Acid/base properties Definitions › Arrhenius › Bronsted-Lowry › Lewis Acid-Base Reactions › Neutralization › Sulfides › Carbonates pH/pOH Scale.
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Acids & BasesUnit 13
Overview
Acid/base properties Definitions
› Arrhenius› Bronsted-Lowry› Lewis
Acid-Base Reactions› Neutralization› Sulfides› Carbonates
pH/pOH Scale Acid/base strength
› Factor affecting› Ka, Kb, Kw
› Percent ionization
Vocabulary› Polyprotic Acids› Amphoteric › Anhydrides
Acids/Bases & Salts› Determine acidity› Calculations
Common Ion Effect Buffers
› Henderson-Hasselbalch Titration
› Indicators› 4 types of curves
Acids 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
Common Acids
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
Acids
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)
Bases 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
Common Bases
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
Definition: Arrhenius
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-
Definition: Bronsted-Lowry
Acid› Substance that is capable of donating a
proton (H+ ion)
Base› Substance that is capable of accepting a
proton (H+ ion)
Examples: Bronsted-Lowry
HC2H3O2 + H2O ↔ C2H3O2- + H3O+
Acids: HC2H3O2 and H3O+
Bases: H2O and C2H3O2-
NH3 + H2O ↔ NH4+ + OH-
Acids: H2O and NH4+
Bases: NH3 and OH-
Notice that water can act as an acid or a base
Bronsted-Lowry
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
Bronsted-Lowry
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
Acid Base Reactions
Neutralization › Salt + water
Sulfides › Salt + sulfide gas
Carbonates › Salt + CO2 + H2O
Neutralization
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)
Sulfides
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 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)
The pH Scale
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
pH Scale
Acids: pH < 7
Neutral: pH = 7
Bases: pH > 7
Increasing [H+] means decreasing pH
Increasing pH means decreasing [H+]
The pOH Scale
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
Comparing pH and pOH
pH + pOH = 14
An acid has a pH of 4, what is the pOH?
4 + pOH = 14pOH = 10
Measuring pH
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…
Strong Acids and Bases
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
H2SO4If acid/base is not on this list, it is a weak acid/base
Weak Acids and Bases
Do not completely ionize in water (weak electrolytes)
Common weak acids:› HF, acids with -COOH group
Common weak bases:› NH3
Factors Affecting Acid Strength
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
Factors Affecting Acid Strength
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
Percent Ionization
Tells us what percent of an acid (or base) is ionized in water › Helps determine the strength of an acid (or
base)
Percent Ionization = ×100
[H+] at equilibrium
Initial Acid Concentration
Percent Ionization (Example)
A 0.035 M solution of HNO2 contains 3.7×10-3 M H+(aq). Calculate the percent ionization.
= = 11%
This means that 11% of the acid will dissociate in water.
3.7×10-3 M0.035 M
Strong Acids
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
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-]
Equilibrium Time!
Weak Acids and Bases
Do not dissociate completely Reversible reactions Need to use equilibrium to solve for
[H+]
K = [Products]
[Reactants]
Acid Dissociation
HA(aq) + H2O(l) H+(aq) + A-(aq)
acid water proton conjugate base
Write the equilibrium expression for the acid dissociation constant, Ka.
][
]][[
HA
AHKa
Base Dissociation
B(aq) + H2O(l) ↔ BH+(aq) + OH-(aq) base water conjugate hydroxide
acid ion
Write the equilibrium expression for the base dissociation constant, Kb.
[ ][ ]
[ ]b
BH OHK
B
Size of K
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
Example: Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step #1: Write the dissociation equation
HC2H3O2 C2H3O2- + H+
Example: Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step #2: ICE
HC2H3O2 C2H3O2- + H+
I
C
E
0.50 0 0
- x +x +x
0.50 - x xx
Example: Weak Acid Equilibrium Problem
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
)50.0()50.0(
))((108.1
25 x
x
xxx
If percent ionization is less than 5%, you can ignore using the quadratic.
3100.3 xx
Example: Weak Acid Equilibrium Problem
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
52.2)100.3log( 3 xpH
You can use the Kb expression to solve for pOH using the same method!
