Answer to Pop Question 1
Recall 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, we were making an assumption here. One of
the conditions that must be satisfied inan 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no longer valid.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? Some will recognize the problem
and come 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!!)