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.

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.

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).