Untitled - Kelly Walsh High School

242 CHEMISTRY FOR THE UTTERLY CONFUSED
have been added. Usually you will find the moles by multiplying the concentration
by the volume of the solution. In a few cases, you may find the number of
moles from the mass and the molar mass.
In the second region of the titration curve, the substance added is the limiting
reagent. As a limiting reagent, it is no longer present to affect the pH. The other
reactant is in excess, and this excess will have an affect upon the pH. The moles
of excess reactant divided by the total volume of the solution (the initial volume
plus the volume of the solution added) gives the concentration. The solution
also contains the products of the reaction. If the excess reactant is either a weak
acid or a weak base, the concentration of the products will be important. A
buffer solution will be present any time you have an excess of a weak acid or
weak base along with its conjugate.
The next point in the titration curve is the equivalence point. At this point,
both the material added and the material originally present are limiting. At
this point, neither of the reactants will be present and therefore will not
affect the pH. If the titration involves a strong acid and a strong base, the pH
at the equivalence point is 7. If the titration involves a weak base, only the
conjugate acid is present to affect the pH. This will require a K a calculation.
If the titration involves a weak acid, only the conjugate base is present to
affect the pH. This will require a K b calculation. The calculation of the conjugate
acid or base will be the moles produced divided by the total volume of
the solution.
The final region of the titration curve is after the equivalence point. In this
region, the material originally present in the container is limiting. The excess
reagent, the material added, will affect the pH. If this excess reactant is a weak
acid or a weak base, this will be a buffer solution.
Let’s use the following example to see how to deal with each of these regions.
This will be a titration of the weak acid with a strong base.
A 100.0 mL sample of 0.150 M nitrous acid (pK a 3.35) was titrated with 0.300 M
sodium hydroxide, NaOH. Determine the pH of the solution after the following
quantities of base have been added to the acid solution: a. 0.00 mL, b. 25.00 mL,
c. 50.00 mL, and d. 55.00 mL.
a. 0.00 mL. Since no NaOH has been added, this is the initial pH section of the
titration curve.
The only substance present is nitrous acid, HNO 2, a weak acid. Since this is a
weak acid, this must be a K a problem. As a K a problem, we can set up a simple
equilibrium problem: