NAME ____________________________
AP NOTES: UNIT 4 (4) APPLICATIONS OF EQUILIBRIUA:
COMMON ION, ACID-BASE BUFFERS
XV-XVIII) Salts to Oxides to Lewis Theory
In the first three sections of this unit we dealt with either a strong acid/strong base system OR calculated equilibria
concentrations for solutions of weak acids (or weak bases) …. At this point we must now turn our gaze upon systems
which have a weak acid (HA) and its salt (NaA) … Techniques developed earlier will be used.
It is very important to recall that the ionic salt is a strong electrolyte (and it will break up completely into its ions) and (!)
the weak acid will dissociate to only a certain point… This means that both dissociations will have a common ion.
XIX) Common Ion Effect
First, a simple demonstration ….. Imagine a beaker with 250 mL of a saturated NaCl(aq) at 21.0°C
NaCl(aq)  Na1+(aq) + Cl1-(aq)
Now consider a concentrated solution of HCl(aq) …. HCl(aq) → H+(aq) + Cl1-(aq) The two solutions share a
common ion …. Cl-1.
According to Le Chatelier’s Principle if the concentrated solution of HCl
were added to the saturated solution of NaCl … what might happen?
*NaCl(s) will precipitate out of solution.
A) Common Ion Effect: *The addition of an ion already involved in the equilibrium reaction, causes a
shift in the equilibrium position…resulting in a decreased solubility of the original salt, due to the
presence of a common ion in solution. Using it allows us to change the solubility of a slightly
soluble salt and Common Ion Effect allows us to change the pH of a solution.
1) Imagine a system with the weak acid
HF(aq) H+(aq) + F-(aq) and to it, add the salt NaF
NaF(s) → Na+(aq) + F-(aq) …. The dissociation of HF will be inhibited, and the
equilibrium position of the HF will be driven to the left, by the common ion fluoride.
I wonder if you folks are realizing that the underlying theme here is Le Chatelier’s principle
whether it is a strong acid/weak acid mixture or common ion effect? This whole Big Idea 6
which states: Any bond or intermolecular attraction, that can be formed, can be broken.
These two processes are in dynamic competition sensitive to initial conditions and
external perturbations … is in operation here, over, and over, and over again…. AND
often has Le Chatelier’s applied.
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B) Acidic solutions and Common Ion
I love what Steven and Susan Zumdahl have written … I am copying it here … for your
benefit. … First, the problem they propose … but take a serious look at the advice they give …
Problem: Calculate [H+], pH and the percent dissociation of HF in an aqueous mixture
containing 1.0 M HF (Ka = 7.2 x 10 -4) and 1.0 M NaF
Advice: “As the aqueous solutions we consider become more complex, it is more important than ever to
be systematic and to focus on the chemistry occurring in the solution before thinking about
mathematical procedures. The way to do this is always to write the major species first and
consider the chemical properties of each one. (p713-14)
ARE YOU LISTENING? Does this sound familiar? What’s in the beaker and
What do the species do?
In this solution containing 1.0 M HF and 1.0 M NaF, there are 5 (or so) major species:
of course there is water H2O (but it is so weak we can ignore it), but there is 4 other
major species:
*HF (dissolved, but still molecular), H3O+, Na+, and FWhich of these other 4 major species can we probably discount … and why?
We can probably ignore *Na+1 since it is the conjugate acid of a strong base and
has virtually no effect on pH.
Thus we must focus on the chemical activity of * HF and F- and H3O+
Now, we want to know about the [H+] so we must concern ourselves with the
ionization and the equilibrium position of the hydrofluoric acid.
Write the ionization of HF in water: * HF(aq) + H2O(ℓ)  H3O+(aq) + F-(aq)
This is an equilibrium issue … regarding a weak acid (and there will be a small
twist with respect to the [F-], because some will be due to the HF ionization, but a
huge amount depends upon the NaF dissociation (which is 100%).
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This common-ion (of F-) will affect the equilibrium position of the acid, and it will
in all likelihood (because of Le Chatelier’s Principle) decrease the expected
production of H+, in solution … and causing the pH to be higher than expected….
Na+ + FHF(aq) + H2O(ℓ)  H3O+(aq) + F-(aq)
Since this is a weak acid equilibrium issue … the first thing to do is to *write the Ka
*Ka = [H+][F-]
[HF]
Since this is a weak acid equilibrium issue … the next expected math procedure is to
create *an ICE table
WHAT????
