1
Lecture 9
November 6, 2002
[Look in Chem Cave for the figures used in this lecture.]
Reading for next week Chapter 11; Chapter 27-1 to 27-3.
Homework Assignment due next Wednesday 10:2,4a,8,11,20,32,39; 27:18
XXXXX
By definition, a buffer solution is a solution of a conjugate acid/base pair that resists
changes in pH. We employ buffers whenever we need to maintain the pH of a solution at a
constant and predetermined level.
A solution most have appreciable amounts of both the conjugate acid and conjugate base in
order to be a buffer.
Since buffers are just solutions of conjugate acid/base pairs, so we can use weak acid Ka's to
calculate the pH of buffer solutions. Because buffers are employed so extensively, a buffer
equation has been developed to simplify buffer calculations.
XXXX
This is called the Henderson-Hasselbalch equation and is derived in your book. The
derivation is based on two assumptions that you need to be aware of.
First we really must have a weak acid. If Ka is larger than 1 x 10-3 then the equation starts
to break down.
Second, as with any weak acid system, if the concentrations of the conjugate acid and
conjugate base are very low, say below 10-4 M, then autoionization of water starts to affect
the system.
If either of the above conditions apply, then we need to use our Systematic Approach for
solving equilibrium from Chapter 9.
But other than these two conditions the equation has very broad and general application.
XXXXX
pH = pKa + log [base form]
[acid form]
=
pKa + log [A-]A[HA]HA
We also have here the equation as a chemist would correct it, which takes into account the
effect of ion strength on buffer solution pH. Of course Jim would take this one step further
and use molality instead of molarity.
2
Notice that the hydrogen ion concentration of a solution containing a weak acid and its
conjugate base is dependent only on the ratio of the molar concentrations of these two
solutes.
Furthermore this ratio is independent of dilution. If we dilute the solution by a factor of 2
we divide both the numerator and denominator by two and the ratio doesn’t change.
Also notice that you can prepare a buffer at a given pH by putting into solution the right
ratio of the weak acid and its conjugate base.
For example.
1. What ratio do we need to prepare a solution with pH = 4.15 using formic acid and its
conjugate base, the formate ion? Formic acid = ant acid
pKa = 3.745 from Appendix G
pH = pKa + log [HCOO-] = 4.15 = 3.745 + log ratio  0.405 = log ratio
[HCOOH]
[HCOO-]
[HCOOH]
=
100.405 = 2.54 = 2.5 to 2 sig figs.
Let's look at the same question the other way around. What is the pH of a solution that is
0.0250 M in formic acid and 0.0625 M in sodium formate?
Note sodium formate is a salt. What do we know about sodium salts? Highly soluble.
So what does 0.0250 M sodium formate mean?
The equilibrium governing the hydrogen ion concentration in this solution is
HCOOH < == > H+ + HCOO-
ratio = 2.5
pH = pKa + log [HCOO-] = 3.75 + log 0.0625
[HCOOH]
0.0250
= 3.75 + .40 = 4.15
Notice that we can make up a buffer at a desired pH by simply combining the correct
number of moles of the weak acid with a soluble salt of the conjugate base.
Now let's see what the pH would be if we correct for ionic strength.
XXXXXXX
 = ½ { [Na+](1)2 + [HCOO-](1)2} = ½ { 0.0625 + 0.0625} = 0.0625
What is gamma for formic acid? 1 not ionized.
3
For formate ion we can interpolate using the Chapter 8 Table on page 154.
formate ion has charge of –1 and size of 350 pm values are
0.81 at  = 0.05 and 0.76 at = 0.1.
XXXXXX
Best way to think about interpolation is to draw a diagram.
0.76 – 0.81 =
0.76 – x 
0.1 – 0.050
0.1 – 0.0625
-0.05 = 0.76 – x  -0.0375 – 0.76 = -x
0.050
0.0375
x = 0.7975 = 0.80
pH = pKa + log [HCOO-]HCOO = 3.75 + log 0.0625x0.80
[HCOOH]
0.0250
= 3.75 + .30 = 4.05
On page 194, Harris discusses the top five reasons why buffers are not made using
calculations but instead using a pH meter. The effect of activity was reason #1. #5 is “You
will probably make an arithmetic error anyway.”
Properties of buffers.
The pH of a buffer solution remains essentially independent of dilution until the
concentrations of the species it contains are decreased to the point where the approximations
used to develop our equations become invalid.
XXXXX
This transparency contrasts the behavior of buffered and unbuffered solutions with dilution.
For each solution, the initial solute concentration was 1.00 M. Notice also that the graph is
essentially a log/log plot.
In a strong acid solution, HCl, the concentration of acid falls steadily as the solution is
diluted. The weak acid HA solution also falls steadily but with a different slope.
