Section 14.1
Nature of Acids and Bases
Arrhenius Definition

Acids produce hydrogen ions in
aqueous solution.
– HCl (aq)  H+(aq) + Cl-(aq)

Bases produce hydroxide ions when
dissolved in water.
– NaOH (aq)  Na+(aq) + OH-(aq)

Limits what can be considered bases (as
we’ll see with other definitions).
Lewis Definition
An acid is an electron pair acceptor.
 A base is an electron pair donor.
 Easy to see if Lewis structures are
drawn:

e- pair
acceptor
e- pair
donor
Bronsted-Lowry Definition
This is the one we’ll focus on!
 An acid is a proton (H+) donor and a
base is a proton acceptor.
 HCl is an acid.
 When it dissolves in water it gives its
proton to water.

H3O+ + Cl-

HCl(g) + H2O(l)

Water is a base since it accepts the H+.
– In the Arrhenius definition water would
not be considered a base!
Conjugate Acid-Base Pairs
General equation
 HA(aq) + H2O(l)
H3O+(aq) + A-(aq)
acid + base
conjugate acid + conjugate
base
 This is an equilibrium.
 Equilibrium favors the side with the weaker
acid and base.
 Refer to the handout for acid and base strength.
 Note: conjugate bases of strong acids are weak,
and conjugate bases of weak acids are strong.
In other words, the stronger the acid/base, the
weaker the conjugate base/acid and vice-versa.

Acid dissociation constant Ka

Recall strong vs. weak acids/bases!
– Strong = essentially completely dissociate
in water (no equilibrium).
– Weak = partially dissociate in water.
Equilibrium is therefore present!
For a weak acid:
HA(aq) + H2O(l)
H3O+(aq) + A-(aq)
 Water is often left out:
HA (aq)
H+(aq) + A-(aq)

Acid dissociation constant Ka


Since weak acids result in equilibrium,
an equilibrium expression can be
written:
HA(aq) + H2O(l)
H3O+(aq) + A-(aq)
Ka = [H3O+][A-]
[HA]
Acid dissociation constant = Ka.
 Note: technically all acids have a Ka
value. Size of Ka indicates strength!

Acid dissociation constant Ka
If water is left out:
 HA(aq)
H+(aq) + A-(aq)


Ka = [H+][A-]
[HA]
Shown without H3O+
Section 14.2
Acid Strength
Types of Acids

Polyprotic Acids- more than 1 acidic
hydrogen (diprotic, triprotic).
– Ex: H2SO4, H3PO4

Oxyacids - Proton is attached to the
oxygen of an ion.
– Ex: H2SO4, HNO3

Organic acids contain the carboxyl
group -COOH (acidic H+ attached to O).
– Ex: CH3COOH (acetic acid)

Generally very weak.
Ka of Polyprotic Acids
The first Ka value is much larger than
the second, third, etc.
 The second Ka value is much larger
than the third Ka value, etc.
 Since the first Ka value is by far the
largest, second, third, etc. Ka values can
be ‘ignored’ and the first Ka value can
usually be used for the acid dissociation
constant of the acid.

Amphoteric









Can behave as either an acid or a base.
Water undergoes autonionization (it
autoionizes).
2H2O(l)
H3O+(aq) + OH-(aq)
KW= [H3O+][OH-]=[H+][OH-]
Notice it’s
-14
At 25ºC KW = 1.0 x10
very small!
Occurs in EVERY aqueous solution.
Neutral solution [H+] = [OH-]= 1.0 x10-7
Acidic solution [H+] > [OH-]
Basic solution [H+] < [OH-]
Amphoteric Continued
Note: although water dissociation
occurs in all solutions and contributes
to [H+], it is small in comparison and
can be ignored when calculating [H+].
 Adding other species to water can also
allow water to act as an acid or a base.

Strong Acids
HBr, HI, HCl, HNO3, H2SO4, HClO4
 Completely dissociated
 [H+] = [HA]
 Ex: What is [H+] in a 0.10M HNO3
solution?