*
Initial
Change
Equil
[HF] mol/L
1.0
1.0 –x
1.0 – x
H+
0
+x
x
F1.0
1.0 +x
1.0 + x
Solution for pH: *Ka = 7.2 x 10-4 = (x)(1.0 +x) ≈ (x)(1.0)
1.0 – x
1.0
* x = 7.2 x 10-4 …. Thus: pH = - log[7.2 x 10-4 ] = 3.14
this change in [HF] or “x” is well below the 5% rule, because the ionization of HF
is inhibited by the presence of a common ion…even though the Ka is “larger than I
would like” at 10-2 and the [HF ] is relatively high, at 1.0 M ... because of the inhibiting
effect of the common ion I can still ignore “x”
Solution for % ionization: *%ionization = [H+] x 100 = 7.2 x 10-4 M x 100 = 0.072%
[HF]0
1.0 M
The pH and % ionization of JUST a solution of 1.0 M HF = a pH of that solution is
1.56, the equilibrium concentration of H+ in a 1.0 M HF solution is 2.7 x 10-2 M, and the
percent dissociation of HF is 2.7%. …. The increase in pH of the solution of this
problem, with NaF present, is obviously due to a lesser value of [H+], which is in all
probability due to the presence of the F- shifting the position of equilibrium to the left.
Get it?
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TRY THIS!
A (buffered) solution contains 0.50 M acetic acid (HC2H3O2 Ka = 1.8 x 10-5) and
0.5 M sodium acetate (NaC2H3O2). Calculate the pH of this solution.
Zumdahl p. 717
1) What are the major species?*HC2H3O2 (wa), H+, Na+, C2H3O2- (conj base) & H2O (very wa/wb)
Which matter?
*
HC2H3O2, H+, C2H3O2-
a) Write the equilibrium expression which includes the most important major species.
* HC2H3O2(aq)  H+(aq) + C2H3O2(aq)b) Next? … It is a weak acid problem… * Write the Ka
*Ka = 1.8 x 10-5 = [H+][ C2H3O2-]
[HC2H3O2]
c) Next? *ICE it … for 20 minutes (giggle)
*
Initial
Change
Equil
2)
HC2H3O2 mol/L
0.50
-x
0.50 –x
Plug and Chug: *1.8 x 10-5 = [x][ 0.50 + x] ≈
[0.50-x]
H+ mol/L
0
+x
x
C2H3O2- mol/L
0.50
0.50 + x
0.50 + x
(x)(0.50)
0.50
*x = 1.8 x 10-5 ….You can check “ignoring x” … with 1.8 x 10-5 x 100 = 3.6 x 10-3
0.50
3) Hence: pH = *4.74
Assignment: Trivedi: 16.1 – 16.5 … These later sections 16.4 and 16.5 get a bit tougher.
XX) Buffered Solutions … Calling all future Chemists / Doctors / Vets / PAs / Biologists!!!! (This is the
SHOW! … We’ve been working up to this … Learn This!!!!)
A) A Buffer or Buffered Solution or pH Buffer: * a solution which resists drastic changes in
its pH when either hydroxide ion or protons (hydronium ions) are added.
….a buffered solution is essentially an expression of some form of common ion effect…..
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1) Buffer solutions achieve their resistance to pH change because of the presence
of an equilibrium (Big Idea 6) between the weak acid HA and its conjugate base A-.
HA ↔ H+ + AWhen some strong acid is added to an equilibrium mixture of the weak acid and its
conjugate base, the equilibrium is shifted to the left, in accordance with Le Chatelier's
principle. Because of this, the hydrogen ion concentration increases by less than the
amount expected for the quantity of strong acid added. Similarly, if strong alkali is
added to the mixture the hydrogen ion concentration decreases by less than the
amount expected for the quantity of alkali added. https://en.wikipedia.org/wiki/Buffer_solution:
(See “How Buffering Works….on the next page)
2) A buffered solution may contain:

*a weak acid and its salt or stated differently … a weak acid and its conjugate
base (e.g. HF and NaF, acetic acid and sodium acetate) Buffers are applications
of the common-ion effect…
OR

*a weak base and its salt (e.g. NH3 and NH4Cl)
a) Thus a buffer contains:

an acidic species to *neutralize excess (OH)- ions

a basic species to *neutralize excess H+ ions
a) For us, the most important of all buffers will be our blood …. We can live only within
a rather narrow range of pH values … yet our bodily functions are producing all sorts
of acids, bases, salts ….
b) I guess the big question is “How does a buffered solution resist the change in pH?
