Acidic & Caustic Remarks

Chm 1311 Lecture for 22 June 1999

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OK...there's lots of things to memorize in Freshman Chemistry up to this point, but we'll soon move to some philosophical items that serve to rationalize what's what. But one explanation you're not scheduled to see 'til much later could usefully be offerred now. It's the importance of dynamic equilibrium to chemical reactions; understand it, and you'll likely see why some reactions proceed in the direction they do.
Speed of
Chemical Reaction Rates is a topic for later, but some broad hints about them will make dynamic equilibrium lots easier to take. For example, it's obvious that chemical reactions can't proceed at all until the reactant molecules encounter one another. And it's not too difficult to understand that when they encounter one another, there's not necessarily a 100% chance that they'll actually do the reaction. After all, they might be colliding at an unreacting end or with insufficient force...whatever.

So the speed with which reactants become products is directly proportional to two factors: the frequency of their encounters and the efficiency of each encounter; double either the frequency or the efficiency, and that reaction goes twice as fast as before.
Frequency of
If you are molecule A in a reaction in which A+B goes to C+D, then how often you encounter B is most important. Double B, and A and B annihilate one another (to produce C+D) twice as quickly. Same thing holds true in reverse; that is, if C+D can get together and unreact, doubling D gives C twice the chances of finding a D target. But if C is doubled as well, there's twice as many missiles finding twice as many targets making four times as many explosions (OK, "reverse reactions" not explosions).
So when there's no C and D yet (the reaction hasn't yet started), there aren't any C and D to get together. A+B is then the only game in town, and the "forward" reaction goes its merry way. But as it does so, fewer and fewer A and B are to be found; each one searches longer to find and react with the other. So the forward progress becomes less frantic.
But while A and B are petering out, C and D are coming into existence and starting to unreact back to A and B! You can see how a balance of continuous reaction might then arise. If both reactions were equally efficient, it's clear that when the reaction appeared to be ½ over, A and B and C and D would all be at the same concentration (i.e., "number density" or mol per volume) and reacting as fast forward (to make C and D) as backward (to reproduce A and B)!
Efficiency of
However there's no reason to expect that A+B encounters would be as efficient as C+D ones. In fact, let's presume that A+B are 4 times as efficient (in reacting to make C+D) as C+D are in unreacting to make A+B. If that's the case, by the time we'd come to equal amounts of all the molecules, A+B although encountering one another as frequently as C+D, would be ahead in that more such encounters would be effective. Now where does balance lie?
Clearly when A+B had lost some more ground...but how much more ground? It's easy; the concentrations of C and D (equal at all times since they're made at the same time), must give four times more frequent encounters to offset the four-fold efficiency of the A+B hits. Aha...from four paragraphs ago, we could get four times as many C+D unreactions if the concentrations of C (or D, they're equal) were twice that of A (or B, assuming they started equal)! Think about that for a moment, and you'll see that dynamic equilibrium will come to that balance when the stuff in the reaction vessel is 1/3 A+B and 2/3 C+D. Unsurprisingly then, a four-fold forward-to-backward reaction efficiency favors the products (by 2:1).

That's dynamic equilibrium in a nutshell: although the forward and reverse reactions just keep on keeping on, that 2:1 preference of products to reactants is assured (if forward is 4× as efficient as reverse). With other efficiency ratios, your mileage may vary, but you'll be able to know it.


The reason for telling you all this now (besides my respect for your intellectual prowess) is that this knowledge can give you a strong hint about how to meddle in the lives of chemical reactions!
If you want to mess with the products' abilities to unreact, surely one of the best things you can do is to deny them frequent or any encounters! How do I do that?!? (You ask.) By getting rid of them as fast as they are formed, if you can. Don't leave them lying around the solution to be picked off in a reverse reaction, and you let the reactants win!

Gas Evolution

For example, suppose one of the products is a gas. That means that it is reasonably insoluble in the solution and prefers to bubble out. Bingo! It's no longer in solution to unreact; so the reaction goes forward with a vengeance. Now if you were of a playful nature, you could trap the gas in a volume above the solution, and as it's concentration there rose, it would become less and less reluctant to be dissolved, but that would be defeating your purpose if you wanted the reaction to go as far to completion as possible. So let the gas escape!
Not surprisingly, Gas Evolution Reactions are a category worth knowing about since the loss of the product gas drives such reactions forward! We saw an example of one in the first lecture when I sprinkled a little MnO2 dust into H2O2, hydrogen peroxide, to speed up its effervesce. It was doing:

2 H2O2 (aq)  arrow right 2 H2O + O2(g)

which it's doing slowly, quietly in your medicine cabinets even as we speak. Your 3% hydrogen peroxide was only that when you bought it; it'll be considerably weaker now. Since it evolves O2(g), it is (among other things) a gas evolution reaction.


Alternately, you could hope a product would become insoluble and solidify, leaving the solution that way. Such a solid settling out of a reaction is called a precipitate (the noun not the verb). Since neither gas nor solid is in the solution phase, those products cannot participate in reverse reactions. So insolubility is a strong motivation to complete reactions in solutions.
We've seen an example of this type of reaction too. It was the Landlord's yellow product which precipitated (verb) as a precipitate (noun) from a solution of Pb(C2H3O2)2 and K2CrO4 giving PbCrO4(s) and what? Answer: a solution with the 2 K+ and 2 C2H3O2- floating around as ions with nothing to do, the lazy beggars. Indeed, their only advantage in the entire operation was that the acetate ion provided soluble Pb2+, and the potassium ion provided soluble CrO42- since all potassium or acetate compounds are water soluble!

