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

last SG · 2010-07-26T23:40:35+00:00

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'A' levels / 'O' levels / Scholarstic Assessment Test

Equal volumes of 0.1mol/dm3 of sodium phosphate and 0.1mol/dm3 of magnesium nitrate were mixed together, to precipitate out magnesium phosphate solid.

a) Determine the limiting reactant.
b) Calculate the molarity of aqueous sodium cations remaining at the end of the reaction.

Ans :

a) Magnesium nitrate.
b) 0.15 mol/dm3

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'A' Level Qn.

for the electrophilic addition of aqueous bromine to propene, why does the Br add to the first carbon atom rather then the second carbon atom?

Here's my guess. Someone correct me if I'm wrong. Bromine enters the first carbon atom rather than the second one due to steric hindrance. If you take observe the structure of propene, the first carbon atom is surrounded by only hydrogen atoms while the second carbon atom is surrounded by 2 alkyl groups. So if bromine enters the second carbon atom, the resulting molecule will be unstable.

Good attempt, but errorneous. Since alkenes are trigonal planar about their sp2 hybridized carbons, there is little steric hinderance for the incoming electrophile.

daniel798 persisted :

since the secondary carbocation is more stable, wouldnt it mean the bromine will add to the second carbon to form a secondary carbocation. then H2O will add to the first carbon atom?

I patiently continued :

No, to generate the secondary carbocation, the Br+ electrophile will be added onto the terminal C atom, and not to the secondary C atom.

When aqueous solvent is used, the propene nucleophile will attack the electrophilic bromine atom (due to inter electron repulsion between the electron-rich alkene and the Br2 molecule, a temporary dipole is induced in the Br2 molecule rendering it electrophilic), consequently forming either a secondary carbocation (major product pathway) or a primary carbocation (minor product pathway).

The water nucleophile (note : although JCs commonly teach the nucleophile being the hydroxide ion nucleophile, but this is technically inaccurate as the molarity of water nucleophile far exceeds the molarity of hydroxide ions, even if the hydroxide ion is a much stronger nucleophile than water) then attacks the carbocation intermediate, consequently losing a proton (to rid itself of the destabilizing positive formal charge on the O atom).

Accordingly, the secondary alcohol would be the major product, and the primary alcohol would be the minor product.

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'A' Level Qn.

Audi asked :

For each of the following reactions A and B,

(i) identify the 2 acids and 2 bases present

(ii) suggest, with reasons, which ion or molecule is the stronger acid, and which is the stronger base.

A : NH3 + H2O <----> NH4+ + OH- Kc =1.8 X 10^-5 mol/dm3

B C6H5O- + CH3CO2H <---> C6H5OH + CH3CO2 Kc=1.3 X 10^5 mol/dm3

Solution :

For A, NH3 is the base, H2O is the acid, NH4+ is NH3's conjugate acid, OH- is H2O's conjugate base. Since position of equilibrium lies to the left, OH- ion is a stronger base than NH3, and NH4+ is a stronger acid than H2O.

For B, phenoxide ion is the base, ethanoic acid is the acid, phenol is phenoxide ion's conjugate acid, and ethanoate ion is ethanoic acid's conjugate base. Since position of equilibrium lies to the right, phenoxide ion is a stronger base than ethanoate ion, and ethanoic acid is a stronger acid than phenol.

As an exercise in Organic Chem, you should also be able to come to the same conclusion for which are the stronger acids & bases, by looking at electronegativities, induction and resonance considerations of the charges (ie. analysis of electronics) of LHS vs RHS species, as an alternative to using Kc values. If you do this correctly, the conclusions should concur with the Kc values.

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'A' Level Qn.

Audi asked :

Calculate the pH of a solution which is 0.40 moldm-3 w.r.t ethanoic acid and 0.20mol dm-3 w/r/tsodium ethanoate

[Ka for ethanoic acid = 1.80 X 10^-5 mol/dm3)

Calculate the change of pH of 1.0 dm3 of the above solution in (a) when 0.050 mol of solid NaOH is added (assume no change in vol)

What is the change of pH when 0.050mol of NaOH is aded to 1.0 dm^3 of pure water?

Solution :

I recommend you do not use the Henderson-Hasselbach equation. Instead just use the formulaic definition of Ka, which (like any Kc) is basically the molarities of RHS over the molarities of the LHS, for the proton dissociation equation of the weak acid.

Do the ICE table for CH3COOH <---> H+ + CH3COO-. You'll notice that for a buffer system, the equilibrium molarities of the weak acid and its conjugate base are approximated back to initial molarities, which means you can skip the ICE table if you wish. Your ICE table should have units as molarities.

