Acids and Acids and BasesBases
Acid/Base Definitions Arrhenius Model
Acids produce hydrogen ions in aqueous solutions
Bases produce hydroxide ions in aqueous solutions
Bronsted-Lowry Model Acids are proton donors Bases are proton acceptors
Lewis Acid Model Acids are electron pair acceptors Bases are electron pair donors
WhoTheory:
Acid=When
Arrhenius increases H+ 1880’s
Brønsted proton donor 1923
Lowry ditto 1923
LewisElectron-pair acceptor
1923
Three definitions of acid
Some Definitions
• Arrhenius acids and bases– Acid: Substance that, when dissolved
in water, increases the concentration of hydrogen ions (protons, H+).
– Base: Substance that, when dissolved in water, increases the concentration of hydroxide ions.
Some Definitions• Brønsted–Lowry: must have both
1. an Acid: Proton donor
and
2. a Base: Proton acceptor
If it can be either…
...it is amphiprotic.
HCO3–
HSO4 –
H2O
What Happens When an Acid Dissolves in Water?
• Water acts as a Brønsted–Lowry base and abstracts a proton (H+) from the acid.
• As a result, the conjugate base of the acid and a hydronium ion are formed.
Movies…
Conjugate Acids and Bases:• From the Latin word conjugare, meaning “to
join together.”• Reactions between acids and bases always
yield their conjugate bases and acids.
Acid and Base Strength
• Strong acids are completely dissociated in water.– Their conjugate bases are
quite weak.
• Weak acids only dissociate partially in water.– Their conjugate bases are
weak bases.
Acid and Base Strength
• Substances with negligible acidity do not dissociate in water.– Their conjugate bases are
exceedingly strong.
Acid DissociationAcid Dissociation
HA H+ + A-
Acid Proton Conjugate
base
][
]][[
HA
AHKa
Alternately, H+ may be written in its hydrated form, H3O+ (hydronium ion)
Ka > 1 = product favored
Dissociation of Strong AcidsDissociation of Strong AcidsStrong acids are assumed to dissociate completely in solution.
Large Ka or small Ka?Reactant favored or product favored?
Dissociation Constants: Strong Dissociation Constants: Strong AcidsAcids
AcidFormul
aConjugate
BaseKa
Perchloric HClO4 ClO4- Very large
Hydriodic HI I- Very large
Hydrobromic HBr Br- Very large
Hydrochloric HCl Cl- Very large
Nitric HNO3 NO3- Very large
Sulfuric H2SO4 HSO4- Very large
Hydronium ion H3O+ H2O 1.0
Dissociation of Weak AcidsDissociation of Weak AcidsWeak acids are assumed to dissociate only slightly (less than 5%) in solution.
Large Ka or small Ka?Reactant favored or product favored?
Dissociation Constants: Weak Dissociation Constants: Weak AcidsAcids
Acid FormulaConjugate Base
Ka
Iodic HIO3 IO3- 1.7 x 10-1
Oxalic H2C2O4 HC2O4- 5.9 x 10-2
Sulfurous H2SO3 HSO3- 1.5 x 10-2
Phosphoric H3PO4 H2PO4- 7.5 x 10-3
Citric H3C6H5O7 H2C6H5O7- 7.1 x 10-4
Nitrous HNO2 NO2- 4.6 x 10-4
Hydrofluoric HF F- 3.5 x 10-4
Formic HCOOH HCOO- 1.8 x 10-4
Benzoic C6H5COOH C6H5COO- 6.5 x 10-5
Acetic CH3COOH CH3COO- 1.8 x 10-5
Carbonic H2CO3 HCO3- 4.3 x 10-7
Hypochlorous HClO ClO- 3.0 x 10-8
Hydrocyanic HCN CN- 4.9 x 10-10
Acid and Base Strength
Acetate is a stronger base than H2O, so the equilibrium favors the left side (K<1).
The stronger base “wins” the proton.
HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2–(aq)
Self-Ionization of WaterSelf-Ionization of Wateraka “autoionization”aka “autoionization”
H2O + H2O H3O+ + OH-
At 25, [H3O+] = [OH-] = 1 x 10-7
The ion-product constant for water at 25 C:
Kw = [H3O+][OH-]
Kw = (1 x 10-7)(1 x 10-7) = 1 x 10-14
Calculating pH, Calculating pH, pOHpOHpH = -log (H3O+)
pOH = -log (OH-)
Relationship between pH and Relationship between pH and pOHpOH pH + pOH = 14
Finding [HFinding [H33OO++], [OH], [OH--] from pH, pOH] from pH, pOH
[H3O+] = 10-pH
[OH-] = 10-pOH
Acidsand
Bases
pH
• In pure water,
Kw = [H3O+] [OH–] = 1.0 10-14
• Because in pure water [H3O+] = [OH-],
[H3O+] = (1.0 10-14)1/2 = 1.0 10-7
Acidsand
Bases
pH• Therefore, in pure water,
pH = –log [H3O+] = –log (1.0 10-7) = 7.00
• An acid has a higher [H3O+] than pure water, so its pH is <7
• A base has a lower [H3O+] than pure water, so its pH is >7.
