MEDICAL BIOCHEMISTRY

Lectures 2:  Acid-Base Concepts

Harry W. Jarrett, Ph.D.






I.	ACIDS AND BASES

	A.	Definitions
	B.	Dissociation of weak electrolytes
	C.	Definition of pK
	D.	The Henderson-Hasselbalch equation
	E.	Titration curves
	F.	Properties of buffers


Suggested Reading:	Lubert Stryer, Biochemistry (Fourth 
Edition), pp.  42-44.


Ionization of Water:  Although they do so very weakly, water molecules 
will dissociate into ions. A proton from one water molecule is transferred to 
the oxygen atom of another water molecule resulting in the formation of an 
hydronium ion and an hydroxide ion.  The equilibrium constant for this 
reaction is represented by the following equation:

			[H+] [OH-] 
	Keq	=	________	=	1.8 x 10-16 M
			  [H2O]

Because the equilibrium constant for this reaction is so low the 
concentration of water is unchanged by ionization.  The molarity of water is 
55.6 M and if we substitute this value in the equation it reduces to the 
following:

	Kw	=	[H+] [OH-]	=	1 x 10-14 M2

The term Kw is the ion product of water and is the product of the molar 
concentrations of the ions in any sample of water.  We can solve this 
equation for the concentration of each ion by taking the square root of Kw.  
The concentration of either ion in any sample of pure water, therefore, is 1 x 
10-7 M.

The ion product expresses an equilibrium for the ionization of water.  An 
increase in the concentration of either ion will result in a decrease in the 
concentration of the other ion because the numerical value for the ion 
product must remain the same.  If, for example, we increase the 
concentration of protons by a factor of 100 to 1 x 10-5 M, the 
concentration of hydroxide ions must decrease to a value of 1 x 10-9 M.  
Under these conditions, the solution has an excess of protons, and the 
solution is called acidic.  Conversely, if we increase the concentration 
of hydroxide ions, the solution is called basic.

It is inconvenient to express the acidity or basicity of a solution using such 
cumbersome numbers.  In order to simplify calculations and graphical 
representations of acidity or basicity, we use a logarithmic scale to define a 
quantity called pH.  By definition, pH is the negative logarithm of the 
hydrogen ion concentration of a solution:

	pH = -log10 [H+]

This is sometimes expressed as the reciprocal of log [H+] or:

	pH =1 / log [H+]

Solution with equal concentrations of protons and hydroxide ions will have a 
pH level of 7 because the proton concentration is 1 x 10-7 M.  Any acidic 
solution will have a pH of less than 7 because the concentration of protons is 
correspondingly increased.  In our other example, the acidic solution with 1 
x 10-5 M [H+] has a pH of 5.  If on the other hand, we increase the 
concentration of hydroxide ions by a factor of 100, the proton concentration 
is correspondingly decreased by the same factor and becomes 1 x 10-9 M.  
Such a solution, therefore, has a pH of 9 and is basic.

We can express the acidity or basicity of solutions with concentrations of 
protons from 1 M (pH = 0) to 1 x 10-14 M (pH = 14).  The 
concentration of protons in a solution can be altered by the addition of acids 
or bases.

Overhead showing Log Scale -- Discuss logarithms

There are a number of definitions of acids and bases.  For our immediate 
purposes, the best one is the Bršnsted definition which states that an acid 
is any substance that is a proton donor.  Conversely, a base is any 
substance that is a proton acceptor.  Hydrochloric acid is a good example 
of an acid.  It ionizes completely in water into a proton and a chloride ion.  
Since HC1 is completely ionized, it is called a strong acid.  Sodium 
hydroxide is a good example of a strong base.  It ionizes completely in water 
to yield a sodium ion and an hydroxide ion.  The hydroxide ion is a proton 
acceptor, reacting with a proton to yield water.  Since in this reaction, 
protons are removed from solution, their concentration goes down and pH of 
the solution increases.

