Creative Crud and Gargle

Chm 1311 Lecture for 23 June 1999

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Since equilibrium reactions can be bullied by adding or removing molecules involved in the equilibrium, we can ensure an equilibrium reaction goes to completion but taking away any product such that the reverse reaction is stopped dead. Since the equilibrium was dynamic, the forward reaction just keeps chugging away producing more and more product until it runs out of the limiting reagent.

The two obvious candidates for this encouraging removal are precipitates and gases since both cease to be part of the solution chemistry! Less obvious encouragement, but still as effective, is the formation of a very weak electrolyte product; while it may well still be part of the solution, it is no part of the ion chemistry! Thus, it can no longer participate in ionic reactions.

Let's look as these possibilities in turn:

Products known to be insoluble will precipitate out of solution and will be a signal to us that reactions that produce them ought to go to completion. Indeed, the weighing of precipitates is a time-honored quantitative chemistry technique for determining the amounts of atoms in the reactants!

The table of solubilities was given in an earlier incarnation of this course as one in a series of other lectures. You should go to that webpage and memorize (sorry 'bout that) which compounds are soluble and which are not. But remember to use your BACK button to return here when you are through. Jump now to a Solubility Table.

Understanding that you can refer back to that table whenever you wish just by clicking on GO FORWARD, let's look at what the table tells us must occur in some chemical reactions.

The compounds in that table are all strong electrolytes. Even the insoluble ones are strong electrolytes not in the sense that they yield lots of ions but rather in the more technical sense that any of the molecules which dissolve will ionize completely in water. So we should really write these species as aqueous ions (e.g., Na+(aq) + NO3-(aq) not NaNO3(aq)) unless they are insoluble; then we'll write them as molecular species (e.g., Ca3(PO3)2(s) not 3 Ca2+(aq) + 2 PO33-(aq))

The usual kind of problem tackled here is a metathetical reaction (double exchange) one product of which exits the solution as a precipitate. An example might be the dissolution of lead(II) acetate and barium chloride in water. They're both soluble and strong electrolytes, so in solutions by themselves they'd be Pb2+ + 2 (C2H5O2)- and Ba2+ + 2 Cl-. Note that one mole of either compound makes 3 moles of ions. Also note I've left the "(aq)" off the ions...too little room.

However, if they're in the same solution, lead(II) chloride will be insoluble and precipitate out:

Pb2+ + 2(C2H5O2)- + Ba2+ + 2Cl-  arrow right PbCl2(s) + Ba2+ + 2(C2H5O2)-

But the "spectator ions," Ba2+ and (C2H5O2)- are just along for the ride. Although Ba2+ certainly can be precipitated, it can't be precipitated by either chloride or acetate ions. Common practice is to write instead the net ionic equation which ignores spectators like:

Pb2+(aq) + 2 Cl-(aq)  arrow right PbCl2(s)

It not only saves space (enough to include the aq!), it focusses on what's really happening.

Now it is certainly possible to rig up such a metathetical reaction where the ions change partners and both precipitate! But it's rare. An example would be adding barium sulfide to tin(IV) sulfate:

2Ba2+ + 2S2- + Sn4+ + 2SO42-  arrow right 2BaSO4(s) + SnS2(s)


While it's certainly true that some gases are very soluble in water (HCl springs to mind), it's also true that it's easy to discourage such solubility and evacuate the dissolved gas, if necessary, just by heating up the solution. Would you want to open a hot soft drink can without a raincoat?

So any solution chemistry which evolves a gas is likely to proceed when the gas bubbles out of solution. What kinds of ionic reactions produces gases? Certainly those which would reverse the dissolution of acid anyhdrides like

HSO3-(aq) + H+(aq)  arrow right SO2(g) + H2O

especially if the concentration of the H+ is high to ensure near complete reaction back to the weak acid, H2SO3, which decomposes into SO2(g) and water. So the best bet would be to use a highly concentrated, strong acid (HCl, say) directly on a hydrogen sulfite salt, maybe NaHSO3. The only caution would be not to use a highly oxidizing strong acid (HClO3 say) or the HSO3- might be oxidized to HSO4- instead, and you'd get the wrong (SO3(g)) anhydride!


Finally, if ionic reactions can produce a soluble molecule that doesn't ionize, the ions it ties up may no longer participate in the chemistry (unless bullied). That too will permit one to predict a chemical reaction.

