H2 and NaOH are produced in the molar ratio of 1:1:2.

This industrial process is used to manufacture 3 different useful products of chlorine, hydrogen and sodium hydroxide.

As to the purpose of the diaphragm :
“In diaphragm cell electrolysis, an asbestos (or polymer-fiber) diaphragm separates a cathode and an anode, preventing the chlorine forming at the anode from re-mixing with the sodium hydroxide and the hydrogen formed at the cathode.” – http://en.wikipedia.org/wiki/Chlorine_production

Hydrogen reacts with chlorine to form hydrogen chloride gas (which undergoes hydrolysis to form hydrochloric acid). Chlorine reacts with sodium hydroxide to form sodium chlorate(I) (at 15 deg C) and sodium chlorate(V) (at 70 deg C), and additionally reforms sodium chloride and water.

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Sun, 08 Nov 2009 10:50:17 +0800

UltimaOnline — Sun, 08 Nov 2009 10:50:17 +0800

‘A’ Level Qn.

1. Solid NaCl and conc. H2SO4 can be used to generate HCl in situ for the reaction with alkenes to produce choloroalkanes.
(a) Explain why 2-chloropropane is formed as the major product instead of 1-cholorpropane.
(b) Will a high yield of 2-iodopropane be obtained by a similar method by replacing NaCl with NaI? explain your answer with appropriate equations.

Solution :

(a) As predicted by Markonikov’s Rule, the reaction pathway involving the more stable secondary carbocation intermediate resulting in the generation of 2-chloropropane, will be the major product. The reaction pathway involving the less stable primary carbocation intermediate resulting in the generation of 1-chloropropane, will be the minor product.

(b) A redox reaction occurs between the iodide ions and sulfuric(VI) acid :

H2SO4 + NaI —> NaHSO4 + HI
8HI + H2SO4 -> 4I2 + H2S + 4H2O

Hence, a lower yield of 2-iodopropane will result.

(Notice that the O.S. of sulfur decreases from +6 to -2, due to the stronger reducing power of iodide ions. If bromide ions were used instead (eg. H2SO4 + NaBr), the O.S. of sulfur decreases only from +6 (H2SO4) to +4 (SO2). The difference in reducing powers of the various halide ions, indicated by their different redox potentials in the Data Booklet (quote these values in the exam!), is explained by the difference of the distance between the positively charged nucleus and the valence shell, for chloride vs bromide vs iodide.)

'A' & 'O' Level Chemistry Qns (A Collection) replied by SBS n SMRT @ Thu, 05 Nov 2009 16:21:16 +0800

SBS n SMRT — Thu, 05 Nov 2009 16:21:16 +0800

‘A’ Level Qn.

The solubility products of some Group II fluorides, at rtp, are given below:

Numerical values of Ksp
BaF2 1.84×10–7
CaF2 3.45×10–11

A student accidentally mixed 25.0 cm3 of 0.100 mol dm–3 CaCl2 solution with 25.0 cm3 of 0.100 mol dm–3 BaCl2 solution in the laboratory. To separate the two metal ions, he added just enough solid KF to precipitate the maximum amount of CaF2 from the mixture, without precipitating BaF2.

(i) Determine the [F–] required for this separation.
(ii) Determine the concentration of Ca2+ remaining in the final solution.

Answers :
(i) 1.92×10–3 mol dm–3
(ii) 9.38×10–6 mol dm–3

Solution (pun intended) Hints :
- Since equal volumes used, molarities are halved.
- Based on Ksp BaF2, work out [F-]. The molarity of fluoride ions must be slightly LESS than this calculated molarity, so as to avoid precipitating BaF2.
- Based on Ksp CaF2, work out [Ca2+]. This works out to be the molarity of aqueous Ca2+ ions, because for any given temperature which remains constant during the reaction, the product of the molarities of the aqueous cations and aqueous anions, each raised to the powers of their stoichiometric coefficients, MUST be equal to (or less than, but never more than) this ionic compound’s Solubility Product (ie. Ksp) at that given temperature. Any additional aqueous ions are precipitated out to allow this value to hold true.

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Thu, 05 Nov 2009 16:20:18 +0800

UltimaOnline — Thu, 05 Nov 2009 16:20:18 +0800

‘A’ Level Qn.

Explain the differences in the stereostructures of the complex ions formed for the cobalt(II) ion when complexed with water ligands, versus with chloride ligands.

