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

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BedokFunland JC's 'A' Levels H2 Chemistry Notes on :

Chemical Bonding (intermolecular interactions) and Proteins (tertiary structure)

Many Singapore JC teachers still subscribe to the outdated idea of regard van der Waals strictly as purely instantaneous dipole - induced dipole between non-polar molecules or parts of molecules. This is outdated, and the correct updated version that Cambridge uses today is : Van der Waals forces are a collective term for all 3 types : instantaneous dipole - induced dipole, permanent dipole - permanent dipole, and permanent dipole - induced dipole.

Increasing number of electrons, and/or increasing size of atom/ion/molecule, and/or increasing polarizability of electron clouds; any and all of these (which are intimately related) will result in greater magnitude of the dipoles (whether permanent or induced) and partial charges, and hence stronger or more extensive van der Waals interactions, stronger instantaneous dipole - induced dipole interactions, stronger permanent dipole - permanent dipole interactions, and stronger permanent dipole - induced dipole interactions, all of these.

For hydrophobic interactions of non-polar or weakly polar 'R' groups (ie. groups not capable of hydrogen bonding or ionic bonding) of amino acid residues of a polypeptide chain in a protein : as hydrogen bonds are significantly stronger, water would rather hydrogen bond amongst themselves (because the stronger the bonds or interactions formed, the more stable the resulting structure/product/species), hence pushing or repelling away non-polar 'R' groups. These non-polar 'R' groups subsequently get pushed until they come within close proximity of each other, whereupon the consequent van der Waals forces of attraction (whether they be permanent dipole - permanent dipole, or permanent dipole - induced dipole, or instantaneous dipole - induced dipole) will hold these non-polar or weakly polar 'R' groups together, contributing to the protein's tertiary structure.

The exam-smart student will include both key phrases "hydrophobic" and "van der Waals" in his answer on the nature of the interaction betwen non-polar or weakly polar 'R' groups in a protein.

Disulfide bridges, or disulfide bonds, are covalent bonds between two S atoms, usually present in proteins as occurring between cysteine amino acid residues of the polypeptide. These are the covalent bonds that are involved in the perming and rebonding of hair. (the term "rebond" means to recreate the disulfide covalent bonds that allow hair to hold its shape, whether it be curly or straight.)

If your hair is curly, the disulfide bonds are cleaved (via adding of hydrogen; a reduction process for sulfur since the OS of the sulfur atoms is decreased; since sulfur is more electronegative than hydrogen), then your hair is straightened, and the disulfide bonds are recreated (hence "rebond") (via removal of hydrogen; an oxidation process for sulfur since the OS of the sulfur atoms are increased; since sulfur and sulfur have equal electronegativity) these newly recreated disulfide bonds hold your hair straight after a rebonding session.

Hydrogen bonds, belong to a class on its own. They are stronger than permanent dipole - permanent dipole forces of attraction, but weaker than covalent bonds. They are therefore neither, but possess some characteristics of both.

Hydrogen bonds have similar strength to ion-dipole attractions (of which there are two types : ion - permanent dipoles are naturally usually stronger than ion - induced dipoles), though (depending on the charge densities of the ions involved) ion - dipole attractions are usually slightly stronger than hydrogen bonds.

An illustrative example of this, is that phenol precipitate or benzoic acid precipitate, dissolves when aqueous sodium hydroxide is added : deprotonation of phenol or benzoic acid, which only had limited hydrogen bonding with water, results in the phenoxide anion or benzoate anion which experiences the somewhat stronger (than hydrogen bonding) ion - (permanent) dipole interactions with water, allowing the precipitate to dissolve. An upgrade in the class of inter-species physical interactions, so to speak, resulting in greater solubility in water.

An interesting case in point, would be whether amino-acids in zwitterionic form have hydrogen bonds or ionic bonds with each other. The answer is : the forces of attraction that results between the R-NH3+ group on one zwitterionic amino acid, and the R-COO- group on the next zwitterionic amino acid, may be regarded as either. Again, the exam-smart candidate would write both, but with brief qualifications or explanations, so the examiner understands, "this candidate really knows his stuff!" and award the full marks.