Plug the Equilibrium molarities into the Ka formula, make molarity of protons [H+] the subject, and hence find pH. (Ans : 4.44)

Do the ICF table for adding the sodium hydroxide. Usually we use moles as units for the ICF table, but since the volume of solution does not change in this instance, we can work directly in molarities.

Your ICF table for CH3COOH + OH- --> CH3COO- should have Initial molarities 0.4, 0.05 and 0.2. Your Final molarities should be 0.35, 0, 0.25.

Plugging these values into your Ka formula (for acidic buffers, you can use Ka of the acid or Kb of the conjugate base), and making molarity of protons the subject, you obtain pH = 4.5986 or 4.6 to 3 sf. Hence the change in pH is approx 0.1586 or 0.159 to 3 sf.

As to the adding of NaOH(s) to pure water (pH 7 at rtp), notice that NaOH is a strong alkali. Hence, no Kb value is required. The pOH is 1.3 hence pH is 12.7, hence change in pH is 5.7.

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'A' level H2 Chem Qn (Acid-Base Equilibria)

Given 150 cm3 of 2.0 mol/dm3 of ammonia solution (pKa of NH4+ = 9.25) at r.t.p., what volume of 0.5 mol/dm3 of hydrochloric acid is needed to prepare a buffer solution with pH = 9.5?

Solution (pun intended) :

Let x be the volume of HCl added to achieve a pH of 9.5

NH3 + H+ ---> NH4+
I (mols) 0.3 | 0.5x | 0
C (mols) -0.5x | -0.5x | +0.5x
F (mols) 0.3 - 0.5x | 0 | 0.5x

[NH3] = (0.3 - 0.5x) / (0.15 + x)
[NH4+] = (0.5x) / (0.15 + x)
At pH 9.5, pOH = 4.5 and [OH-] = 10^-4.5

Hydrolysis of NH3 : NH3 + H2O <---> NH4+ + OH-

Kb = ( [NH4+] [OH-] ) / [NH3]
1.7783 x 10^-5 = [(0.5x) / (0.15 + x)] [10^-4.5] / [(0.3 - 0.5x) / (0.15 + x)]
x = 0.21596 dm3 = 216 cm3 (to 3 sf)

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'A' level H2 Chem Qn (Acid-Base Equilibria)

Q1. Which of the following is true for pure (ie. non-acidic, non-alkaline) water at (i) 274 K? (ii) 299 K?
a) pH = 7 b) pH < 7 c) pH > 7 d) pOH = 7 e) pOH < 7 f) pOH > 7

Q2. Is a solution with pH = 7 considered acidic, neutral or alkaline, at (i) 274 K? (ii) 299 K?

Answers :

Q1.(i) - pH > 7 and pOH > 7
Q1.(ii) - pH < 7 and pOH < 7
Q2.(i) - acidic
Q2.(ii) - alkaline

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‘A’ Level H2 Chem : Notes on Acid-Base nature of Salts.

Just consider each of the species present in the following salts, to determine whether they are acidic or basic.

Example #1 : sodium chloride salt.
Example #2 : sodium ethanoate salt.
Example #3 : ammonium chloride salt.
Example #4 : ammonium ethanoate salt.

For salt #1, Na+ low charge density, doesn’t undergo hydrolysis. Cl- low charge density, doesn’t undergo hydrolysis. Hence NaCl is neutral.

For salt #2, Na+ low charge density, doesn’t undergo hydrolysis. CH3COO- not very stable (somewhat stabilized by resonance, but still not as stable as say, Cl-, because O is smaller than Cl, therefore higher charge density), hence in a bid to stabilize itself, it will abstract ( “pluck” or “grab” ) a proton from water. Hence we say it undergoes “hydrolysis” (ie. chemical reaction with water), to generate OH- ions. Hence CH3COO-Na+ is alkaline, because molarity of OH- > molarity of protons.

For salt #3 NH4+ is acidic, because N has a positive formal charge ( central dogma of chemistry is “charge is destabilizing” ) and will try to lose a proton in a bid to lose the positive formal charge to stabilize itself. Hence protons from NH4+ dissociated into water, molarity of protons increases. Cl- low charge density, doesn’t undergo hydrolysis. Therefore, NH4+Cl- is acidic.