Solution :
The hexahydroxocobalt(II) ion has an octahedral stereostructure, while the tetrachlorocobaltate(II) ion has a tetrahedral stereostructure. This is due to the steric considerations : the O atoms (of water ligands) are significantly smaller than the Cl- ion ligands; hence (six) water ligands can pack closer to the Co2+ ion (to give an octahedral stereostructure), while there is not enough space to pack six Cl- ion ligands to give a similar octahedral structure.

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Mon, 12 Oct 2009 12:50:17 +0800

UltimaOnline — Mon, 12 Oct 2009 12:50:17 +0800

‘A’ Level Qn.

Suggest why calcium sulfate(VI) decomposes at a higher temperature than calcium carbonate.

Solution :

From the lower thermal decomposition temperature, we may deduce that the electron charge clouds on the carbonate anion is more easily polarized and distorted by the dipositive Ca2+ cation, as compared to the sulfate(VI) anion.
(Further explanation, beyond the requirements of the H2 syllabus)
The reason for the greater polarizability and distortability of the carbonate anion, in comparison with the sulfate(VI) anion, is due to the greater excess electron density on the oxygen atoms for the carbonate anion, over the sulfate(VI) anion. Notice that the resonance hybrid sees a -1/2 charge on each O atom for the sulfate(VI) anion (ie. dinegative anionic charge dispersed by resonance over 4 oxygen atoms), in contrast to a -2/3 charge on each O atom for the carbonate anion (ie. dinegative anionic charge dispersed by resonance over only 3 oxygen atoms).

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Thu, 08 Oct 2009 07:06:46 +0800

UltimaOnline — Thu, 08 Oct 2009 07:06:46 +0800

‘O’ & ‘A’ Level Qn.

A 0.1246g sample of a compound of Cr and Cl was dissolved in water. All of the chloride ions were then captured by silver ion in the form of AgCl ppt. A sample mass of 0.3383g of AgCl(s) was obtained. Determine the empirical formula of the compound of Cr and Cl.

Answer :
CrCl3

Hints :
Since the valency of chlorine when ionically bonded with metals is already known, let the cationic charge of Cr be x, ie. empirical formula be CrClx. Calculate moles of CrClx in terms of x. Calculate moles of Cl- present (precipitated as AgCl). Form an equation in x, solve for x.

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Tue, 06 Oct 2009 15:37:22 +0800

UltimaOnline — Tue, 06 Oct 2009 15:37:22 +0800

‘A’ Level Qn.

Given that the Ka of ethanoic acid is 1.8×10^-5, calculate the pH at equivalence point in the titration of 0.1 mol/dm3 of ethanoic acid with 0.1 mol/dm3 of sodium hydroxide.

Answer :
pH = 8.72

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Sun, 04 Oct 2009 12:38:20 +0800

UltimaOnline — Sun, 04 Oct 2009 12:38:20 +0800

‘O’ & ‘A’ Level Qn.

“Aqueous iodine” actually refers to the triiodide ion, as molecular iodine isn’t soluble in water.
10cm3 of a solution of iodate(V) ions of unknown molarity, was mixed with excess iodide ions. Both iodate(V) ions and iodide ions reacted with each other (in a redox reaction) to generate molecular iodine. The iodine generated required 30cm3 of 0.1 mol/dm3 of thiosulfate ions for complete redox reaction.
a) Write all half-equations and balanced redox equations relevant to the problem above.
b) Calculate the molarity of the iodate(V) solution used.
c) Explain why molecular iodine is insoluble, and why the triiodide ion is soluble, in water.
d) Draw the full displayed structural formulae of the iodate(V) ion, the thiosulfate ion, the tetrathionate ion and the triiodide ion.
e) State the ionic geometry of the triiodide ion as predicted by VSEPR model; and calculate the oxidation states for the individual sulfur atoms in the thiosulfate and tetrathionate ions.

Solution :
0.05 mol/dm3
Permanent dipole – induced dipole is weaker than hydrogen bonding; ion-dipole interactions stronger than hydrogen bonding.
Triiodide ion is linear; OS of sulfur atoms are +4 and 0 in thiosulfate, and +5, 0, 0, +5 in tetrathionate.

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Fri, 02 Oct 2009 15:52:43 +0800

UltimaOnline — Fri, 02 Oct 2009 15:52:43 +0800

‘A’ Level Qn.

limywv asked :
“How to draw NO2 correctly? Guys, this question has been a major bugbear for me for quite some time now…I’ve searched the internet but no dot-and-cross diagrams were offered… can anyone help me on this?”

Solution (plus my comments).

A couple years ago, my tuition student was just about the only one in her cohort who could correctly draw the structure of NO2 in a major exam. My point is (not about boasting, but about) most JC teachers don’t focus (not blaming them though, they have it tough enough) on teaching the relevant, fundamental and crucial aspects of Chemistry that empower students to understand, appreciate and enjoy the Science as an Art. JC students most often fail to allow themselves to do the above (not blaming them though, they have it tough enough), and instead blindly memorize their way through JC.

Specifically (for this question), I refer to the concept of formal charges and resonance (that is ill-taught in JCs), without an understand of these, drawing structures become senseless and hopeless.

To answer your question :

Central atom is N (with a +ve charge), single (and dative) bonded to an O atom (with a -ve charge), and double bonded to a (neutral) O atom. The N atom thus has 3 bond pairs and an unpaired electron, the (-ve charged) O atom has 3 lone pairs and 1 bond pair, the (neutral) O atom has 2 lone pairs and 2 bond pairs.

The electron geometry is trigonal planar, the molecular geometry is v-shape or bent or non-linear.

The resonance hybrid sees each of both N to O bonds as having bond lengths & bond strengths equivalent to 1.5 bonds. The charge on each O atom is an averaged -1/2 charge.

For ‘A’ levels exams, it will suffice to draw a single resonance contributor.

limywv replied with :
“so it’s O=N—>O? with the Natom having 7 electrons while the 2 O atoms have full octet structures? So is this the reason why they can form diamers occassionally?”

I replied :
Yes, that’s correct on both counts (N atom has 7 electrons in terms of an octet, and 4 electrons in terms of formal charge). If the question tasks you to draw NO2, it might also require you to draw N2O4 in the same qn. The unpaired electron in one NO2 molecule combines with another unpaired electron in another NO2 molecule to form N2O4.

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Fri, 25 Sep 2009 13:55:51 +0800

UltimaOnline — Fri, 25 Sep 2009 13:55:51 +0800

‘A’ Level Qn.

By reference to the images provided on this Wikipedia URL ( http://en.wikipedia.org/wiki/Phenolthalein ), draw the mechanisms involved for the 4 structures of the phenolphthalein molecule ( IUPAC name 3,3-bis(4-hydroxyphenyl)isobenzofuran-1(3H)-one ) as the aqueous environment progresses with increasing pH from 0 to 12.

http://en.wikipedia.org/wiki/Phenolthalein

I’ll give some tips to help you along :
(but you should attempt to solve it by yourself first).

1 to 2
deprotonation of carboxylic; lone pair on carboxylate attacks trigonal planar carbocation.

2 to 3
deprotonation of both phenolic groups; lone pair on phenoxide forms pi bond; pi electrons in ring shifts down through ring to form pi bond with sp3 carbon; sigma bond cleaves to become lone pair on carboxylate.

3 to 4
hydroxide nucleophile attacks sp2 carbon outside rings; pi bond outside ring shifts into ring to regenerate the phenoxide.

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Thu, 24 Sep 2009 22:54:03 +0800

UltimaOnline — Thu, 24 Sep 2009 22:54:03 +0800

‘A’ & ‘O’ Level Qn.

On heating, Group I metal nitrate(V) compounds decompose giving the metal nitrate(III) and oxygen gas, while Group II metal nitrate(V) compounds decompose giving the metal oxide, nitrogen dioxide and oxygen gas. 15.35g of a mixture of sodium nitrate(V) and magnesium nitrate(V) was heated in a fume cupboard until no more gases were evolved. The water soluble part of the residue was dissolved in 1dm3 of water to give solution A. 10cm3 of solution A was mixed with 20cm3 of 0.02mol/dm3 acidified potassium manganate(VII) solution, giving solution B. The excess potassium manganate(VII) in 15cm3 of solution B required 6cm3 of 0.05mol/dm3 of sodium ethandioate solution for complete reaction.

a) Write all half-equations, balanced redox equations, and thermal decomposition equations relevant to the problem above.

b) Calculate the sample masses of sodium nitrate(V) and magnesium nitrate(V) in the mixture.

c) Draw the full displayed structural formulae of
sodium nitrate(III), magnesium nitrate(V), potassium manganate(VII) and sodium ethandioate.