An acid is also defined as an electron acceptor whereas a base can be 
defined as an electron donor.  This is the Lewis definition and electron 
acceptors are Lewis acids whereas electron donors are Lewis bases.  We will 
have an occasion later in the course to utilize this definition.

In biological systems, we deal primarily with weak acids and weak bases.  
These are compounds that ionize only partially in solution.  A good example 
of weak acid is acetic acid which ionizes to yield a proton and an acetate ion.  
The degree of ionization of this compound can be expressed by the following 
equilibrium equations:

			[H+] [CH3COO-] 
	Keq	=	_____________	=	1.76 x 10-5 M
			   [CH3COOH]

This value is constant (and Molar units).  Again such a small number is 
cumbersome to deal with mathematically; it is more convenient to use its 
logarithm.  By analogy with pH, we can define the pK of an acid or bases as 
the negative logarithm of its dissociation constant as an acid:

	pKa = -log Ka

Weak acids differ quite widely in strength of ionization as an acid or in pKa.  
For any acid or base, the pKa determines the pH of a solution when the 
compound is dissolved in water.  We can derive an equation that 
illustrates this relationship in the following manner:

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

Where A- is the conjugate base of the acid and HA is the undissociated 
acid.  We can rearrange this equation to:

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

This rearranges to:

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

We have already defined -log [H+] as pH and -log Ka as pK so the equation 
becomes:
						[A-] 
	pH	=	pKa	+	log   _______
						[HA]

The Henderson-Hasselbach equation relates the pH of any acid solution 
with pKa and the relative concentration of the conjugate base of the acid 
and the undissociated acid.  From this equation we can determine that the 
pKa of any acid is the pH at which the concentrations of conjugate base and 
undissociated acid are equivalent since the ratio is 1 and the log of 1 is 
0.  The term cancels out and pH = pKa.

The equivalence is reached when we titrate a weak acid with a strong base 
such as NaOH.  If we carry out such a titration and plot the pH of the solution 
as a function of the equivalents of base added, we see that the acid is 
completely neutralized after addition of one equivalent and the pH of the 
solution is close to neutrality.  This is the end point of the titration.  The 
concentrations of [A-] and [HA] are equal midway between the start of the 
titration and the end point, or after the addition of half an equivalent of base.  
For acetic acid, the pH of the solution at this stage is 4.75 which is the pKa 
for this acid.  For any acid, its pKa is always the pH at this point in the 
titration.

Slide 7 -- D. Voet and J.G. Voet (1990) Biochemistry Fig. 2-11, pp 39 -- Phosphate titration

The curve for acetic acid is relatively simple because it has only one 
ionizable proton.  Other acids can donate multiple protons and their titration 
curves are more complicated.  An example is the titration of phosphoric acid.  
Complete titration of phosphoric acid requires three equivalents of base and 
there are three end points.  For each phase of the titration, there are three pH 
values at which the concentration of conjugate base and undissociated acid 
are equal, and there are, therefore, three pKa values for phosphoric acid.

I put this slide up to make you aware that the graphs you see may not all be arranged the same -- 
sometimes equivalents of base or acid are shown on either the x-axis or the y-axis.  You must be 
careful since this will affect how you interpret the figure.

Slide 8 -- C.K. Mathews and K.E. van Holden (1990) Biochemistry Fig. 2.19, pp 46 -- 
titration of formate and ammonium

Also, many of you may have noticed that we have been treating ammonia as a weak acid when in 
fact it is a basic compound.  That is what the "a" in pKa is all about -- it says we are considering the 
substance to be an acid.  We can write a dissociation for ammonium ion as follows:

	NH4+		--->		NH3	+	H+
And symbolize this as:
	HA+		--->		A	+	H+
just as we did for weak acids.  As long as we always write these as an acid dissociation, we can use 
the exact same equation, the Henderson-Hasselbach equation, to describe both weak acids and weak 
bases.

