I find the following equation to be good practice for my students (both 'O' levels and 'A' levels). Note that this is a simple, non-ionic, non-redox equation.

 

You know how you learnt at 'O' levels how nitrogen dioxide contributes to acid rain? Well, try writing a balanced equation for the reaction. You might find it helpful to be told that one of the two products is dinitrogen monoxide (for 'A' level students, also draw the Kekule structure or displayed structural formula of dinitrogen monoxide, showing all lone pairs and formal charges).

 

Note : you can easily google the balanced equation, but the point of this question is to teach students the correct (ie. systematic) way to balance this and other equations, instead of the haphazard trial and error method.

 

I suggest fellow educators (teachers and/or tutors) use this equation to help your students learn the correct (ie. systematic) way of balancing (simple, non-ionic, non-redox) equations.

 

I think there is a typo right? should be nitrogen monoxide and not dinitrogen monoxide right?

cos i google the equation and it happens otherwise.

Hi, no there is no typo. The purpose of this exercise is in balancing. The chemistry of oxides of nitrogen is interesting with many possible pathways. When nitrogen dioxide, for instance, is dissolved in water, depending on temperature, pressure, etc, a variety of products may be formed including nitrous acid, nitric acid, nitrogen monoxide, etc.

 

For the purpose of this question, assume (ie. if it is given in the question itself, that) you obtain dinitrogen monoxide and nitric acid. Balancing this equation can be educational for students. Only 2 out of 10 of my students could balance this equation, before I taught them the systematic way to balance it.

 

Try it?

 

Hint : remember how you used 'x' and 'y' algebraic variables when balancing the combustion of any hydrocarbon of unknown formula? (ie. CxHy + (???)O2 ---> (???) CO2 + (???) H2O). Do the same for our nitrogen dioxide question.

 

UltimaOnline's Solution :

 

NO2 + H2O ---> N2O + HNO3

 

Let coefficient of H2O be x.

 

NO2 + xH2O ---> N2O + HNO3

 

Then coefficient of HNO3 becomes 2x.

 

NO2 + xH2O ---> N2O + 2xHNO3

 

Let coefficient of N2O be y.

 

NO2 + xH2O ---> yN2O + 2xHNO3

 

Then coefficient of NO2 becomes 2x+2y.

 

(2x+2y)NO2 + xH2O ---> yN2O + 2xHNO3

 

Looking at oxygen, we have :

 

2(2x+2y) + x = y + 6x

 

Simplifying, we get :

 

x = 3y

 

Notice that x is the larger value here, and y the smaller value.

 

Hence, let y be the smallest possible integer, ie. 1.

 

Consequently, y = 1, x = 3.

 

Substituting these values, we obtain the balaced equation :

(8)NO2 + (3)H2O ---> (1)N2O + (6)HNO3

 

 

This is the systematic way of balancing equations that I teach to my students. I always remind my students, "Algebra is your friend who is here to make your life easier, so use it whenever you can in Chemistry calculations!"

 

Here are some 'O' level questions (but 'A' level students should also try them for your own revision and practice too!) for students' practice and enjoyment :

 

 

1)   

A sample of 0.43g of an organic compound containing only carbon, hydrogen and oxygen, was burnt in excess oxygen. The combustion produced 1.10g of carbon dioxide and 0.45g of water.

                a) Calculate the empirical formula of the compound.

                b) Given that this compound has a relative molecular mass of 250g, deduce its molecular formula.

 

Ans : C15 H30 O3

 

 

 

 

2)  20cm³ of a gaseous hydrocarbon was mixed with 100cm³ of oxygen so that the hydrocarbon was completely burnt. The volume of gas remaining at the end of the combustion was 70cm³. After passing through soda lime, the volume was reduced to 10cm³. All gases were measured at r.t.p. Determine the formula of the hydrocarbon.