Answers :
Sample mass of NaNO3 = 3.4g
Sample mass of Mg(NO3)2 = 11.95g

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Thu, 24 Sep 2009 22:47:27 +0800

UltimaOnline — Thu, 24 Sep 2009 22:47:27 +0800

‘A’ & ‘O’ Level Qn.

50cm3 of aqueous sulfuric(VI) acid of unknown molarity was mixed with 75cm3 of 0.45mol/dm3 of aqueous potassium hydroxide. 55cm3 of the resulting alkaline solution A was mixed with 80cm3 of 0.35mol/dm3 of phosphoric(V) acid. 90cm3 of the new resulting acidic solution B required 39.97cm3 of aqueous sodium carbonate of unknown molarity, for complete neutralization. In a separate experiment, 625cm3 of the same aqueous sulfuric(VI) acid solution required 192.3cm3 of the same aqueous sodium carbonate solution for complete neutralization.

a) Calculate the molarities of the sulfuric(VI) acid solution and the sodium carbonate solution.

b) Draw the full displayed structural formulae of sulfuric(VI) acid, potassium hydroxide, phosphoric(V) acid, sodium carbonate, and name all salts present in solution A and solution B.

Answers :
[H2SO4] = 0.2 mol/dm3
[Na2CO3] = 0.65 mol/dm3

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Thu, 10 Sep 2009 23:17:43 +0800

UltimaOnline — Thu, 10 Sep 2009 23:17:43 +0800

‘A’ Level Qn.

lianamaster asked :
Q1) What relative amounts of 1-bromobutane and 2-bromobutane are actually formed when butane is brominated?
Q2) If all C-H bonds had the same homolytic dissociation energy, what would be the relative amounts of 1-bromobutane and 2-bromobutane produced in the bromination of butane?

My reply / the Solution :

For Q1.

The reactivity (radical intermediate stability) factor depends on temperature, but bromination is a lot more selective than chlorination.

At 125 deg C, a tertiary radical is formed 1600x faster, and a secondary radical 82x faster, compared to primary radical.

You needn’t memorize these values; the exam question will provide the relevant values if they require you to factor this in.

Relative amount of 1-bromobutane :
Number of H atoms substitutable x reactivity (radical stability)
= 6×1 = 6

Percentage yield = 6/(328+6) = 1.80%

Relative amount of 2-bromobutane :
Number of H atoms substitutable x reactivity (radical stability)
= 4×82 = 328

Percentage yield = 6/(328+6) = 98.2%

For Q2.

Assuming all C-H bonds had the same homolytic dissociation energy, in other words, ignoring the reactivity (radical stability) factor, simply count the number of hydrogens substitutable to obtain the two isomeric monobrominated products :

Relative amount of 1-bromobutane :
Number of H atoms substitutable
= 6

Percentage yield = 6/(6+4) = 60%

Relative amount of 2-bromobutane :
Number of H atoms substitutable
= 4

Percentage yield = 4/(6+4) = 40%

youphoria queried :
is q1 an a level qn? i have never seen it before. particularly regarding “reactivity (radical stability)” factor. only qn 2 seen before.

My reply :

It’s actually University level Organic Chemistry.

At ‘A’ levels, the exam question will usually allow you to explain and apply only the simpler factor of number of hydrogens substitutable.

If the exam question at ‘A’ levels requires you to factor in primary, secondary, tertiary radical stability (or equivalently, dissociation energies of primary, secondary, tertiary C-H bonds), the question will clue you in with relevant data (a brief explanation, together with relative rates of free radical substitution involving primary, secondary, tertiary alkyl radical intermediates, eg. 5.0 > 3.8 > 1.0 for chlorination at 298K).

For the 2007 ‘A’ level H2 Chemistry exam, for instance, the question required the candidate to “describe, explain and calculate based on any ONE factor that determines isomeric product distribution of monochlorination of butane” (where the two factors are of course, number of hydrogens substitutable, and stability of alkyl radical intermediates).

So don’t worry too much about this.

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Tue, 08 Sep 2009 23:31:55 +0800

UltimaOnline — Tue, 08 Sep 2009 23:31:55 +0800

‘A’ Level Notes.

