Biochemistry 304 2014 student edition acids, bases and p h

1. Acids, Bases and pH
Student Edition 5/23/13 Version
Pharm. 304
Biochemistry
Fall 2014
Dr. Brad Chazotte
213 Maddox Hall
chazotte@campbell.edu
Web Site:
http://www.campbell.edu/faculty/chazotte
Original material only ©2004-14 B. Chazotte

2. Goals
•Review the ionization of water, Keq and Kw
•Review the concept of an acid dissociation constant, Ka
•Understand the concept of pH and a pH scale
•Review the Henderson-Hasselbalch equation & its use
•Be able to calculate the pH of a weak acid
•Understand buffers
•Review titration curves for monoprotic & polyprotic acids
•Review the effect of pH on protein solubility and enzyme function
•Review the concept of, and the calculation of, ionic strength

3. Ionization of Water
Water has a slight tendency to undergo a
reversible ionization.
H2O  H+ + OH-
Since free protons do not exist in solution one
writes:
H2O  H3O+ + OH-
(H3O+, a hydronium ion)
Proton jumping: the proton can rapidly jump
from one water molecule to the next in one of the
fastest reactions in solutions.

4. Keq and Kw
A + B  C + D Keq = [C] [D]
[A][B]
[] concentration approximates the activity coefficient
FOR WATER:
H2O  H+ + OH- Keq = [H+] [OH-]
[H2O]
In pure water at 25 ºC
[H2O] =55.5 M i.e., essentially constant compared to ions
Therefore we can write :
(55.5 M) (Keq) = [H+] [OH-] = Kw (ion product of
water)
Kw = 1 * 10-14 @ 25.0 ºC

5. Lehninger 2001 Table 4.2
Table: The pH Scale
As defined by
Sorensen
pH = -log [H+]
pH + pOH = 14
pH is a shorthand
way of designating
the hydrogen ion
activity of a solution
The “p” in pH designates the negative logarithm
The ion product of
water is the basis for
the pH scale
Sum of pH and pOH
always =14
Voet, Voet & Pratt 2013 Fig. 2-16

6. pH’s of
Some
Aqueous
Fluids
Lehninger 2000, Figure 4.13
>pH 7 basic
[H+] < [OH-]
pH 7 neutral
[H+] = [OH-]
<pH 7 acidic
[H+] > [OH-]

7. Acids & Bases - Definitions
Brønsted and Lowry:
Brønsted Acid - a substance that donates protons (hydrogen ions)
Brønsted Base - a substance that accepts protons (hydrogen ions)
When a Brønsted acid loses a proton a Brønsted base is produced.
In this context the original acid loses a proton and produces a base,
These are referred to as a conjugate pair, e.g. HA and A-.
A strong acid or base is one that ionizes almost 100% in aqueous
solution.

8. Ka (Dissociation Constant)
Ka numerically describes the strength of an acid.
K = [H3O+] [A-]
[HA] [H2O]
since [water] = 55.5 M in dilute solution, [H2O] ≈ constant
Ka = K [H2O] = [H+] [A-]
[HA]
Strong acids Ka >> 1 Weak acids Ka < 1

9. Henderson- Hasselbalch Equation
Shows the relationship of a solution’s pH and the concentration
of an acid and its conjugate base in solution.
[H+] = Ka ([HA] / [A-]) (1)
pH = - log Ka - log ([HA] / [A-]) since pH = -log [H+] (2)
pH = pKa + log ([A-] / [HA]) since pK = -log K (3)
HA = proton donor A- = proton acceptor
When [HA] = [A-] then log ([A-] / [HA]) = 0
And pH = pKa !
But the H-H equation cannot be used for strong acid or base.

10. Henderson-Hasselbalch Calculation
Calculate the pKa of lactic acid given that the concentration
of lactic acid is 0.010 M, the concentration of lactate is
0.087 M and the pH is 4.80.
[lactate]
pH = pKa + log [lactic acid]
[lactate]
pKa = pH - log [lactic acid]
0.087 M
pKa = 4.80 - log 0.010 M = 4.80 - log 8.7 = 4.80 - 0.94
pKa =3.9

11. Extent of Ionization
Relationship of pH and pKa
Weak acids % Compound Ionized
Weak Bases
pH = pKa ~50%
pH = pKa + 1 ~90%
pH = pKa + 2 ~99%
pH = pKa + 3 ~99.9%
pH = pKa + 4 ~99.99%
pH = pKa ~50%
pH = pKa - 1 ~90%
pH = pKa - 2 ~99%
pH = pKa - 3 ~99.9%
pH = pKa - 4 ~99.99%
Cairns “Essentials of Pharmaceutical Chemistry”2008 p.22

12. pH of a Weak Acid Solution
Calculation I
Problem: A weak acid HA is 0.1% ionized (dissociated) in a
0.2 M solution.
A)What is the equilibrium constant for the acid’s
dissociation?
B)What is the pH of the solution?

