




Acid/base properties
Definitions

› Polyprotic Acids
› Amphoteric
› Anhydrides
› Arrhenius
› Bronsted-Lowry
› Lewis

› Neutralization
› Sulfides
› Carbonates


Acid-Base Reactions
pH/pOH Scale
Acid/base strength
› Factor affecting
› K a, K b , K w
› Percent ionization
Vocabulary
Acids/Bases & Salts
› Determine acidity
› Calculations
Common Ion Effect
Buffers
› Henderson-Hasselbalch

Titration
› Indicators
› 4 types of curves
Sour taste
 React with active metals to produce hydrogen gas
 Change the color of acid-base indicators
 React with bases to produce salt and water
 Conduct an electric current (electrolytes)
 Turn litmus paper red


Sulfuric Acid
› Car batteries; production of metals, paints, dyes,
detergents

Nitric Acid
› Explosives, pharmaceuticals, rubber, plastics, dyes

Phosphoric acid
› Soda, fertilizers, animal feed, detergents

Hydrochloric Acid
› Stomach acid, cleaning metals, found in hardware stores
(muriatic acid)

Acetic Acid
› Vinegar, food supplements, fungicide

Citric Acid
› Fruit juices

Binary acids
› Contain only two different elements
› Name as “hydro - ic acid”
› Example: HCl (hydrochloric acid)

Oxyacids
› Acid consisting of hydrogen and a polyatomic
anion that contains oxygen (oxyanion)
› To name, drop ending of polyatomic ion and ad
“- ic acid”
› Example: HNO3 (nitric acid)
Taste bitter
 Feel slippery
 Change the color of
acid-base indicators
 React with bases to
produce salt and
water
 Conduct an electric
current (electrolytes)
 Turn litmus paper blue


Ammonium hydroxide
› Household cleaners, window cleaner

Ammonia
› (Gas) inhalant to revive unconscious person

Sodium bicarbonate (baking soda)
› Acid neutralizers in acid spills
› Antacids for upset stomachs

Sodium hydroxide
› Drain cleaner (drano), oven cleaner, production of
soap

Magnesium hydroxide
› Antacids, milk of magnesia, laxatives

Acid
› Substance that ionizes in water and
produces H+ ions
› Example: HCl  H+ + Cl
Base
› Substance that ionizes in water and
produces OH- ions
› Example: NaOH  Na+ + OH-

Acid
› Substance that is capable of donating a
proton (H+ ion)

Base
› Substance that is capable of accepting a
proton (H+ ion)
HC2H3O2 + H2O ↔ C2H3O2- + H3O+
Acids: HC2H3O2 and H3O+
Bases: H2O and C2H3O2NH3 + H2O ↔ NH4+ + OHAcids: H2O and NH4+
Bases: NH3 and OHNotice that water can act as an acid or a base

Conjugate Pair – a BL acid/base
pair(one with H+ and one without H+)
› Examples:
 HC2H3O2 and C2H3O2 H3O+ and H2O
 H2O and OH NH4+ and NH3

The more easily a substance gives up a
proton, the less easily the conjugate base
accepts a proton (and vice versa)
› The stronger the acid, the weaker the conjugate
base
› The stronger the base, the weaker the
conjugate acid

Neutralization
› Salt + water

Sulfides
› Salt + sulfide gas

Carbonates
› Salt + CO2 + H2O

Solution of an acid and solution of a base are mixed
Products have no characteristics of either the acid
or the base

Acid + Base (metal hydroxide)  salt + water

› Salt comes from cation of base and anion of acid
HY + XOH  XY + H2O
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)

Acid reacts with a sulfide
Gaseous product (H2S) has a foul odor (rotten eggs)

Acid + metal sulfide  salt + hydrogen sulfide

› Salt comes from cation of sulfide and anion of acid
HY + XS  XY + H2S
HCl(aq) + Na2S(aq)  NaCl(aq) + H2S(g)

Carbonates and bicarbonates react with
acids
HY + XHCO3  XY + H2CO3
H2CO3 is not stable so breaks into H2O and CO2
Then HY + XHCO3  XY + H2O + CO2
HCl(aq) + NaHCO3(aq)  NaCl(aq) + H2CO3 (aq)
HCl(aq) + NaHCO3(aq)  NaCl(aq) + H2O(l) + CO2(g)

pH scale: measures concentration of
hydrogen ions in solution
pH = -log[H+]

therefore
[H+] = 10-pH
Example: What is the pH of a solution with a
[H+] of 1.4×10-5?
pH = -log[1.4×10-5] = 4.9
Acids: pH < 7
Neutral: pH = 7
Bases: pH > 7
Increasing [H+] means
decreasing pH
Increasing pH means
decreasing [H+]

pOH scale: measures concentration of
hydroxide ions in solution
pOH = -log[OH-] therefore [OH-] = 10-pOH