For salt #4 Both cation and anion of the salt undergoes hydrolysis. NH4+ is clearly acidic, and CH3COO- is clearly basic/alkaline (an alkaline solution is what results when a base undergo hydrolysis). So overall, is the salt NH4+CH3COO- more acidic or more basic/alkaline? To determine this, you must compare the Ka value of NH4+ with the Kb value of CH3COO-. Remember that the magnitude of a Ka value tells you how strong the acid is. Similarly, the magnitude of a Kb vaule tells you how strong the base is. Accordingly, if the Ka value of the acidic cation is larger in magnitude than the Kb of the basic anion, then ammonium ethanoate is overall acidic. If the Kb value of the basic anion is larger in magnitude than the Ka of the acidic cation, then ammonium ethanoate is overall basic/alkaline.

Additional notes on acidic salts (covered in the H2 Chem syllabus under Inorganic Chemistry) :

Aluminium salts are usually acidic. This is due to the high charge density of the tripositive Al3+ cation, which consequently has a high propensity to accept dative co-ordinate covalent bonds from water ligands to form hexxaaquoaluminium complex ions. In such a complex ion, electron density is withdrawn inductively from the O-H bonds in the water ligands, through the O-Al bonds and towards the highly charged Al3+ ion. Consequently the O-H bonds in the water ligands weaken, and protons dissociate as a result.

AlCl3(s) + 6H2O -> [Al(H2O)6]3+(aq) + 3Cl(aq)
[Al(H2O)6]3+(aq) -> [Al(H2O)5)OH)]2+ + H+(aq)

Accordingly, magnesium salts, eg. MgCl2, are also acidic, but much less so compared to the aluminium salts (since the Al3+ ion is tripositive, and thus of a higher charge density, compared to the dipositive Mg2+ ion).

Also note that due to the diagonal relationship in the periodic table indicating similar charge densities between the smaller Be2+ and the larger Al3+ ions, beryllium and aluminium salts are approximately equally acidic.

The solutions of BeCl2 and AlCl3 salts have an approximate pH of 3. The solution of MgCl2 has an approximate pH of 6.5.

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Topic : Acid-Base Equilibria

Regarding the Henderson-Hasselbalch equation, it's not that you shouldn't use it, it's that you need never use it. You can of course, use it if you want. But it's *never* compulsory to use it.

Units for Molarity (ie. molar concentration) = M = mol/dm3.

0.2 moles of sodium benzoate is dissolved in 10 mL of 0.1 M benzoic acid solution. Calculate the pH. (Kb of benzoate ion = 1.5x10^-10).

Solution (pun intended) :

Assuming you used solid sodium benzoate and the volume of solution didn't change, then we have a buffer solution consisting of weak acid benzoic and its conjugate base benzoate. Since it is a buffer system, to calculate pH we can consider either the hydrolysis of the base (ie. Kb of benzoate), or the proton dissociation of the acid (ie. Ka of benzoic acid).

Let's consider the hydrolysis of the benzoate ion.

Benzoate ion + water <--> Benzoic acid + hydroxide ion
I (molarity) 20 | - | 0.1 | 0
C (molarity) -x | - | +x | +x
E (molarity) 20-x | - | 0.1+x | x

Since x is much smaller than initial molarities of both weak acid and its conjugate base (since this is a buffer solution), hence equilibrium molarities are approximated back to initial molarities

Kb = [benzoic acid] [hydroxide ion] / [benzoate ion]
1.5 x 10^-10 = [0.1] [OH-] / [20]
[OH-] = 3 x 10^-8
pOH = 7.52
pH = 6.48

Also, note that because the only reaction between benzoic acid (an acid) and benzoate ion (a base) is a Bronsted-Lowry acid-base proton transfer reaction (don't say "neutralization reaction" that's 'O' level terminology that shouldn't be used at JC/Poly level). In such a reaction, a proton is transferred from the Bronsted acid to the Bronsted base.

benzoic acid + benzoate ---> benzoate ion + benzoic acid.

As you can see, the final moles (and hence the molarities) of benzoic acid and benzoate ion do not change. If it helps you to understand, you can take it that BOTH the weak acid AND its conjugate base both undergo hydrolysis to an equal extent, and hence cancel out each other's effect.

And since the moles and molarities (of both weak acid and conjugate base) do not change, we have the accurate values for molarities of both the weak acid and its conjugate base, and therefore by using either the Kb (of the conjugate base) or Ka (of the weak acid), we are considered to already have taken everything (ie. proton dissociation from the weak acid AND hydrolysis of the conjugate base) in our calculations. This is because Ka and Kb are related by Kw.

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'A' Level H2 Chem Qns. Topic : Organic Chemistry mechanisms.

These questions test whether the candidate is able to apply fundamental concepts of OC (Organic Chem) to deduce mechanisms involving (either as reactants or products) compounds not usually given much attention to, in the H2 Chem syllabus. Examples of such compounds are ethers and epoxides.