13. pH of a Weak Acid Solution
Calculation II
“A” HA  H+ A-
Start 0.2M 0 0
Change -(0.1% of 0.2M)
= -(2 x 10-4M) +(2 x 10-4M) +(2 x 10-4M)
Equilibrium 0.2M -(2 x 10-4M) (2 x 10-4M) (2 x 10-4M)
[H+] [A-] (2 x 10-4M) x (2 x 10-4M)
Ka = [HA] = (0.2 M - 2 x 10-4M)
4 x 10-8
Ka = 1.998 x 10-1 = 2 x 10-7
Segal 1975

14. pH of a Weak Acid Solution
Calculation III
“B”
pH = log (1 / [H+])
= log (1 / [2 x 10-4])
= log (5000)
= 3.7

15. Acid-Base Neutralization
Calculation Problem
How many ml of 0.025 M H2SO4 are required to neutralize 525 ml of
0.06 M KOH? Setup:
1 # moles H+ (equivalents) req. = # moles OH- (equivalents) present
2 Liters x normality (N) = # equivalents
3 Litersacid x Nacid = litersbase x Nbase
1 H2SO4 = 0.025 M = 0.05 N (two hydrogens per mole sulfuric)
Litersacid x 0.05 M = 0.525 litersbase x 0.06 M
0.63 litersacid = (0.525 liters x 0.06 M) / 0.05 M

16. Acids, Bases, Salts & Solutions
Types of salts and solutions formed when acid and base combine
Strong acid + Strong base → Neutral salt HCl + NaOH → Na+Cl- + H2O
Strong acid + Weak base → Acidic salt HCl + NH3 → NH4
+ Cl-
Weak acid + Strong base → Basic salt CH3COOH+ NaOH → CH3COO- Na+ + H2O
Weak acid + Weak base → Neutral salt CH3COOH + NH4OH → NH4
+ CH3COO- + H2O
Cairns “Essentials of Pharmaceutical Chemistry”2008 p11
NH4
+ Cl- ↔ NH4
+ Cl-
NH4
+ + H2O ↔ NH3 + H3O+
Example of strong acid and weak
base from above → acidic salt
This is why:
Some Examples of Drugs Formulated as Salts
Diphenhydramine hydrochloride (Benadryl)
Naproxen sodium (Aleve)
Cetirizine hydrochloride (Zyrtec)
Morphine sulfate
Oxycodone Hydrochloride (Oxycontin)

17. Buffers
The maintenance of a relatively constant pH is critically
important to most biological systems. pH changes can effect the
structure, function, and/or reactivity of biomolecules.
One pH unit is equivalent to a 10-fold change in H+ ion
concentration.
A buffer (system) resists changes in solution pH by changes in
the concentration of the buffers’ acid and conjugate base (HA
and A-). This can be illustrated in the titration curve of a weak
acid. A system’s maximum buffering “capacity” is at or near
the pKa of the molecule with a range typically ± 1 pH.

18. Titration
Curve
Examples
Acetic Acid,
Phosphate &
Ammonia
Voet. Voet & Pratt, 2013 Fig 2.17
HA  H+ + A-
Ka = [H+] [A-]
[HA]
CH3COOH
CH3COO-
5.76
3.76
4.76
0 50 100% titrated

19. Titration of a Polyprotic Acid
e.g., H3PO4
Voet. Voet & Pratt, 2013 Fig 2.18
A polyprotic
acid has a pKa
for each
ionization or
ionizable
group.
The ionization of
one group creates
an electrostatic
charge that raises
the pKa of the
subsequent
ionization

20. A Buffer System: Acetic Acid / Acetate
Lehninger 2000, Figure 4.17
Adding H+ ions drives
the equilibrium to acetic
acid taking up the added
H+ ions while adding OH-
ions drive the equilibrium
to acetate.
Sum of the buffer
components does not
change, only their ratio.

21. The Bicarbonate Buffer System
(Lungs & Blood)
Lehninger 2000, page 105
(Voet et al. Rx #1)
(Voet et al. Rx #2)

22. pH and Solubility of a Protein
Matthews et al 1999 Figure 2.21
At a protein’s isoelectric
point, called pI, the sum of
the positive charges equals
that of the negative charges
leaving no net charge
At the pI a protein is
least soluble.

23. pH Optima of Three Enzymes
Lehninger 2000, Figure 4.19

24. Ionic Strength
A measure of the total concentration of ions in solution.
The more charged the ion the more it is counted
μ = ½ Σ ciZi
2 where ci = concentration of ith species
and Zi = the charge on the ith species
Why care? The greater the μ of a solution the higher the
charge in the “ionic atmosphere” around an ion and the
less net charge so the less attraction between any given
cation and anion.
Relates thermodynamic activity, ai, to concentration [A].
aA = [A] γA (see also extended Debye-Hückel Equation)

25. Sample Calculation of Ionic Strength
 = ½ ((0.090))
 = 0.045
What is the ionic strength of a 0.015 M solution of CaCl2 ?
The equation is  = ½ ciZ i
2
The reaction is CaCl2  Ca2+ + 2 Cl-
The concentrations are [0.015] + [0.030]
 = ½ ((0.015 x 22) + (0.030 x -12))
 = ½ ((0.060) + (0.030))

26. The Effect of Ionic Strength
Matthews et al 1999 Figure 2.22

27. End of Lecture