NAME ____________________________
AP NOTES: UNIT 4 (5) APPLICATIONS OF EQUILIBRIUA:
WORKING A BUFFER PROBLEM
XX) Intro to buffers to buffer capacity….
XXI) Just a review re: The characteristics of a buffer ….
(brought to you by Nivaldo Tro p. 755)
A) A weak acid by itself, even though it partially ionizes to form some of its conjugate base, does NOT
contain sufficient base to be a buffer. Similarly, a weak base by itself, even though it partially
ionizes water to form some of its conjugate acid, does not contain sufficient acid to be a buffer.
1) A buffer must contain significant amounts of both a weak acid & its conjugate base (or vice versa)
2) The presence of the weak acid and its conjugate base (for instance) in equilibrium with each
other will resist pH change.
3) The weak acid of the buffer will neutralize added base
4) The conjugate base neutralizes added acid
B) Consider the simple buffer made by dissolving acetic acid (ethanoic acid) (HC2H3O2) and
sodium acetate (NaC2H3O2) in water.
1) Add a strong base such as NaOH
NaOH(aq) + HC2H3O2(aq) → H2O(l) + NaC2H3O2(aq)
As long as the amount of added NaOH is less than the amount of acetic acid the extra
NaOH will be neutralized and the change in pH is small.
2) Add a strong acid such as HCl
HCl(aq) + NaC2H3O2(aq)
→ NaCl(aq) + HC2H3O2(aq)
As long as the amount of added HCl is less than the amount of NaC2H3O2(aq) in solution
the buffer neutralizes the added HCl and the resulting pH change is small
C) Solution to a buffer problem may be determined in TWO ways
Using the equilibrium solution approach …
After accounting for the stoichiometric issues,
one could write a reaction, write a Ka
expression, run an ICE table, plug and chug
into the Ka expression and then the pH equation
Using the Henderson-Hasselbalch equation
the Ka is fairly small
AND WHEN
the initial [acid] or [base]
is/are fairly concentrated
and at least 100 to 1,000
larger than the Ka itself
Use this when you feel you can use the
“x” is small approximation & you are
given a Ka [so you can find a pKa]…
The “check” for “x” is a bit
tautological with H.H. eq … as you
must calculate the [H+] from the pH
you calculated, in order to run the
check. Don’t worry, I’ll show you….
Check out Brown and LeMay
p. 733 the comment section for
exercise 17.4 … It makes a
terrific case for when to use
H.H. eq … and when not to.
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D) Preparing a Buffer
Imagine you need to make a buffer solution of a specific pH … You need to calculate the
amounts of acid and its conjugate base
e.g.) How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer
with a pH of 9.00? The Kb of NH3 = 1.8 x 10-5 (assume the addition of the NH4Cl does not change
the volume of the solution) B&L p.733
Think about this: You want to know how much NH4+ needs to be used.
There’s an equilibrium to consider:
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
1.8 x 10-5 = [NH4+][OH-]
[NH3]
You have the pH … that can give you the pOH …. that can get you the [OH-] …
pH + pOH = 14.00 ….
9.00 + pOH = 14.00, pOH = 5.00
pOH = - log[OH-] ….. 5.00 = -log[x] ….
x = 1.0 x 10-5 M
Since Kb is small and the common ion of ammonium is present … you can assume that
the concentration of ammonia is essentially, its initial concentration
1.8 x 10-5 =[x][1x10-5]
[0.10]
x = 0.18 M
So, in order to have a buffer with a pH = 9.00, the [NH4+] must = 0.18 M
M = moles
L of sol’n
E) Given:
0.18 M = mol
2L
=
0.36 mol of NH4Cl
Acid HNO2
HClO
-4
Ka
4.5 x10 3.0 x 10-8
1) Which of the acids would be more suitable for use in a solution buffered at pH = 7.0?
*ans: Since the best buffer is when the pH = pKa , take the antilog of the Ka values
The best buffer would be made with HClO (hypochlorous acid)
2) What other substance(s) would be needed to make the buffer?
* You would need a salt containing ClO-, such as KClO, or NaClO & water!