Draw the mechanisms for :

1) Propene + aqueous bromine; followed by alcoholic KOH (heat under reflux). Product : 1,2-epoxypropane.

2) Ethoxyethane + hydroiodic acid (heat under reflux). Immediate products : ethanol and iodoethane. Final products (with excess HI) : iodoethane and water.

3) Epoxyethane + aqueous acid (heat under reflux). Products : ethan-1,2-diol.

4) Epoxyethane + methylammonium chloride (heat under reflux). Products : 2-hydroxy-N-methyl-ethylamine.

5) 1,2-epoxypropane + ethyl magnesium bromide (Grignard reagent), followed by hydrolysis. Hint : alkyl addition occurs at the less sterically hindered carbon atom. Product : pentan-2-ol

Solution :

1) Propene nucleophile attacks Br+ electrophile; H2O nucleophile attacks secondary carbocation; proton lost to rid O+ of +ve formal charge; basic OH- ion abstracts proton from hydroxy group of 1-bromo-propan-2-ol; -ve formal charged O atom nucleophilically attacks (ie. donates dative bond to) electrophilic C atom (bonded to the electronegative halogen) without violating C's octect; thanks to the halogen which leaves as a halide ion (halogens are good leaving groups because halide ions are stable, due to their large ionic radius and therefore low charge density).

2) Immediate products : O atom of epoxide is protonated, giving rise to a +ve formal charge on O atom and a strong +ve partial charge on the adjacent C atom, rendering it electrophilic. Halide ion nucleophile then attacks the electrophilic C atom, causing heterolytic cleavage of the O-C bond, ridding the O atom of its +ve formal charge.

Final products : the ethanol further reacts with hydroiodic acid as follows : the hydroxy group is protonated, upgrading a poor leaving group (because hydroxide ion is unstable) to an excellent leaving group (because water is stable). Iodide ion nucleophile attacks electrophilic C atom and water is eliminated.

3) O atom of epoxy group protonated; water nucleophile attacks electrophilic C atom; O atom loses proton to rid itself of the +ve formal charge.

4) O atom of epoxy group protonated; methylamine nucleophile attacks electrophilic C atom; N atom loses proton to rid itself of the +ve formal charge.

5) Grignard reagent (ie. carbon / alkyl nucleophile) attacks the less sterically hindered C atom (ie. the 1st C atom not the 2nd). The C-O bond heterolytically cleaves to generate the pent-2-oxide ion, which abstracts a proton from water (ie. hydrolysis) to generate the final pentan-2-ol product.

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'A' levels / Challenging 'O' levels

Chemistry Calculations Qn.

Titanium(IV) oxide TiO2, is heated in hydrogen has to give water and a new titanium oxide. TixOy. if 1.598g of TiO2 produced 1.438g of TixOy, what is the formula of the new oxide?

Solution :

No of moles of Ti before redox = No of moles of Ti after redox

this implies :
1 (No of moles of TiO2) = x (No of moles of TixOy)

this implies :
1 (1.598 / 79.9) = x (1.438 / (47.9x + 16y))

this implies :
x / y = 2 / 3

Therefore, formula of titanium oxide = Ti2O3

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'A' Levels / Challenging 'O' Levels

Chemistry Calculations Qn.

1) A mixture of MgSO4.7H2O and CuSO4.5H2O is heated until a mixture of anhydrous salts is formed. if 5.0g of the mixture give 3.0g of the anhydrous salts, calculate the percentage by mass of MgSO4.7H2O in the mixture.

2) A solution of acid contains 7.30g of HOOC(CH2)nCOOH per dm^3 of solution. 20.0cm^3 of this acid was titrated against NaOH, using phenolpthalein indicator. in the titration, 25.0cm^3 of the alkali was used. the sodium hydroxide contained 1.36g of hydroxide ion per dm^3. Calculate the value of n in the molecular formula and therefore name this acid.

Solution :

Q1. Let x and y be the moles of CuSO4 and MgSO4 present. You can write up to 3 mathematical equations.

?x + ?y = 2g
and
?x + ?y = 3g
and
?x + ?y = 5g

Choose any 2 of these equations, and solve simultaneous equations for x and y.

Q2. From mass concentration of OH- solution, find molarity of OH- solution. Hence find moles of OH- ions used in titration. Hence find moles of protons (ie H+ ions) neutralized. Hence find moles of the (diprotic) dicarboxylic acid present. Hence find molarity of the (diprotic) dicarboxylic acid. Find molar mass of the (diprotic) dicarboxylic acid, in terms of algebraic variable n. Hence find (continuing in terms of algebraic variable n) mass concentration of the (diprotic) dicarboxylic acid. Therefore equate this mass concentration value to 7.30g/dm3. The acid is hexane-1,6-dioic acid, or simply hexanedioic acid.