Example: What is the [OH-] of a solution with
a pOH of 6.2?
[OH-] = 10-6.2 = 6.3×10-7
pH + pOH = 14

An acid has a pH of 4, what is the pOH?
4 + pOH = 14
pOH = 10

pH meter
› Electrodes measure [H+]

Acid-base indicators
› Change color in presence of acid or
base (or certain pH ranges)
› Litmus paper, phenolphthalein,
cabbage juice, methyl orange, thymol
blue…

Completely ionize in solution (strong electrolytes)
Strong Acids
Strong Bases
HCl
Group 1 metals + OH
HBr
(LiOH, NaOH, KOH,…)
HI
HClO3
Heavy group 2 metals + OH
HClO4
Ca(OH)2, Sr(OH)2, Ba(OH)2
HNO3
H2SO4
If acid/base is not on this list, it is a weak acid/base

Do not completely ionize in water (weak
electrolytes)

Common weak acids:
› HF, acids with -COOH group

Common weak bases:
› NH3
1.
Electronegativity of element bonded to H
› Binary acids
› More electronegative bond = stronger acid
› Example: HCl stronger than HBr
2.
Bond Strength
› Stronger bonds do not allow hydrogen to dissociate as
easily
› Reason why HF is not a strong acid (F is most
electronegative, but H-F bond is strongest bond)
3.
Stability of Conjugate base
› More stable the conjugate base, the stronger the acid
4.
For polyatomic ions, the more electronegative the
nonmetal, the stronger the acid (when comparing
acids with same number of O atoms)
› Example: HClO3 is stronger than HBrO3
5.
For polyatomic ions, when nonmetal is the same,
the more O atoms, the stronger the acid
› Example: HClO3 is stronger than HClO2

Tells us what percent of an acid (or
base) is ionized in water
› Helps determine the strength of an acid (or
base)
Percent Ionization =
[H+] at equilibrium
Initial Acid
Concentration
×100

A 0.035 M solution of HNO2 contains
3.7×10-3 M H+(aq). Calculate the
percent ionization.
=
3.7×10-3 M
0.035 M
= 11%
This means that 11% of the acid will dissociate in water.
HA(aq) + H2O(l)  H+(aq) + A-(aq)
acid



water
proton conjugate base
Strong acids dissociate completely
The dissociation is not reversible
The acid is the only significant source of H+ ions, so
pH can be calculated directly from the [H+]
› Example: A 0.20 M solution of HNO3 has an [H+] of 0.20 M
› pH = -log[H+]



Strong bases dissociate completely
The dissociation is not reversible
The base is the only significant source of OH- ions,
so pOH can be calculated directly from the [OH-]
› Example: A 0.30 M solution of NaOH has a [OH-] of 0.30 M
› pOH = -log[OH-]
Do not dissociate completely
 Reversible reactions
 Need to use equilibrium to solve for [H+]

K=
[Products]
[Reactants]
HA(aq) + H2O(l)  H+(aq) + A-(aq)
acid
water
proton conjugate base

Write the equilibrium expression for the
acid dissociation constant, Ka.


[ H ][ A ]
Ka 
[ HA]
B(aq) + H2O(l) ↔ BH+(aq) + OH-(aq)
base

water
conjugate hydroxide
acid
ion
Write the equilibrium expression for the
base dissociation constant, Kb.


[ BH ][OH ]
Kb 
[ B]
The greater the Ka, the stronger the acid
 The smaller the Ka, the weaker the acid

The greater the Kb, the stronger the base
 The smaller the Kb, the weaker the base


What is the pH of a 0.50 M solution of
acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
HC2H3O2  C2H3O2- + H+

Step #1: Write the dissociation equation

What is the pH of a 0.50 M solution of
acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?

Step #2: ICE
I
C
E
HC2H3O2  C2H3O2- + H+
0.50
-x
0.50 - x
0
0
+x
+x
x
x

What is the pH of a 0.50 M solution of
acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?

Step #3: Set up the equilibrium expression
2
( x)(x)
x
1.8 x10 

(0.50  x) (0.50)
5
3
x  3.0 x10
If percent ionization is less than 5%, you can ignore using the quadratic.

