REVIEW OF FRESHMAN CHEMISTRY:
pH, pK, buffers, Henderson-Hasselbalch Equation (see also Stryer pp. 73-74)
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
Example:
H
+
+
–
A
R-COOH
H
+
+
RCOO
–
OR
+
HA
Example:
H
+
+
+
R-NH3
A
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 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.
[H + ][A − ]
Ka =
[HA]
1
1 [A − ]
=
•
[H + ] Ka [ HA]
[A− ]
pH = pKa + log
[HA]
The last equation is the Henderson-Hasselbalch Equation.
pH/pK/buffer review & practice problems, page 1 of 3
(Henderson-Hasselbalch Equation)
[A − ]
pH = pKa + log
[HA]
Note that when pH = pKa, [A–] = [HA]. At this condition, the “buffering capacity” of the buffer will
be the greatest. That is, a given change in concentration of base or acid will result in the smallest ∆pH.
pH titration
10
9
pH
8
7
6
5
0
0.2
0.4
0.6
0.8
1
Equivalents hydroxide
Upon addition of 0.5 equivalents of base, 1/2 of the acid will be converted to the conjugate acid,
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.)
+
+
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?
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 of that protein) in the uncharged (neutral) form at pH
4.5. What is the pKa of that specific His residue in that protein?
pH/pK/buffer review & practice problems, page 2 of 3
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. 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.
3. You need to know the ionization properties of the functional group of His to answer this question!
(Abbreviate the imidazole group in question as Im, so neutral form is Im, the conjugate base, and
+
charged form is Im , the conjugate acid.) 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 "typical" pKa for a "generic" His imidazole group in proteins as given in Table
3.1 in Berg, Timoczko & Stryer.)
pH/pK/buffer review & practice problems, page 3 of 3