HNO3  H+ + NO3*Mole ratio is 1:1 for HNO3:H+
*Thus [H+] = 0.10M
Weak Acids
Ka will be small.
 It will be an equilibrium problem to
find [H+].
 Determine whether most of the H+ will
come from the acid or the water.
 Compare Ka and Kw.

– Whichever is larger is the one that will
donate more H+ (this is usually the acid).

Rest is just like last chapter.
Weak Acids
Because you’re dealing with weak
acids, it can be assumed that the acid
won’t dissociate much.
 Lets you make assumptions and
simplify terms when solving for x. (If
you forget and don’t simplify that’s
OK- you will just need to use the
quadratic formula).
 Then calculate the [H+].

Weak Acids
For weak acids, the concentrations of
H+ and A- may be found if both the
initial concentration of the acid and Ka
are known.
 Ex: What is the [H+] in 0.300M acetic
acid solution? Ka = 1.8 x 10-5
 Remember- this is an equilibrium
problem! Set up your ICE table, then
solve for x.

– Answer: x = [H+] = 2.3 x 10-3 M
Practice Problem #1

Pg. 675 #55:
A 0.0560g sample of acetic acid is
added to enough water to make
50.00mL of solution. Calculate the [H+],
[CH3COO-], and [CH3COOH]. The Ka
for acetic acid = 1.8 x 10-5.
Summary: Strong vs. Weak
Remember- strong acids: [HA] = [H+]
and weak acids = equilibrium problem
to find [H+].
 Otherwise, once the [H+] is known, pH
problems are the same.

Section 1 Homework

Pg. 672 # 19(a&c), 21, 24, 28, 29, 32
pH Scale


pH= -log[H+] or pH = -log [H3O+]
pH of pure water = -log(1.0 x 10-7) = 7.00
– In other words, this is neutral.
– pH < 7.00 is acidic; pH > 7.00 is basic
– Notice that as pH decreases, [H+] increases


Sig figs: only the digits after the decimal
place of a pH are significant
[H+] = 1.0 x 10-8 pH= 8.00 2 sig figs
pH Scale Continued

The pH of a solution can be estimated if
the [H+] is known.
– Look at the exponent to estimate the pH
– Ex: [H+] = 1 x 10-5; pH = 5

pOH can also be calculated:
 pOH = -log[OH-]
[H+]
100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14
pH
0
1
Acidic
14 13
3
11
5
7 9
Neutral
9
7 5
10-14 10-13 10-11 10-9
11
3
13
14
Basic
1
0
pOH
10-7 10-5 10-3 10-1 100
[OH-]
Important Relationships
KW = 1.0 x 10-14 = [H+][OH-]
 pKW = 14.00 = pH + pOH
 pH = -log[H+]
 pOH = -log[OH-]
+
 [H ],[OH ],pH and pOH
Given any one of these we can find
the other three!
 There are also often multiple ways to
correctly solve these problems!

Example #1
The [H+] = 1.00 x 10-5 M. What is the
pOH for this solution? Note: if [H+] is
given, no need to
 Step 1: Find pH.
worry if acid is
 Step 2: Then use
strong or weak.
14 = pH + pOH to solve for pOH.
 Can you think of another approach?
 Find [OH-] using: 1.0x10-14 = [H+][OH-].
Then use: pOH = -log[OH-].
Answer: Either way you get pOH = 9.000

Example #2
Calculate the pH of a 1.00M solution of HF,
whose Ka = 7.2x10-4.
[H+] is not given,
(1) Small Ka, so it’s a weak acid.
so now we need
to decide if the
(2) Major species: HF and H2O.
acid is strong or
+
(3) Which provides the most H ions?
weak, and then
Ka = 7.2 x 10-4 and Kw = 1.0 x 10-14
solve for [H+].
So HF provides more H+ ions.
(4) Equilibrium we’re investigating:
HF (aq)
H+ (aq) + F- (aq)
So, Ka = 7.2 x 10-4 = [H+][F-]/[HF]