 

 

Ans : C3 H6

 

 

 

 

3) A mixture of MgSO₄.7H₂O and CuSO₄.5H₂O is heated until a mixture of the anhydrous salts, is obtained. If 5.0g of the hydrated mixture when heated gives 3.0g of the anhydrous salts, calculate the % by mass of CuSO₄.5H₂O in the initial hydrated mixture.

 

Ans : 73.9%

 

 

 

 

 

4)  When Fe and Fe³⁺ are mixed together, a reaction occurs in which Fe²⁺ is produced. What is the ratio of Fe to Fe³⁺ required to produce equal moles of Fe²⁺ and Fe³⁺ when the reaction is complete?

 

Ans : (For fun's sake, I'll leave it to you to work out the answer. You can post your answer to check with me if you like).

 

 

 

 

 

Enjoy!

 

               

 

Like all the others above, this is a question that is good practice for both 'O' level and 'A' level students. For 'O' level students, if you're not yet familiar with balancing half-equations (usually Pure Chem 'O' level students are familiar but Sci Chem 'O' level students are not; also familiarity and extent of exposure also varies from school to school), I'll give you an example so you can do the Redox question at the bottom of this post :

 

In 'O' level Biology, placing a piece of potato (or piece of animal liver) into a test tube of hydrogen peroxide will result in vigorous effervescence. What is happening, is that the potato cell (and all animal cells, especially the liver since one of its main functions is that of detoxification - hydrogen peroxide is toxic) contains the enzyme catalase / peroxidase (alternatively, use inorganic catalyst manganese(VI) oxide), which catalyzes the decomposition reaction for hydrogen peroxide into water and oxygen gas.

 

So from the above, you should know that hydrogen peroxide can act as a reducing agent and kena oxidized, or it can act as an oxidizing agent and kena reduced.

 

Because reduction means (among other things) the removal of oxygen, reducing H2O2 gives you H2O. Because oxidation means (among other things) the removal of hydrogen, oxidizing H2O2 gives you O2 gas. This is how you remember the products of the reduction and oxidation of H2O2.

 

Here is how you balance half-equations :

 

[reduction]

H2O2 --> H2O

To balance oxygen,

H2O2 --> 2H2O

To balance hydrogen,

2H+ + H2O2 --> 2H2O

To balance charges,

2H+ + 2e- + H2O2 --> 2H2O

 

[oxidation]

H2O2 --> O2

To balance hydrogen,

H2O2 --> O2 + 2H+

To balance charges,

H2O2 --> O2 + 2H+ + 2e-

 

[Overall Balanced Redox]

Cancelling away the common protons and electrons on both sides, we have :

2 H2O2 --> 2H2O + O2

This is the decomposition of hydrogen peroxide into water and oxygen gas.

 

Note that for the question below, the two Redox reactions are :

a) nitrogen dioxide (reducing agent) reacting with hydrogen peroxide (oxidizing agent)

b) manganate(VII) ions (oxidizing agent) reacting with hydrogen peroxide (reducing agent).

 

A final note : if there is sufficient acidity (ie. molarity of protons), purple MnO4- ions will be reduced to colourless Mn2+ ions. Otherwise (in neutral or alkaline conditions), MnO4- ions will only be reduced to brown ppt of MnO2(s). Compare the two - the former sees a reduction of Oxidation State (O.S.) from +7 to +2, while the latter only sees a reduction of O.S. from +7 to +4. Hence, we usually acidify MnO4- ions because we want it to have a stronger oxidizing efficacy or power.