Example 2: Weak Acid Equilibrium - Solving for Ka
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
Example 2: Weak Acid Equilibrium - Solving for Ka
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- + H+
I
C
E
0.10 0 0
4.2×10-3
Example 2: Weak Acid Equilibrium - Solving for Ka
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
E
0.10 0 0
4.2×10-34.2×10-3
4.2×10-34.2×10-34.2×10-3
0.0096
Example 2: Weak Acid Equilibrium - Solving for Ka
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
Rule of Thumb
The larger the value of Ka, the stronger the acid
(Self-) Auto-ionization of Water
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-
Auto-ionization of Water
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
Auto-ionization of Water
1.4×10-14 = [H+][OH-]
In basic solutions, [OH-] > [H+]
In acidic solutions, [H+] > [OH-]
In neutral solutions, [H+] = [OH-]
Auto-ionization of Water
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
Relating pKw to pKa and pKb
Acid or base dissociation constants are sometimes expressed as pKa and Kb. › pKa = –logKa
› pKb = -logKb
pKw = 14 = pKa+ pKb
Polyprotic Acids
Acids with more than one ionizable H+ ion
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….
H2SO3(aq) ↔ H+(aq) + HSO3-(aq) Ka1 = 1.7×10-2
HSO3-(aq) ↔ H+(aq) + SO3
-2(aq) Ka2 = 6.4×10-8
Polyprotic Acids
To calculate the overall K for the reaction, treat it as a multi-step equilibrium
Overall reaction…
H2SO3(aq) ↔ H+(aq) + HSO3-(aq) Ka1 = 1.7×10-
2
HSO3-(aq) ↔ H+(aq) + SO3
-2(aq) Ka2 = 6.4×10-
8
H2SO3(aq) ↔ 2H+(aq) + SO3-2(aq) Ka = Ka1 ×
Ka2
Ka = (1.7×10-2)(6.4×10-8) = 1.1×10-9
Polyprotic Acids
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
Polyprotic Acids
Monoprotic acid = 1 ionizable H+
Diprotic acid = 2 ionizable H +
Triprotic acid = 3 ionizable H +
Etc…
Amphoteric (amphiprotic) Substances
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
Anhydrides
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-
Acids Bases and Salts#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.
Strong Acid + Strong Base
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
Strong Acid + Weak Base
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
Weak Acid + Strong Base
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 C2H3O2- ions
react to produce OH- in solution
C2H3O2-(aq) + H2O HC2H3O2 + OH-
Solution Is basic
Weak Acid + Weak Base
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)
Vocabulary
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
Acid Base Salt Calculations
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
Example (Acid Base Salt)
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 #1: We will use Kb because the solution will be basic. First find the Kb for HC2H3O2.
C2H3O2- + H2O HC2H3O2 + OH-
Example (Acid Base Salt)
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-
Kw = KaKb
1.0 × 10-14 = (1.8x10-5)(Kb)
Kb = 5.6 × 10-10
Example (Acid Base Salt)
Step #2: Write the equilibrium expression.
C2H3O2- + H2O HC2H3O2 + OH-
Kb = [HC2H3O2][OH-]
[C2H3O2-]
Example: Weak Acid Equilibrium Problem
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: ICEC2H3O2
- HC2H3O2 OH-
I
C
E
0.10 0 0
- x +x +x
0.10 - x xx
Example (Acid Base Salt)
Step #4: Solve for OH-.
C2H3O2- + H2O HC2H3O2 + OH-
5.6×10-10 = = = 7.5×10-6
[x][x]
[0.10-x]
[x][x]
[0.10]
Example (Acid Base Salt)
Step #4: Solve for OH-.
C2H3O2- + H2O HC2H3O2 + OH-
x = 7.5×10-6 = [OH-]
Example (Acid Base Salt)
Step #5: Solve for pH.
[OH-] = 7.5×10-6
pOH = -log[OH-] = -log[7.5×10-6]pOH = 5.1
14 = pH + pOH14 = pH + 5.1
pH = 8.9
Common Ion Effect
Use LeChatlier’s Principle to determine how the addition of a substance will affect the equilibrium
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
HC2H3O2 C2H3O2 - + H+
Buffers
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)
Buffers
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
Henderson-Hasselbalch Equation
Reminder: pKa = –logKa and pKb = -logKb
][
][log
][
][log
acid
basepK
HA
ApKpH aa
][
][log
][
][log
base
acidpK
B
BHpKpOH bb
Given on AP Cheat Sheet!
Example (Henderson-Hasselbalch)
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.
][
][log
][
][log
acid
basepK
HA
ApKpH aa
pH = -log(1.8×10-5) + log(0.50/0.20) pH = 4.7 + 0.4 = 5.1
Example 2(Henderson-Hasselbalch)
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.
