Page 297 - Modern Analytical Chemistry

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280 Modern Analytical Chemistry

approach will make use of the equilibrium calculations described in Chapter 6. We
also will learn how to sketch a good approximation to any titration curve using only
a limited number of simple calculations.

Titrating Strong Acids and Strong Bases For our first titration curve let’s consider
the titration of 50.0 mL of 0.100 M HCl with 0.200 M NaOH. For the reaction of a
strong base with a strong acid the only equilibrium reaction of importance is
+
–
H 3 O (aq)+OH (aq) t 2H 2 O(l) 9.1
The first task in constructing the titration curve is to calculate the volume of NaOH
needed to reach the equivalence point. At the equivalence point we know from re-
action 9.1 that
Moles HCl = moles NaOH

or
M a V a = M b V b
where the subscript ‘a’ indicates the acid, HCl, and the subscript ‘b’ indicates the
base, NaOH. The volume of NaOH needed to reach the equivalence point, there-
fore, is
50
0
MV (.100 M)( .0 mL)
aa
25
V eq = V b = = = .0 mL
0
M b (.200 M)
Before the equivalence point, HCl is present in excess and the pH is determined by
the concentration of excess HCl. Initially the solution is 0.100 M in HCl, which,
since HCl is a strong acid, means that the pH is
+
pH = –log[H 3 O ] = –log[HCl] = –log(0.100) = 1.00
–1
14
The equilibrium constant for reaction 9.1 is (K w ) , or 1.00 ´10 . Since this is such
a large value we can treat reaction 9.1 as though it goes to completion. After adding
10.0 mL of NaOH, therefore, the concentration of excess HCl is
moles excess HCl MV b b
aa -M V
[HCl] = =
total volume V a +V b
0
0
)
0
50
( .100 M )( . mL - ( .200 M )( . mL )
10
0
= = . 0 050 M
0
0
50 . mL +10 . mL
giving a pH of 1.30.
At the equivalence point the moles of HCl and the moles of NaOH are equal.
Since neither the acid nor the base is in excess, the pH is determined by the dissoci-
ation of water.
–
+
+ 2
K w = 1.00 ´10 –14 =[H 3 O ][OH ]=[H 3 O ]
–7
+
[H 3 O ] = 1.00 ´10 M
Thus, the pH at the equivalence point is 7.00.
Finally, for volumes of NaOH greater than the equivalence point volume, the
–
pH is determined by the concentration of excess OH . For example, after adding
–
30.0 mL of titrant the concentration of OH is