Chem 438 - Introductory Biochemistry - Spring 2013
How does water interact with non-polar molecules?
- Recall the solution of charged ions such as sodium.
- The problem here is that in order to dissolve in water a
non-polar molecule must disrupt a series of H-bonds and no new
bonds of equal strength are substituted.
- Thus water tends to exclude non-polar substances.
When we forcefully disperse a non-polar substance into water
then the water must form a cage around the molecule to maximize
H-bonds for each water molecule. [overhead 7-51, VV]
- Now an additional problem arises-the waters hydrating the
non-polar group are "locked" in place - they can't
easily flip about because there is no interior bond for substitution!
- Thus the entropy of these water molecules is greatly
reduced. So the insolubility of non-polar groups in water
has both enthalpic and entropic factors!
- Note that many non-polar molecules may dissolve in water
by the formation of micelles as in Fig. 2.9, p 33 of Text. [slide]
public domain image via Wikipedia Creative Commons
Finally, recall that water is a good nucleophile and so will
participate in many chemical reactions-readily hydrolyzes esters,
amides, anhydrides etc.
Ionization of Water, pH & Buffers
Dissociation of water molecules: One aspect of water we have yet to talk about is its dissociation or ionization. In normal aqueous solution there is a certain probability that a hydrogen nucleus (a proton) can exchange between two hydrogen bonded molecules:
(Of course the hydronium ion, H3O+, will be associated with additional water molecules as well through H-bonding. For simplicity we will just write H+, with the understanding that it refers in fact to hydrated hydronium ions in aqueous solution. ) Note the reaction is not highly favored, in neutral solution (no excess H+or OH-) there will only be 10-7 molar hydronium ions, in other words only about 2 of every billion water molecules will be protonated!
For aqueous solution [H+][OH-]= 10-14;
- pH = -log [H+]. Remember that a lo pH means a high concentration of protons.
- More generally 'p' = -log[ ], so pOH = -log[OH-] and by analogy, pKa = -logKa
- Note also, that if we take logs of both sides of the equation: [H+][OH-]= 10-14, we get pH + pOH = 14.
Brönsted acid: we will be using the Brönsted definition for acids and bases:
- an acid is a proton donor
- a base is a proton acceptor.
- Recall the corollary that acids and bases therefore exist as conjugate acid base pairs. Note that when an acid by this definition gives up a proton it becomes a base, since the reverse reaction would be accepting a proton:
- Thus the acetic acid in the first reaction becomes its conjugate base acetate ion, while the base, hydroxide ion, becomes its conjugate acid, water. In the reverse reaction the nomenclature also reverses. Note that a molecule such as water can be both an acid, donating a proton to become its conjugate base hydroxide ion, or it can be a base, accepting a proton to become its conjugate acid, a hydronium ion.
pH & Buffers
The strengths (ability to donate protons) of acids vary considerably.
- For the general acid HA we can write:
HA + H2O
A– + H3O+
Ka = [H3O+] [A–] / [HA]
- Where Ka is the acid dissociation constant.
(Note that the definition of Ka is based on
the Brønsted definition.) Values of Ka can vary tremendously (1015 to 10-60) -
after all anything with at least one proton can be considered
an acid under some circumstances with this definition. The common
definition of a strong acid is an acid which dissociates completely
in a 1 M solution. The common strong acids in aqueous solution,
such as sulfuric, nitric and hydrochloric acids have Ka values (for the first dissociation in the case of sulfuric) of
102 to 109. Thus they all dissociate completely
(first dissociation only for sulfuric) in aqueous solution, though
they will have different strengths in some other solvents. Most
common organic acids are weak in aqueous solution, having Ka values of 10-5 to 10-15. Note that whether
an acid is strong or weak is dependent on the solvent system!
Strong acids have weak conjugate bases, and vice-versa.
- For reactions involving a strong acid or base we can assume,
for practical purposes, that all of the strong acid or base added
to a mixture will react until the base or acid originally present
in solution is completely consumed. (Of course this is an approximation,
all reactions actually approach an equilibrium condition, so
that, in theory, there is always some reactant and some product
present.) For example, if we start with a solution containing
0.100 mole of acetic acid and add 0.050 moles of sodium hydroxide
the resulting mixture will contain 0.050 moles acetic acid, 0.050
moles sodium acetate and 0.000 moles sodium hydroxide (actually
about 10-10 moles, which is 0.000 for our thousandths
place significant figure calculation).
The equilibrium equation for a mixture of a weak acid and its
conjugate base can be rewritten by taking logs of both sides and
rearranging to give the Henderson-Hasselbalch equation: pH = pKa + log [A-]/[HA]
We frequently represent the reaction of an acid with a base
as a titration curve (Fig. 2.16).
understand these curves and be able to label them for axis, percent
dissociation at beginning, middle and end, buffer region, end-point,
and how to find pKa. An exercise to help you to review titrations curves is available.
© R. A. Paselk 2010;
Last modified 4 February 2013