Pb2+ + 2 C2H3O2- + 2 K+ + CrO42-  arrow right PbCrO4(s) + 2 K+ + 2 C2H3O2-

carries those couch potato ions on both sides of the arrow even though nothing happens to them.

& Net Ionic

Chemist don't call those unreactive counter ions "couch potatos." We call them "spectators." And we usually ignore them unless we need to calculate weights of compounds of which they're part. So the reaction above could have been reduced to what we call the net ionic reaction, or

Pb2+ + CrO42-  arrow right PbCrO4(s)
which has the advantage of telling us it didn't matter what soluble ions we used; we could as easily added plumbous nitrate and sodium chromate. The results would have been the same! Net Ionic Reactions are a neat shorthand.
On the
Other Hand
We could just use reactions which produces molecules sublimely disinterested in reverse reaction regardless of how often they encounter one another. In other words, products with lousy reaction efficiency are proof against unreaction, right?
The way to ensure such complete reaction is to produce a molecule so stable that it cannot engage in a reverse reaction without much prodding. For example, the copper bound up in cupric oxide (CuO) ore is not going to leap out as a free metal since the oxide is quite stable. But even Bronze Age civilizations had discovered that feeding this ore a compound (charcoal) which wanted that oxygen even more desperately would do the trick:

2 CuO(s) + C(s)  arrow right 2 Cu(l) + CO2(g)

but in the absence of that prodding, CuO ore would be quite stable.


In aqueous solution there is just such a compound, available in great abundance; it is the water molecule itself. Often chemical reactions yield water as a product, and its reluctance to react makes it a "good leaving group", shorthand for "a stable product." So even though it encounters other molecules in solution relentlessly, its relative unreactivity makes it a desirable product in the sense that it won't compete against the reactions which produced it.
Acid The best examples of water as a leaving group are the acid-base reactions. Svante Arrhenius, he who postulated the existence of ions, proposed that acids have all the properties they do as a consequence of making H3O+, hydronium ions in aqueous solution. Well, actually he didn't know that protons in solution have such an enormous charge gradient (no electrons to protect them from other molecules), that they latch onto the nearest available water to make H3O+. Arrhenius thought it was H+(aq) that was actually doing the acid trick, and to this day, us chemists usually get lazy and use H+(aq) as a shorthand for the actual hydronium ion. Forgive us when we so sin. So you'll often see

HNO3(aq)  arrow right H+(aq) + NO3-
Base Arrhenius also proposed the OH-(aq) ion as the active ingredient in caustic compounds. There he happened to be on firmer ground. While, no doubt, that negative charge attracts a "solvation shell" of water molecules, as does any aqueous ion, it never binds with any one water as does the proton. The water molecules in solvation shells are constantly exchanging with bulk waters. So the only complaint one might level at OH-(aq) is that the charge really ought to be shown on the oxygen, but a century of tradition rejects writing this hydroxide ion as HO-(aq). Pity.

So a common base reaction in water would be:

NaOH(aq)  arrow right Na+(aq) + OH-(aq)

vs. Weak

The nitric acid (HNO3) and sodium hydroxide (NaOH) base above are examples of a strong acid and base, respectively. Like strong electrolyte, that means that they ionize in aqueous solution completely; in contrast, a weak acid or base ionizes poorly in aqueous solution. If HNO3 and NaOH react with one another, the nitrate and sodium ions are (again) just spectators to the reaction, and the resultant net ionic reaction:

H+(aq) + OH-  arrow right H2

is the generic reaction for all strong acids with strong bases. It is called a neutralization reaction since, with stoichiometrically balanced amounts, the hydronium and hydroxide ions not only neutralize one another electrically (making neutral water) but, more to the point, they neutralize one another's corrosive properties!
Weak Acids
and Bases
Very, very few acids or bases are strong; they were profiled in the previous lecture. The vast majority are weak, reluctant to give up their acid protons or basic hydroxides even assuming that they are soluble in water. So the dynamic equilibrium for the ionization of a weak acid, like citric acid (H3C6H5O7, where the 3 bold H atoms are the acidic protons),

H3C6H5O7  arrow right H+(aq) + H2C6H5O7-

really lies quite far toward the reactants. The same is true of the weak base, ammonia,

NH3(aq) + H2O  arrow right NH4+(aq) + OH-(aq)

which also "favors" the reactants.
Conjugates This means (getting ahead of the textbook once again) that dynamic equilibrium would suggest that the anion of a weak acid

H2C6H5O7- + H2O  arrow right H3C6H5O7 + OH-

is basic! While the cation of a weak base

NH4+(aq) + H2O  arrow right NH3(aq) + H3O+(aq)

is acidic! Such reversed behavior from the salts of weak acids are referred to as the acid's "conjugate base." Ditto the "conjugate acid" of a weak base is its counter ion.

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Last modified 22 June 1999. Chris Parr