How does a buffer solution work?”
i) Recall that buffers simply contain weak acids or bases and their respective
conjugates delivered into the solution, as ionic salts and…
ii) Since acids react with bases and not with other acids, while bases react with
acids but not with other bases, only a solution which contains both an acid
and a base can resist the effect of both kinds of addition by reaction.
This leads us to using a solution with a weak acid and its conjugate base
or a weak base and its conjugate acid.
373
Here “IT” is ... in a nutshell
ii) When a small amount of OH– is added to the buffer, the OH– reacts with HA to
produce A– and water.
But the [HA]/[ A–] ratio remains more or less constant,
so the pH is not significantly changed.
When a small amount of H+ is added to the buffer, *X– is consumed to
produce HX (the molecular weak acid).
http://blog.sellsiusrealestate.com/wp-content/nutshell1.jpg
Once again, the [HA]/[ A–] ratio is more or less constant,
so the pH does not change significantly
C) How Buffering Works …. Not, In a Nutshell
(Zumdahl p. 718-9)
1) The essential key is a buffered solution contains *relatively large quantities of a
weak acid (HA) and its conjugate base (A-) …
When hydroxide ions are added to the solution, the weak acid (which is the best source of
protons) reacts:
OH-(aq) + HA(aq) A-(aq) + H2O(l) … The result is that the OH- are NOT allowed
to accumulate but are replaced by the weak conjugate base (A- ions)
a) The stability of the pH under these conditions can be grasped by re-arranging
the equilibrium expression for HA
Ka = [H+][A-]
[HA]
…. becomes
[H+] = Ka[HA]
[A-]
Perhaps the above helps you see that the equilibrium concentration of
H+ (and thus the pH) is determined by the ration of [HA]
[A-]
When OH- ions are added, from some outside source to the buffered
solution, the HA is converted to A- …. and the ratio decreases ….
BUT!!!!! When the amounts of HA and A- originally present are
very large, compared with the amount of added OH-, the change in the
[HA] ratio is rather small … thus pH remains essentially constant.
[A-]
Similar reasoning works, when extra H+ is added to a buffered solution of a
weak acid and a salt of its conjugate base. A- has a high affinity for H+, so
when H+ is added, the molecular weak acid is re-constituted.
H+(aq) + A-(aq) HA(aq) Thus free H+ do NOT accumulate.
374
Again, there will be a change in the ratio of [HA] but when the initial
[A-]
concentrations are rather large relative to the added H+ … then the pH remains
relatively stable.
Summary: *An example of a buffer is a solution *made by mixing a weak acid
and a salt containing the conjugate base of that acid.
A buffer has the ability to *resist changes in pH
TRY THIS! Which of these could represent buffer solutions? (There are 5 answers) Remember, you want
solutions of weak acids (or bases) and a salt of that acid (or base)
….As a rule for our course, buffers do NOT have strong acids or strong bases. Thus a buffer
must be a mixture of a weak acid, (or conjugate acid) & a weak base (or conjugate base)
a) ___ H3PO4(aq) & KH2PO4(aq)
b) ___ HNO2(aq) /& NaC2H3O2(aq)
c) ___ CH3COOH(aq) & NaC2H3O2(aq)
d) ___ H2SO4(aq) & K2SO4(aq)
ans: *a,c, g (amphiprotic), h, i, (alkaline buffer)
e) ___ just Cu(OH)2(aq)
f) ___ just NaC2H3O2(aq)
g) ___NaHCO3(aq) & Na2CO3(aq) (tricky)
h) ___H3BO3(aq) & Na3BO3(aq)
i) ___NH3(aq) (or NH4OH(aq)) & NH4Cl(aq)
j) ___ HClO2(aq) & NaOH(aq)
TRY THIS! Assume you wished to make a buffer, using HClO(aq) as the first component. Suggest a formula
of the second component. * Possibilities: NaClO, KClO
TRY THIS! Assume you wished to make a buffer, using HCN(aq) as the first component. Suggest a formula
of the second component. * Possibilities: NaCN, KCN
375
2) Extra! Extra!
pKa Rears Its Head Again!!!