The buffered solution resists changes in pH during dilution until the concentrations become
low.
XXXXX
This graphs shows what are known as titration curves. The curves follow the pH of
solutions that start out with 50 mL of a 0.1000 M acid solution.
As sodium hydroxide is added (x axis) the H+ in solution reacts to form water. For a strong
acid the reaction is directly with the H+ in solution. For weaker acids the hydroxide also
reacts with HA the undissociated acid according to the equation
4
HA + OH- < == > HOH + AYou can see that once we are into the region that resists change in pH that adding either
base (going to the right) or acid (going to the left) does not change the pH appreciably.
This is our definition of a buffer, a solution that resists changes in pH whether we add acid
or base to the solution.
We see that the slope of this curve is flattest at the point where the titration is half over.
This is where [HA] = [A-]. From the Henderson-Hasselbalch equation we see that this
means
pH = pKa + log [base]/[acid] = pKa + log 1 = pKa
Fairly flat from pKa ± 1 pH unit.
The curves also illustrate that we would use different conjugate acid/base pairs in order to
buffer at different pH values.
XXXXXXX
Here is a list of buffering systems that cover a wide range of pH values.
We noted that the curves are flattest at the point where pH = pKa. This leads to the concept
of buffering capacity.
The ability of a buffer to prevent a significant change in pH is directly related to the total
concentration of the buffering species as well as to their concentration ratio.
XXXXX
The buffer capacity of a solution is defined as the number of moles of a strong acid or a
strong base that causes 1.00 L of the buffer to undergo a 1.00-unit change in pH.
XXXXX
This graph shows how buffer capacity varies with the ratio of the conjugate base/acid pair.
Buffer capacity falls off moderately rapidly as the concentration ratio of acid to conjugate
base becomes larger or smaller than unity.
For this reason, the pKa of the acid chosen for a given application should lie within 1 unit of
the desired pH in order for the buffer to have a reasonable capacity.
You can also see that if you know you have a solution in which hydrogen ions are going to
be generated, then you might want to start the buffer on the base side of the pKa.
XXXXX
You are all familiar with Alkalinity by now. We define alkalinity as the capacity of
natural water to react with H+ to reach a pH of 4.5.
5
H+ reacts with substances that are bases and in natural water this involves the following
reactions almost exclusively.
XXXX
H+ + OH- <= => HOH
H+ + HCO3- <= => HOH + CO2
2 H+ + CO32- <= => HOH + CO2
so alkalinity is approximately equal to the sum of the hydroxide, hydrogen carbonate and
two times the carbonate ion concentrations.
Alkalinity  [OH-] + [HCO3-] + 2 [CO32-]
One way alkalinity is expressed as mmol H+ needed to neutralize the base in liter of
sample.
As you know, geologists express alkalinity in mg of CaCO3 /Liter as if all the base in the
sample comes from calcium carbonate.
We can determine the relationship between these two measures using the stoichiometry
from the following chemical equation.
2 H+ + CO32- <= => HOH + CO2
So
1.000 mmol H+ x 1 mmol CO32- x 1 mmol CaCO3 x 100.087 mg CaCO3
1 Liter
2 mmol H+
1 mmol CO321 mmol CaCO3
= 50.04 mg CaCO3
So 1.00 mmol H+ used in the titration measures the equivalent of 50.0 mg CaCO3.
The next step in our examination of acids and bases is to consider acids that have more
than one proton to donate in acid/base reactions.
COOH
An example is citric acid which has the structure HOOC-CH2CCH2COOH
OH
And the formula C5H8O7.
There are three carboxylic acid groups in the molecule, each with a different Ka. Citric
acid is a triprotic acid.
XXXXX
6
Harris lists values for pKa1 = 3.128, for pKa2 = 4.761, and for pKa3 = 6.396, that
correspond to the three equations.
H3Citrate < === > H+ + H2CitrateH2Citrate- < === > H+ + HCitrate2HCitrate2- < === > H+ + Citrate3All of these equilibria occur simultaneously. In Chapter 10 you will read about detailed
equations that apply to specific equilibrium circumstances.
But we can usually simplify problems by identifying the important species in the chemical
system.
Remember, you can always approach these problems by using the systematic approach for
solving equilibrium given in Chapter 9.
XXXXX
We have already encountered one diprotic acid – the carbonate system.
H2CO3 <= => H+ + HCO3-
Ka1 = 4.45 x 10-7
pKa1 = 6.352
HCO3- <= => H+ + CO32-
Ka2 = 4.69 x 10-11
pKa2 = 10.329
Now let’s think through what these equations tell us.
What is the pH of a solution that has its carbonic acid concentration equal to its hydrogen
carbonate concentration?
