Lecture Notes from CHM 1341
27 June 1996

Solution Chemistry

The overwhelming advantage of solution chemistry is the ease with which well calibrated quantities of substances can be dispensed and mixed to pursue both qualitative and quantitative analysis. Additional advantages come from potential products immiscible in the reaction solution (solids, gases, and incompatible liquids) and thus forced into virtually automatic separation from it. But it is the soluble components which can be diluted to any degree upon which we focus.

Since a liquid conforms to the shape of its container, it is volume which is the most easily measured. So the number of moles of substance per unit volume of solution is a key conversion quantity. Standard nomenclature assigns "solute" to the substance dissolved, "solvent" to the substance dissolving it, and "solution" to their mixture. But, in reality, once mixed, who has dissolved whom doesn't much matter. All that's critical is that there's some substance to be delivered, and it's now packaged in convenient form to be delivered drop by drop...virtually continuously for most purposes.

The measure used most often in solution chemistry is M, the number of moles of solute per liter of solution (not solvent). And this measure of concentration of a substance, S, is designated as [S] with the molecular formula for the substance placed between square brackets and quoted in units of M (mol/L). Since its small controlled quantities which are so handily measured this way, it's good to note that the same value of mol/L is also obtained as mmol/mL, millimoles per milliliter. A mL is also 1 cc or cubic centimeter, if you work through the volume conversions and the original definition of a liter as 1 cubic decimeter.

While M or molar concentration compares moles to volume, density compares mass to volume; so the two are intimately related via the molar mass. For example, the density of methanol (methyl alcohol) is 0.796 g/cc. So we could, should be so inclined, compute the molar concentration of methanol in itself as a solvent! It would be

                       0.796 g methanol   1 mmol methanol
molar conc. of CH3OH = ---------------- * ----------------
                         1 cc methanol    0.032 g methanol


                     = 24.9 mmol/cc = 24.9 mol/L pure CH3OH

The same trick with water shows its pure concentration to be 55.5 mol/L, or 55.5 M.

But concentrations of pure solvent aren't usually of interest. Instead, we're usually after concentrations of dissolved molecules or ions in aqueous solution, water being the near perfect solvent for many substances...and non-toxic too! And the standard use for molar concentrations is in titrations.

A titration involves reacting a substance dissolved in one solution with a known volume of a know strength (concentration) of substance in some "standard" solution. It's just "standard" because you know everything about what's in it. The reaction might be any of the kinds we've mentioned before: acid/base, redox, or precipitation. And the key to the titration is knowing when all of the unknown has been reacted by the known substance. So acid-base reactions rely upon sudden changes of acid or base concentration (signalled by color-changing "indicator" molecules or electronic means). The "endpoint" of a redox reaction might be signalled by sudden change in a molecule less interested than the unknown in being redoxed (as it were). Likewise for a precipitation, the indicator would be a substance which fails to compete for the standard until no further unknown remains to be precipitated.

This chapter focusses upon concentration calculations in acid/base reactions and strong acid/strong base at that. Under those conditions, we expect a quantitative (complete) neutralization of the unknown base by the standard acid, or vice versa. Actually, standard bases are a little unreliable; that's because they're always exposed to the carbon dioxide of the air which makes the (admittedly weak) carbonic acid in aqueous solutions. It's enough, however, to make the shelf life of "standard bases" notoriously short. Bottom line: you need a standard base solution? Make it fresh!

So the net ionic equation for all strong acid/strong base reactions is the same:

     H3O+(aq) + OH-(aq)  ----->  2 H2O(l)

The obvious advantage here is that hydronium ion neutralizes hydroxide ion one-to-one. You might think you're home free because of that, but think again. Polyprotic acids (more than one acidic proton) and/or polyhydroxyl bases (more than one hydroxide ion) mean that the ratio of acid to the hydronium or base to the hydroxide isn't one-to-one. So stoichiometric conversions will be required in those cases.

It is true that in polyprotic acids, for example, as acidic protons are removed, those remaining represent weaker and weaker acids...after all, the anion is becoming steadily more negative and thus more attractive to those remaining protons. Still, to render an example, let us suppose that the three acidic protons of phosphoric acid (the acid in Coca Cola) can be neutralized by the hydroxyls of barium hydroxide:

2 H3PO4(aq) + 3 Ba(OH)2(aq)  ----->  6 H2O(l) + 3 Ba2+(aq) + 2 PO43-(aq)

In all honesty, although the barium ion is a spectator, the phosphate is an "actor" since the HPO₄^2-(aq) ion is extremely weakly acidic! So I have to fudge the case here (or pick some polyprotic organic acid instead) to make the point.

And the point is (the envelope, please), that one must incorporate those stoichiometric factors in acid/base reactions, if they're not one-to-one. So if you were asked how many mL of 0.1 M phosphoric acid would it take to neutralize 15 mL of 0.2 M barium hydroxide (assuming both were strong), you'd solve it as follows:

                                  0.2 mmol Ba(OH)2     2 mmol OH-
# moles OH-(aq) = 15 mL Ba(OH)2 * ---------------  * -------------- = 6 mmol
                                    mL Ba(OH)2       1 mmol Ba(OH)2


                                  1 mmol H+    1 mmol H3PO4   1 mL H3PO4
# mL H3PO4(aq) = 6 mmol OH-(aq) * ---------  * ------------ * -----------
                                  1 mmol OH-     3 mmol H+    0.1 mmol H3PO4

               = 20 mL

where we've managed to keep mmol and mL throughout since we know mol/L = mmol/mL.

Chris Parr
University of Texas at Dallas
Programs in Chemistry, Room BE3.506
P.O. Box 830688  M/S BE2.6 (for snailmail)
Richardson, TX  75083-0688

   Voice: (214) 883-2485
     Fax: (214) 883-2925
     BBS: (214) 883-2168 (HST) or -2932 (V.32bis)
Internet: parr@utdallas.edu  (Click on that address to send Chris e-mail.)