Experiment 6: Standardization of a Solution Using
a Primary Standard
and
Experiment 7: Determination of Concentration by
Acid-Base Titration
Background: A neutralization reaction results when a strong acid reacts with a strong base and
is represented by the following net-ionic equation:
HOH HO+-]]aqgg+ aq " (^2) ]lg
The equilibrium constant for this reaction at room temperature is approximately 10^14. The mag-
nitude of the equilibrium constant shows that the reaction proceeds essentially to the product
side, using up nearly all of the ion present as the limiting reactant. If both H+and OH–are pre-
sent in equal quantities as reactants, the resulting solution will be neutral, because water is the
only product of the reaction. When weak acids are titrated with strong bases, the resulting solu-
tion will be basic because the conjugate base of the weak acid will form. When weak bases are
titrated with strong acids, the resulting solution will be acidic because the conjugate acid of the
weak base will form.
In Part I of this experiment a titration will be performed by titrating a standardized solution of
HCl with a NaOH solution whose concentration is not known. One cannot make a standardized
solution of NaOH directly by weighing out an exact amount of NaOH and diluting it with dis-
tilled water. This is because solid NaOH absorbs both H 2 O and CO 2. The titration will allow the
concentration of the OH–to be calculated accurately. The equivalence point is the point in the
titration when enough titrant has been added to react exactly with the substance in solution be-
ing titrated and will be determined by using an indicator. The end point, the point in the titra-
tion at which the indicator changes color, will be very close to 7. At this point, a drop of acid or
base added to the solution will change the pH by several pH units.
In Part II, a weak acid is titrated with the strong base from Part I and the equivalence point will
be somewhat higher than 7 (8 – 9). By titrating a known amount of solid acid with the stan-
dardized NaOH, it will be possible to determine the number of moles of H+the acid furnished.
From this information, one can obtain the equivalent mass (EMa) of the acid:
EM
moles of H
grams acid
a= +
However, the equivalent mass of the acid may or may not be the same as the molecular mass of
the acid since some acids produce more than one mole of H+per mole of acid. In order to find
the molecular mass from the EMa, the molecular formula is required.
Part III: AP Chemistry Laboratory Experiments