Assignment: Get to Trivedi and attack 16.10 for a full presentation
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XXII) Calculating pH changes in a buffer solution
A) There are 2 parts to calculating the small change in pH which occurs in a buffer solution, when an
acid or base is added to the buffer.
The Stoichiometric Calculation in
which you must calculate the extent to
which the addition changes the relative
amounts of acid and conjugate base.
The Equilibrium Calculation in
which you must calculate the pH based
on the new amounts of acid and
conjugate base
It is best to complete the stoichiometric
calculations first, prior to the equilibrium
calculation.
The calculation of the pH can be
performed by using either the
Henderson-Hasselbalch equation
OR by working a full equilibrium
problem.
This could include something like an
ICE table … though it is not equivalent
to an ICE table. It is simply an
attempt to track the stoichiometric
changes … Often ≈0.00 mol for the
amount of H+ because the amount is
so small compared to the amounts of
A- and HA
The calculation of the pH can be
performed by using either the
Henderson-Hasselbalch equation
OR by working a full equilibrium
problem.
Recall that weak acids ionize only to a
small extent and that the presence of the
common ion further suppresses the
ionization. The amount of H+, of
course, is not exactly zero … as will be
seen by completing the equilibrium part
of the calculation.
In summary:
Adding a small amount of strong acid to a buffer coverts a stoichiometric amount of the base to the
conjugate acid and decreases the pH of the buffer (adding acid decreases pH just as one would expect.)
H+
+
A-
→
added strong acid conjugate base
already part of buffer
HA
molecular weak acid, which binds up H+, removing it from solution
Adding a small amount of strong base to a buffer converts a stoichiometric amount of the acid to the
conjugate base and increases the pH of the buffer (adding base increased the pH just as one would
expect)
OH- + HA
→ H2O + Aadded strong base
weak acid
already part of buffer
conjugate base of the weak acid
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B) Addition of Strong Acids or Bases to Buffers
Brown and LeMay text gives you sound advice …
1) Consider the acid-base neutralization reaction and determine its effect on [HA] and [A-]
This is a limiting reagent calculation … and you will need a bit o’ stoichiometry.
2) Us the calculated values of [HA] and [A-] as well as the Ka to calculate [H+]. This is the
equilibrium calculation and can be done with an ICE table OR with Henderson-Hasselbalch
equation if the concentrations of the weak acid-base pair are very large compared to the
Ka for the acid (p.735)
3) Calculate pH using pH = - log[H+]
e.g.) A buffer is made by adding 0.300 mol CH3COOH (Ka = 1.8 x 10-5) & 0.300 CH3COONa
to enough water to make 1.000 L of solution. The pH of the buffer is 4.74
Calculate the pH of this solution after 5.0 mL of 4.0 M NaOH(aq) is added
 Stoichiometric calculation: First we must determine how the added (OH)- affects the
concentration of the acid.
neutralizes
CH3COOH + NaOH
→
CH3COO- + Na+ + H2O
This can be tricky … the volumes are changing, thus the molarities change. Work
with moles and then convert to molarity by dividing by the TOTAL volume of the
resulting solution.
What’s in the beaker? … 0.300 mol CH3COOH and 0.300 mol CH3COONa
What is added?
0.020 mol NaOH
M = moles/L 4.0 M = x/0.005L
Something like an ICE table might help track the stoichiometric changes….
Before
Addition
The Addition
CH3COOH
0.300 mol
-0.020
+
OH0.020 mol
→ CH3COO- +
0.300 mol
-0.020 mol
+0.020 mol
-----
0.00 mol
0.320 mol
-----
H2 O
-----
because it is neutralized by the
1:1 rato with the added base
After the
Addition
0.280
limiting reagent and
consumed completely
 Equilibrium calculation: CH3COOH(aq) ↔ H+(aq) + CH3COO-(aq)
Using the remaining concentrations from the above table, we can use the
Henderson-Hasselbalch equation to solve for pH. HOWEVER… since the equation
demands that we use the molar concentrations of the conjugate base and the acid,
we must determine the NEW molarities … recall that we began with 1.000L of buffer
and added another 0.005 L of NaOH … making a new volume of 1.005 Liters…
Hence the new [CH3COOH] = 0.280/1.005 and the new [CH3COO-] = 0.320/1.005L
= 0.279 M
and
0.318 M
keep going…
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NOW you are ready to plug and chug into the Henderson-Hasselbalch equation.
