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

last SG · 2010-07-26T23:40:35+00:00

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A couple of BedokFunland JC Buffer Riddles

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?

Final answer :

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.

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A buffer solution is titrated with NaOH(aq). Upon addition of 3 volumes of NaOH(aq), the buffer solution becomes most effective. Equivalence point is attained when 8 volumes of NaOH(aq) is added.

a) Calculate the ratio of the molarities of both members of the conjugate acid-base pair of the buffer solution, before the NaOH(aq) is added.

b) Given that the original pH of the buffer solution (ie. before NaOH(aq) is added) is 4.16, calculate the proton dissociation constant of the acidic component of the buffer.

Solution :

a) At maximum buffer capacity :

HA + OH- ---> A- + H2O

Initial (mol)         8y            3y              x-8y           n.a.
Change (mol)   -3y           -3y              +3y            n.a
Final (mol)         5y             0               x-5y            n.a.

At equivalence point :

HA + OH- ---> A- + H2O

Initial (mol)         8y            8y              x-8y          n.a.
Change (mol)   -8y           -8y              +8y          n.a.
Final (mol)          0              0                x              n.a.

First, complete the ICF table for equivalence point. Next, complete the ICF table for maximum buffer capacity, bearing in mind that the initial moles of HA and A- are the same (for both tables).

Since at maximum buffer capacity, [HA] = [A-], this implies 5y = x-5y and hence x = 10y.

Substituting x = 10y into the initial moles of A-, we have moles of A- = 2y.

Therefore, the ratio of the amounts of both members of the conjugate acid-base pair present in the buffer (before NaOH(aq) is added), is 8y HA : 2y A-, ie. 4 HA : 1 A-

b) Using the Henderson-Hasselbalch equation, we have

pH = pKa + log ( [base] / [acid] )

4.16 = pKa + log (1/4)

pKa = 4.762

Ka = 1.73 x 10^-5

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AJC_P2_2013

1(a)(iii) Is there any other solvent that is suitable besides the one given in the solution?

Also we required to know the functions of ether?

HCJC_P2_2013

1(b) How do we know that the precipitation of barium sulfate is exothermic and sodium sulfate is not?

ACJC lah, not AJC.

Grignard reagents, like LiAlH4, must be anhydrous and kept in dry ether, because they are sources of C- and H- strong Lewis bases, which are strong nucleophiles and strong Bronsted-Lowry bases, and hence will readily abstract protons H+ from moist air (which will result in your expensive Grignard and LiAlH4 reagents being wasted, and worse, pose a safety flammable explosive hazard due to the H2 generated for LiAlH4) if not kept safely solvated in a dry solvent.

You need to know that ether is an excellent solvent to use because it is inert (it is aprotic due to a lack of a partial positive H atom and thus non-acidic; and it is non-nucleophilic and non-basic as it's already bonded to 2 R groups which are non-viable leaving groups) and it's a good balance of polar (the partial negative charged O atom) and non-polar (the hydrophobic alkyl groups) and although it can't donate H bonds, it can receive H bonds, and consequently it is an excellent inert solvent which can dissolve almost all organic molecules (both polar and non-polar, both protic and aprotic), including phenol and phenylamine.

HCJC_P2_2013 - Wah biangz eh, all sodium salts are soluble lah (everything covered in O levels is examinable at A levels!). So obviously no sodium ppt forms. BaSO4 is famously insoluble (O levels!), and (as part of the H2 syllabus, you should know this mah) obviously ionic precipitations are exothermic due to the formation of (large no. of strong ionic bonds holding together the) giant ionic lattice structure that you see (with your naked eye) as a precipitate.

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Topic : atomic radius

Hi! This is from singapore cambridge nov a level exam year 2010,

Paper 1 qn 1

Qn: which element has the largest atomic radius?

So suggested answer is potassium, as the valence electrons of the atom of K experience the weakest nuclear charge effect among the 4 options above

And Kr is the atom which valence electrons experience the strongest attraction to the nucleus, as nuclear charge effect increases across the same period while shielding effect remain approximately constant across the same period, hence effective nuclear charge increases across the period

My question is , why not Kr ?

From Data booklet :

Atomic radii /nm :

Na : 0.186

Ar : 0.192

Why cant we extrapolate this observation and apply it for period 4?