Redox potentials refer to reduction potentials (if a reduction half-equation is written) or oxidation potentials (if an oxidation half-equation is written).

As long as the equation is given as reversible (ie. equilibrium double half arrow notation used), it means you can reduce one species (eg. Fe3+) to a reduced state (eg. Fe2+), and you can also oxidize the reduced state species (eg Fe2+) to the oxidized state species (eg. Fe3+).

Hence, for such reversible equations, you can reverse the half equations and the standard potentials.

When written as a reduction half-equation, the redox potential is called the (standard) reduction potential (if standard conditions). When written as an oxidation half-equation, the redox potential is called the (standard) oxidation potential (if standard conditions).

The positive/negative sign of the redox potential is reversed when re-writing a reduction half-equation as an oxidation half-equation, or vice-versa.

Whichever way (reduction or oxidation) you choose to write it, the more positive the standard redox potential, the more the position of equilibrium lies to the right. The more negative the standard redox potential, the more the position lies to the left.

You can Google/Wikipedia the technical physics definition of redox/reduction/oxidation potentials (which is not required for ‘A’ level Chemistry). Technical physics definition aside, it is helpful for you to think of the meaning of “potential” as “tendency” or “inclination” or “propensity”.

Hence, the Reduction Potential numerical value (including positive/negative sign) describes the potential or tendency/inclination/propensity for a species to be reduced. The more positive the value, the more the position of equilibrium lies to the right (ie. the reactant species eg. Ag+ prefers to be reduced).

(Note : due to formatting problems, the word equation is given instead)
Eg. Silver cation + electron = (ie. left-to-right arrow) = Silver atom ( + 0.80V )
( + 0.80V means Silver cation likes to be reduced to Silver atom)
which can be also be written as :
Silver atom = (ie. left-to-right arrow) = Silver cation + electron ( – 0.80V )
(Indeed, – 0.80V means Silver atom doesn’t like to be oxidized to Silver cation)

Likewise, the Oxidation Potential numerical value (including positive/negative sign) describes the potential or tendency/inclination/propensity for a species to be oxidized. The more positive the value, the more the position of equilibrium lies to the right. (ie. the reactant species prefers to be oxidized).

(Note : due to formatting problems, the word equation is given instead)
Eg. Potassium atom = (ie. left-to-right arrow) = Potassium cation + electron ( + 2.92V )
( + 2.92V means Potassium atom loves to be oxidized to Potassium cation)
which can be also be written as :
Potassium cation + electron = (ie. left-to-right arrow) = Potassium atom ( – 2.92V )
( – 2.92V means Potassium cation hates to be reduced to Potassium atom)

Standard cell potential (which must be positive to be feasible, ie. for the current of a Galvanic/Voltaic cell to flow) is calculated as :

Oxidation potential at Anode + Reduction potential at Cathode

Reduction potential at Cathode – Reduction potential at Anode

As long as standard cell potential is positive, the reaction is feasible, and the current in the Galvanic/Voltaic cell will flow.

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Sun, 06 Sep 2009 12:33:57 +0800

UltimaOnline — Sun, 06 Sep 2009 12:33:57 +0800

‘O’ & ‘A’ Level Qn.

a) Is zinc hydroxide soluble in water? How about in excess OH-(aq)?
b) If we heat NaOH, do we get Na2O and water? How about Zn(OH)2?

Solution :

Zn(OH)2 is an insoluble ppt, but can form a soluble complex ion of tetrahydroxozincate(II) ion [Zn(OH-)4]2-(aq), in excess OH-(aq). The complex ion is soluble because it is capable of ion-dipole interactions with water, unlike Zn(OH)2(s).

NaOH is stable (as Na+ has relatively low charge density), and will not lose water to form Na2O. When excess water is evaporated away, you will get white solid NaOH(s).

On the other hand, Zn2+ has relatively higher charge density, is more unstable and hence Zn(OH)2 will lose water when heated to form ZnO(s).

Zinc oxide is yellow when hot, white when cold; the yellow colour arises from interactions in the crystal lattice, and not due to d-d* transition, since Zn2+ has no partially filled d-orbitals and doesn’t qualify as a transition metal.

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Sat, 05 Sep 2009 12:02:04 +0800

UltimaOnline — Sat, 05 Sep 2009 12:02:04 +0800

‘O’ & ‘A’ Level Qn.