][
][log
][
][log
acid
basepK
HA
ApKpH aa
pH = -log(1.8×10-5) + log(0.20/0.20) pH = 4.7 + 0 = 4.7
Choosing a Buffer
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.
Weak AcidFormula
of the acidExample of a salt of the
weak acid Hydrofluoric HF KF – Potassium fluoride
Formic HCOOH KHCOO – Potassium formate
Benzoic C6H5COOH NaC6H5COO – Sodium benzoate
Acetic CH3COOH NaH3COO – Sodium acetate
Carbonic H2CO3 NaHCO3 - Sodium bicarbonate
Propanoic HC3H5O2 NaC3H5O2 - Sodium propanoate
Hydrocyanic HCN KCN - potassium cyanide
The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH)
Acid/Salt Buffering Pairs
The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO3)
BaseFormula of the base
Example of a salt of the weak acid
Ammonia NH3 NH4Cl - ammonium chloride
Methylamine
CH3NH2 CH3NH3Cl – methylammonium
chloride
Ethylamine C2H5NH2 C2H5NH3NO3 - ethylammonium
nitrate
Aniline C6H5NH2 C6H5NH3Cl – aniline hydrochloride
Pyridine C5H5N C5H5NHCl – pyridine hydrochloride
Base/Salt Buffering Pairs
Buffer Solutions in ActionExample 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
Buffer Solutions in Action (#1)
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 0.50
- 0.10
0.10 0.600
+ 0.10- 0.10 New conc. for acid and salt
Buffer Solutions in Action (#1)
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+
][
][log
HA
ApKpH a
pH = -log(1.8×10-5) + log(0.60/0.10) pH = 4.7 + 0.8 = 5.5
Buffer Solutions in Action (#2)
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 0.20
- 0.05
0.45 0.250
+ 0.05- 0.05 New conc. for acid and salt
Buffer Solutions in Action (#2)
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+
][
][log
HA
ApKpH a
pH = -log(1.8×10-5) + log(0.45/0.25) pH = 4.7 + 0.3 = 5.0
Titration 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 Curves
Titration Curve graph of the pH as acid or base is added to the solution
Indicators
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 Transition pH range High pH color
Gentian violet (Methyl violet 10B) yellow 0.0–2.0 blue-violet
Leucomalachite green (first transition) yellow 0.0–2.0 green
Leucomalachite green (second transition) green 11.6–14 colorless
Thymol blue (first transition) red 1.2–2.8 yellow
Thymol blue (second transition) yellow 8.0–9.6 blue
Methyl yellow red 2.9–4.0 yellow
Bromophenol blue yellow 3.0–4.6 purple
Congo red blue-violet 3.0–5.0 red
Methyl orange red 3.1–4.4 orange
Bromocresol green yellow 3.8–5.4 blue
Methyl red red 4.4–6.2 yellow
Methyl red red 4.5–5.2 green
Azolitmin red 4.5–8.3 blue
Bromocresol purple yellow 5.2–6.8 purple
Bromothymol blue yellow 6.0–7.6 blue
Phenol red yellow 6.8–8.4 red
Neutral red red 6.8–8.0 yellow
Naphtholphthalein colorless to reddish 7.3–8.7 greenish to blue
Cresol Red yellow 7.2–8.8 reddish-purple
Phenolphthalein colorless 8.3–10.0 fuchsia
Thymolphthalein colorless 9.3–10.5 blue
Alizarine Yellow R yellow 10.2–12.0 red
Litmus red 4.5-8.3 blue
Indicators
Selection of Indicators
Titration
1. Strong acid-strong base titration
2. Weak acid-strong base titration
3. Strong acid-weak base titration
4. Polyprotic acid-strong base titration
Strong Acid - Strong Base
1
2
3
4
5
6
7
8
9
10
11
12
13
0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00
m illiliters NaOH (0.10 M)
pH
A solution that is 0.10 M HCl is titrated with 0.10 M NaOH
Endpoint is at pH 7
Weak Acid - Strong Base
1
2
3
4
5
6
7
8
9
10
11
12
13
0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00m illiliters NaOH (0.10 M)
pH
A solution that is 0.10 M CH3COOH is titrated with 0.10 M NaOH
Endpoint is above pH 7
Strong Acid – Weak Base
1
2
3
4
5
6
7
8
9
10
11
12
13
0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00
m illiliters NH3 (0.10 M)
pH
A solution that is 0.10 M HCl is titrated with 0.10 M NH3
Endpoint is below pH 7
Polyprotic Acids
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
Titration Calculations
RememberM1V1 = 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)
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