From Wikipedia :

In order of atomic radius, covalent radius, van da waals radius :

Na : empirical: 186 pm, 166±9 pm, 227 pm

K : empirical: 227 pm, 203±12 pm, 275 pm

In order of covalent radius, van da waals radius :

Ar : 106±10 pm, 188 pm

Kr : 116±4 pm, 202 pm

Thank you (:

https://en.m.wikipedia.org/wiki/Sodium

https://en.m.wikipedia.org/wiki/Potassium

https://en.m.wikipedia.org/wiki/Argon

https://en.m.wikipedia.org/wiki/Krypton

It's not a fair comparison, because K (being a metal) doesn't have van der Waals radius, while Kr (being unreactive) *only* has van der Waals radius. You could say it's Cambridge's fault for setting this problematic question, but for A level purposes, since in the H2 syllabus, you focus only on the concept of increasing electronegativity and ionization energy across a period, instead of the different types of atomic radii (specifically covalent radius versus van da waals radius) which isn't really in the H2 syllabus, thus Cambridge (reasonably) expects you to just apply the concept of increasing electronegativity and ionization energy across a period, and that's it. Don't need to care about covalent radius versus van der Waals.

Other than atomic radius, Cambridge can test you on comparing the *ionic* radii of iso-electronic species, eg. K+ vs Cl- (both isoelectronic with Ar atom). That's about it. No covalent vs van der Waals radius.

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Hi all, would like to check on an MCQ from 2013.

Which gases that go into the bottom of the blast furnace undergo a reaction during the extraction of iron?

1. Carbon dioxide 2. Nitrogen 3. Oxygen

A. 1 only B. 1, 2 and 3 C. 1 and 3 only D. 3 only

Would you include nitrogen (forming nitrogen oxides) as your answer?

Again, it's Cambridge's fault for setting ambiguous qns like these (every year there'll be a couple of such qns, in which different school teachers will disagree with each other and/or with Cambridge, and teach their students different answers) at both O levels and (especially) A levels.

The most correct answer (regardless of whether it's the answer Cambridge wanted, which is always debatable for such qns), is that all 3 gases do indeed react at the high temperatures present in blast furnances (whether for extraction of iron from its ore, or other purposes).

As nitrogen (the most abundant gas in the atmosphere) is inevitably present in the air used by blast furnaces (air is afterall, the cheapest source of oxygen, and obviously all industries are profit-oriented and want to make as much money as possible by reducing costs), at such high temperatures of blast furnaces, it's inevitable that various oxides of nitrogen (NOx) are generated. Such oxides of nitrogen are toxic to humans, and are known to be an industrial medical health hazard for blast furnace workers.

While the O level syllabus doesn't specifically include mention of this health hazard suffered by blast furnace workers, nonetheless (assuming you believe Cambridge wants you to including nitrogen as the correct answer) at O levels and A levels, students are arguably expected to put 2 and 2 together (presence of nitrogen and oxygen, and at high temperatures) and conclude oxides of nitrogen may (and are) being produced in blast furnaces, albeit as undesirable (and medically hazardous) side-products.

Yup, I'm a tutor and I raised this question to check the consensus; its one of those that could be argued either way.

Appreciated Jurongresident's detailed reply, apologies if he spent the extra time to do it thinking I'm a student.

For what it's worth, the publishers (cambridge?) suggested answer is only carbon dioxide and oxygen, but I thought a student shouldn't be faulted to include nitrogen also.

Accounts of the need for NOX control in blast furnace has been documented, within the secondary textbook; we might also link it to the chapter on Air and Air pollution.

Alas, its one of those ambiguous questions for tutors and students to take note.

TYS publishers are not certainly not Cambridge. TYS publishers pay MOE for the right to publish the questions (the answers are by the various authors hired by the publishers), as the copyrights are co-owned by Cambridge and MOE. The official Cambridge mark schemes are (by law, due to a legal agreement between MOE and Cambridge) confidential and cannot be revealed to the public. It is fairly common (especially for A levels, when it becomes even more complicated) for all the different TYS publications available on the market to all give different answers for an ambiguous, challenging or problematic, open-ended Singapore A level exam question (every year there's a couple of such qns). This happens even for MCQs, but naturally to a lesser extent.