 

Without further ado / adieu, here is the long awaited question that is guaranteed to delight both 'O' level and 'A' level students alike :

 

 

 

 

Ans :

1a(i,ii,iii,iv,v) Final Ans is 46%

1b) Final Ans is +2

[For 'A' Level Students only]

 

 

 My "Loch Ness Monster" Classic Question 

 

 

A bubble of gas (initial diameter 1.585 cm; mass 2.0973 x 10-2 g) that contains twice as much oxygen as it does carbon dioxide, emerges from a photosynthesizing aquatic plant (there’s a plesiosaur reptile aka “Loch Ness monster” right next to the plant, btw) at the bottom of a lake, where the temperature is really cold and the pressure is 6.4 times greater than at the lake’s surface; and the bubble rises until it reaches the surface of the lake (which happens to be at sea level, and at that moment is equivalent to room temperature), and bursts “pop!”.

i) Calculate the diameter (in cm) of the bubble just before it bursts.

ii) Calculate the body temperature (in °C) of the plesiosaur at the time the bubble emerges.

iii) Calculate the diameter (in cm) of this gas bubble if it were instantaneously teleported to the core of the sun (600 thousand times hotter than room temperature and 340,000 million times more crushing than Earth’s sea level atmospheric pressure), and assuming it somehow remains miraculously intact (although at the temperature and pressure, atomic nuclei are being forcibly fused together, so trust me when I say you wouldn't want to be hanging around there)

 

 

 

 

Final Ans to (i) is 3.0 cm.

Final Ans to (ii) is 8 deg C.
Final Ans to (iii) is 0.0159 cm.

'A' Level Qn (Combining Acid-Base Equilibria with Solubility Equilibria)

 

 

Calculate the molarity of NH₃ (aq) needed to initiate the precipitation of Fe(OH)₂ (s) from a 0.003 mol/dm³ solution of FeCl₂ (aq). Given K_sp Fe(OH)₂ = 1.6 X 10^-14 and K_b NH₃ = 1.8 x 10^-5.                  

                                                                                                                                    

Ans : [NH₃] > 2.6 X 10^-6 mol/dm³

 

'A' Level Qns (Hess Law and Energetics)

Calculate the enthalpy change for the reaction N₂O₃ (g) + N₂O₅ (s) ®

2 N₂O₄ (g), given :

NO(g) + NO₂(g) ® N₂O₃(g)                            ΔH_r = -39.8 kJ/mol

NO(g) + NO₂(g) + O₂(g) ® N₂O₅(g)              ΔH_r = -112.5 kJ/mol

2NO₂(g) ® N₂O₄(g)                                        ΔH_r = -57.2 kJ/mol

2NO(g) + O₂(g) ® 2NO₂(g)                            ΔH_r = -114.2 kJ/mol

N₂O₅(g) ® N₂O₅(s)                                        ΔH_r = -54.1 kJ/mol

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This Acid-Base Equilibria qn, asked by "contemporarydancer" attempting a JC prelim paper, is an excellent distinction question.

http://www.sgforums.com/forums/2297/topics/325645

 

Reproduced here :

 

 

 

Qn : 25 cm3 of Phosphoric Acid H₃PO₃ of initial concentration 0.1M was titrated against NaOH, of concentration 0.1M.

Given the following equations and corresponding Ka values:-

H₃PO₃      =     H₂PO₃^-    +    H⁺               Ka₁ = 1x 10^-2
H₂PO₃^-     =     HPO₃^2-     +    H⁺               Ka₂ = 2.5 x10^-7

Sketch the shape of the pH curve during this titration, labelling significant pH values and significant volumes of NaOH on the axes of the curve.

 

 

Solution Guide (by UltimaOnline) :

H₃PO₃ is more correctly named (for common name; for systematic name, work out the oxidation state of phosphorus in the acids – phosphoric(???) acid) "phosphorous acid" (rather than "phosphoric acid", H₃PO₄). Unlike phosphoric acid, a triprotic acid, phosphorous acid is indeed a diprotic acid (hence Ka1 and Ka2) - the proton bonded to the phosphorus atom is non-acidic (ie. cannot be dissociated).

"Equivalence point" vs “end point”. “Equivalence point” refers to the exact stoichiometric neutralization point per each acidic proton. "End point" of a titration occurs when the (relevant) indicator used changes colour. The end point approximates the equivalence point.

There are two equivalence points, and hence two end points. This is because phosphorous acid is a diprotic acid (ie. one mole of acid gives two moles of protons; aka dibasic acid).