You will notice that in each one of these curves there are areas that are 
relatively flat.  This occurs at pH values close to the pK of the acid.  These 
are regions at which the solution is buffered.  By this we mean that there is 
relatively small change in pH as base is added to the acid.  A buffer is a 
system that resists change in some parameter.  In this case it is a 
change in pH.  In other systems, a buffer may resist change in some other 
parameter such as the concentration of a metal ion.  The system is resistant 
to a change in pH in this area because it consists of a mixture of 
undissociated acid and the products of dissociation.  As we titrate the 
solution with base, dissociated protons react with OH- ions to yield water.  
Since we are removing one of the products of the reaction, more acetic acid 
dissociates to replace these protons to preserve the equilibrium.  The result 
is that there  is only a relatively small decrease in the proton concentration 
so the pH does not change appreciably.  Conversely, if we titrate in the 
opposite direction (i.e. by the addition of strong acid) acetate ions (the 
conjugate base) react with protons to form undissociated acid.  Again the 
concentration of protons does not increase appreciably so the pH remains 
relatively constant.  It is only when we approach the end point of the 
titration that there is an appreciable decrease in the proton concentration 
and a sharp rise in pH.  At this stage of the titration most of the 
undissociated acid has been consumed and there is little buffering 
capacity.  In any other buffer system, maximum buffering capacity is 
in the pH range of 1 pH unit on either side of the pKa.  At the 
extremes of this range, 10% or less of the buffering capacity remains.  On the 
acid side, 90% of the buffer pair is undissociated acid.  On the basic side, 
90% of the buffer pair is completely dissociated acid.  For acetic acid, this 
range is from pH 3.75 to 5.75.  Maximum buffering capacity, however, is at pH 
4.75 or the pKa.

As shown by the Henderson-Hasselbach equation, the pH of a buffer 
system is determined solely by the ratio of conjugate base to undissociated 
acid.  The pH is independent of the total concentration of the dissolved 
solutes in a reasonably dilute solution.  The capacity of a buffer, however, 
is determined by the total concentration of solutes.  Increasing the 
concentration of the buffer pair increases the capacity of the buffer system 
to resist change in pH.  Coca-Cola for example contains an extremely large 
amount of phosphoric acid in addition to its other ingredients.  Its pH, 
however, is only slightly acid due to the presence of a high concentration of 
the conjugate base of phosphoric acid.

Open System Buffering in the Blood

Buffers are extremely important in biological systems because most life is 
possible only within a very limited range of pH.  For most systems this is 
close to neutrality although some metabolic processes and chemical 
reactions occur outside this pH range.  Reactions occur in the stomach, for 
example, at a pH of 2 which is equivalent to 0.01 M HCl.  The pH of blood 
remains fairly constant at 7.4.  This is important for respiration since CO2 
cannot be removed from tissues below pH 7.3 and at pH values greater than 
7.7, CO2 cannot pass from the blood to the lungs.

There are two major buffering systems in the blood:
1.	Hemoglobin -- large amount and oxygenated form releases protons 
while deoxygenated form binds protons.  I won't cover this.

2.	More important is the CO2/HCO3- system.  This is a complex, open 
system buffer.  The relevant equilibria are:

	    k			      K1	           Ka1	         Ka2
CO2(gas)  Û  CO2(dissolved) + H2O  Û  H2CO3  Û  H+ + HCO3-  Û  H+ + CO3-2
(carbon dioxide)		   (carbonic acid)   (bicarbonate)   (carbonate)

How do you do pH calculations with this monster?  First of all, you simplify 
it.  All of the constants are known:

[CO2]dissolved  =  k(PCO2)	Where PCO2 is the partial 
pressure in mm Hg

k	=	3.01 x 10-5
K1	=	5 x 10-3
Ka1	=	1.58 x 10-4	(pKa1 = 3.8)
Ka2	=	5.6 x 10-11	(pKa2 = 10.25)

Since effective buffering only occurs near a pKa, how does this maintain 
blood pH = 7.4?  Because it is an open system and CO2 is removed by 
breathing.