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When calculations on TM are done, when do we include the hydrated H2O or the anhydrous salt? Eg 2013 P2 Qn 2(c)(iii) we use the Mr of anhydrous salt

2013 Paper 2
Q3(a)(i) can reagent A be Na metal instead of NaOH?
Q4(a)(i) When we compare between main group and TM can we use NC increase across the group and SE remains relatively constant to compare ENC and from there explain the atomic radii? Since all the elements are in the same period?

2013 Paper 3
Q1(c)(i) When we draw the mechanism do we use equilibrium arrow? Sincs that is given in the question?

2012 Paper 2
Q2(a)(ii) Can we say that nitrogen containing hydrocarbon is combusted to produce NO gas?
Q2(c)(i) How do we know that delta G is supposed to be measured in joules?
Q5(a)(ii) Do we have to describe the mechanism by steps like nucleophilic attack and regeneration of catalyst in the mechanism?

2013 P2 Qn 2(c)(iii) - If the question doesn't specify "hydrated", you've to use the molar mass of the anhydrous form.

2013 Paper 2 Q3(a)(i) - You've only a small chance that Cambridge might show you mercy and give you that mark, for 3 reasons. 1stly, Cambridge is testing to see if the student knows that phenol, being more acidic than alcohols, can be deprotonated just by the use of NaOH(aq) alone. 2ndly, NaOH is a lot cheaper and less hazardous than Na. You should always only use the cheaper and safer reagent, if both reagents have the same effect. 3rdly, beyond the H2 syllabus, Na being such a strong reducing agent, can actually (under certain circumstances or with certain co-reagents) reduce the benzene ring (eg. Birch reduction).

Q4(a)(i) - Acronyms are not acceptable by Cambridge in the A levels. Don't over-use them even in your own self-revision. No, your answer is relevant only for period 2 and 3 elements excluding d block metals. For this question, you need to specify that 3d orbitals (being more diffused) provide relatively poor shielding for the 4s orbitals electrons, which are consequently more strongly electrostatically attracted to the positively charged nucleus, of which Fe and Cu have 6 and 9 more protons compared to Ca.

2013 Paper 3 Q1(c)(i) - Cambridge won't give a damn either way.

2012 Paper 2 Q2(a)(ii) - You won't get any marks for your answer. Because in the 1st place by definition of hydrocarbon, there is no nitrogen present (immediately zero marks for this part of the question liao). Next, even if some organic fuels contained nitrogen (eg. amine group, nitrile group, amide group, imine group, etc), that isn't the main mechanism (which Cambridge is testing you for) by which oxides of nitrogen are formed. It's the atmospheric N2 which reacts with atmospheric O2 at high temperatures that is being tested here. Be sure you mention the strong NN triple bond, the high activation energy involved, and the high temperatures present in combustion engines.

Q2(c)(i) - If before today you didn't know, after today you know liao. Yes you're supposed to have memorized this in the H2 syllabus lah!

Q5(a)(ii) - As long as you draw out the full mechanism correctly (yes including regeneration of the catalyst), you don't need to describe anything in words.

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A level 2014 paper 1 qn 29

Isnt treating a c=c double bond with aq bromine will result it to be a halohydrin ( a br bonded to one C, OH bonded to other C).

Treat Br to the phenol group will result it attached to the ortho position

So wouldn't the max number be 3 br atoms?

If diluted soln, then 3. If concentrated soln, maximum 4 (Br2 added across double bond) is possible. Cambridge knew many students would think as you did, hence they deliberately (and too kindly) left out 3 as a possible option.

But qn state it is aqueous bromine, not concentrated

Then, can i form dihalide from c=c by using conc br2, instead of X2 in ccl4, dark

Aqueous can be concentrated aqueous. Different from liquid. Yes you can obtain some dihalide (previous qn stated *maximum possible*, which could be a minor product, not necessarily major), but it won't be 100% yield, so you won't get marks for using this method. You should still specify X2 in CCl4 when asked.

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A level 2009 paper 2 qn 2g)

How do we know this zn2+ +2oh -> zinc hydroxide ?, ( cant find this in cs toh study guide): )

And why cant or can, instead write this hydrated [Zn(H2O)6]2+ form?

Do we need to know/memorise strength of ligands so that we would know which ligand displace which ligand?