Since phosphorous acid is a diprotic acid, there will hence be 2 buffer regions (ie. regions in which any protons or hydroxide ions added will be immediately removed from solution by the conjugate base and conjugate acid respectively, hence the 'buffering' effect). As to the pH of these buffer regions (or to be precise the pH of the exact points of maximum buffer capacity for the 2 regions), use pKa₁ and pKa₂ to determine.

Remember that an acid buffer is most effective (ie. at its maximum buffer capacity) when molarity of acid = molarity of conjugate base / salt. Understand that this also means the pH of maximum buffer capacity occurs midpoint(s) of the neutralization process, before the equivalence point(s) is/are reached.

Hence you should be able to work out the volumes of NaOH for the 2 points of maximum buffer capacity, which given that the question is kind enough to have the molarities of phosphorous acid and sodium hydroxide as being equal, makes your job much easier.

To determine the pH at 1st and 2nd equivalence point, you need 2 tables :

a) ICF (Initial Change Final) table for 1st equivalence point, in moles.

b) ICF (Initial Change Final) table for 2nd equivalence point, in moles.

To determine the volume of NaOH required for both equivalence points, use stoichiometry (ie. looking at the balanced chemical equation(s); in this context, for neutralization of the first and second acidic protons).

The important and tricky PitTrap here (which many students fall into and lose their precious marks and hence also their chance for a distinction grade), is when calculating the molarity of the salt (or acid, or base), you have to be very careful to take into consideration and correctly add up the total volumes of both the acid and alkali used (including or epsecially for 2nd equivalence point, where for the final volume of solution you must add up total NaOH used, including for previous equivalence point!).

 

 

 

Calculating the pH of the 2nd equivalence point.

 

Notice that for Na₂HPO₃(aq) of which the anion is the HPO₃^2-(aq) aka monohydrogen phosphite ion (ie. the twice deprotonated phosphorous acid), because Na⁺ has no affinity to covalent bond with (or remove from solution) OH^- ions, while HPO₃^2- has significant affinity to covalent bond with (or remove from solution) H⁺ ions aka protons, the salts Na₂HPO₃ is basic/alkaline upon hydrolysis. (Most students find it easier to think of it summarily and remember it as "strong base + weak acid = basic salt").

Hence, using Kb₂ expression and value (you are given Ka₂ value; hence using the relationship between Ka, Kb and Kw, you can determine the Kb₂ value), work out the molarity of hydroxide ions, hence pOH, and therefore finally pH.

 

Kb2 = ([OH-][H2PO3 -])/[HPO3 2-]

((1x10^-14)/(2.5x10^-7)) = ([OH-]^2)/((25/1000)x0.1)/(75/1000))

(4x10^-8) = ([OH-]^2) / (0.03333)

[OH-]^2 = 1.333 x 10^-9

[OH-] = 3.651x10^-5

pOH = 4.438

pH = 9.562 = 9.56 (to 3 sig fig)

 

 

 

 

Calculating the pH of the 1st equivalence point.

Now, the first deprotonated phosphorous acid, NaH₂PO₃(aq) of which the anion is H₂PO₃^-(aq) aka dihydrogen phosphite ion, is a little trickier. One the one hand, it may protonate itself (ie. remove protons from solution to become phosphorous acid again) and hence be basic. On the other hand, it may serve as a source of protons itself (since its second acidic proton may dissociate, to become monohydrogen phosphite ion) and hence be acidic.

To determine which, compare the relevant Ka value against the relevant Kb value. You should be able to figure out it's the Ka₂ value against the Kb₁ value (tip : focus on the protonation and deprotonation of the ion being discussed, the dihydrogen phosphite ion).

Because the Ka₂ (deprotonation of ion as an acid) value is significantly greater than the Kb₁ (protonation of ion as a base) value, you can conclude that the dihydrogen phosphite ion is acidic rather than basic.