What is the average distance between two air molecules in front of your face? (assume you’re at sea level and at room temperature.)

Answer :
3.42 X 10-7 cm
or
3.42×10-8 dm
or
3.42×10-9 m

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Sat, 05 Sep 2009 11:57:39 +0800

UltimaOnline — Sat, 05 Sep 2009 11:57:39 +0800

‘A’ Level Qn.

_{~~

Hydroxylamine NH2OH is a toxic substance to
many aquatic organisms, and it has chemical properties similar to
ammonia. For example, it reacts with hydrochloric acid to form the
salt hydroxylammonium chloride NH3OHCl.

(a) In an experiment, 25.0cm^{3 sample of the hydroxylammonium
chloride solution was titrated against a solution of 0.100M sodium
hydroxide, and 14.40 cm}3 of sodium hydroxide was needed for
complete neutralisation.

(i) With the aid of an eqn, determine if the nature of the end
point of titration is neutral, basic or acidic.

(ii) Calculate the PH value at the equivalence point of the
titration. The base dissociate value, Kb of hydroxylamine is
1.10×10^-8 mol/dm3.}_~

Points to note :

(Don’t worry about the counterbalancing ion Cl-, its just a spectator ion; it doesn’t undergo hydrolysis much because of its low charge density and stability from its strong ion-dipole interactions with water).

The HONH3+ ion is obviously acidic, as it has an available proton to donate, and no lone pairs available to accept protons. It’s a relatively weak acid though, not a strong one, just like NH4+ losing protons to become NH3 (since HONH2, like NH3, will again have the tendency to accept protons back, and become HOHN3+ or NH4+ again; to properly understand why, we have to look at both conjugate species, charge densities, stabilities, solute-solvent interactions and thermodynamics; not the focus of your question here. Suffice that students familiar with NH3 and NH4+ system are aware this is a weak base and its weak* conjugate acid, and extrapolate its similiarities to the HONH2 and HONH3+ system).

(*Many students just memorize without understanding “weak base = strong conjugate acid” without understanding, which is unfortunate (that most Sg students just memorize without understanding and appreciating what they study). I make sure my tuition students understand why, and therefore the limitations of applying such ideas. For instance, understanding why Markonikov’s Rule usually works (ie. stability of carbocation intermediate species), will enable you to recognize and understand when it actually fails (ie. when the major product is actually the anti-Markonikov product). Back to the “weak base = strong conjugate acid” idea, it’s definitely true but to a relative degree, based on relative stabilities. But the fact is, while NH3 is a weak base, NH4+ certainly isn’t a strong acid. If you truly understand Chemistry, rather than blindly memorizing, this fact shouldn’t confuse you at all.)

Titrating HONH3+ (again, you can ignore the Cl- counterbalancing spectator ion) against NaOH, a strong alkali, we end up with HONH2 (ie. deprotonated conjugate base form of HONH3+) at equivalence point (and spectator ions are Na+Cl-, ignore these harmless buggers), and HONH2 is obviously basic (due to an available lone pair on N; rather than the less available lone pair on O), as its 2 remaining protons are not at all acidic at this stage/form (since the conjugate base of HONH2, which is HONH-, is exceedingly unstable), and it is a lone pair available (on the N atom) to accept a proton from water.

Hence at equivalence point, the species present HONH2, will undergo hydrolysis (ie. reaction with water), in a Bronsted-Lowry acid-base proton transfer reaction, proton transferred from water to HONH2, to generate HONH3+ and OH-.

Therefore, your Initial Change Equilibrium (ICE) table will focus on to what extent (as specified by the Kb value), hydrolysis will occur to generate how much OH- ions; so you can calcuate molarity of OH- ions, and thus pOH, and therefore pH.

As a general guideline for acid-base equilibria questions, do an ICF (Initial Change Final) table first, in number of moles; then do an ICE (Initial Change Equilibrium) table next, in molarities. Bear in mind when applying Ka or Kb expression, it is always in molarities of all species, and take care to factor in changing volumes at different stages (eg. 1st equivalence point, 2nd equivalence point, etc) when calculating molarities.

Also note that the hydroxy group of HONH2 / HONH3+ is neither strongly acidic nor basic. Although there can be reaction pathways to forcefully protonate or deprotonate the hydroxy group (after being done with the amine group), but in water, no significant hydrolysis of the hydroxy group will occur. Treat this hydroxy group as you would an alcohol – neither strongly acidic or basic; inert as far as this titration question is concerned.