Example #2
(5) Set up ICE table:
HF (aq)
H+ (aq) + F- (aq)
I 1.00
0
0
C -x
+x
+x
E 1.00-x
x
x
Simplification: since Ka is
7.2 x 10-4 = (x)(x) small, can assume the initial
[HF] won’t change noticeably,
(1.00-x) so at equilibrium [HF] ≈ 1.00.
So: 7.2 x 10-4 = x2 Solve for x: x = 0.027
1.00
Example #2
(5) Verify assumption was OK- x divided by the
initial concentration of HF must be less than
or equal to 5%:
0.027 x 100% = 2.7%
1.00
Since 2.7% < 5%, the assumption that x was
small enough to be neglected was valid.
(6) Calculate [H+] and then the pH:
[H+] = x = 0.027M
pH = -log(0.027) = 1.57
Practice Problem #1

Calculate the pH of a 0.100M solution of
hypochlorous acid (HOCl). The Ka
value = 3.5 x 10-8.

[H+] = 5.9 x 10-5M, so pH = 4.23
A mixture of Weak Acids
 The
process is the same.
 Determine the major species.
 The stronger will predominate
(whichever has the largest Ka
value).
 Doubt you’ll see this on the AP
exam, but just in case!
Example




Calculate the pH of a solution that contains
1.00MHCN whose Ka = 6.2x10-10 and
5.00MHNO2 whose Ka = 4.0x10-4.
Approach: same as before, now just need to
consider THREE K values (the two above and
water).
Since Ka for HNO2 is much larger than Kw
and the Ka for HCN, this is the only one that
needs to be used for finding [H+].
[H+] = 4.5 x 10-2, so pH = 1.35
AP Practice Question
A 0.1M solution of acetic acid
(CH3COOH) has a pH of about:
a)1
b)3
c)7
d)10
You can answer this question without
doing any math!
AP Practice Question
What is the ionization constant, Ka, for a
weak monoprotic acid if a 0.30M solution
has a pH of 4.0?
a) 3.3 x 10-8
b) 4.7 x 10-2
c) 1.7 x 10-6
d) 3.0 x 10-4
Solve without using a calculator:
-Estimate [H+]: pH = 4.0, so [H+] = 1 x 10-4
-Ka = (1 x 10-4)2/0.30  ÷1 = 1 x 10-8 so Ka
should be a little bigger
When Kw Matters…
Percent Dissociation

=

% dissociation = 0.42%
[H+] at
equilibrium
amount dissociated (M) x 100
initial concentration (M)
 Example: Calculate the % dissociation
of 1.00 M acetic acid, Ka = 1.8 x 10-5.
 Approach is the same as weak acid
problems! Solve like an equilibrium
problem for the necessary concentrations,
then calculate % dissociation.
Sections 2-3 Homework Problems
Pg. 673 #17, 22, 40, 43, 47, 51, 57
Section 14.7
Polyprotic Acids
Polyprotic acids
Always dissociate stepwise.
+
 The first H comes off much easier than
the second.
 Ka for the first step is much bigger than
Ka for the second, the second is bigger
than the third, etc.
– More difficult to lose the next H+ because
the negative charge increases.
 Denoted Ka1, Ka2, Ka3.

Polyprotic acid

H2CO3
H+ + HCO3- Ka1= 4.3 x 10-7

HCO3-
H+ + CO3-2
Ka2= 4.3 x 10-10
Conjugate base in first step is the acid in
second.
 In calculations we can normally ignore
the second, third, etc. dissociation.

Sulfuric Acid is Special
In the first step it is a strong acid.
– No Ka value given- complete
dissociation.
 Second step is a weak acid.
– Ka2 = 1.2 x 10-2
– Small, but not always small enough to
ignore.
– If the initial concentration of H2SO4 is
low enough, the second H+ impacts pH!

Sulfuric Acid Example
Calculate the pH of a 0.0100M H2SO4
solution. Ka2 = 0.012.
 The first acidic proton (H+) dissociates
completely (H2SO4 is strong at first):
H2SO4  H+ + HSO4Thus: [H2SO4] = [H+] = [HSO4-] = 0.0100M
 Now, we need to consider the second
acidic proton.