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A levels H2 Chem
Topic : Organic Chem

For (C6H5)CH=CHCH3,
when this compound undergoes nitration, which position does the propene direct the NO2+?

also when this molecule undergoes hydration in H3PO4, where does the H+ and the OH- add since both carbon atoms have 1 R group. When adding to an alkene, does length of R group affect the product or only the number of R groups affect the product?

when an alkyl halide undergoes elimination, under sodium hydroxide in alcoholic medium,
CH3CH2X -----> CH2=CH2 + HX

then,

HX + OH- -----> H2O + X-

why does HX react with OH- ion when NaOH is in alcohoic medium, so there aren't any OH- ions present?

also when reacting alkyl halide with ammonia in alcoholic medium,
CH3CH2X + NH3(alc) -----> CH3CH2NH2 + HX

why doesn't HX react with NH3 here?

Regard the alkene group as an alkyl group (by induction, it is neither electron donating nor withdrawing; by resonance it can be both electron donating and withdrawing). In other words, a weakly activating, ortho-para directing substituent.

Accordingly, the product of the electrophilic aromatic subtitution is generated as the NO2+ electrophile substitutes away a proton on either the para position (since a propenyl group presents significant steric hinderance) or the ortho position (since there are two ortho positions versus one para position). For A levels, you need to give both ortho and para isomeric products.

During hydration of the alkene double bond, the more stable carbocation intermediate will be the carbocation atom (ie. C with +ve formal charge) bonded directly to the benzene ring, because the benzene ring can stabilize the carbocation by both induction/hyperconjugation and resonance, while the terminal methyl group can stabilize the carbocation by induction/hyperconjugation only.

Therefore, the major alcohol product will have the OH group bonded to the C atom next to the benzene ring, since a more stable intermediate means lower activation energy means faster rate of reaction means more of *that* product formed, which we identify by use of the label "major product".

As for length of alkyl chain, yes the greater the length, the greater its electron-donating by induction capacity. Although this is not usually tested at H2 Chem level, so just FYI.

The equation you quoted for elimination of HX from alkyl halides using alcoholic OH- ions, is misleading. If you know the mechanism (I encourage all H2 students to approach OC (Organic Chem) from an understanding-of-mechanism instead of blind-route-memorization approach), then you would realize that the equations given are misleading.

The OH- ion in alcoholic solvent has a greater propensity to behave as a base rather than a nucleophile (since the alcoholic solvent affords less hydrogen bonding stabilization to the OH- ion compared to aqueous water solvent). Understand that bases accept protons in a bid to stabilize themselves. And so the OH- ion under alcholic solvent functions as a base (and carries out elimination) rather than as a nucleophile (to carry out substitution).

A Note on Nucleophilicity versus Basicity : Comparing F- with I-, you should be able to articulate that F- is a stronger base (since higher charge density = more unstable = more desperate to protonate itself) while I- is a stronger nucleophile (since greater ionic radius = electron charge clouds more polarizable = more willing to donate dative bonds to electrophiles).

Mechanism of Elimination of HX :
The alcoholic OH- ion abstracts ("plucks") a beta proton; the sigma bond between the beta carbon and beta proton becomes a pi bond between the alpha and beta protons; without violating the alpha carbon's octet thanks to halogen which leaves as a halide ion (remember that halogens are good leaving groups because halide ions are stable because of large ionic radius and thus low charge density; of the halogens only F is a poor leaving group because F- is unstable due to high charge density).

Accordingly, if you wish to represent the elimination reaction with an equation (although the full mechanism is always preferred).

CH3CH2X(alc) + OH-(alc) -----> CH2=CH2(alc) + H2O(alc) + NaX(alc)

Write the above equation, and *not* the two step equation you quoted (ignore your JC's notes, don't use or buy or study *any* JC's lecture notes, every JC's notes are different and many are often filled with nonsense, just use CS Toh's A level Study Guide instead for the A levels).

Similarly, based on the mechanism, the nucleophilic attack of ammonia on alkyl halides should be written as :

CH3CH2X(alc) + NH3(alc) -----> CH3CH2NH3+X-(alc)
or
CH3CH2X(alc) + NH3(alc) -----> CH3CH2NH3+(alc) + X-(alc)

Note that the positive formal charge is on the N atom on the ammonium salt, because the N atom has 4 bond pairs and 0 lone pairs, but N is in Grp V. To obtain the amine (ie. the conjugate base form), simply deprotonate the ammonium salt using aqueous NaOH or KOH.

