Page 178 - Modern Analytical Chemistry

1400-CH06 9/9/99 7:40 AM Page 161

Chapter 6 Equilibrium Chemistry 161

The second equilibrium reaction is the dissociation of water, which is an obvious
yet easily disregarded reaction
–
+
2H 2 O(l) t H 3 O (aq)+OH (aq)
+
–
K w =[H 3 O ][OH ] = 1.00 ´10 –14 6.36
–
–
+
Counting unknowns, we find four ([HF], [F ], [H 3 O ], and [OH ]). To solve this
problem, therefore, we need to write two additional equations involving these un-
knowns. These equations are a mass balance equation
–
C HF =[HF]+[F ] 6.37
and a charge balance equation
–
+
–
[H 3O ]=[F ] + [OH ] 6.38
We now have four equations (6.35, 6.36, 6.37, and 6.38) and four unknowns
+
–
–
([HF], [F ], [H 3 O ], and [OH ]) and are ready to solve the problem. Before doing
so, however, we will simplify the algebra by making two reasonable assumptions.
First, since HF is a weak acid, we expect the solution to be acidic; thus it is reason-
able to assume that
+
–
[H 3 O ] >> [OH ]
simplifying the charge balance equation (6.38) to
–
+
[H 3 O ]=[F ] 6.39
Second, since HF is a weak acid we expect that very little dissociation occurs, and
–
[HF] >> [F ]
Thus, the mass balance equation (6.36) simplifies to
C HF = [HF] 6.40

For this exercise we will accept our assumptions if the error introduced by each as-
sumption is less than ±5%.
Substituting equations 6.39 and 6.40 into the equilibrium constant expression
for the dissociation of HF (equation 6.35) and solving for the concentration of
+
H 3O gives us
+
+
[ HO ][ HO ]
3
3
K a =
C HF
[HO + ] = KC ( .68 ´ 10 –4 )( . ) = . 26 ´ –2 M
10
aHF =
10
3

Before accepting this answer, we must verify that our assumptions are acceptable.
–
+
The first assumption was that the [OH ] is significantly smaller than the [H 3 O ]. To
–
calculate the concentration of OH we use the K w expression (6.36)
K w . 100 ´10 –14 –13
–
[OH ] = + = –2 = 38 M
. ´10
. ´10
[HO ] 26
3
–
Clearly this assumption is reasonable. The second assumption was that the [F ] is
significantly smaller than the [HF]. From equation 6.39 we have
–2
–
[F ] = 2.6 ´10 M