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When a strong acid or a strong base is added to water, it nearly
completely dissociates into its ion constituents
because it has a p*K*
_{a} or p*K*
_{b} less
than zero. For
example, a solution of
H_{2}SO_{4} in water contains mostly H^{+} and
SO_{4}
^{2-}, and almost no H_{2}SO_{4} is left
undissolved.
This makes calculating the pH of a strong acid or strong base solution
exceedingly
simple--*the concentration
of acid equals the concentration of H ^{+}.* Recall that pH is
computed by taking the negative log of of [H

Calculating the pH of weak acid and weak base solutions is much more complicated than the above case--weak acids and bases do not completely dissociate in aqueous solution but are in equilibrium with their dissociated forms. Therefore, we must apply what we know about equilibria to solve these types of problems. For example, let's calculate the pH of a 0.10 M solution of acetic acid in water. To do this, we first write down the equilibrium involved and the expression for the equilibrium constant:

Figure %: Note that H^{+} represents H_{3}O^{+} in the equilibrium
constant expression.

Next, you should compile a table of values for the concentration of all
species involved in the
equilibrium. We already know that the initial concentration, [
]_{o}, of acetic acid is 0.10
M and that the initial concentration of H^{+} is 10^{-7} (since
the solvent is neutral water).
Even though there is
an initial concentration of H^{+} in solution, it is so small compared
to the amount
produced by the acid that it is usually ignored. From the stoichiometry of
the reaction, one mole of
H^{+} and one mole of acetate (Ac^{-}) are produced for
each mole of acetic
acid dissociated. Therefore, if we denote the amount of acetic acid
dissociated as x, the final
concentrations of H^{+} and Ac^{-} are both x, and the final
concentration of HAc in
solution is 0.10 M - x. This data is summarized in :

Figure %: Table of equilibrium values for the dissociation of acetic acid

After you have compiled the table of values, you can
substitute the equilibrium
concentration values for each species into your expression for
*K*
_{a} as shown
below:

You should note that the above equation is a quadratic
equation in x and
therefore requires use of the quadratic equation to solve for x. If,
however, we can make the
assumption that [HAc]_{o} - x = [HAc]_{o}, the
equation becomes much
easier to solve. We can do this if HAc is a weak enough acid that it
dissociates very little, and the change in [HAc] is negligible.

Is this assumption valid? Solving the quadratic equation
using
the quadratic formula gives a pH of 2.88. The difference of 0.01 pH unit
is small enough to be insignificant, so the
assumption is valid in this case and will certainly save you some
time on a test. We can make the approximation [HA]_{o} - x =
[HA]_{o} so long as x is less than 5% of the initial concentration of
HA.
X will be greater than 5% of
[HA]_{o} with stronger weak acids at low concentrations. Consider these
guidelines when you decide whether or not to make the approximation: you can
simplify the quadratic equation if the solution is
more concentrated than 0.01 M and the p*K*
_{a} of the acid is
greater than 3.

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