CH3CH2NH3+X-(aq) + Na+OH-(aq) -----> CH3CH2NH2(aq) + H2O(aq) + Na+X-(aq)
or
CH3CH2NH3X(aq) + NaOH(aq) -----> CH3CH2NH2(aq) + H2O(aq) + NaX(aq)

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'A' Level H2 Chem.

Topic : Balancing Redox Eqns

Q1.
Fe2+ + MnO4- => Mn2+ + Fe3+

Write the reduction half-equation :
MnO4- ---> Mn2+

Balance the heteroatoms, ie. Manganese. Already balanced.

Balance the O atoms.
MnO4- ---> Mn2+ + 4H2O

Balance the H atoms by adding protons.
8H+ + MnO4- ---> Mn2+ + 4H2O

Balance the charges by adding electrons.
8H+ + MnO4- + 5e- ---> Mn2+ + 4H2O

Write the oxidation half-equation :
Fe2+ ---> Fe3+

Balance the heteroatoms, ie. Iron. Already balanced.

Balance the O atoms. No need.

Balance the H atoms. No need.

Balance the charges.
Fe2+ ---> Fe3+ + e-

To obtain lowest common multiple for electrons,
multiply oxidation half-equation by 5
5Fe2+ ---> 5Fe3+ + 5e-

Add LHS + LHS, RHS + RHS for both half-equations :

5Fe2+ ---> 5Fe3+ + 5e-
+
8H+ + MnO4- + 5e- ---> Mn2+ + 4H2O
=
5Fe2+ + 8H+ + MnO4- ---> 5Fe3+ + Mn2+ + 4H2O

Q2.
Cr2O7 2- + C2O4 2- => Cr3+ + CO2

Write down the reduction half-equation.
Cr2O7 2- ---> Cr3+

Balance the heteroatoms, ie. Chromium.
Cr2O7 2- ---> 2Cr3+

Balance the O atoms.
Cr2O7 2- ---> 2Cr3+ + 7H2O

Balance the H atoms.
14H+ + Cr2O7 2- ---> 2Cr3+ + 7H2O

Balance the charges.
6e- + 14H+ + Cr2O7 2- ---> 2Cr3+ + 7H2O

Write the oxidation half-equation.
C2O4 2- ---> CO2

Balance the heteroatoms, ie. Carbon.
C2O4 2- ---> 2CO2

Balance the O atoms. Already balanced.

Balance the H atoms. None present.

Balance the charges.
C2O4 2- ---> 2CO2 + 2e-

To obtain lowest common multiple for electrons,
multiply oxidation half-equation by 3
3C2O4 2- ---> 6CO2 + 6e-

Add LHS + LHS, RHS + RHS for both half-equations :

3C2O4 2- ---> 6CO2 + 6e-
+
6e- + 14H+ + Cr2O7 2- ---> 2Cr3+ + 7H2O
=
3C2O4 2- + 14H+ + Cr2O7 2- ---> 6CO2 + 2Cr3+ + 7H2O

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'A' Level H2/H1 Chem

Challenging 'O' level Chem

Topic : Chem Calculations

A mixture of methane and ethane with combined sample mass of 13.43g is completely burned in oxygen. If the total sample mass of CO2 & H2O produced is 64.84g, calculate the percentage by volume and percentage by mass, of the two alkanes in the mixture.

Solution :

CH4 + 2 O2 ---> CO2 + 2 H2O
C2H6 + 7/2 O2 ---> 2 CO2 + 3 H2O

Let x and y be the moles of CH4 and C2H6 burnt.

x(12+4) + y(12+12+6) = 13.43g ---------- (equation 1)

and

x(12+16+16) + 2x(2+16) + 2y(12+16+16) + 3y(2+16) = 64.84g ---------- (equation 2)

From equation 1,
16x + 30y = 13.43
x = (13.43 - 30y)/16

Substitute into equation 2,

44(13.43 - 30y)/16 + 36(13.43 - 30y)/16 + 88y + 54y = 64.84g
44(13.43 - 30y) + 36(13.43 - 30y) + 16(88)y + 16(54)y = 16(64.84)
590.92 - 1320y + 483.48 - 1080y + 1408y + 864y = 1037.44
-128y = -36.96

y = 0.28875 moles of C2H6

x = 0.29797 moles of CH4

Percentage by volume of CH4 = percentage by moles of CH4 (assuming constant temperature and pressure) = moles of CH4 / total moles of both gases
= 0.29797 / (0.28875 + 0.29797) = 50.8%
Percentage by volume of C2H6 = 100% - percentage by volume of CH4 = 49.2%

Percentage by mass of CH4 = sample mass of CH4 / total sample mass of both gases
= 4.76752 / (4.76752 + 8.6625) = 35.5%
Percentage by mass of C2H6 = 100% - percentage by mass of CH4 = 64.5%

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'A' Level H2 Chemistry

Topic : Solubility (Ionic) Equilibria

Given that Ksp of silver sulfate is 1.4 x 10^-5, what is the sample mass of solid silver nitrate to be added to
250cm3 of saturated silver sulfate solution, such that the mass solubility of silver sulfate is reduced to 0.3121 g/dm3?

