acids, bases and ph student edition 5/23/13 version pharm. 304 biochemistry fall 2014 dr. brad...
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Acids, Bases and pH
Student Edition 5/23/13 Version
Pharm. 304 Biochemistry
Fall 2014
Dr. Brad Chazotte 213 Maddox Hall
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