E.g. Zn (OH)2 + NH3 -> [Zn (NH3)4 ]2+ + OH-

(Sorry didnt balance and didnt write state symbol )

Is the NH3 ligand always stronger than H2O ligand? As it displaces H2O ligand from cu2+ complex

In absence of OH-(aq), you write the hydrated [Zn(H2O)6]2+ form. With OH-(aq), you get Zn(OH)2(s) ppt.

No need to memorize ligand strength or ligand field strength (google if you don't know the difference). Just apply common chemistry sense (eg. ligand strength, nucleophilic strength and Bronsted-Lowry basicity strength, all parallel each other because all 3 are functions of Lews basicity strength, with some exceptions you needn't care about at A levels), and most importantly, follow the clues given in the question. If there's a colour change upon adding a reagent, it implies ligand substitution aka ligand exchange has occurred.

On a related note, whether the coordination complex generated has 4 or 6 ligands, is a complex matter (pun intended), and the underlying reasons cannot be fully covered at A levels. For A level purposes, try to familiarize (no need to blindly memorize 100+ coordination complexes, their geometries and colours, etc) yourself with the dozen or so complexes more commonly encountered at A levels. For this purpose, Chang Kim Seng & Jeanne Tan's book, and George Chong's book, will be more useful than CS Toh's book (which is summarized lecture notes). Your own JC should have given you a list as well.

Tips : if the ligands are small and neutral, eg, H2O ligands, there will usually be 6 ligands, ie. octahedral geometry. If the ligands are large, eg. Cl ligands, there will usually be only 4 ligands, ie. tetrahedral geometry, due to steric hindrance or van der Waals repulsions. If the ligands are small but have high anionic charge densities, eg. OH- ligands, there will usually be only 4 ligands, ie. tetrahedral geometry, due to anionic electrostatic repulsions.

The notable special case (which the H2 syllabus specifies you need to memorize), is the deep-blue coloured and with square planar geometry, tetraamminecopper(II) ionic coordination complex, [Cu(NH3)4]2+. The reasons why Cu2+ only accepts 4 NH3 ligands, and why the ionic complex's geometry is square planar instead of tetrahedral, is because of the Jahn–Teller effect and other reasons too complex to be dealt with at A level purposes. For A level purposes, other than the few commonly encountered coordination complexes you need to memorize (their colours and geometries), simply follow the clues which will be provided by Cambridge in the question.

Yes, NH3 ligand is stronger than H2O ligand, but don't forget molarity (ie. concentration) also determines the position of equilibrium in the Kstab or Kf equation (google it), and therefore determines which ligands substitute or exchange away which ligands. In other words, a weaker ligand with a higher molarity, can substitute away a stronger ligand with a lower molarity. Which is why the Cambridge question will specify, "added in excess" to ensure the ligand substitution occurs, due to the molarity affecting the position of equilibrium.

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What affects Nucleophilic and electrophilic strength?

Is it How e- deficient an ephile is, and how delta + positive charge on the nucleophile?

Does the electronegativity of an electrophile play part too?

Nucleophilicity and Electrophilicity is affected by several factors :

- Magnitude of charges and charge densities, whether formal charges or partial charges, which is affected by electronegativity (eg. is the negative formal charge on a C atom or O atom?), % s orbital character in hybridized orbital (affecting charge density), atomic/ionic radius, shielding effect, induction and resonance.

- Availability of vacant orbitals on electrophile (addition-elimination mechanism requires lower Ea than SN1 & SN2 aliphatic nucleophilic substitutions, which is one reason why in Periodicity of H2 syllabus, SiCl4 readily undergoes hydrolysis, while CCl4 is hydrolysis resistant; the availability of 3d orbitals in Si (but not C) creates the potential for addition-elimination hydrolysis mechanisms with lower activation energy).

- Availability of lone pair on nucleophile, which is affected by electronegativity, % s orbital character in hybridized orbital (affecting lone pair availability), atomic/ionic radius, shielding effect, induction and resonance.

- Steric hindrance by bulky groups (either on electrophile reducing electrophilicity, or on nucleophile reducing nucleophilicity and increasing Bronsted-Lowry basicity instead) around the electrophilic or nucleophilic atom.