Solution :

Let sample mass of AgNO3(s) added be y grams.
=> No of moles of AgNO3(s) added = (y/170) mol
=> [Ag+] added = (y/170)/(250/1000) = 0.023529y mol/dm3

Since molar solubility of Ag2SO4(s) reduced to (0.3121/312.1) = 1x10^-3 mol/dm3

Ag2SO4(s) <---> 2Ag+(aq) + SO4 2-(aq)
[ 2(1x10^-3)+ 0.023529y ] [1x10^-3]

=> [ 2(1x10^-3)+ 0.023529y ] x [1x10^-3] = 1.4x10^-5
=> (5.5363x10^-4)y^2 + (9.4118x10^-5)y + (-0.013996) = 0
=> y = 4.9437g = 4.94g (3 sf)

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'A' Level H2/H1 Chemistry
Challenging 'O' Level Chemistry
Topic : Redox Calculations

5 cm3 of 0.10 mol dm-3 of potassium manganate (VII) reacts with 12.0cm3 of nitrogen monoxide at r.t.p. A colourless solution containing a brown precipitate of manganese (IV) oxide, MnO2 was produced. Determine the final oxidation number of nitrogen in the compound and hence suggest a possible formula of the nitrogen-containing product of the reaction.

Solution :

When MnO4- is reduced to MnO2, only 3e- are transferred per mole of MnO4- used.
5x10^-3 moles of MnO4- implies that 5x10^-4 x 3 moles of electrons are removed from 12/24000 moles of NO.
This implies that 5x10^-4 moles of NO loses 1.5x10^-3 moles of e-.
This implies that each mole of NO loses 3 moles of e-.
Hence, new O.S. = = (+2) + (+3) = +5

Therefore, the nitrogen-containing product of the redox reaction could be either dinitrogen pentoxide N2O5, or the uninegative nitrate(V) ion NO3-.

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Drbiology posted :

Hey guys! Here's a question I can't seem to do. It's in redox topic:

The average oxidation number of the 2 sulfur atoms in sodium thiosulfate, Na2S2O3, is +2.

When solid sulfur containing the radioactive isotope ^35 S is boiled with sodium sulfite, Na2^32SO3, a radioactive sample of sodium thiosulfate is produced. Adding HCl(aq) to this sample causes all the ^35 S to precipitate as sulfur. The resulting solution contains non-radioactive sulfite ions.

i) Draw the structure formula of the sulfite ion.

ii) Use your answer to (i) and the reactions described to suggest a structure for the thiosulfate ion produced. On your structure, label any radioactive atom.

iii) Deduce the oxidation number for each sulfur atom in the thiosulfate ion.

I hope that experts in this forum could help. This assignment is due tmrw btw. thanks in advance! :D

Drbiology, first a warning : the concepts required to properly and fully understand this aspect of chemistry, include formal charges, oxidation states, resonance contributors and resonance hybrids, are almost never properly taught in the JC. Even if you were to ask your JC teacher about these concepts, more often than not, he/she will say, "it's not in the syllabus, you don't need to know, don't ask so much."

It's true that it's not strictly required by the syllabus (by which reasoning, this question is an unfair question and therefore you should skip this qn, no?), but without understanding about formal charges, oxidation states, resonance contributors and resonance hybrids, JC students can never have a correct fundamental understanding of chemistry, which is unfortunate.

If you're interested, you can always try asking your JC teacher, or your own private tuition teacher (if you have one), as well as Google and Wikipedia these out yourself (this is afterall, the Age of the Internet, where "All Knowledge is Available to All Mankind").

Solutions :
(i)
Latin name : sulfite ion
Stock name : sulfate(IV) ion
Kekule structure (all 3 resonance contributors as well as the resonance hybrid shown)
http://en.wikipedia.org/wiki/Sulfite

(ii)
Latin name / Stock name : thiosulfate ion
Kekule structure (the resonance hybrid shown; the resonance contributors are simply any two of the three O atoms taking turns to be the singly-bonded, negatively formal charged O atoms. Note that Ionic Charge = Sum of Formal Charges.)
http://en.wikipedia.org/wiki/Thiosulfate

Radioactive S atom is the terminal (ie. non-central) S atom, of the thiosulfate ion.

