Page 159 - Modern Analytical Chemistry

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142 Modern Analytical Chemistry

–
+
2–
HCO 3 (aq)+H 2 O(l) t H 3 O (aq)+CO 3 (aq) 6.8
–
–
HCO 3 (aq)+H 2 O(l) t OH (aq)+H 2 CO 3 (aq) 6.9
amphiprotic A species that can serve as both a proton donor and a proton acceptor is called am-
A species capable of acting as both an phiprotic. Whether an amphiprotic species behaves as an acid or as a base depends
acid and a base.
on the equilibrium constants for the two competing reactions. For bicarbonate, the
acid dissociation constant for reaction 6.8
K a2 = 4.69 ´10 –11
is smaller than the base dissociation constant for reaction 6.9.
K b2 = 2.25 ´10 –8
Since bicarbonate is a stronger base than it is an acid (k b2 > a2 ), we expect that
k
–
aqueous solutions of HCO 3 will be basic.
Dissociation of Water Water is an amphiprotic solvent in that it can serve as an
acid or a base. An interesting feature of an amphiprotic solvent is that it is capable
of reacting with itself as an acid and a base.
+
–
H 2 O(l)+H 2 O(l) t H 3 O (aq)+OH (aq)
The equilibrium constant for this reaction is called water’s dissociation con-
stant, K w ,
–
+
K w =[H 3 O ][OH ] 6.10
which has a value of 1.0000 ´10 –14 at a temperature of 24 °C. The value of K w varies
substantially with temperature. For example, at 20 °C, K w is 6.809 ´10 –15 , but at
30 °C K w is 1.469 ´10 –14 . At the standard state temperature of 25 °C, K w is
1.008 ´10 –14 , which is sufficiently close to 1.00 ´10 –14 that the latter value can be
used with negligible error.

The pH Scale An important consequence of equation 6.10 is that the concentra-
+
+
–
–
tions of H 3 O and OH are related. If we know [H 3 O ] for a solution, then [OH ]
can be calculated using equation 6.10.
EXAMPLE 6.2
+
–5
–
What is the [OH ] if the [H 3O ] is 6.12 ´10 M?
SOLUTION
K w . 100 ´10 –14
[OH – ] = + = –5 = . 163 ´10 –10
[HO ] . 612 ´10
3

pH Equation 6.10 also allows us to develop a pH scale that indicates the acidity of a so-
–
+
+
Defined as pH = –log[H 3 O ]. lution. When the concentrations of H 3 O and OH are equal, a solution is neither
acidic nor basic; that is, the solution is neutral. Letting
–
+
[H 3 O ] = [OH ]
and substituting into equation 6.10 leaves us with
+ 2
K w =[H 3 O ] = 1.00 ´10 –14