Also, in case the exam question requires you to calculate initial pH, you need to be able to calculate the initial molarity of HONH3+.

(a) In an experiment, 25.0cm3 sample of the hydroxylammonium chloride solution was titrated against a solution of 0.100M sodium hydroxide, and 14.40 cm3 of sodium hydroxide was needed for complete neutralisation. ~

Find the number of moles of OH- ions required for complete neutralization.

This is thus the number of moles of the weak monoprotic HONH3+ acid present.

Using the formula Molarity = Moles / Volume (dm3), you can find the molarity of the species HONH3+ at initial.

For equivalence point, the no. of moles of the species HONH2 is exactly the same as the no. of moles of HONH3+ at initial, but the volume will be different (since you added NaOH solution), hence the molarity of HONH2 at equivalence point is certainly different.

Analogy (for a diprotic acid) to understand why no. of moles of the species is always the same, just not molarity :
You begun with 10 teddy bears with 2 arms (representing protons) each. During titration neutralization, you’re chopping the arms off the teddy bears. At 1st equivalence point, there are still 10 teddy bears, only now they’re all one-armed. At 2nd equivalence point, there are still 10 teddy bears, only now they all have no arms left. Bear in mind while the number (of moles) of the teddy bears remain the same throughout, the molarity of the teddy bears keep changing, as the volume of the solution keeps increasing (as you keep adding the alkali).

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Thu, 03 Sep 2009 11:44:49 +0800

UltimaOnline — Thu, 03 Sep 2009 11:44:49 +0800

‘A’ Level Qn.

—-—-—-—-—-—-—-
With reference to the data booklet predict and describe the observations when FeCl3 (aq) is added to the following solutions. In each case, calculate the electromotive force or standard cell potential and write balanced redox equations.
i) Mn2(SO4)3
ii) CuI
How am I supposed to know what will happen? Help needed thanks!
—-—-—-—-—-—-—-

List of Standard Redox Potentials : http://infinity.usanethosting.com/Tuition/Stand…

Standard Cell Potential (ElectroMotive Force) = Standard Reduction Potential at Cathode + Standard Oxidation Potential at Anode.

If the cell potential is positive, then the redox reaction is feasible, and the current will flow (for galvanic/voltaic cells).

Note that for a standard solution of FeCl3 (for this question asked),
Standard Reduction potential for Fe3+ to Fe2+ is + 0.77V
Standard Oxidation potential for 2Cl- to Cl2 is – 1.36V

For experiment (i)
Standard Reduction potential for Mn3+ to Mn2+ is + 1.49V
Standard Reduction potential for SO4 2- to SO2 is + 0.17V
Standard Oxidation potential for 2SO4 2- to S2O8 2- is – 2.01V
Conclusion :
Under standard conditions, Redox reaction between Mn3+ and Cl- (to produce Mn2+ and Cl2) is feasible.

For experiment (ii)
Standard Reduction potential for Cu+ to Cu is + 0.52V
Standard Oxidation potential for Cu+ to Cu2+ is – 0.15V
Standard Oxidation potential for 2I- to I2 is – 0.54V
Conclusion :
Under standard conditions, Redox reaction between Fe3+ and both Cu+ and I- (to produce Fe2+, Cu2+ and I2) is feasible.

(Note : there are of course other species to consider as possible products of additional possible redox reactions, but unless the exam question specifies the redox potentials for such, otherwise based on the Data Booklet given in the exams, the above are the only directly relevant considerations).

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Thu, 27 Aug 2009 19:12:44 +0800

UltimaOnline — Thu, 27 Aug 2009 19:12:44 +0800

‘A’ & ‘O’ Level Qn.

125cm3 of a solution of phosphoric(V) acid were mixed with 75cm3 of 0.25mol/dm3 of barium hydroxide. A 50cm3 aliquot of the resultant mixture required 7.5g of sodium carbonate for exact neutralization, in which a gas was produced.

(a) Calculate the volume of the gas produced, at (i) r.t.p, and at (ii) s.t.p.

(b) Describe and explain (with equations) how the identity of this gas may be ascertained.