The interaction between R-NH3+ and R-COO-, may be regarded as hydrogen bonding, because the interaction is between a partially positively charged hydrogen on R-NH3+ and a formally negatively charged oxygen on R-COO-. But because R-NH3+ may also be regarded as cationic (though the positive formal charge is on N, not H; more on that in a while) and the R-COO- group may be regarded as anionic; hence one may argue ionic bonding occurs between zwitterionic amino acids. Either way, this explains why amino acids are least soluble in water, at their isoelectric point.

Related to the above discussion, is whether ammonium cations and hydroxide anions experience hydrogen bonding or ion-dipole attractions with water. Both are acceptable (some chemists would label such as the former, while other chemists would label such as the latter), and the exam-smart student will again, write both, but with brief qualifications and explanations.

Technically, NH4+ experiences hydrogen bonding rather than ion-dipole interactions with water. This is because :
1) the tetrahedral geometry of the NH4+ ion will result in water molecule physically interacting with the partially positively charged H atoms, rather than the N atom.
2) the positively formally charged N atom would be even more strongly electron-withdrawing by induction, resulting in a larger magnitude of a partial positive charge on the H atoms, resulting in stronger hydrogen bonding with water (compared to say, ammonia with water).

The hydroxide ion, consists of a partially positively charged H atom, and a formally negatively charged O atom. The H atom therefore experiences hydrogen bonding with water molecules, while the O atom experiences what may be regarded as ion - permanent dipole interactions with water, or what may be regarded as especially strong hydrogen bonding with water.

For hydrogen bonding to occur, there must be :
#1 - a partially positively charged H atom; the magnitude of the partial positive charge must be significant enough, this occurs naturally when H is covalently bonded to electronegative F, O or N atoms.
#2 - a partially or formally negatively charged F, O or N atom with at least one available lone pair; because hydrogen bonding is mainly electrostatic in nature, hence there needs to be a partial or formal negative charge on the F, O or N atom for hydrogen bonding to occur with the partially positively charged H atom. Because hydrogen bonding is also somewhat covalent in nature, there needs to be at least one available lone pair for donation of the hydrogen bond (which may be viewed as almost dative in nature, but not quite, since H atoms only has one 1S orbital and cannot violate its duplet, ie. a H atom can have at most only one lone pair or one bond pair at any time).

Therefore, the negatively formally charged O atom in hydroxide ion experiences particularly strong hydrogen bonding, which accordingly may arguably also be labelled as an ion - permanent dipole interaction.

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

Warning :

Being exam smart, means to write both possible answers if the question is ambiguous, or the context or concepts involved, allow for such. This will greatly increase the exam candidate's chances of successfully scoring the allocated marks. However, the exam smart student *must* qualify and explain (at least briefly) why both answers may be argued to be correct or relevant. If this qualification or explanation is not done, the examiner/marker may think the student is trying to smoke/bluff his way through, by giving multiple answers to a single question; which will result in the student losing the marks.

Example of an ambiguous question : "What is the product formed when the following compound is reacted with KMnO4(aq)?".

The exam smart candidate would write : "Dear Examiner Sir/Mdm, because the question failed to specify umambiguously the conditions (eg. temperature) at which the reaction is carried out, therefore I will explore both alternative possibilites as follows."

"If hot, concentrated, acidified KMnO4(aq) is used, oxidative cleavage of the alkene double bond would occur, and the structure of the product would be... If cold, dilute, alkaline KmNO4(aq) is used, then two OH groups would be added across the alkene double bond, and the structure of the product would be..."

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'A' Level Qn.

From Wikipedia :

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

Snake wine (蛇酒, pinyin: shéjǐu; rượu rắn in Vietnamese) is an alcoholic beverage produced by infusing whole snakes in rice wine or grain alcohol. The drink was first recorded to have been consumed in China during the Western Zhou dynasty and considered an important curative and believed to reinvigorate a person according to Traditional Chinese medicine^[1]. It can be found in China, Vietnam and throughout Southeast Asia.

The snakes, preferably venomous ones, are not usually preserved for their meat but to have their "essence" and snake poison dissolved in the liquor.

Click here for photos.

'A' Level Qn :

By the use of relevant diagrams and/or equations, suggest why drinking venomous snake wines will not usually poison the drinker.

