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

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

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.

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

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

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-]

Acids & Bases - DefinitionsBrønsted 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.

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] ≈ constantKa = K [H2O] = [H+] [A-]

[HA]

Strong acids Ka >> 1 Weak acids Ka < 1

Henderson- Hasselbalch EquationShows 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.

Henderson-Hasselbalch CalculationCalculate 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 MpKa = 4.80 - log 0.010 M = 4.80 - log 8.7 = 4.80 - 0.94

pKa =3.9

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

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?

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

pH of a Weak Acid Solution Calculation III

“B”

pH = log (1 / [H+])

= log (1 / [2 x 10-4])

= log (5000)

= 3.7

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

Acids, Bases, Salts & SolutionsTypes 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 saltThis is why:

Some Examples of Drugs Formulated as SaltsDiphenhydramine hydrochloride (Benadryl)Naproxen sodium (Aleve)Cetirizine hydrochloride (Zyrtec)Morphine sulfateOxycodone Hydrochloride (Oxycontin)

BuffersThe 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.

Titration Curve

ExamplesAcetic Acid, Phosphate

& Ammonia

Voet. Voet & Pratt, 2013 Fig 2.17

HA H+ + A-

Ka = [H+] [A-] [HA]

CH3COOH

CH3COO-5.76

3.764.76

0 50 100% titrated

Titration of a Polyprotic Acide.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

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.

The Bicarbonate Buffer System (Lungs & Blood)

Lehninger 2000, page 105

(Voet et al. Rx #1)

(Voet et al. Rx #2)

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.

pH Optima of Three Enzymes

Lehninger 2000, Figure 4.19

Ionic StrengthA 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)

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))

The Effect of Ionic Strength

Matthews et al 1999 Figure 2.22

End of Lecture