(c) By calculating the molarity of phosphoric(V) acid used, calculate the sample mass of solid phosphoric(V) acid that needs to be dissolved in 125cm3 of pure water to prepare this molarity.

(d) Draw the structural formulae for (i) phosphoric(V) acid, (ii) barium hydroxide, and (iii) sodium carbonate.

Answers :
(c) The sample mass of solid phosphoric(V) acid that needs to be dissolved is 19.7g (to 3 significant figures).
Note : getting© correct means you already have parts (a) and (b) correct. Check Wikipedia for (d).

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Sat, 22 Aug 2009 23:27:27 +0800

UltimaOnline — Sat, 22 Aug 2009 23:27:27 +0800

‘A’ Level Organic Chem Qn.

Qn : How do you differentiate between electron withdrawing groups and electron releasing groups (or benzene ring activator/deactivator)?

Look for positive formal charges (eg. on N of the NO2 group) or positive partial charges (eg. on C of COOH group) to identify electron withdrawing by induction effect (and where possible because of pi bonds present, eg. NO2 group, electron withdrawing by resonance effect as well).

Look for lone pairs which may be delocalized into the benzene ring by resonance; a resonance contributor sees a lone pair (on the atom bonded to the benzene ring) becoming a pi bond with the benzene ring, with a negative formal charge spread out over the ortho and para C atoms of the benzene ring. Viewed from another perspective, we say for instance, that the p orbitals of the halogen (eg. chlorine of chlorobenzene) overlaps with the pi orbitals of the benzene ring, giving the C-X (eg. C-Cl) bond partial double bond character.

Groups which donate electrons inductively (and/or by hyperconjugation{1}), as well as groups which donate electrons by resonance, are both ortho-para directing, though interestingly for different reasons{2}. Such groups activate the benzene ring towards electrophilic aromatic substitution, as they donate electrons into pi system of the benzene ring, making the pi electrons of benzene a stronger nucleophile.

Groups which withdraw electrons by induction or by resonance, are meta directing, and deactivate the benzene ring against electrophilic aromatic substitution, as they withdraw electrons away from the pi system of the benzene ring, making the pi electrons of benzene a weaker nucleophile.

Note :
{1} and {2} are not required to be understood at ‘A’ levels. Those interested to learn more may like to check out Wikipedia as well as the following image files (from Paula Yurkanis Bruice’s Organic Chemistry).

http://infinity.usanethosting.com/Tuition/Organ…
http://infinity.usanethosting.com/Tuition/Organ…
http://infinity.usanethosting.com/Tuition/Organ…
http://infinity.usanethosting.com/Tuition/Organ…

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Thu, 13 Aug 2009 01:41:02 +0800

UltimaOnline — Thu, 13 Aug 2009 01:41:02 +0800

Australian Science Olympiads - Physics, Chemistry, Biology.

http://www.aso.edu.au/www/index.cfm?itemid=31

The Chemistry (and I expect Physics as well) is close enough to the local H2 'A' level syllabus, but not the Biology though.

Suitable for use at 'A' levels (and equivalent, eg. NUS High Diploma, International Baccalaurate, Integrated Programs, etc) and possibly for 'O' levels (for those who fancy a challenge).

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Tue, 21 Jul 2009 22:47:02 +0800

UltimaOnline — Tue, 21 Jul 2009 22:47:02 +0800

'A' Level Qn.

An ester Q, was refluxed with aqueous sodium hydroxide and the resulting mixture then distilled. The distillate was subjected to hot, acidified KMnO4 which decolourized. The residue in the distillation flask, upon acidification, gave a white ppt.

Which of these could be Q?

A) CH3CO2C6H5

B) C6H5CO2CH3

C) CH3CO2CH2C6H5

D) CH3CH2CO2CH2C6H5

Answer (highlight with mouse) :

'A' & 'O' Level Chemistry Qns (A Collection) replied by UltimaOnline @ Tue, 21 Jul 2009 22:44:15 +0800

UltimaOnline — Tue, 21 Jul 2009 22:44:15 +0800

'A' Level Qn.

A sparingly soluble calcium salt ionizes according to the equation :

Ca3X2(s) <---> 3Ca2+(aq) + 2X3-(aq)

If the solubility product Ksp of Ca3X2 is S, what is the molarity of Ca2+(aq) at equilibrium?

Answer :

3((S/108)^1/5) which may be simplified to (9S/4)^1/5

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