Know This! Buffers function best when the pKa of the weak acid used is close to the desired
working range of the buffer. So, the optimal buffering system has a pKa value close to the
desired pH of the solution. Or another way to look at this, the optimal buffering will
occur when the [A-]/[HA] approaches 1 …. This ratio (approaching 1) suggests that there is
adequate base and acid to deal with additions of stronger acids or bases to the system ..and
resist changes to the pH.
Thus, when the concentrations of A- and HA are equal, the [H+] is equal to Ka,
(or equivalently the –log forms are equal, hence: pH = pKa)
Since pKa is a property of the weak acid used in the buffer, the pH is controlled by
manipulating the proportion of weak base (A-) and weak acid (HA) in solution.
http://chemcollective.org/activities/tutorials/buffers/buffers4
e.g ) Suppose you wish to construct a buffer with an acid HA, which has a pKa = 5
In order to have a system as close to [A-] = [HA] as possible, so that [A-]/[HA] =1
the best pH for the buffer would be *5
more later….
…. But let’s first see how pKa is re-introduced to our present conversation.
3) This aforementioned form of the acid dissociation equilibrium expression is often useful
for calculating [H+] … it is a bit of a “trick of the trade” … (I shudder to think I wrote that)
The pH of the buffer is related to Ka and to the relative concentrations of the acid and base.
Hence … to calculate the [H+] …. and thus use that to calculate a pH ….we can use:
[H+] = Ka[HA] For instance: Calculate the [H+] in a buffered solution containing
[A-] 0.10 M HF (Ka = 7.2 x 10-4) and 0.30 M NaF
[H+] = 7.2 x 10-4 (0.10) = 2.4 x 10-4 M
(0.30)
However, as written previously … pKa is often preferred…so….
376
4) Another useful form of [H+] = Ka[HA] can be obtained, by taking the negative log
[A-] of both sides…
-log[H+] = - log (Ka) – log [HA]
[A-]
That is
[A−]
pH = pKa + log ([HA]
) =
𝒃𝒂𝒔𝒆]
pH= pKa + log ([𝒄𝒐𝒏𝒋
)
[𝒂𝒄𝒊𝒅]
𝒃𝒂𝒔𝒆]
a) pH = pKa + log ([𝒄𝒐𝒏𝒋
) is called the Henderson-Hasselbalch equation
[𝒂𝒄𝒊𝒅]
It is on your reference tables!!! So, expect to use it.
b) It is the -log form of the Ka and it is useful for calculating the pH of solutions
when the ratio [HA]/[A-] is known….that means it’s useful for buffers….
i) For a particular buffering system of a conjugate acid-base pair, all solutions
which have the same ratio [A-]/[HA] will have the same pH.
e.g.) a buffered solution containing 5.0 M HC2H3O2 and 3.0 M NaC2H3O2
will have the same pH as one containing
0.050 M HC2H3O2 & 0.030 M NaC2H3O2
ii) Note that this equation uses the equilibrium concentrations of
the acid and conjugate base.
However, when the Ka is sufficiently small (when the equilibrium
concentration of the un-dissociated acid is close to the initial concentration…
or rather, when “x” is negligible), then we can use the initial values of the
acid and base concentrations to get a good estimate of pH.
http://smithstudents.com/AP_Resources_files/SG17bufferKsp_AP.pdf
5) So let me zip over to: http://chemcollective.org/activities/tutorials/buffers/buffers4
Keep the Henderson-Hasselbalch equation in mind and let’s use it …
377
So, at this point, there are two ways for calculating the pH of a buffer system

You can use: The Ka & the equilibrium expression …& then find pH
using pH = - log [H+]
AND/OR

You can use the Henderson-Hasselbalch equation (under the right conditions)
pH = pKa + log [conj base]
[acid]
recall that pKa= - log Ka
Let’s look at one problem, solved in these two different ways ….
Calculate the pH of a solution containing 0.75 M lactic acid (Ka = 1.4 x 10-4) and
0.25 M sodium lactate. Lactic acid (HC3H5O3) is a common constituent of biologic
systems. For example, it is found in milk and is present in human muscle tissue during
exertion. (Zumdahl p. 720)
What’s in the beaker? * HC3H5O3, Na+, C3H5O3-, H2O
Which are the 2 most important species? * HC3H5O3 and C3H5O3- WHY? Na+ has
virtually no ability to affect pH and water is a far weaker acid than lactic acid … check
out the Ka value ….
