How To Predict the Outcome of Acid–Base Reactions
How To Predict the Outcome of Acid–Base Reactions
❒The following
table gives the approximate pKa values for a range of representative compounds.
❒While you
probably will not be expected to memorize all of the pKa values in the table ,
it is a good idea to begin to learn the general order of acidity and basicity
for some of the common acids and bases.
❒The examples
given the table are representative of their class or functional group. For
example, acetic acid has a pKa = 4.75, and carboxylic acids generally have pKa values
near this value (in the range pKa = 3–5). Ethyl alcohol is given as an example of
an alcohol, and alcohols generally have pKa values near that of ethyl alcohol
(in the pKa range 15–18), and so on. There are exceptions, of course, and we
shall learn what these exceptions are as we go on.
❒By learning the
relative scale of acidity of common acids now, you will be able to predict whether
or not an acid–base reaction will occur as written.
❒The general
principle to apply is this: acid–base reactions always favor the formation of
the weaker acid and the weaker base.
❒The reason for
this is that the outcome of an acid–base reaction is determined by the position
of an equilibrium. Acid–base reactions are said, therefore, to be under equilibrium
control, and reactions under equilibrium control always favor the formation of
the most stable (lowest potential energy) species. The weaker acid and weaker base
are more stable (lower in potential energy) than the stronger acid and stronger
base.
❒Using this
principle, we can predict that a carboxylic acid (RCO2H) will react
with aqueous NaOH in the following way because the reaction will lead to the
formation of the weaker acid (H2O) and weaker base (RCO2-):
❒Because there
is a large difference in the value of the pKa of the two acids, the position of
equilibrium will greatly favor the formation of the products. In instances like
these we commonly show the reaction with a one-way arrow even though the
reaction is an equilibrium.
Solved problem
Probem (1): Consider
the mixing of an aqueous solution of phenol, C6H5OH (see
Table above), and NaOH. What acid–base reaction, if any, would take place?
Strategy:
Consider the
relative acidities of the reactant (phenol) and of the acid that might be
formed (water) by a proton transfer to the base (the hydroxide ion).
Answer:
The following
reaction would take place because it would lead to the formation of a weaker acid
(water) from the stronger acid (phenol). It would also lead to the formation of
a weaker base, C6H5ONa, from the stronger base, NaOH.
Probem (2): Using
Table above, explain why the acid–base reaction that takes place between NaH
(as source of CH- ions) and CH3OH is:
Answer:
A hydride ion is a very strong base, being the
conjugate base of H2 (a very weak acid, pKa = 35). Hydride will remove
the most acidic proton from CH3OH.
Although CH3OH
is not given in the Table above, we can compare it to CH3CH2OH,
a similar alcohol whose hydroxyl group pKa is 16, far more acidic than any
proton attached to a carbon without a functional group (e.g., a proton of CH3CH3,
which has pKa = 50).
Because the
proton attached to the oxygen is much more acidic, it is removed
preferentially.
Water Solubility as the Result of Salt Formation
❒Although acetic
acid and other carboxylic acids containing fewer than five carbon atoms are
soluble in water, many other carboxylic acids of higher molecular weight are
not appreciably soluble in water.
❒Because of
their acidity, however, water-insoluble carboxylic acids dissolve in aqueous sodium
hydroxide; they do so by reacting to form water-soluble sodium salts:
❒We can also
predict that an amine will react with aqueous hydrochloric acid in the
following
way:
❒While
methylamine and most amines of low molecular weight are very soluble in water,
amines with higher molecular weights, such as aniline (C6H5NH2),
have limited water solubility. However, these water-insoluble amines dissolve
readily in hydrochloric acid because the acid–base reactions convert them into
soluble salts:
No comments