What is the pH of a 0.50 M solution of
acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?

Step #5: Solve for pH
3
pH   log(3.0 x10 )  2.52
You can use the Kb expression to solve for pOH using the same method!

A student prepared a 0.10 M solution of formic acid
(HCOOH) and measured its pH. The pH at 25°C
was found to be 2.38. Calculate the Ka for formic
acid at this temperature.

Step #1: Solve for [H+] from pH
[H+] = 10-2.38 = 4.2×10-3 M


A student prepared a 0.10 M solution of formic acid (HCOOH)
and measured its pH. The pH at 25°C was found to be 2.38.
Calculate the Ka for formic acid at this temperature.
Step #2: Set up ICE table
HCOOH(aq)  HCOO- +
I
C
E
0.10
0
H+
0
4.2×10-3


A student prepared a 0.10 M solution of formic acid (HCOOH)
and measured its pH. The pH at 25°C was found to be 2.38.
Calculate the Ka for formic acid at this temperature.
Step #3: Use stoichiometry to complete table
HCOOH(aq)  HCOO-
+
H+
I
C
0.10
0
0
4.2×10-3
4.2×10-3
4.2×10-3
E
0.0096
4.2×10-3
4.2×10-3

A student prepared a 0.10 M solution of formic acid
(HCOOH) and measured its pH. The pH at 25°C
was found to be 2.38. Calculate the Ka for formic
acid at this temperature.

Step #4: Solve for Ka using equilibrium
expression
Ka = 1.8×10-4
The larger the value of Ka, the
stronger the acid

According to Bronsted Lowry, H2O can act as
either an acid or a base

Auto-ionization: One water molecule can donate
a proton to another water molecule
› Extremely rapid reaction and no molecule remains ionized
for long
› At room temperature 1 out of every 109 molecule are
ionized at a given instant
› Water is a nonelectrolyte and consists almost entirely of
H2O molecules
H2O(l) + H2O(l) ↔ H3O+ + OH-


H2O(l) + H2O(l) ↔ H3O+ + OH-
Auto-ionization of water is an equilibrium
process (use Kw - ion product constant)
Kw = [H3O+][OH-]
Also written as Kw = [H+][OH-]

At 25°C, Kw =1.4×10-14
1.4×10-14 = [H+][OH-]

In basic solutions, [OH-] > [H+]

In acidic solutions, [H+] > [OH-]

In neutral solutions, [H+] = [OH-]
1.4×10-14 = [H+][OH-]

If the concentration of one ion is known, you can
solve for the concentration of the other ion
Example: Calculate the concentration of H+ in a
solution in which the concentration of OH- is
0.010M.
1.4×10-14 = [H+][0.010]
[H+] = 1.0×10-12 M
 Acid
or base dissociation constants
are sometimes expressed as pKa
and Kb.
› pKa = –logKa
› pKb = -logKb
pKw = 14 = pKa+ pKb

Acids with more than one ionizable H+ ion
H2SO3(aq) ↔ H+(aq) + HSO3-(aq)
HSO3-(aq) ↔ H+(aq) + SO3-2(aq)
Ka1 = 1.7×10-2
Ka2 = 6.4×10-8
The acid-dissociation constants are Ka1, Ka2,
etc…
 The first proton is most easily removed

› As protons are removed, it becomes more and
more difficult to remove protons
› Ka1>Ka2>Ka3….

To calculate the overall K for the reaction,
treat it as a multi-step equilibrium
H2SO3(aq) ↔ H+(aq) + HSO3-(aq)
HSO3-(aq) ↔ H+(aq) + SO3-2(aq)
Ka1 = 1.7×10-2
Ka2 = 6.4×10-8
Overall reaction…
H2SO3(aq) ↔ 2H+(aq) + SO3-2(aq)
Ka = (1.7×10-2)(6.4×10-8) = 1.1×10-9
Ka = Ka1 × Ka2

Strong acids (ex H2SO4) completely ionize
with the first step
› pH can be calculated by treating the acid as if it
were a monoprotic acid (one ionizable
hydrogen)

If Ka values differ by a factor of 103 or more,
acids can be treated as monoprotic
Monoprotic acid = 1 ionizable H+
 Diprotic acid = 2 ionizable H +
 Triprotic acid = 3 ionizable H +
 Etc…