Sulfuric Acid Example
The second acidic proton comes from
HSO4-, which is a weak acid. So this is
treated like an equilibrium problem:
HSO4H+ + SO4-2
I 0.0100
0.0100
0
C -x
+x
+x
E 0.0100-x
0.0100+x
x
 Plug into Ka2 expression:
0.012 = (x)(0.0100+x)/(0.0100-x)

Sulfuric Acid Example





We can try to use the simplification that x
is negligible with respect to HSO4- and H+:
0.012 = (x)(0.0100)/(0.0100),
however the value for x = 0.012, which
does not make sense. Thus the
simplification is not valid!
Use quadratic formula to solve!
x = 0.0045
[H+] = 0.0100+0.0045 = 0.0145M
pH = 1.84
Notice that most Ka1 values are significantly
larger than the Ka2 values, and thus typically
do not impact the [H+] or the pH. Typically,
exceptions are sulfuric acid and oxalic acid.
Section 7 Homework
Pg. 676 #93, 96, 97, 98
Lesson Essential Question:
How do calculations with
acids differ from calculations
with bases?
Section 14.6
Bases
Bases
The OH- is a strong base.
 Hydroxides of the alkali metals are
strong bases because they dissociate
completely when dissolved.
 The hydroxides of alkaline earth metals
(Ca(OH)2, etc.) are also strong, but they
don’t dissolve well in water.
 Used as antacids because [OH ] can’t
build up.

Bases
without OH
Bases are proton acceptors:
 NH3 + H2O
NH4+ + OH It is the lone pair on N that accepts the
proton.
• Many weak bases contain N
 Use Kb instead:
 B(aq) + H2O(l)
HB+(aq) + OH- (aq)


Kb = [HB+][OH- ]
[B]
Strength of Bases
 Hydroxides
are strong.
 Others are weak.
 Smaller Kb = weaker base.
Strong Base Example
Calculate the pH of a 0.050M
solution NaOH.
Treat it like a strong acid
problem, just use pOH.
pOH = 1.3; pH = 14 – 1.3 = 12.7
Weak Base Example
Calculate the pH for a 1.0M
solution of CH3NH2, whose
Kb=1.8x10-5.
Treat it like a weak acid problem,
just with OH- instead of H+.
[OH-] = 0.016M; pOH = 1.8;
pH = 12.2
Another Important Relationship
For conjugate acid-base pairs,
Ka x Kb = Kw.
This can be particularly useful if
you’re dealing with a weak base and
only the Ka value of the conjugate acid
is known.
The Kb for the base can be calculated
and the equilibrium concentrations
can then be found.
Example
0.400M of the acetate ion is
dissolved in solution. What is the
pH of the solution? Ka = 1.8x10-5
for acetic acid.
Write the equilibrium equation.
Then find Kb for acetate and solve.
[OH-] = 1.5x10-5 M; pH = 9.18
Section 14.8
Acid-Base Properties of Salts
Salts Can Affect pH
Salts are ionic compounds.
 Most salts are soluble in water, and
completely dissociate.
 Depending upon the ions making up
the salt, they COULD affect the pH
once dissociated.

Neutral Salts
Salts whose ions could react with water
to produce a strong acid or base DO
NOT effect the pH.
– Ex: Add NaCl to water. Two possible
reactions:
Strong acids
Na+ + H2O
NaOH + H+ and bases
dissociate
Cl + H2O
HCl + OH
completely
 Both possible reactions favor the
reactants, meaning no H+ or OH- is
produced, so pH remains neutral.

Acidic Salts
If the cation of a salt reacts with water
to produce a weak base and H+, the
solution is acidic.
 Ex: NH4Cl has two possible reactions:
 We already know Cl- has no effect.
 NH4+
NH3 + H+
 NH3 is a weak base, so equilibrium does
NOT favor the reactants; thus the
formation of H+ is significant, and pH is
effected.