Answer :

Elmhurt College's Virtual ChemBook by Charles E Ophardt

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'A' Level Qn - Kinetics.

1. Do we include a catalyst when writing down the rate equation?

2.Explain the following observations as fully as you can.

When ammonia is passed over a platinum gauze, the rate of decomposition into nitrogen and hydrogen is independent of the partial pressure of ammonia. However, at very low pressure, the rate is directly proportional to the partial pressure of ammonia.

1. Yes. Since the rate of reaction is affected by the presence and molarity of a catalyst. For any given reaction, the numerical value of the rate constant k is said to be affected only by temperature. This is true. But when a catalyst is employed, the reaction pathway involves an alternate mechanism with a lower activation energy (at the rate determining elementary step), which means to say that this may be considered a different reaction altogether, and hence has a different (a much larger) rate constant k numerical value.

2. At sufficiently high pressure, the catalyst (ie. platinum) is saturated with its substrate (ie. ammonia, the excess reactant), and hence altering (eg. further increasing) the partial pressure (or molarity) of ammonia (ie. the excess reactant) does not alter the rate of reaction, ie. the reaction is zero order with respect to ammonia. At lower partial pressures (or molarities) of ammonia, the limiting reactant is the ammonia itself, and increasing partial pressure (or molarity) of ammonia (ie. the limiting reactant) will result in an increase in rate of reaction, ie. the reaction is first order with respect to ammonia.

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'A' Level Qn

Can iron(III) iodide exist in aqueous solution? Explain.

aka "The Legend of the Mythical FeI3(aq)"

Solution :

No, because Fe3+ would oxidize I- to I2, and be itself reduced to Fe2+. Or you can think of it as I- would reduce Fe3+ to Fe2+, and be itself oxidized to I2.

Always quote relevant Data Booklet values in your answer. Under standard conditions, ie. 1 mol/dm3 of Fe3+/Fe2+ and I2/2I- , the standard reduction potential of Fe3+ to Fe2+ is +0.77V; while the standard oxidation potential of 2I- to I2 is -0.54V.

Hence, under standard molarities, we have cell potential = reduction potential @ cathode + oxidation potential @ anode (alternatively, cell potential = reduction potential @ cathode - reduction potential @ anode) = +0.77V + -0.54V = +0.23V, which proves that our conjecture as stated in our opening paragraph above holds true.

Moreover, notice that the stoichiometry or mole ratio of I- to Fe3+ in a solution of FeI3(aq) is not the required (ie. for our calculations to hold true) 2 : 1 ratio, but is actually 3 : 1.

This means that as predicted by Le Chatelier's principle, having a greater molarity (ie. molar concentration) of I- will shift position of equilibrium to the left, meaning that the oxidation potential of our I- to I2 is now even more positive than before, which results in our cell potential being even positive than +0.23V, which in effect means all the more I- will react with Fe3+ in a redox reaction to generate I2 and Fe2+.

[Reduction] 2Fe3+ + 2e- ---> 2Fe2+

[Oxidation] 2I- --> I2 + 2e-

[Balanced Redox] 2Fe3+ + 2I- --> I2 + 2Fe2+

So we have 2 moles of FeI3 reacting to generate one mole of I2 and 2 moles of Fe2+ and there are 4 moles of I- left over, which is equivalent to having 2 moles of FeI2(aq), which can indeed exist happily ever after in aqueous solution, unlike the non-existent fairy-tale mythical FeI3(aq).

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'A' Level Qn.

You have 2 solutions of Cu2+, both having equal molarities of 1mol/dm3 Cu2+(aq). In one solution, you add 2 mol/dm3 of methylamine. In the other solution, you add 1mol/dm3 of ethylenediamine (aka ethane-1,2-diamine). In which solution do you expect to have a larger Kc value (to be precise : stability constant aka formation constant) for the formation of the complexed Cu2+(aq) ions? Explain.

Solution :

Although the enthalpy of Cu-N bond formation is approximately the same for both ligands (or put in another way, both ligands' donor N atoms have approximately equal donor power towards Cu2+), however there will be a significantly greater molarity of the ethylenediamine copper(II) complex ions, over the dimethylamine copper(II) complex ions.