Substances that can act as either acids
or bases
› Example: H2O
 Can give up H+ to become OH  Can receive H + to become H3O +
› Example: H2PO4 Can give up H + to become HPO4-2
 Can receive H + to become H3PO4

Acid Anhydride: combines with water to
form an acid
› CO2 + H2O  H2CO3
› SO3 + H2O  H + + HSO4
Base Anhydride: combines with water to
form a base
› CaO + H2O  Ca(OH)2
› Na2O + H2O  2Na+ + 2OH-
#1
If a salt is composed of the conjugates of a
strong acid and strong base, the solution will be
neutral.
#2
If a salt is composed of the conjugates of a weak
base and a strong acid, its solution will be acidic.
#3
If a salt is composed of the conjugates of a strong
base and a weak acid, its solution will be basic.
#4
If a salt is composed of the conjugates of a weak
base and a weak acid, the pH of its solution will
depend on the relative strengths of the conjugate
acid and base of the specific ions in the salt.

If a salt is composed of the conjugates of a
strong acid and strong base, the solution
will be neutral.
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)

The Na+ and Cl- ions do not further ionize in
water

If a salt is composed of the conjugates of a
strong acid and weak base, the solution will
be acidic.
HCl(aq) + NH3(aq)  NH4Cl(s)

The NH4Cl ionizes in water to produce some
H+ ions (the Cl - ions do not react in water)
NH4Cl(s)  NH4+(aq) + Cl-(aq)
NH4+ + H2O  NH3+ + H+
Solution
Is acidic

If a salt is composed of the conjugates of a weak
acid and strong base, the solution will be basic.
NaOH(aq) + HC2H3O2(aq)  NaC2H3O2(s) + H2O(l)

The NaC2H3O2 ionizes in water
NaC2H3O2(s)  Na+(aq) + C2H3O2-(aq)

The Na+ ions do not react at all, but the C2H3O2ions react to produce OH- in solution
C2H3O2-(aq) + H2O  HC2H3O2 + OH- Solution
Is basic

If a salt is composed of the conjugates of a
weak acid and weak base, the pH of the
solution will depend on the relative
strengths of the conjugate acid and base
of the specific ions in the salt.
› The ion with the larger equilibrium constant (Ka or
Kb) will have the greater influence on pH)

Hydrolysis
› When the ions react with water to produce
hydrogen or hydroxide ions in solution
 Example: Strong Acid/Weak Base
 Example: Strong Acid/Strong Base
Solve similar to weak acid/weak base
problems
 Solve for Ka or Kb depending on if the salt
makes an acidic (Ka) or basic (Kb) solution

› Use the ion of the strong acid or strong base for
equilibrium equation

What is the pH of a 0.10 M solution of
acetic acid, NaC2H3O2. The Ka of
HC2H3O2 is 1.8 x 10-5 .
C2H3O2- + H2O  HC2H3O2 + OH
Step #1: We will use Kb because the
solution will be basic. First find the Kb for
HC2H3O2.

What is the pH of a 0.10 M solution of
acetic acid, NaC2H3O2. The Ka of
HC2H3O2 is 1.8 x 10-5 .
C2H3O2- + H2O  HC2H3O2 + OHKw = K aKb
1.0 × 10-14 = (1.8x10-5)(Kb)
Kb = 5.6 × 10-10

Step #2: Write the equilibrium expression.
C2H3O2- + H2O  HC2H3O2 + OH-
Kb =
[HC2H3O2][OH-]
[C2H3O2-]

What is the pH of a 0.10 M solution of
acetic acid, NaC2H3O2. The Ka of
HC2H3O2 is 1.8 x 10-5 .

Step #3: ICE
I
C
E
C2H3O20.10
-x
0.10 - x
HC2H3O2
OH-
0
0
+x
+x
x
x

Step #4: Solve for OH-.
C2H3O2- + H2O  HC2H3O2 + OH-
5.6×10-10 =
[x][x]
[0.10-x]
=
[x][x]
[0.10]
= 7.5×10-6

Step #4: Solve for OH-.
C2H3O2- + H2O  HC2H3O2 + OH-
x = 7.5×10-6 = [OH-]

Step #5: Solve for pH.
[OH-] = 7.5×10-6
pOH = -log[OH-] = -log[7.5×10-6]
pOH = 5.1
14 = pH + pOH
14 = pH + 5.1
pH = 8.9

Use LeChatlier’s Principle to determine how the
addition of a substance will affect the equilibrium
HC2H3O2  C2H3O2 - + H+