(iii) Oxidation Number (ON) = Oxidation State (OS) = Formal Charge + Electronegativity consideration.
OS of the central S atom = (0) + (+4) = +4
OS of the terminal S atom = (0) + (0) = 0
(Notice that for electronegativity consideration, since S and S are 'own people', or as we say in chinese : "zi ji ren" (mandarin) or "kaki lang" (hokkien) or "kaki nang" (teochew) or "ji gei yan" (cantonese), meaning both atoms having the same electronegativity since both are of the same element, hence electronegativity consideration is zero between the two S atoms.)

(Technically, there is a slight difference between OS and ON, but for 'A' level purposes they are considered exactly the same thing.)

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Iodine and chlorine react together to form compound X (ICln).

When 0.0010 mol of X was reacted with an excess of KI(aq), all of its iodine was converted into I2. The iodine liberated required 40.0 cm3 of 0.10 mol/dm3 sodium thiosulfate, Na2S2O3, for complete reaction.

Calculate the value of n in ICln.

Identify the reactants and products :

Reactants : ICln, KI

Products : I2, KCl

Let the coefficient of ICln be 1.

To balance the chlorine, coefficient of KCl must be n.

To balance the potassium, coefficient of KI must be n.

To balance the iodine, coefficient of I2 must be (n+1)/2

Accordingly, write the balanced equation (using coefficients above) for the reaction between ICln and KI to generate I2 and KCl.

By stoichiometry of the balanced redox equation (between iodine and thiosulfate), we note that each mole of ICln generates 2 moles of I2.

By stoichiometry of the balanced reaction between ICln and KI, this implies (n+1)/2 = 2, hence n = 3

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Organic Chemistry (Synthesis problem)

The common name for the active ingredient in Panadol, is paracetamol. Other names, more descriptive of its structure, are : N-(para-hydroxyphenyl)ethanamide, or para-ethanoylaminophenol.

Q1. Based on the names given above, draw the structure of paracetamol.

Ans :

Click here.

Q2. From the starting material of phenol, propose a synthesis pathway and draw the mechanisms for this pathway, to generate paracetamol.

Ans :

Click here for synthesis outline.

From phenol, use dilute nitric(V) acid and dilute sulfuric(VI) acid to carry out mono-nitration. Use fractional distillation to separate the ortho and para isomers.

From para-nitrophenol, reflux with Sn (or Zn or Fe) and excess concentrated HCl(aq), followed by NaOH(aq), to reduce the nitro group to amine. Alternatively, hydrogen with nickel catalyst at 300 deg C may be used for the reduction.

From para-aminophenol, use limited amounts of (ie. as limiting reactant) ethanoic anhydride (or for H2 Chem studens not familiar with acid anhydries, you may use ethanolyl chloride instead) to carry out nucleophilic acyl substitution to generate paracetamol. Notice that the (phenyl) amine group is more nucleophilic than the (phenolic) hydroxy group, because nitrogen is less electronegative than oxygen, and hence more willing to donate dative bonds during nucleophilic attack.

Mechanism wise, H2 Chem students should be able to draw the mechanism for every step of the pathway, except for the reduction of nitro group to amine, and possibly the last step (the mechanism for nucleophilic acyl substitution is addition-elimination, usually taught only to H3 students). But H2 Chem students are encouraged to self-learn this addition-elimination mechanism, because nucleophilic acyl substitution is responsible for organic chem reactions that H2 Chem students should be familiar with, including generating esters, hydrolyzing esters, generating amides, hydrolyzing amides, etc.

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ElectroChem Qn.

What happens when you mix aqueous tin(II) chloride and acidified aqueous H2O2?

Solution :

There are 5 possible half-equations, 2 reduction half-equations and 3 oxidation half-equations :

H2O2 can be reduced to H2O
Sn2+ can be reduced to Sn

H2O2 can be oxidized to O2
Sn2+ can be oxidized to Sn4+
Cl- can be oxidized to Cl2

Accordingly, there are 6 possible balanced redox equations. Calculate all 6 cell potentials, and select all redox reactions whose cell potentials are positive, and furthermore highlight the redox reaction with the most positive cell potential.

If time is short, and you wish to give only one, most feasible redox equation, here's the shortcut : select the reduction half-equation with the most positive reduction potential, and select the oxidation half-equation with the most positive oxidation potential, add these potentials up to give you the cell potential of the most feasible redox reaction.

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