pH = pKa + log [conj base]
[acid]
Thus:
recall that pKa= -log Ka
pH = 4.74 + log [0.318]
[0.279]
= 4.80
Notice that the pH of the problem is 4.74 and the pKa is also 4.74 …
Hence, due to the buffering ability of the solution, the pH changed slightly
from 4.74 to 4.80, with the addition of a small amount of a concentrated,
strong base.
NOW COMPARE
Question: What would be the pH of a 1.000 L of water, to which 5.0 mL of 4.0 M NaOH(aq)
were added?
MOH- = moles
L
Msol’n = moles
L
4.0 M = moles thus, 0.020 moles of NaOH were added
0.0050 L
to the water
Msol’n = 0.020 mol NaOH = 0.020 M solution
1.005 L
Thus pOH = - log [0.020]
pH = 14.00 – 1.70
= 1.70
Thus the pH of the new solution = 12.30
Without any buffering ability, the pH of pure water changed from 7.0 to 12.30
… a change in 5.30 pH units, with the addition of the same amount of the
strong concentrated base, as was added in the earlier buffer problem. One
natural conclusion then is that buffering ability matters, when the goal is
to resist changes in pH.
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TRY THIS!
Which of these statements is true?
1) If you were to add strong acid or base to a buffer, the pH will never change
2) In order to do calculations in which strong acid or base is added to a buffer, you only need
to use the Henderson-Hasselbalch equation.
3) Strong bases react with strong acids, but not with weak acids.
4) If you were to add a strong acid or base to a buffer, the buffer’s pKa or pKb will change.
5) In order to do calculations in which strong acid or base is added to a buffer, you need to
calculate the amounts of substances from the neutralization reaction and then equilibrate.
ans:*5 some might choose “4”, but pKa tells us how strong an acid is … not its extent of
ionization. Hence, the remaining acid is a strong as it always was although its molar
concentration in the buffer may change slightly with the addition of a strong acid or base.
TRY THIS!
A friend comes to you with the following problem:
Calculate the change in pH of a buffer made with 6.50 x 10-4M HOCl (Ka = 3.0 x 10-5) and 7.5 x 10-4 M
NaOCl, when 5.0 mL of 1.50 M HCl is added to the buffer solution.
Would you advise your friend to solve the problem using;
the equilibrium solution approach (run the stoichiometry changes, write an equilibrium
reaction equation, write the Ka expression, pH equation etc…)
OR by
the Henderson-Hasselbalch equation.
Defend your reasoning: * I would urge my friend to use the first approach and to avoid use of the
Henderson-Hasselbalch equation. While the equation is simpler, the first approach always works, while
the Henderson-Hasselbalch equation only works well, when the buffer concentrations of the acid and
conjugate base are relatively high, and at least 1,000 times larger than the Ka. Given the numbers in the
problem, these conditions are NOT met. The Ka is NOT significantly smaller than the concentrations of
the buffer components (There is just about a 10 times difference). We cannot assume then that the
concentrations are quite large relative to the dissociated amount of acid. “x” cannot be ignored and
using the Henderson-Hasselbalch equation, assumes that “x” can be ignored.
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TRY THIS!
What is the pH of a solution that is 0.50 M in propanoic acid and 0.40 M in
sodium propanoate. (Ka for propanoic acid = 1.3 x 10-5)
1)
2)
3)
4)
0.097
0.47
4.98
4.79
Hint 1: * Take a look at the value of
the Ka and the initial concentrations
… There are huge differences,
greater than 1,000 times difference.
Hint 2: * You need to calculate the
pH … think Henderson-Hasselbalch
eq.
ans: choice 4
Ka for HNO2 is 4.5 x 10-4, calculate the pH of a buffer solution made by mixing
0.225 mol of HNO2 and 0.450 mol of NaNO2 in enough water to make 0.400 liter of solution.