The addition of the acetate ion (adding soluble
salt) causes the equilibrium to shift to the LEFT
The hydrogen concentration will DECREASE
The acetic acid ionizes less with the addition of the
acetate ion than it would alone in solution

A solution with a very stable pH
› You can add an acid or base to a buffer solution
without it greatly affecting the pH of the solution

Consists a weak acid-base conjugate pair
› Usually a weak acid or base with the salt of that
acid or base
› Resists changes in addition of strong acid or
base

Example: blood (pH of 7.4)

Buffer Capacity – the amount of acid or
base the buffer can neutralize before
the pH begins to change to an
appreciable degree

pH Range (of a buffer) – the range over
which the buffer acts effectively
 [ A ] 
 [base] 
  pKa  log

pH  pKa  log
 [acid] 
 [ HA] 
 [ BH  ] 
 [acid] 
  pKb  log

pOH  pKb  log
 [base] 
 [ B] 

Reminder: pKa = –logKa and pKb = -logKb
Given on AP Cheat Sheet!

What is the pH of a buffer solution with
concentrations of 0.20M HC2H3O2 and 0.50 M
C2H3O2-? The acid dissociation constant for
HC2H3O2 is 1.8×10-5.
 [ A ] 
 [base] 
  pKa  log

pH  pKa  log
 [acid] 
 [ HA] 
pH = -log(1.8×10-5) + log(0.50/0.20)
pH = 4.7 + 0.4 = 5.1

What is the pH of a buffer solution with
concentrations of 0.20M HC2H3O2 and 0.20 M
C2H3O2-? The acid dissociation constant for
HC2H3O2 is 1.8×10-5.
 [ A ] 
 [base] 
  pKa  log

pH  pKa  log
 [acid] 
 [ HA] 
pH = -log(1.8×10-5) + log(0.20/0.20)
pH = 4.7 + 0 = 4.7

When concentrations of acid and
conjugate base are the same, pH = pKa
(and pOH = pKb)

When you want to prepare a buffer with
a desired pH, choose an acid with a pKa
close to the desired pH.
The salt will contain the anion of the acid, and
the cation of a strong base (NaOH, KOH)
Weak Acid
Formula
of the acid
Hydrofluoric
HF
Formic
HCOOH
Benzoic
C6H5COOH
Acetic
Carbonic
Propanoic
Hydrocyanic
CH3COOH
H2CO3
HC3H5O2
HCN
Example of a salt of the
weak acid
KF – Potassium fluoride
KHCOO – Potassium formate
NaC6H5COO – Sodium benzoate
NaH3COO – Sodium acetate
NaHCO3 - Sodium bicarbonate
NaC3H5O2 - Sodium propanoate
KCN - potassium cyanide
The salt will contain the cation of the base, and
the anion of a strong acid (HCl, HNO3)
Formula of
the base
Example of a salt of the weak
acid
NH3
NH4Cl - ammonium chloride
Methylamine
CH3NH2
CH3NH3Cl – methylammonium chloride
Ethylamine
C2H5NH2
C2H5NH3NO3 - ethylammonium nitrate
Aniline
C6H5NH2
C6H5NH3Cl – aniline hydrochloride
Base
Ammonia
Pyridine
C5H5N
C5H5NHCl – pyridine hydrochloride
Example of a buffer of HC2H3O2 with NaC2H3O2
HC2H3O2 ↔ C2H3O2- + H+

When adding a strong acid to a buffer solution, the
base acts to neutralize the acid (ex: add HCl)
NaC2H3O2 + HCl ↔ HC2H3O2 + NaCl
Na+ + C2H3O2- + H+ + Cl- ↔ HC2H3O2 + Na+ + Cl-

When adding a strong base to a buffer solution,
the acid acts to neutralize the base (ex: add KOH)
HC2H3O2 + KOH ↔ KC2H3O2 + H2O
HC2H3O2 + K+ + OH- ↔ K+ + C2H3O2- + H2O

What is the pH of a buffer solution with
concentrations of 0.20M HC2H3O2 and 0.50 M
C2H3O2- after the addition of 0.10 M KOH? The acid
dissociation constant for HC2H3O2 is 1.8×10-5.
The strong base is neutralized by the acid:
HC2H3O2 + KOH  C2H3O2 - + H+
0.20
- 0.10
0.10
0.10
- 0.10
0
0.50
+ 0.10
0.60
New conc. for
acid and salt