- Viability and stability of leaving group upon elimination (in nucleophilic aliphatic and acyl substitutions ; for electrophilic aromatic substitutions the incoming electrophilic substitutes away a proton, to restore aromaticity and resonance stabilization energy), which is affected by electronegativity (eg. is the negative formal charge on a C atom or O atom?), % s orbital character in hybridized orbital (affecting stabilization of charge), atomic/ionic radius, shielding effect, induction and resonance.

- Endothermicity of bond dissociation enthalpy and its activation energy, or (in other words) the strength of the covalent bond required to be cleaved (be sure to quote Data Booklet values whenever relevant), which is in turn affected by % s orbital character in hybridized orbital (affecting strength of covalent bond, eg. sp2-sp3 bonds stronger than sp3-sp3 bonds), atomic/ionic radius and period of element (effectiveness of overlap, whether head-on for sigma bonds or side-on for pi bonds, is greatest between 2nd electron shell orbitals and other 2nd electron shell orbitals, and much less effective between 2nd and 3rd/4th/5th electron shell orbitals, or 3rd/4th/5th electron shell orbitals and other 3rd/4th/5th electron shell orbitals, regardless of whether hybridized or not), electronegavity & induction (increasing polarity of covalent bond increases it's ionic character and thus strengthens it, which is why CO triple bond is stronger than NN triple bond in Chemical Bonding, or why Al3+ and Mg2+ salts are acidic in Periodicity and Acid-base Equilibria, due to decreasing polarity and thus weakening of O-H bond in water ligands) and resonance (the C-X bond in aryl & vinyl halides have partial double-bond character in their resonance hybrid, hence aryl & vinyl halides are resistant to hydrolysis).

- % s orbital vs p orbital character of hybridized orbitals, as mentioned above, affects availability of lone pair on Lewis base (ie. nucleophile and/or Bronsted-Lowry base; eg. why amines are more nucleophilic & basic over imines, and imines over nitriles; for amides the resonance factor takes precedence), affects strength of covalent bond required to be cleaved (and hence activaiton energy required), and affects stability of charge (both in nucleophile / Bronsted-Lowry base, as well as in the leaving group after elimination).

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That 2009 qn,

So the ammonia ligand replaced oh ligand?

So zinc hydroxide is soluble?

Probably sparingly?

From 2009 paper 2 qn 2g - Eh jialat lah, you really don't understand coordination complex equilibria, izzit? You made 2 errors right here : 1st of all, in this question OH- isn't a ligand at all. It's a counter ion which precipitates out Zn2+ as Zn(OH)2(s). *If* you had added excess NaOH(aq), giving a sufficiently high molarity of OH-, then the excess OH- can start to function as ligands, and that's how you get the tetrahydroxozincate(II) coordination complex.

But in this qn, you did *not* add excess NaOH(aq), rather you added excess NH3(aq), giving a sufficiently high molarity of NH3, then the excess NH3 can start to function as ligands, and that's how you get the tetraamminezinc(II) coordination complex.

All 'insoluble' salts are actually 'sparingly soluble'. But most importantly, you need to be able to explain to Cambridge *why and how* the 'sparingly soluble' metal hydroxide solid precipitate, eg. Cu(OH)2(s) dissolves when adding in excess a solution containing a suitable ligand, eg. NH3(aq), in terms of shifting of positions of equilibria across 2 (or more) reaction equations.

Eqn 1 : NH3(aq) + H2O(l) ---> NH4+(aq) + OH-(aq)

Eqn 2 : Cu2+(aq) + 2OH-(aq) ---> Cu(OH)2(s)

Eqn 3 : Cu2+(aq) + 4NH3(aq) ---> [Cu(NH3)4]2+(aq)

When ammonia solution is added to copper(II) sulfate solution, the hydrolysis of ammonia (Eqn 1) generates OH-(aq), and when Qsp (aka Ionic Product) > Ksp (aka Solubility Product), position of equilibrium in Eqn 2 shifts to the RHS, generating Cu(OH)2(s) precipitate. When excess ammonia solution is added, the position of equilibrium in Eqn 3 shifts to the RHS, resulting in a decrease in molarity of Cu2+(aq), which causes the position of equilibrium in Eqn 2 to shift to the LHS, which causes the copper(II) hydroxide precipitate to dissolve.

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