Download REVIEW OF FRESHMAN CHEMISTRY: pH, pK, buffers, Henderson

REVIEW OF FRESHMAN CHEMISTRY:
pH, pK, buffers, Henderson-Hasselbalch Equation (see also Lehninger Principles of
Biochemistry pp. 95-105; pp. 123-127 cover ionization properties of amino acids.)
Biological reactions occur in aqueous environments, or in environments defined by the active centers
of enzymes which themselves exist in aqueous environments. The active centers of enzymes have
functional groups that participate in the processes of catalysis, and the chemical nature of those
functional groups is often a function of the pH. The importance of pH in the functional properties of
proteins and enzymes is well known.
Dissociation of weak acids
+
Consider the simple monoprotic acid HA (or alternatively, HA ):
HA
H
Example:
+
+
–
A
R-COOH
H
+
+
RCOO
–
OR
+
HA
Example:
H
+
+
A
+
R-NH3
H
+
+ R-NH2
The form with the associated proton is the conjugate acid form.
The form after dissociation of the proton (the form without the proton) is the conjugate base.
The equilibrium DISSOCIATION constant is called Ka. (The subscript "a" stands for "acid"
dissociation, not an association equilibrium. Biochemists often simply use the term "pK".) The higher
the Ka (the greater the tendency of the acid to DISSOCIATE/DONATE its proton), the stronger the
+
acid. Remember that pH = – log[H ] and pKa = – logKa. Thus the lower the pKa, the stronger the
acid.
Ka =
[H + ][A − ]
[HA]
1
1 [A − ]
=
•
[H + ] Ka [ HA]
pH = pKa + log
[A− ]
[HA]
The last equation is the Henderson-Hasselbalch Equation.
pH/pK/buffer review & practice problems, page 1 of 4
(Henderson-Hasselbalch Equation)
[A − ]
pH = pKa + log
[HA]
Look again at the equilibrium dissociation reaction and remember LeChatelier's Principle:
+
–
HA
H
+
A
+
The higher the [H ] (i.e., the lower the pH), the more the equilibrium shifts to the left, so the more
conjugate acid will be present. Conversely, the lower the [H+] (i.e., the higher the pH), the more the
equilibrium shifts to the right, so the more conjugate base will be present. The exact ratio of base to acid
depends on the pH and the pKa (Henderson-Hasselbalch Equation).
Note that when pH = pKa, [A–] = [HA], so [base] /[ acid] ratio = 1/1 ([base] = [acid]). When pH
= pKa, the “buffering capacity” of the mixture will be the greatest. That is, a given change in
–
concentration of base (OH ) or acid (H+) will result in the smallest change in pH. If the pH is ABOVE
the pKa, there will be more base than acid in the solution, i.e., the [base] /[ acid] ratio will be greater
than 1/1. If the pH is BELOW the pKa, there will be more acid than base, i.e., the [base] /[ acid]
ratio will be less than 1/1. When you do calculations, always stop at the end to think about whether
your answer makes sense.
Consider the following titration curve:
pH titration
10
9
pH
8
7
6
5
0
0.2
0.4
0.6
0.8
Equivalents hydroxide
pH/pK/buffer review & practice problems, page 2 of 4
1
Upon addition of 0.5 equivalents of base, 1/2 of the acid will be converted to the conjugate base,
such that [A–] = [HA], and pH = pKa. This condition will allow the largest concentration of both
species, and therefore the maximum buffering effect.
PRACTICE PROBLEMS (Work these without looking at the answers, which are on the last page.)
NOTE ESPECIALLY PROBLEM 2B, how to go from the ratio of base/acid (Henderson-Hasselbalch
calculation) to the fraction of the total (acid + base) that's in the form of the acid (or what fraction
is in the form of the base). In fact, anytime you calculate a RATIO and get a number, STATE the
(understood) "1" in the denominator. If ratio is 0.4, state it as 0.4 / 1. Also, start by writing chemical
equations to describe acid dissociation reactions, with CHARGE BALANCE.