Of course, the ultimate example of this is the most important reaction we'll study in this course. It is the net ionic equation that governs the neutralization of all strong acids by strong bases. Those neutralizations produce "a salt and water." The salt is irrelevant (unless it precipitates!) and can be left out of the focal equation thus:

H+(aq) + OH-(aq)  arrow right H2O

Chemists speak of water as "a good leaving group" when they mean to convey that its stability and weak electrolyte character conspire to prevent it from engaging in any significant reverse reaction.


Yes, I snuck a fast one in on you above when I cautioned against inadvertantly "oxidizing" HSO3- and ending up with the wrong acid anhydride. I had trusted that you'd read ahead to the section on Oxidation/Reduction and knew what I was talking about.

OK, if you didn't do that, read the following and then go back to that gas evolution example newly infused with understanding. :)

OXIDATION The name obviously has something to do with oxygen; it pays homage to the voracious appetite of oxygen (and all the non-noble gas elements in the upper right-hand corner of the periodic table) for electrons. Remember that the oxide anion is O2-, implying that oxygen will steal up to 2 electrons from some hapless atom or molecule that can't hold onto them.

So oxidation is just that theft: it is the loss of electrons, not necessarily to oxygen, mind you. Atoms, molecules, or even ions which have been ripped off of one or more of their electrons are said to have been oxidized, and the thief is said to have been the "oxidizing agent".

In the rusting of copper that changes the bright metallic luster to a soft, beautiful patina of pastelle green, say that seen on the dome of the Multipurpose Building (when are they going to rename that?) is a classic example of oxidation (of the copper) as in

2 Cu(s) + O2(g)  arrow right 2 CuO(s)

REDUCTION But if electrons (2) leave the copper, they have to go somewhere; so that reaction is also an example of the antioxidation of O atoms. But nobody calls it by such a convoluted name. Instead the name given to the addition of electrons to atoms, molecules, or ions is called "reduction."

The word originally was applied to the method of reducing a metal ore to its metal. (Alchemists were forever "reducing" things to their "essences" and created a lot of chemistry into the bargain.) So undoing that reaction above would constituted REDUCTION of copper ore to metallic copper, the trick that ushered in the Bronze Age. That reaction, known by prehistoric man, was

CuO(s) + CO(g)  arrow right Cu(s) + CO2(g)

wherein copper ions gain back their 2 electrons and are "reduced." But where did they come from? Not oxygen! It would never be that generous. So instead the electrons came from (the only thing that's left) the carbon! So here the carbon had to have been oxidized.


Were CO ionic (it most definitely is not!), we'd imagine the O to be the oxide anion O2- which would make carbon the C2+ cation in order to retain the neutrality of the CO molecule. That was the windup; now here comes the pitch ...

In figuring oxidation/reduction reactions (abbreviated "redox"), we assign oxidation numbers which consist of the charges we'd expect to find on all of the atoms of the molecules as if they were all ionic...even the ones that aren't!

Not surprisingly, you have to memorize the Table of Common Oxidation Numbers. OK, groan and get it over with. But this table, from another course's lectures, is at least clever enough to direct you to the correct answer as soon as it's practical. You go through the rules, stopping when they've produced an unambiguous answer for the oxidation state of an atom within a molecule or ion. After reading those Oxidation Number Rules, use your BACK button to return here.

So the oxidation numbers of the atoms in H2PO42- would be +1, +4, and -2, for H, P, and all four O atoms. We arrive at that by knowing that they have to add up to -2, the charge on the dihydrogen phosphate ion, which permits us to determine P's oxidation number by difference between that -2 and the sum of the others. The others are easy; H must be H+ because P and O are nonmetals, and O must be O2- since P isn't F!

So 2(+1) + X + 4(-2) = -2 can only be solved if X=+4, phosphorous's oxidation state in this compound. It could be something else in another compound, say H2PO32-, dihydrogen phosphite ion; there it's +2 for the P atom.

(Spectator ions are even easier; since nothing changes them, we can just give the WHOLE ion an oxidation number equal to its charge.)

Oxidation increases the oxidation number, making it more positive as electrons are stolen from the atom. Reduction decreases the oxidation number making it more negative as the atom steals electrons from its neighbors.

THE trick to all "redox" reactions is to remember that since the electrons have to go somewhere, oxidation never occurs without reduction!

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Last modified 5 June 2000. Chris Parr