There are primarily 2 reasons for this phenomenon, which is known as the Chelate Effect.

The 1st reason, is that of Proximity. Once one end of a bidentate ligand (eg. ethylenediamine) is coordinated, the other end is forced to be in close proximity to the central metal ion. Only a rotation of the ligand is required for further coordination, which makes it far more likely that this would occur. In contrast, just because a monodentate ligand (eg. methylamine) is coordinated, has no direct effect on the chance or rate of another monodentate ligand being coordinated as well.

The 2nd reason, is that of Thermodynamics, specifically Entropy. With the monodentate methylamine ligand, we have :

2CH3NH2 + [Cu(H2O)6]2+ --> [Cu(CH3NH2)2(H2O)4]2+ + 2H2O

Notice we have 3 species on the left, and 3 species on the right. Almost no change in entropy.

With the bidentate ethylenediamine ligand, we have :

NH2CH2CH2NH2 + [Cu(H2O)6]2+ --> [Cu(NH2CH2CH2NH2)(H2O)4]2+ + 2H2O

Notice we have 2 species on the left, and 3 species on the right. A significant, positive, favourable change in entropy.

Hence, with the bidentate (or in general, multidentate or polydentate ligands) ligands, compared to monodentate ligands, we observe a more positive (ie. favourable) change in entropy, which in effect resuts in a more negative (ie. favourable) change in Gibbs Free Energy.

The two reasons above, explain why the Kc (stability constant or formation constant) value for multidentate or polydentate ligands, are larger than the corresponding Kc value for monodentate ligands, ceteris paribus (ie. all else being equal).

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'A' Level Qn.

What type of reaction is this?

ROH + RCOOH --> RCOOR + 2HO

Solution :

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

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

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

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

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

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

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'A' Level Qn.

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

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

Solution :

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Either way, you obtain your calcium carbonate white precipitate.

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'A' Level Qn - Chemical Equilibria.

Question :

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

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

Answers :

157.5g and 2.667mol

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

Guidance or Notes on Equilibria :

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

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

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

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

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

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

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

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

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

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‘A’ Level Qn : Acid-Base Equilibria

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

Answer :
2.20×10-4 (to 3 sig fig)

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'O' Levels / 'A' Levels Qn : Qualitative Analysis.

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

and

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

Here's the Solution (pun!)

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

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

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

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

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

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

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

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

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

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

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

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

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On the QA test for nitrate(V) ions :

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

Al ---> Al3+ + 3e-

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

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

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

NH4 + OH- ---> NH3 + H2O

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

NH3 + H2O ---> NH4 + OH-

The hydroxide ions turn red litmus paper blue.

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'A' Level Qn.

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

A) Hydrogen bonding

B) Permanent dipole - permanent dipole attraction

C) Ion - permanent dipole attraction

D) None of the above

Answer :

D) None of the above.

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

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

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'A' Level Qn.

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

Solution :

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

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

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

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

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'A' Level Qn : Acid-base Equilibria.

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

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

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

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

Answers :
a) 10.5

b) 7.93

c) 4.55

d) (graph)

e) 8.30

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

h) 3N2H4 --> 4NH3 + N2

i) -171.3 kJ/mol

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'A' Level Qn : Hess Law.

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

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

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

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

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

Given :

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

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

Answers :
Qn2) +90.25 kJ/mol

Qn3) +172.5 kJ/mol

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'A' Level Qn : Redox & Structures

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

Answers :

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

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

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'A' Level H2 Notes on "Chemical Bonding" and "Solubility of Grp II compounds"

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

will compounds with hydrogen bonding have pd pd also?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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'A' Level Qn : Acid-Base Equilibria.

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

Answer :

2.20 x 10^-4 moles

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'A' Level Qn : Acid-Base Equilibria.

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

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

Answers :

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

b) Buffer : initial 8.9, final 8.9

Base : initial 10.95, final 10.43

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

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'A' Level Qn : Acid-Base Equilibria

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

Answer : 1.864 x 10^-3 mol/dm3

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'A' Level Qn : Acid-Base Equilibria

Calculate the pKa of dihydrogen monoxide at 298K.

Answer :

15.7

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