+
7
+
1. For a solution whose pH is 6.0, what is [H ]? If [H ] is 5 x 10– M, what is the pH? (Use a simple
scientific calculator if necessary to do log problems.)
2. A. For a weak acid such as the R group carboxyl group of Glu or Asp in a protein, if the pKa of that
specific residue's carboxyl group is exactly 4.0, in an environment in which the pH = 5.0, what
–
would be the ratio of base to acid ( [COO ]/[COOH] )?
B. For that same carboxyl group, what fraction, or what percentage, of the total (population of all
the molecules in solution) is present in the form of the ACID (COOH) at pH 5.0? What fraction
–
of the total is present in the form of the BASE (COO ) at pH 5?
3. Suppose that about 1% of the molecules of a particular protein in solution have the imidazole group
of a specific His residue (say it's residue #20 in the amino acid sequence of that protein) in the
uncharged (neutral) form at pH 4.5. What is the pKa of that specific His residue in that protein?
(You need to know the ionization properties of the functional group of His to answer this question -look at the ionization equilibrium for His in Fig. 5-12, p. 125, of Lehninger Principles. All that
concerns you is the ionization of the R group, the imidazole group, which for this amino acid is the
middle pKa. The pKa of the imidazole group of the free amino acid histidine is about 6.0, but the
exact pKa values of amino acid functional groups in proteins vary somewhat, depending on the
specific environment around that residue in that protein; you're being asked to calculate this specific
His residue's pKa.)
pH/pK/buffer review & practice problems, page 3 of 4
SOLUTIONS TO PRACTICE PROBLEMS
+
1. pH = – log[H ] (Remember, the log is the exponent.)
pH
+
If pH = 6.0, [H ] =10–
6.0
= 10–
M
7
+
7
(Use a calculator for this.) If [H ] = 5 x 10– M, pH = – log(5 x 10– ) = 6.3
–
+
2. acid dissocation reaction for a carboxyl group: (R)-COOH ↔ (R)-COO + H
A. Use the Henderson-Hasselbalch Equation whenever you see a problem involving the relationship
of pH, pK, and base/acid ratio (or fraction of the total group that's in form of the base, or of the
acid).
pH = pKa + log([base]/[acid])
log([base]/[acid]) = pH – pKa = 5.0-4.0 = 1.0
1
([base]/[acid]) = 10 = 10/1 (There's 10 times as much of the carboxyl group in the form of the
base as in the form of the acid when the pH is 1 unit above the pKa.)
–
B. At pH 5.0, the COOH/COO ratio is 10/1. We can do this problem most simply by using
–
proportions; the total of the carboxyl group is COOH + COO = 10 + 1 = 11. The fraction of the
total in the form of the acid (COOH) is
–
(COOH) / total = (COOH) / (COOH + COO ) = 1/11 = 0.091
9.1% of the molecules in solution have that carboxyl group in its acid form.
–
–
–
COO / total = COO / (COOH + COO ) = 10/11 = 0.909
Alternatively, if you've already calculated that acid/total = 0.091, then
base/total = 1 – (acid/total) = 1 – (0.091) = 0.909. Either way,
90.9% of the molecules in solution have that carboxyl group in its base form.
3. We'll abbreviate the imidazole group in question as Im, so neutral form is Im, the conjugate base,
+
and charged form is Im , the conjugate acid.
+
+
Im ↔ Im + H
Use the Henderson-Hasselbalch Equation again.
"About 1% neutral" means that base (Im) / total = about 0.01 or 1/100. Thus base/acid ratio would
be base/(total – base) = 1 / (100-1) = 1/99.
pH = pKa + log([base]/[acid])
pKa = pH – log([base]/[acid]) = 4.5 – log (1/99) = 4.5 – (– 2.0) = 4.5 + 2.0 = 6.5
The pKa of that specific His residue in that protein must be about 6.5. (Note that this is slightly
different from the pKa of the R group of the free amino acid histidine.)
pH/pK/buffer review & practice problems, page 4 of 4