First of all, CO2 not in solution and not in bicarbonate or carbonate is neutral 
and so should be eliminated.  We can accomplish this by defining a new Ka1':

	  [H+] [HCO3-]
Ka1'   =   __________  =  K1 x Ka1  =  (5 x 10-3)(1.58 x 10-4)	=  7.9 x 10-7
	 [CO2]dissolved

pKa1'	=	6.1, so getting closer.

Since we usually know the partial pressure of CO2, a more "user friendly" 
form of this equations is:

		[H+] [HCO3-]		[H+] [HCO3-]
Ka1'	=	__________	=	_______________	=  7.9 x 10-7
		  k  (PCO2)		(3.01 x 10-5) PCO2

Thus, the relevant Henderson-Hasselbach equation is:

		   [HCO3-]		  [HCO3-]
pH  =  6.1  +  log  _______  =  6.1  +  log  _______ 
		     [CO2]		(3.01 x 10-5) PCO2

In normal arterial blood, these are:

		 24 x 10-3 M
pH  =  6.1  +  log  __________  =  7.4
		 1.2 x 10-3 M

The body can maintain this pH because CO2 in the lungs equilibrates with 
alveolar air and so maintains the CO2 and thus pH fairly constant.  So, if 
metabolism or diet add H+ to the blood, it combines with HCO3- to make 
ultimately CO2 which is then removed by the lungs.  If these processes add 
OH-, this consumes H+ (from H2CO3) and makes HCO3- from CO2 but this CO2 
is replaced during breathing and so pH changes very little.

Solving pH Problems

The Henderson-Hasselbach equation has may uses:

1.  The pH of a solution can be calculated if the pK and the ratio of conjugate 
base and undissociated acid are given.

Example 1:  Calculate the pH of a solution containing 100 ml of 1 M 
phosphoric acid to which has been added 60 ml of 1 N NaOH.
a.  Calculate meq acid and conjugate base:
	 Acid = 100 meq total - 60 meq = 40 meq
	Conjugate base = 60 meq
b.  Calculate ratio of conjugate base to undissociated acid:
	 Ratio = 60/40 = 6/4.
c.  Use henderson-Hasselbach equation to calculate pH:
	pH = pK + log 6/4 = 2.1 + .18 = 2.28

Example 2:  Calculate the pH after titrating with an additional 50 ml of 1 N 
NaOH.
	a.  Conjugate base = 10 meq®¯¯¯ Undissociated acid = 90 meq
	b.  pH = pK + log 10/90 = 7.2 + (-.95) = 6.25

Example 3:  Calculate the concentration of conjugate base and undissociated 
acid in a 0.1 M solution of phosphoric acid at pH 7.4.

a.  Calculate ratio of base to acid:
	pH  = pK  + log B/A
	7.4 = 7.2 + log B/A
	log B/A = .2
	B/A = 1.6
	B = 1.6 A
	B + A = 0.1 M
	1.6 A + A = 0.1 M
	2.6 A = 0.1 M
	A = 0.1/2.6 = 0.04M
	B = 1.6 A = 0.06 M

Example 4:  Calculate the pH of a solution containing 8 meq of sodium 
dihydrogen phosphate and 2 meq of disodium monohydrogen phosphate.

a.  Identify pK:
	These ionic species exist between the first and second end point of 
the titration of phosphoric acid.  The second pK (7.2) is the appropriate one.

b.  pH = 7.2 log 2/8 = 7.2 + (-.6) = 6.6

Hints for solving pH problems:

1.  Know the nomenclature and the structures of the ionic forms that 
predominate at a particular pH value.
2.  Visualize or actually draw a rough titration curve of the compound in 
question, including pK values and end points.
3.  Use the appropriate pK value needed to solve the problem.
4.  Calculate meq of base form and acid form of the buffer pair.
5.  Insert information into the Henderson-Hasselbach equation and make 
calculations.
6.  If pH and pK are given, calculate relative amounts of conjugate base and 
conjugate acid using Henderson-Hasselbach equation 

If you are still not clear about pH, you might try the pH tutorial available on-line from Jefferson University.