Basic Salts
If the anion of a salt reacts with water to
produce a weak acid and OH-, the
solution is basic.
 Ex: NaF has two possible reactions:
 We already know Na+ has no effect.
 F- + H2O
HF + OH HF is a weak acid, so equilibrium does
NOT favor the reactants; thus the
formation of OH- is significant, and pH
is effected.

Acidic & Basic Salts



It is possible that a salt could contain a
cation that forms a weak base and an
anion that forms a weak acid.
If this is the case, Ka and Kb must be
compared.
The larger value dictates the pH. If Ka is
larger, the solution is acidic. If Kb is larger,
the solution is basic. If both are equal, the
pH is neutral.
– Ex: NH4CN in water: Ka & Kb = 5.6 x 10-10
Practice
Given the following salts, predict what
the pH would be if they were added to
water (acidic, basic, or neutral).
1)KNO3
2)Na2CO3
3)NH4Br
neutral
basic
(note: HBr = strong) acidic
4)NH4F (note: HF = weak)
Comparison of Ka and Kb needed.
Sample Problem:
Calculate the pH of a 0.10M NH4Cl
solution. Kb = 1.8 x 10-5 for NH3.
 Only NH4+ contributes to pH change:
NH4+
NH3 + H+
 Need Ka because NH4+ is an acid!
 Kw = Ka x Kb for a conjugate acid/base
pair.
 So: Ka = (1.0x10-14)/(1.8x10-5)
Ka = 5.6x10-10

Sample Problem:
Then, treat calculations like an
equilibrium problem:
NH4+
NH3 + H+
I 0.10
0
0
C -x
+x +x
E 0.10-x
x
x
 5.6 x 10-10 = (x)(x)
0.10-x
5.6 x 10-10 = x2/0.10 => x = 7.5x10-6

Sample Problem:
Simplification: 0.10 – x ≈ 0.10
5.6 x 10-10 = x2/0.10 => x = 7.5x10-6
 Verify:
7.5x10-6 x 100 = 0.0075%
0.10
 Verification is acceptable.
 [H+] = 7.5x10-6 M
 pH = 5.12

Section 8 Homework
Pg. 676 #99, 101, 104, 111
Section 15.2-15.3
Buffers
Buffers

Solutions that maintain a constant pH
when an acid or base is added.
– In other words, change in pH is resisted.
One species present can consume/react
with H+ and another species present
can consume OH-.
 Very useful and important in many
ways, especially for life.

– Blood contains buffers to maintain a constant
pH!
Buffers

Typically achieved by mixing a weak
acid with its conjugate base, or a weak
base with its conjugate acid.
– The conjugate acid or base is usually added as a
salt.

Examples:
– Acetic acid mixed with sodium acetate.
– Ammonia mixed with ammonium chloride.
• Notice these species will not effect the pH.
Acetic Acid & Sodium Acetate
HC2H3O2  H+ + C2H3O2 Sodium acetate completely dissolves.
 Addition of acetate ion shifts
equilibrium to the left.
 At this point there is a lot of acetic acid
and the acetate ion.
 Acetic acid (and H+) can neutralize any
base added, and the acetate ion can
neutralize any acid added.

Ammonia & Ammonium
 NH3 + H2O 
NH4+ + OHChloride

Ammonium chloride completely dissolves.
Addition of ammonium ion shifts
equilibrium to the left.
 At this point there is a lot of ammonia
and the ammonium ion.
 Ammonia (and OH-) can neutralize any
acid added, and the ammonium ion can
neutralize any base added.

Why Weak Acids & Bases?




