**Problem : **

What is the pH of a solution of 0.36 M HCl, 0.62 M NaOH, and 0.15 M
HNO_{3}?

Hydrochloric acid and nitric acid are strong acids, and sodium hydroxide is
a
strong base; these all dissociate completely. The total [H^{+}]
from the two acids is 0.51 M and [OH^{-}]
from NaOH is
0.62 M. Therefore, 0.51 moles per liter of H^{+} will react with
0.51 moles per liter of
OH^{-} to form water. That leaves a 0.11 M NaOH solution. The
pOH
of a 0.11 M NaOH
solution is 0.96 pOH units, and the pH is 13.04 pH units.

**Problem : **

What percent of formic acid (HCOOH) is dissociated in a 0.1 M solution of
formic acid? The
*K*_{a} of formic acid is 1.77 x 10^{-4}.

**Problem : **

What happens to the pH of a 0.1 M solution of formic acid when enough HCl
(g) is added to make
the solution 0.01 M in HCl?

Note that in this problem the volume of the solution does not change so you
do not have to recalculate
the concentration of formic acid. The only change to the problem is that
there is now an initial
concentration of H^{+} in solution:

As you can see, the addition of a strong acid can, by Le Chatlier's
principle, cause a weak acid not to dissociate. An exact solution for this
problem does show
a small dissociation of
formic acid that is insignificant.

**Problem : **

What is the pH of a 0.001 M solution of H_{2}SO_{4}?
HSO_{4}^{-} has a p*K*_{a} of 1.2 x 10^{-
2}.

To solve this problem, you must first note that sulfuric acid's first
deprotonation is as a strong acid,
so we have a concentration of 0.001 M H^{+} to start and 0.001 M
hydrogen sulfate.
Because hydrogen sulfate is a weak acid, this problem becomes very similar
to the last one (see
). The calculated value for pH is 2.73 pH units
(note that it is lower
than 3 due to the dissociation of hydrogen sulfate).