What is the pH of a buffer solution with
concentrations of 0.20M HC2H3O2 and 0.50 M
C2H3O2- after the addition of 0.10 M KOH? The acid
dissociation constant for HC2H3O2 is 1.8×10-5.
HC2H3O2 + KOH  C2H3O2 - + H+
 [ A ] 

pH  pKa  log
 [ HA] 
pH = -log(1.8×10-5) + log(0.60/0.10)
pH = 4.7 + 0.8 = 5.5

What is the pH of a buffer solution with
concentrations of 0.20M HC2H3O2 and 0.50 M
C2H3O2- after the addition of 0.05 M HBr? The acid
dissociation constant for HC2H3O2 is 1.8×10-5.
The strong acid is neutralized by the base:
C2H3O2 - + HBr  HC2H3O2 + Br-
0.50
- 0.05
0.45
0.05
- 0.05
0
0.20
+ 0.05
0.25
New conc. for
acid and salt

What is the pH of a buffer solution with
concentrations of 0.20M HC2H3O2 and 0.50 M
C2H3O2- after the addition of 0.10 M KOH? The acid
dissociation constant for HC2H3O2 is 1.8×10-5.
HC2H3O2 + KOH  C2H3O2 - + H+
 [ A ] 

pH  pKa  log
 [ HA] 
pH = -log(1.8×10-5) + log(0.45/0.25)
pH = 4.7 + 0.3 = 5.0

A base of known
concentration is slowly
added to an acid of
unknown concentration (or
vice versa) to reach
neutralization

Indicators are added to the
solution that change color to
signal the equivalence point
› Equivalence point – point at
which stoichiometrically
equivalent quantities of acid
and base have been brought
together - neutralization
 Titration
Curve
graph of the pH
as acid or base
is added to the
solution

Indicators are added to the solution that
change color to signal the equivalence
point
› Equivalence point – point at which
stoichiometrically equivalent quantities of acid
and base have been brought together –
neutralization

Different indicators change at different pH
levels
Indicator
Low pH color
Gentian violet (Methyl violet 10B)
yellow
0.0–2.0
blue-violet
Leucomalachite green (first transition)
Leucomalachite green (second transition)
yellow
green
0.0–2.0
11.6–14
green
colorless
Thymol blue (first transition)
red
1.2–2.8
yellow
Thymol blue (second transition)
yellow
8.0–9.6
blue
Methyl yellow
Bromophenol blue
Congo red
Methyl orange
Bromocresol green
Methyl red
Methyl red
Azolitmin
Bromocresol purple
Bromothymol blue
Phenol red
Neutral red
Naphtholphthalein
Cresol Red
Phenolphthalein
Thymolphthalein
Alizarine Yellow R
Litmus
red
yellow
blue-violet
red
yellow
red
red
red
yellow
yellow
yellow
red
colorless to reddish
yellow
colorless
colorless
yellow
red
Transition pH range High pH color
2.9–4.0
3.0–4.6
3.0–5.0
3.1–4.4
3.8–5.4
4.4–6.2
4.5–5.2
4.5–8.3
5.2–6.8
6.0–7.6
6.8–8.4
6.8–8.0
7.3–8.7
7.2–8.8
8.3–10.0
9.3–10.5
10.2–12.0
4.5-8.3
yellow
purple
red
orange
blue
yellow
green
blue
purple
blue
red
yellow
greenish to blue
reddish-purple
fuchsia
blue
red
blue
1.
Strong acid-strong base titration
2.
Weak acid-strong base titration
3.
Strong acid-weak base titration
4.
Polyprotic acid-strong base titration
13
12
11
10
9
pH
8
7
Endpoint is
at pH 7
A solution that is
0.10 M HCl is
titrated with
0.10 M NaOH
6
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NaOH (0.10 M)
30.00
35.00
40.00
45.00
13
12
11
10
9
pH
8
Endpoint is
above pH 7
7
6
A solution that is
0.10 M CH3COOH
is titrated with
0.10 M NaOH
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NaOH (0.10 M)
30.00
35.00
40.00
45.00
13
12
11
10
9
pH
8
7
6
5
Endpoint is
below pH 7
A solution that is
0.10 M HCl is
titrated with
0.10 M NH3
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NH3 (0.10 M)
30.00
35.00
40.00
45.00

The titration curve
will have as many
bumps as there
are hydrogen ions
to give up

This curve has 2
bumps so it
represents a
diprotic acid

Remember
M1V1 = M2V2

When finding the concentration or volume
at which an acid and base neutralize each
other, we use the same calculation
MaVa = MbVb
(a = acid and b=base)