Page 181 - Modern Analytical Chemistry

1400-CH06 9/9/99 7:41 AM Page 164

164 Modern Analytical Chemistry

pH of 0.1 M HL Finding the pH of a solution of alanine is more complicated than
+
–
that for H 2 L or L because we must consider two equilibrium reactions involving
HL. Alanine is an amphiprotic species, behaving as an acid
–
+
HL(aq)+H 2 O(l) t H 3 O (aq)+L (aq)
and a base
+
–
HL(aq)+H 2 O(l) t OH (aq)+H 2 L (aq)
As always, we must also consider the dissociation of water
+
–
2H 2 O(l) t H 3 O (aq)+OH (aq)
+
+
–
–
This leaves us with five unknowns ([H 2 L ], [HL], [L ], [H 3 O ], and [OH ]), for
which we need five equations. These equations are K a2 and K b2 for HL,
–
+
L
[ HO ][ ]
3
K a2 =
[ HL]
–
+
K w [ OH ][ H L ]
2
K b2 = =
K a1 [ HL]
the K w equation,
+
–
K w =[H 3 O ][OH ]
a mass balance equation on HL,
–
+
C HL =[H 2 L ]+[HL]+[L ]
and a charge balance equation
–
+
+
–
[H 2 L ]+[H 3 O ] = [OH ]+[L ]
From the ladder diagram it appears that we may safely assume that the concentra-
+
–
tions of H 2 L and L are significantly smaller than that for HL, allowing us to sim-
plify the mass balance equation to
C HL = [HL]
+
Next we solve K b2 for [H 2 L ]
K w [HL ] [HL ][HO + ] C HL [HO + ]
3
3
[HL + ] = = =
2
K a1 [OH – ] K a1 K a1
–
and K a2 for [L ]
[HL ] a K C HL
2
a K 2
–
L
[] = + = +
[HO ] [HO ]
3
3
Substituting these equations, along with the equation for K w , into the charge bal-
ance equation gives us
+
[
C HL HO ] + K w K C HL2
a
3
+ [ HO ] = + + +
3
K a1 [ HO ] [ HO ]
3
3
which simplifies to
æ C HL ö 1
[HO + ] ç + 1 = (K w +K C HL )
÷
3
a2
èK a1 ø [HO + ]
3
KC HL +K w
a2
2
[HO + ] =
3
C
( HL /K a1 ) + 1
KK a2 C HL +K K w
a1
a1
[HO + ] = 6.41
3
C HL +K a1