What is the reaction equation:
* HC3H5O3(aq)  H+(aq) + C3H5O3- (aq)
What is the equilibrium expression?
*Ka = [H+][ C3H5O3-]
[HC3H5O3]
* 1.4 x 10 -4 = [H+][ C3H5O3-]
[HC3H5O3]
Since the [acid] and the [conj base] are relatively large at 0.75 M and 0.25 M respectively,
we can just rearrange the Ka expression and plug and chug these values to find H+
1.4 x 10 -4 = [H+][ C3H5O3-]
[HC3H5O3]
becomes, when re-arranged for H+
[H+] = 1.4 x 10-4 [HC3H5O3]
[C3H5O3-]
and to find pH: pH = - log [H+] or
or
= 1.4 x10-4 (0.75 M) = 4.2 x 10-4M
(0.25 M)
pH = - log (4.2 x 10-4) = 3.38
378
NOW, USING THE HENDERSON-HASSELBALCH EQUATION….
pH = pKa + log [conj base]
[acid]
recall: pKa = - log Ka
pH = 3.85 + log (0.25 M) = 3.38
(0.75 M)
The theorist in one, loves the first approach … The math-lord in one, honors the latter.
Assignment: Trivedi: 16.6 – 16.7 (16.7 is quite important … do not fail to look this over)
Try it again … using a buffer made with a base and its conjugate acid … Let’s look at both approaches.
We can assume, when the concentrations are relatively large, that we can ignore “x” but we will need to
run an ICE table…. We need the ICE table to find the [OH-] … so that we can use the [OH-] to solve
for the pOH and thus pH … This sort of problem is nowhere nearly as direct as the first problem.
First, let’s look at using the Kb equilibrium expression.
A buffered solution contains 0.25 M NH3 (Kb = 1.8 x 10-5) and 0.40 M NH4Cl. Calculate the pH
of the solution. (Zumdahl p. 721)
What is in the beaker? * NH3, NH4+, H2O, ClWhich really matter? *NH3 and NH4+, since Cl- has virtually no effect on pH , being a weak
conjugate base of a strong acid … and water is a very, very weak base … NH3 beats it every time…
What is the reaction equation?
What is the Kb expression?
* NH3(aq) + H2O(l) 
NH4+(aq) + OH-(aq)
1.8 x 10-5 = [NH4+][OH-]
[NH3]
Run the ICE table….
[NH3]
Initial
0.25
Change
-x
Equil
0.25-x
[NH4+] [OH-]
0.40
0
+x
+x
0.40+x
x
1.8 x 10-5 = (0.40)(x)
(0.25) (ignoring “x”) …. thus
1.1 x10-5 = [OH-]
(You can run the check for the 5% rule ….)
pOH = - log (1.1 x 10-5) = 4.95
thus … pH = 14.00 = 4.95 = 9.05
379
NOW, USING THE HENDERSON-HASSELBALCH EQUATION….
pH = pKa + log [conj base]
[acid]
Ka = Kw/Kb … so
recall: pKa = - log Ka and that Kw = (Ka) (Kb)
Ka = 1 x 10-14/1.8 x 10-5 = 5.6 x 10-10
pKa = -log (5.6 x10-10)
= 9.25
pH = pKa + log [conj base] thus: pH = 9.25 + log(0.25 M)
[acid]
0.40 M)
= 9.25 -0.20 = 9.05
Assignment: Trivedi: 16.23 Questions 1,2,3,4,6,8,9,10,12 AND
16.24 Questions 1,2,3,4,6,7,8,9,16,18
D) Buffer Capacity and pH Range (new vocabulary)…
1) Buffer capacity is the * amount of acid or base that can be neutralized by the buffer
before there is significant pH change.
a) Buffer capacity depends on the * concentrations of the components of the buffer.
i) The greater the concentrations of the conjugate acid-base pair, the greater the
buffer capacity.
2) The pH range of a buffer is the pH range over which it is an effective buffer.
a) The pH range of a buffer is generally within one pH unit of the pKa of the
buffering agent
380