Answer to Pop Question 1
(Chemistry 1046--Fall 1996)

     I had a total of 10 responses to the Pop Question.  Only two
people recognized the problem, and they only gave the first approximation 
solution given below.  No one dug a little deeper to consider the effect 
of Le Chatelier's principle.

     Recall, I said when figuring pH of a strong acid, the acid
completely dissociates.  A 0.01 M HCl solution forms 0.1 M Cl- and
0.1 M H+.  But remember, I pointed out that we were making an
assumption here.  One of the conditions that must be satisfied in
an equilibrium is that the number of positively charged species
must equal the number of negatively charged species.  So

      H+  =   Cl-  +  OH-

But in acid solution, even an acid as weak as pH 6, the OH- is
going to be small relative to the Cl-, and so can be neglected. 
But when you get very low concentrations of HCl, this assumption is
no longer valid.  In other words you must take into account the H+
produced from the ionization of water as well as that from the
ionization of HCl.  

So how do you do that?  A couple of people recognized the problem
and came close to the answer by suggesting that you add the two
together:

  H+ from HCl  =  10exp-8

  H+ from H2O  =  10exp-7

  10exp-8  +  10exp-7  =  1.1 *  10exp-7

 -log(1.1*10exp-7) =  6.96    (To two significant figures in the
                              mantissa).

But now LeChatelier's principle comes into play.  As you add
additional hydrogen ion from the HCl, the equilibrium of the water
dissociation shifts toward the formation of water (to lower the H+
again).  So how can we take that into account?

Go back to our charge balance equation:

   H+  =  Cl-  +   OH-

and recognized the OH- = Kw/H+

Therefore   H+  =  Cl-  +   Kw/H+

or (H+)^2  =  Cl-(H+)  +  Kw

or  (H+)^2   -  10exp-8 (H+)  -  10exp-14.

Solving for H+ using the quadratic formula gives

   [+10exp-8 +/- sq.rt.((10exp-8)^2 + 4(10exp-14))]/2

   =    [+10exp-8 +/- sq.rt.(4.01 * 10exp-14)]/2

   =   [+10exp-8 + 2.0025 * 10exp-7]/2

   =  2.1025 * 10exp-7/2  =  1.05125 * 10exp-7

  pH  = -log(1.05125*10exp-7) = 6.97829, or round off to 6.98.


Answer:  6.98

(So note, those who neglected the LeChatelier shift came close at
6.96.  The more important lesson, though is for you to recognize
that you can't start at pH 7, add acid, and expect the solution to
become more basic!!)