Week 5 Buffer and Binding Problems:
For the following problems you may need to use appendix D in the back of the textbook (pp 1022-1023) to find Ka
values.
1. What does buffer capacity mean? What is the typical maximum pH range that a buffer having equal amounts
of acid and base will be effective?
The buffer capacity refers to the amount of acid and base the buffer can react with before the pH
starts to change significantly. The buffer capacity will be directly related to the concentration of
the buffer. Two buffers with the same exact acid-to-base ratio but having different concentrations
will buffer in the same range but the more concentrated buffer will be able to absorb more acid or
base. The buffering range of a buffer is typically  one pH unit from the pKa of the acid.
2. Phosphoric acid (H3PO4) has three different pKa values. Show the equilibrium reactions which account for the
loss of these three acidic hydrogens and assign the appropriate pK a to each equilibrium.
H3PO4
-H
pKa = 2.12
H3PO41-
-H
pKa = 7.21
HPO42-
-H
pKa = 12.38
PO43-
3. What molar concentration of HCO3- (bicarbonate anion) can exist in arterial blood at a pH of 7.35 when the
molar concentration of CO2 is 22 mM? Assume that all of the CO2 is present as H2CO3. Remember that the
units of mM are mmoles per liter and not mmole per mL.
pKa = 6.37
pKa = 10.25
H2CO3
H+ + HCO3CO2 + H2O
H+ + CO32To answer this question you need to use the first ionixation of carbonic acid which will allow you
to relate the concentration of carbonic acid (dissolved CO 2) and bicarbonate ion (HCO3-). The
Henderson-Hasselbach equation can be used to determine the HCO 3-/H2CO3 ratio and then
“plug-in” the known H2CO3 concentration of 22 mM.
pH = pKa + log
[ HCO3-]
[H2CO3]
7.35 = 6.37 + log
0.98 = log
[ HCO3-]
[H2CO3]
[ HCO3-]
[H2CO3]
9.55 =
[ HCO3-]
[H2CO3]
9.55 =
[ HCO 3-]
22mM
[HCO3-] = 210.1 mM
4. For an experiment involving what happens to the growth of a particular fungus in a slightly acidic medium, a
biochemist needs 250 mL of an acetate buffer with a pH of 5.12. The buffer solution has to be able to hold
the pH to within  0.10 pH unit of 5.12 even if 0.01 mol of either H+ or OH- enters the solution. What is the
minimum number of grams of acetic acid and of sodium acetate dihydrate that must be used to prepare the
250 mL of buffer? [NOTE: This is the problem we did in class]
What do we know? Want 250 mL of acetate buffer at pH = 5.12 that will not change by more than
0.1 pH unit after the addition of 0.01 mol of either acid or base.
What needs to be looked up? The pKa of acetic acid is 4.74.
To solve the problem, you must first determine the acid/base ratio of the buffer by using the
Henderson-Hasselbach equation.
5.12 = 4.74 + log
0.38 = log
2.40 =
[base]
[acid ]
[base]
[acid ]
[base]
[acid ]
2.40 [acid] = [base]
Next, determine the base/acid ratio after the maximim amount of acid of base has been added
(0.01 mol) and the pH has changed by the maximim allowed amount (0.1 unit).
5.02 = 4.74 + log
0.28 = log
1.90 =
[base  0.01]
[acid  0.01]
[base  0.01]
[acid  0.01]
[base  0.01]
[acid  0.01]
Since the initial [base] = 2.40[acid], this can be substituted into the above equation for the base
concentration. That gives you one equation and one unknown. The equation can be solved for
the initial acid concentration.
1.90[acid] + 0.019 = [base] – 0.01
1.90[acid] + 0.019 = 2.40[acid] – 0.01
0.029 = 0.50[acid]
0.058 = [acid]
(0.058 moles of acid is the minimum number of moles of acid needed in the initial buffer)
The number of moles of base required in the initial buffer will be 2.40 [0.058] = 0.139 moles.
Converting these amounts to grams finishes answering the question. MW acetic acid = 60g/mol
and 0.058 moles of acid = 3.48 g. MW sodium acetate = 82 g/mol and 0.139 moles of base =
11.40 g.
5. An analytical procedure for separating certain components of a dietary supplement requires that they be
dissolved in a buffered solution having a pH of 3.80. This buffer has to be able to handle the addition of 0.005
mol of either H+ or OH- without changing its pH by more than 0.050 unit.
(a) Using appendix D in your textbook, determine the best acid (and its sodium salt) to use in preparing the
above buffer.
The best choice of acid would be formic acid (HCHO 2) which has a Ka of 1.8 x 10-4 and pKa of
3.74. You want to choose the acid of your buffer to have a pKa close to the pH where the
solution needs to be buffered. The conjugate base would be sodium formate (NaCHO 2).
(b) Calculate the minimum number of grams of pure acid and the pure sodium salt needed to make 750 mL
of the buffer.
pH = pKa + log
[ NaCHO 2]
[ HCHO 2]
3.8 = 3.74 + log
0.06 = log
1.15 =
[ NaCHO 2]
[ HCHO 2]
[ NaCHO 2]
[ HCHO 2]
[ NaCHO 2]
[ HCHO 2]
1.15[HCHO2] = [NaCHO2]
The allowable pH fluctuation is 0.05 pH unit after addition of 0.005 mol of either acid or base.
The new base/acid ratio will be:
3.75 = 3.74 + log
0.01 = log
1.02 =
[ NaCHO 2 - 0.005]
[ HCHO 2  0.005]
[ NaCHO 2 - 0.005]
[ HCHO 2  0.005]
[ NaCHO 2 - 0.005]
[ HCHO 2  0.005]
1.02[HCHO2] + 0.0051 = [NaCHO2] – 0.005
1.02[HCHO2] + 0.0051 = 1.15[HCHO2] – 0.005
0.0101 = 0.13[HCHO2]
0.078 = [HCHO2]
Therefore, 0.078 moles of formic acid and 0.089 moles of sodium formate will be needed in the
750 mL of buffer solution. Using the molecular weight of each component, this translates into
3.59 g of formic acid and 6.05 grams of sodium formate.
(c) What is the concentration of this buffer?
Buffer concentration equals the sum of the concentrations of acid and base present.
0.078moles
= 0.104 M in formic acid
0.750 L
0.089moles
= 0.119 M in sodium formate
0.750 L
The buffer concentration is 0.223 M.
6. Think about the receptors responsible for causing muscles to contract and relax. Is this a system where you
want the ligands to bind tightly or loosely to their receptors? Briefly explain.
For muscle contraction, you would like the biological response (muscle contraction) to be able to
reverse quickly. This is accomplished by loose binding between the receptor and ligand. The
equlilbrium between the bound receptor-ligand complex will lie to the side of unbound ligand
when there is loose binding.
7. Hormones are typically present in very low concentrations. Should hormones bind tightly or loosely to their
receptors? Will hormones typically have high or low KD values?
Hormones bind tightly to their receptors and have a low KD value.