Hence, to determine pH at the 1st equivalence point, after working out the ICF table and upon having determined the molarity of the salt NaH₂PO₃(aq), we hence use the Ka₂ expression and value to work out the molarity of protons, and hence pH.

 

To refresh your memory,

Ka = ( [molarity of protons] x [molarity of conjugate base] ) / [molarity of conjugate acid]

Kb = ( [molarity of hydroxide ions] x [molarity of conjugate acid] ) / [molarity of conjugate base]

 

So to summarize, this is a good discriminating (distinction) question, in which you have to decide whether the 1st-deprotonated phosphorous acid aka dihydrogen phosphite ion, is acidic or basic (upon hydrolysis), based on the relevant Ka and Kb values. Knowing which, will enable you to (decide whether to use Ka or Kb expression to) calculate the pH at the 1st equivalence point. The twice-deprotonated phosphorous acid aka monohydrogen phosphite ion, on the other hand, is very obviously basic (upon hydrolysis) (since the proton bonded to phosphorus cannot dissociate and hence isn't acidic), and hence you must use the Kb expression to calculate [OH^-], then pOH and finally pH, at the 2nd equivalence point.

 

And remember, the more you've the passion to understand, enjoy and love Chemistry (for what it *truly* is, and not just memorizing blindly as instructed dogmatically by your JC lecturers/tutors, or for the sake of paper chase exams), the more you will inevitably excel in it!

 

'A' Levels Qn.

 

Based on the Data Booklet, identify 4 reducing agents that, under standard conditions, will reduce Cr3+ to Cr2+ (without further reduction to Cr). Include all relevant reduction potentials in your answer.

 

Ans :
1 - Fe to Fe2+
2 - Ni and NH3 to hexaaminenickel(II) ion.
3 - Co and NH3 to hexaaminecobalt(II) ion.
4 - Fe(OH)2 to Fe(OH)3

Draw the (Kekule or Lewis or dot-&-cross) structures of

a) triiodide ion, I3 -

b) HF2 -

c) HBr2 -

 

 

 

 

Ans :

 

a) Nucleophilic I- ion attacks instantaneous dipole (delta +ve) of one of the I atoms in the I-I molecule; the attacked iodine atom still retains 3 lone pairs in addition to the newly formed dative bond from the iodide ion. The negative formal charge hence shifts to the central I atom.  

electron geometry - trigonal bipyramidal (ie. hybridization sp3d)

molecular geometry - linear (I-I-I, with 3 lone pairs & 2 bond pairs about central I atom)

Note : in the 'A' level H2 exam, the VSEPR geometry asked is always the molecular geometry; but you should (ie. it is easiest and smartest to) first mentally figure out (don't memorize blindly, which is boring and unreliable) the electron geometry first (based on number of electron pairs, don't care lone pair or bond pair), then considering how many are lone pairs and how many are bond pairs, figure out the molecular geometry.

 

b) [ F- ~ H-F ]-

where F- ion is hydrogen bonded (~ represents hydrogen bond) to the H (which is delta +ve) which is covalently bonded to F.

 

c) [ Br- ~> H-Br   <--->   Br-H <~ Br- ]-

where nucleophilic Br- ion attacks (~> and <~ represent electron flow mechanism curly arrows) the H (which is delta +ve) which is covalently bonded to Br; as H can only have a maximum valence duplet, the covalent bond pair becomes a lone on the covalently bonded Br which consequently becomes a leaving group Br- ion. The opposite occurs in the other resonance contributor (<---> represents resonance). Considering the 2 resonance contributors, the H actually has 2 half-bonds, still amounting to a stable duplet.)

 

The difference in structures between HF2 - and HBr2 - can be attributed to the following :

1) Bond enthalpy of HF is much stronger than H-Br; H-F bond won't cleave readily.

2) Only HF is capable of strong hydrogen bonding; HBr is not.

3) Br- is a much more stable leaving group than F-, due to much larger ionic radius.