You probably noticed that only weak acids
and bases are used for buffer solutions.
Why is this so?
Consider the strong acid HCl and adding
NaCl to it.
HCl completely dissociates; no equilibrium:
HCl  H+ + ClAddition of NaCl doesn’t cause any shift in
equilibrium, and only acid (H+) remains.
This could only neutralize the addition of a
base, not the addition of an acid.
Why Weak Acids & Bases?
 So…
EQUILIBRIUM THAT
EXISTS FOR WEAK ACIDS AND
BASES IS KEY FOR
DEVELOPING A SUCCESSFUL
BUFFER!
 It’s the only way for both acid &
base to exist together!
pH & pOH of Buffers

Henderson-Hasselbalch equation can be
used to find pH or pOH of buffers.
– Must know Ka or Kb and initial concentrations
of the acid/conjugate base or base/conjugate
acid.
pH = pKa + log([A-]/[HA])
 pOH = pKb + log([HB+]/[B])

– Note: pKa or pKb means you take –log of the K
value.
pH & pOH of Buffers
pH = pKa + log([A-]/[HA])
 pOH = pKb + log([HB+]/[B])

– Also notice above that the desired pH or pOH
can be ‘fine-tuned’ by adjusting the ratio of
weak acid/base to its conjugate base/acid.
– Ex: the more A- you add, and the less HA you
add, the higher the buffer pH will be.

Can also find pH as we’ve been doing.
– Set up as an equilibrium problem and solve.
AP Practice Question
The following questions refer to aqueous
solutions containing 1:1 mole ratios of the
following pairs of substances. Assume all
concentrations are 1M.
a) NH3 and H3CCOOH (acetic acid)
b) KOH and NH3
c) HCl and KCl
d) H3PO4 and KH2PO4
e) NH3 and NH4Cl
AP Practice Question Cont.
The solution with the highest pH.
a) NH3 and H3CCOOH (acetic acid)
b) KOH and NH3
c) HCl and KCl
d) H3PO4 and KH2PO4
e) NH3 and NH4Cl
AP Practice Question Cont.
The solution with the lowest pH.
a) NH3 and H3CCOOH (acetic acid)
b) KOH and NH3
c) HCl and KCl
d) H3PO4 and KH2PO4
e) NH3 and NH4Cl
AP Practice Question Cont.
The solution with the pH closest to
neutral.
a) NH3 and H3CCOOH (acetic acid)
b) KOH and NH3
c) HCl and KCl
d) H3PO4 and KH2PO4
e) NH3 and NH4Cl
AP Practice Question Cont.
A buffer at an alkaline pH.
a) NH3 and H3CCOOH (acetic acid)
b) KOH and NH3
c) HCl and KCl
d) H3PO4 and KH2PO4
e) NH3 and NH4Cl
AP Practice Question Cont.
A buffer at an acidic pH.
a) NH3 and H3CCOOH (acetic acid)
b) KOH and NH3
c) HCl and KCl
d) H3PO4 and KH2PO4
e) NH3 and NH4Cl
Buffer Calculation Example
What is the pH of a solution that has
2.00mol NH3 and 3.00mol NH4Cl in 1.00L
of solution? The Kb for NH3 = 1.81 x 10-5.
Henderson-Hasselbalch:
pOH = -log(1.81 x 10-5) + log(3.00/2.00)
pOH = 4.918, so pH = 14 – 4.918 = 9.082
* Can verify answer by solving as an
equilibrium problem. Write equilibrium for
NH3 producing NH4+ since you’re given Kb.
Buffer Calculation Practice
What is the pH of a solution of 0.75M
lactic acid (Ka = 1.4 x 10-4) and 0.25M
sodium lactate? Lactic acid (HC3H5O3) is
often a component of of biologic systems
(such as milk and muscle tissue during
exercise).
Use Henderson-Hasselbalch; answer:
pH = 3.38
Homework

Pg. 740 # 21 & 23
Buffer Capacity
More concentrated acid-base pairs =
more acid or base can be neutralized =
less change in pH.
 This is the buffer capacity: how well a
buffer solution can maintain a constant
pH.

Making Buffers
To make a buffer with an acidic pH:
weak acid + salt containing conjugate
base.
 To make a buffer with a basic pH:
weak base + salt containing conjugate
acid.
 When preparing a buffer solution you
must choose an acid whose pKa value is
close to the desired pH.

