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N-4
In Part III, you will experimentally determine the molar enthalpy of neutralization of a strong acid (HCl,
hydrochloric acid) and a weak acid (CH3COOH, acetic acid). From this you will calculate the
enthalpy of dissociation of acetic acid.
Neutralization of any strong acid with a strong base can be represented by the net ionic equation:
H3O
+
(aq)
+ OH
-
(aq)
2 H2O
(l)
(13)
This reaction only involves bond formation and thus is an exothermic reaction (?H = negative). The
heat evolved by the reaction depends on the amount of acid and base that react but not on the type of
acid or base, provided that they are both strong electrolytes, that is, they both dissociate
completely in an aqueous solution and do not form a precipitate. This measurement illustrates why in
previous experiments you were always cautioned, when mixing strong acids and strong bases together,
to do so slowly so that the heat of neutralization could dissipate.
The second neutralization reaction you will study involves acetic acid (CH3COOH). The molar enthalpy
of neutralization of a weak acid is generally different than that of a strong acid neutralization. By
definition, a weak acid is not fully dissociated (typically much less than 5% dissociated, depending on
the concentration). For instance, a 1.00 M CH3COOH solution is only 0.42% dissociated in solution.
Why does the heat of neutralization of a weak acid differ from that of a strong acid? We can answer
this question by considering the following reactions:
Weak Acid Dissociation: CH3COOH
(aq)
+ H2O
(l)
H3O
+
(aq)
+ CH3COO
-
(aq)
?H
diss
(14)
Neutralization Reaction: H3O
+
(aq)
+ OH
-
(aq)
2 H2O
(l)
?H
neutralization
(15)
Overall Reaction: CH3COOH
(aq)
+ OH
-
(aq)
CH3COO
-
(aq)
+ H2O
(l)
?H
overall
(16)
Since the overall reaction is the sum of the weak acid dissociation and neutralization reaction, then by
Hess’s Law, ?H
overall
= ?H
diss
+ ?H
neut
. Two factors determine the magnitude and sign of the
enthalpy of dissociation of the weak acid (?H
diss
):