Henderson-Hasselbalch equation
pH = pKa + log [base]/[acid]
Which is the base? Which is the acid?
What will be the pH of a solution if its hydrogen carbonate concentration equals its
carbonate concentration?
Now, more difficult. If the pH of a carbonate solution is 12, what can you say about the
concentration of each of the carbonate species?
What if the pH is 9? 6? 3?
XXXX
Now look at this diagram and see if it confirms our answers.
XXXXX
In Chapter 11 Harris is going to present summaries that look like this. We see that this
present most of the same information and leads us to the same conclusions.
7
Another approach takes these diagrams and turns them on their side. These similar
constructs are called ladder diagrams.
A ladder diagram is a visual tool for evaluating systems at equilibrium.
When used properly a ladder diagram can be used to (1) estimate the approximate
composition of a system at equilibrium, (2) determine the reactions likely to occur when
several reagents are combined and (3) evaluate what will happen when a change occurs in
system conditions.
Let’s consider a simple system first. Hydrogen fluoride in aqueous solution is a weak acid
called hydrofluoric acid with Ka = 6.8 x 10-4.
Remember our term weak acid means an acid that holds on to its hydrogen tightly, not that
it is unreactive. Indeed, HF is a very powerful reagent that etches glass.
XXXXX
So we have
HF(aq) <= => H+ + F-
Ka = 6.8 x 10-4
If we write the Henderson-Hasselbalch equation for this system we get
pH = pKa + log [F-]/[HF]
where pKa = 3.17
We know that if [F-] = [HF] then pH = pKa.
We also know that if [F-] > [HF] then the log term is positive and the pH > pKa, i.e. the
solution is more basic.
When [F-] < [HF] then the log term is negative and pH < pKa, and the solution is acidic.
XXXXXX
The ladder diagram for this system consists of a vertical scale of pH values oriented so
that smaller (more acidic) pH levels are at the bottom and larger (more basic) pH levels
are at the top.
A horizontal line is drawn at a pH equal to the pKa of the system. In this case at pH =
3.17. This horizontal line separates the solution into regions where each of the two
conjugate forms of HF predominate.
Because this is a log diagram there are points 1 log unit from the pKa that are 10/1 and
1/10, 2 units are 100/1 and 1/100, etc.
8
By referring to the diagram we can see that at a pH of 2.5, hydrofluoric acid will exist
predominately as HF.
If we add sufficient base to the solution such that the pH increases to 4.5, the predominate
form becomes F-.
Thus you can see that the ladder diagram is just a visual form of the HendersonHasselbalch equation.
XXXXXX
Here is a second ladder diagram that contains information about both hydrofluoric acid
and ammonia.
From this ladder diagram we see that if the pH is less than 3.17, the predominate species
are HF and NH4+; for pH values between 3.17 and 9.24 the predominate species are F- and
NH4+; and above pH 9.24 the predominate species are F- and NH3.
Ladder diagrams are particularly useful for evaluating the reactivity of acids and bases.
An acid and a base can not coexist if their respective areas of predominance do not
overlap.
XXXXX
If we mix together solutions of NH3 and HF, the reaction
HF(aq) + NH3(aq) <= => NH4+ + Foccurs because the predominance areas for HF and NH3 do not overlap, but those of
ammonium ion and fluoride ion do overlap.
We can see that this is a reasonable conclusion by adding together the three equilibria
involved
HF(aq) <= => H+ + F-
Ka = 6.8 x 10-4
NH3(aq) + HOH <= => NH4+ + OH-
Kb = 1.75 x 10-5
H+ + OH-
K = 1/Kw = 1.00 x 1014
<= => HOH
HF(aq) + NH3(aq) <= => NH4+ + F-
K = Ka*Kb*K = 1.19 x 106
What does the size of K tell you?
Since the equilibrium constant is significantly greater than 1, the reaction’s equilibrium
position lies far to the right.
9
This conclusion is general and applies to all ladder diagrams. If two substances are
mixed that do not have a common position on the ladder, then they will react until one of
them is used up to form species that can coexist on the ladder.
Here is another example:
Predict the pH and composition of a solution prepared by adding 0.090 mol of HF to 0.040
mol of NH3.
We see from the ladder diagram that these two substances cannot coexist.
Since HF is present in excess and the reaction between HF and NH3 is favorable, the NH3
will react to form NH4+ consuming HF. At equilibrium, essentially no NH3 remains.
Amount of NH4+ formed is 0.040 mol which also forms 0.040 mol of F-.
This consumes 0.040 mol of HF leaving 0.090 mol – 0.040 mol = 0.050 mol HF.
According to the ladder diagram, for this system, a pH of 3.17 results when there is an
equal amount of HF and F-. Since we have more HF than F-, the pH will be slightly less
than 3.17.