– Look at exponent for easy estimation!
AP Practice Question
What is a solution with an initial H3PO2
concentration of 1M and an initial
KH2PO2 concentration of 1M?
a) solution with pH > 7, which is a buffer
b) solution with pH < 7, which isn’t a buffer
c) solution with pH < 7, which is a buffer
d) solution with pH > 7, which isn’t a buffer
AP Practice Question
What is a solution with an initial KCOOH
concentration of 1M and an initial
KH2PO2 concentration of 1M?
a) solution with pH > 7, which is a buffer
b) solution with pH < 7, which isn’t a buffer
c) solution with pH < 7, which is a buffer
d) solution with pH > 7, which isn’t a buffer
AP Practice Question
Using the following information, choose the
best answer for preparing a pH = 8 buffer.
H3PO4: Ka = 7.2 x 10-3
H2PO4-: Ka = 6.3 x 10-8
HPO4-2: Ka = 4.2 x 10-13
a) K2HPO4 + KH2PO4
b) H3PO4
c) K2HPO4 + K3PO4
d) K3PO4
Adding Acid/Base to a Buffer
Calculate the pH of a solution when 0.10mol of
gaseous HCl is added to 1.0L of a solution
containing 0.25M NH3 and 0.40M NH4Cl.
Acid
is being added, so the base NH3 will react
with it.
When an acid reacts with a base, the reaction
goes to completion: all dissociated H+ reacts
with NH3.
Use stoichiometry to calculate moles of species
left after the reaction.
Adding Acid/Base to a Buffer
Calculate the pH of a solution when 0.10mol of
gaseous HCl is added to 1.0L of a solution
containing 0.25M NH3 and 0.40M NH4Cl.
Set
up an ‘IF’ table (for a reaction that goes to
completion- initial and final values).
– Don’t have to do this; IF table just helps set up your
given information to solve the problem.
NH3
+ H+

I 0.25mol
0.10mol
F
NH4+
0.40mol
Must be in
moles for
stoichiometry
!
Adding Acid/Base to a Buffer
Calculate the pH of a solution when 0.10mol of
gaseous HCl is added to 1.0L of a solution
containing 0.25M NH3 and 0.40M NH4Cl.
Once
you have the initial moles, determine
which reactant is limiting (usually the added
strong acid or base). Then find moles of all
species remaining using stoichiometry.
NH3
+ H+

NH4+
I 0.25mol
0.10mol LR 0.40mol
F 0.15mol
0mol
0.50mol
Adding Acid/Base to a Buffer
Calculate the pH of a solution when 0.10mol of
gaseous HCl is added to 1.0L of a solution
containing 0.25M NH3 and 0.40M NH4Cl.
Use
final mole values to find final
concentrations: [NH3] = 0.15M; [NH4+] = 0.50M
Then use final concentrations of acid and base
left in the Henderson-Hasselbalch equation.
Note: Kb for NH3 = 1.8 x 10-5
pOH = -log(1.8 x 10-5) + log(0.50/0.15) = 5.27
pH = 14 – 5.27 = 8.73
pH of original buffer solution = 9.05. Not
much change!
Adding Acid/Base to a Buffer
Calculate the pH of a solution when 0.10mol of
gaseous HCl is added to 1.0L of a solution
containing 0.25M NH3 and 0.40M NH4Cl.
Alternative
calculation: instead of using Kb, you
could also have used Ka.
Ka = (1 x 10-14)/(1.8 x 10-5) = 5.6 x 10-10
pH = -log(5.6 x 10-10) + log(0.15/0.50) = 8.73
Note the change in the log taken of the ratio!
Note also that the pH is the same!
Practice
What is the pH of a solution containing 5.00M
HC2H3O2 and 5.00M NaC2H3O2 when 0.010mol
of HCl gas is added to 1.0L of this solution?
Ka = 1.8 x 10-5 for acetic acid.
*Reaction: C2H3O2- + H+  HC2H3O2
*Set up IF table! Find final concentrations.
LR = H+ added
Final concentrations: [C2H3O2-] = 4.99M and
[HC2H3O2] = 5.01M
Final pH = 4.74
Homework

Pg. 741 #27 & 29 (d) only