1)
2)
3)
4)
5.72
3.51
3.65
2.98
ans: choice 3
Assignment: Get to the Trivedi Flash: 16.8, and 16.9
XXIII) Human blood as a buffer
I have been speaking of our blood as a buffer and we are now at the point where I will qualify that
statement to a point. Our blood approximates a physiological buffer (some of the strict definition
parameters are stretched) …And, maintenance of our blood pH is also greatly aided by the proper
functioning of our heart, lungs and kidneys. It would be irresponsible of the biologist in me (whom
I can normally keep quiet with cookies), to ignore the role of the heart, lungs and kidneys. It is not
all Le Chatelier’s and blood chemistry … but heck, it is ‘pert near fine example of a buffer system.
In human blood, pH is held between 7.36 and 7.42. This nearly constant pH is maintained by a buffer
system. An important buffer in blood is a mixture of carbonic acid (H2CO3 …a weak acid) and the
hydrogen carbonate ion (a.k.a. bicarbonate (HCO3)- ion) ….the weak conjugate base of carbonic acid.
When the blood pH is allowed to drop below below 7.36 or so, the condition called acidosis exists
When the blood pH is allowed to rise above 7.42, the condition of alkalosis exists.
Activities such as exercise decrease blood pH. For instance, CO2 and H+ are produced during the
breakdown of glucose during exercise, and are removed from the muscle via the blood. The production
and removal of CO2 and H+ together with the use and transport of O2, cause chemical changes in the
blood, which can lead to a decrease in pH… unless other mechanisms are present to offset the increase
in acid.
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As the [H+] increases it is reacted with bicarbonate ion, dissolved in the water of blood, producing
molecular carbonic acid.
H+(aq) + HCO3-(aq) ↔ H2CO3(aq) This decreases the [H+], raising the pH
In a non-acid-base reaction, carbonic acid also dissociates rapidly to produce water
and carbon dioxide, which is expelled from the body via the lungs.
H2CO3(aq)
↔
CO2(g) + H2O(l)
Putting the two together:
an acid-base reaction
H+(aq) + HCO3-(aq) ↔
H2CO3(aq) ↔ CO2(g) + H2O(l)
a non-acid-base reaction
http://www.chemistry.wustl.edu/~edudev/LabTutorials/Buffer/Buffer.html
In terms of the Henderson-Hasselbalch equation, the pH ultimately depends upon the ratio of
bicarbonate ion to carbon dioxide.
pH = pKa –log [CO2]
[HCO3-]
This ratio remains relatively constant, because the concentrations of both buffer components
(HCO3- and CO2) are very large, compared to the amount of H+ added to the blood during normal
activities and moderate exercise. When H+ is added to the blood as a result of metabolic processes,
the amount of HCO3- (relative to the amount of CO2) decreases; however, the amount of the change is
tiny compared to the amount of HCO3- present in the blood. This optimal buffering occurs when the pH
is within approximately 1 pH unit from the pKa value for the buffering system, i.e., when the pH is
between 5.1 and 7.1.
However, the normal blood pH of 7.4 is outside the optimal buffering range; therefore, the addition of
protons to the blood due to strenuous exercise may be too great for the buffer alone to effectively control
the pH of the blood. When this happens, other organs must help control the amounts of CO2 and
HCO3- in the blood. The lungs remove excess CO2 from the blood (helping to raise the pH via shifts in
the equilibria in Equation 10), and the kidneys remove excess HCO3- from the body (helping to lower
the pH). The lungs' removal of CO2 from the blood is somewhat impeded during exercise when the
heart rate is very rapid; the blood is pumped through the capillaries very quickly, and so there is little
time in the lungs for carbon dioxide to be exchanged for oxygen. http://www.chemistry.wustl.edu/~edudev/LabTutorials/Buffer/Buffer.html
Additional Reading:
Q/A on Blood and Buffer: http://www.newton.dep.anl.gov/askasci/mole00/mole00656.htm
Ocean Acidification & pH: http://www.pmel.noaa.gov/co2/story/A+primer+on+pH
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