POLYPROTIC BUFFERS
So, I'm not ready to let go of this intermediate form example just yet. My question for you about it now
is this: Is it a buffer? Will it resist meaningful efforts to change its pH through addition of acid or base?
You might be tempted to say yes...but you'd be wrong. (The answer is no!) As we have seen, (and as
Harris preaches) there is very little of either the conjugate base or of the conjugate acid of HCO3- present
in that solution. A buffer requires a significant (read: intentional!) concentration of both an acid and its
conjugate base be simultaneously present in the solution. Another way of looking at it is that the pH
(8.34) is not within one unit of either pKa1 (6.351) or pKa2 (10.329). At those pH's, you would have a
mix of (nearly) equal amounts of both a weak acid and its conjugate base, and thus a buffer. You can't
make a decent pH 8.34 buffer using carbonic acid, though...even if you mix equal amounts of H2CO3
and Na2CO3. Why? Well, because they would react with each other to form mostly HCO3-. It would be
equivalent to dissolving just NaHCO3- to begin with.
However, you can make a buffer with a polyprotic acid. Moreover, in most cases all of the same rules
and provisos apply...you can use the Henderson-Hasselbalch equation and approximations, and you will
generally actually prepare the buffer by mixing a strong acid or base with just one form of the
polyprotic. The new wrinkle that can arise, and that you may have to worry about, comes about when
the pKa you want to use in the buffer is very close to another pKa of the polyprotic. We'll return to that
situation, and polyprotic buffers more generally, once we learn about alpha plots.
FRACTIONAL COMPOSITION EQUATIONS ("α FRACTIONS" AND "α PLOTS")
Professor Kuwata introduced the concepts of fractional composition and principal species for a
monoprotic acid. Our goal is to now extend that notion to a polyprotic system. Recall that
Concentration of species present in specified form
αspecific form ≡
Total concentration of species in all forms
So, for example, for a triprotic acid H3A, we define four α fractions:
Moles present as H 3 A
[H 3 A]
αo = α H 3 A =
=
Total moles of A present (in all forms)
[H 3 A] + [H 2 A − ] + [HA 2 − ] + [A 3− ]
Note: Subscript 0 indicates the form with zero protons removed from the fully protonated form.
[H 2 A − ]
Moles present as H 2 A −
α1 = α H A − =
=
2
Total moles of A present (in all forms)
[H 3 A] + [H 2 A − ] + [HA 2 − ] + [A 3− ]
α2 = α HA 2−
Moles present as HA 2−
[HA 2 − ]
=
=
Total moles of A present (in all forms)
[H 3 A] + [H 2 A − ] + [HA 2 − ] + [A 3− ]
α3 = α A 3−
Moles present as A 3−
[ A 3− ]
=
=
Total moles of A present (in all forms)
[H 3 A] + [H 2 A − ] + [HA 2 − ] + [A 3− ]
These fractions must clearly sum to one: αo + α1 + α2 + α3 = 1
It turns out (though it seems overly optimistic to hope for) that exact, analytic expressions for these can
be put together: mostly because the denominator can be replaced with a formal concentration. Here's
how we go about doing this for the triprotic case:
The key to remember here is don't write a charge balance; instead, use a mass balance on A:
F = [H3A] + [H2A-] + [HA2-] + [A3-]
Next, write each of the acid dissociation equilibria and their corresponding equilibrium expressions:
[H 2 A − ][H + ]
+
H3A
H2A + H
Ka1 =
[H 3 A]
-
H2A
2-
+
HA + H
Ka2
[HA 2 − ][H + ]
=
[H 2 A − ]
[H 2 A − ][H + ]
HA
A +H
Ka3 =
[ A 3− ]
To get an analytic expression for αo, we solve these in sequence to eliminate the deprotonated forms in
favor of the fully protonated form:
K
[H 2 A − ][H + ]
Ka1 =
⇒ [H2A-] = [H3A] a+1
[H 3 A]
[H ]
2-
Ka2
[HA 2 − ][H + ]
=
[H 2 A − ]
Ka3
⇒
[HA2-] = [H2A-]
[H 2 A − ][H + ]
=
[ A 3− ]
⇒
3-
+
Ka2
K
Ka2
K K
= [H3A] a+1
= [H3A] a1 + a22
+
+
[H ]
[H ] [H ]
[H ]
[A3-] = [HA2-]
K a3
K K K
= [H3A] a1 +a 2 3 a 3
+
[H ]
[H ]
Now, we put each of these into the mass balance equation, to get an equation in only F and the fully
protonated form (here, H3A):
F = [H3A] + [H2A-] + [HA2-] + [A3-]
K
K K
K K K
= [H3A] + [H3A] a+1 + [H3A] a1 + a22 + [H3A] a1 +a 2 3 a 3
[H ]
[H ]
[H ]
K
K K
K K K ⎞
⎛
= [H3A] ⎜⎜1 + a+1 + a1 + a22 + a1 +a 2 3 a 3 ⎟⎟
[H ]
⎠
⎝ [H ] [H ]
Plug this into the αo expression and (optionally) simplify:
αo = α H 3 A
K
K K
K K K ⎞
⎛
[H 3 A]
=
= ⎜⎜1 + a+1 + a1 + a22 + a1 +a 2 3 a 3 ⎟⎟
F
[H ]
⎝ [H ] [H ]
⎠
−1
−1
⎛ [H + ]3 + K a1[H + ]2 + K a1K a 2 [H + ] + K a1K a 2 K a 3 ⎞
⎟⎟
= ⎜⎜
[H + ]3
⎠
⎝
+ 3
⎞
⎛
[H ]
⎟⎟
= ⎜⎜ + 3
+ 2
+
⎝ [ H ] + K a1 [ H ] + K a1 K a 2 [ H ] + K a1 K a 2 K a 3 ⎠
You can then get α1 by solving each of the equilibrium expressions for H2A- and following an analogous
procedure, and so on for the other α's. Harris provides the generalized result that is obtained for any
polyprotic acid in the margin on p.192; but we want you to know how to derive any given alpha.
« Demo and discussion: α fraction vs. pH for carbonic acid, and for two hypothetical acids with pKa1 &
pKa2 closer together (1 and 0.5 units apart).»