Other “p” Scales
• The “p” in pH tells us to take the negative log of the quantity (in this case, hydronium ions).
• Some similar examples are– pOH –log [OH-]
– pKw –log Kw
If you know one, you know them
all:
[H+][OH-]pH
pOH
pH and pOH Calculations
H + O H -
pH pO H
[O H -] = 1 x 10 - 1 4
[H + ]
[H + ] = 1 x 10 - 1 4
[O H -]
p O H = 14 - p H
p H = 14 - p O H
pOH
= -l
og[O
H- ]
pH =
-log
[H+
]
[OH
- ] = 1
0-pO
H
[H+
] = 1
0-pH
pH pH ScaleScale
You need to know
the acidity of common
substances!
Acidsand
Bases
How Do We Measure pH?
Litmus paper• “Red” paper turns
blue above ~pH = 8• “Blue” paper turns
red below ~pH = 5An indicator
• Compound that changes color in solution.
Acidsand
Bases
Strong Acids• You will recall that the seven
strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4.
• These are strong electrolytes and exist totally as ions in aqueous solution.
• For the monoprotic strong acids,
[H3O+] = [acid].
Acidsand
Bases
Strong Bases
• Strong bases are the soluble hydroxides, which are the alkali metal (NaOH, KOH)and heavier alkaline earth metal hydroxides (Ca(OH)2, Sr(OH)2, and Ba(OH)2).
• Again, these substances dissociate completely in aqueous solution.
[OH-] = [hydroxide added].
Acidsand
Bases
Dissociation Constants• For a generalized acid dissociation,
the equilibrium expression is
• This equilibrium constant is called the acid-dissociation constant, Ka.
Acidsand
Bases
Calculating Ka from the pH
The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature.
To calculate Ka, we need all equilibrium concentrations.
We can find [H3O+], which is the same as [HCOO−], from the pH.
Acidsand
Bases
Calculating Ka from the pH
pH = –log [H3O+]
– 2.38 = log [H3O+]
10-2.38 = 10log [H3O+] = [H3O+]
4.2 10-3 = [H3O+] = [HCOO–]
Acidsand
Bases
Calculating Ka from pH
In table form:
[HCOOH], M [H3O+], M [HCOO−], M
Initially 0.10 0 0
Change –4.2 10-3 +4.2 10-
3
+4.2 10-3
At Equilibrium
0.10 – 4.2 10-3
= 0.0958 = 0.10
4.2 10-3 4.2 10 - 3
Acidsand
Bases
Calculating Ka from pH
[4.2 10-3] [4.2 10-3][0.10]
Ka =
= 1.8 10-4
A Weak Acid Equilibrium A Weak Acid Equilibrium ProblemProblem
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+
A Weak Acid Equilibrium A Weak Acid Equilibrium ProblemProblem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-
5 ?Step #2: ICE it!
HC2H3O2 C2H3O2- + H+
II
CC
EE
0.50 0 0
- x +x +x
0.50 - x xx
A Weak Acid Equilibrium A Weak Acid Equilibrium ProblemProblem
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 law of mass action
HC2H3O2 C2H3O2- + H+
0.50 - x xxEE
)50.0()50.0(
))((108.1
25 x
x
xxx
A Weak Acid Equilibrium A Weak Acid Equilibrium ProblemProblem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-
5 ?Step #4: Solve for x, which is also [H+]
HC2H3O2 C2H3O2- + H+
0.50 - x xxEE
)50.0(108.1
25 x
x [H+] = 3.0 x 10-3 M
A Weak Acid Equilibrium A Weak Acid Equilibrium ProblemProblem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-
5 ?Step #5: Convert [H+] to pH
HC2H3O2 C2H3O2- + H+
0.50 - x xxEE
52.4)100.3log( 5 xpH
Acidsand
Bases
Calculating Percent Ionization
In the example:
[A-]eq = [H3O+]eq = 4.2 10-3 M
[A-]eq + [HCOOH]eq = [HCOOH]initial = 0.10 M
Acidsand
Bases
Calculating pH from Ka
Calculate the pH of a 0.30 M solution of acetic acid, C2H3O2H, at 25°C.
Ka for acetic acid at 25°C is 1.8 10-5.Is acetic acid more or less ionized than
formic acid (Ka=1.8 x 10-4)?
Acidsand
Bases
Calculating pH from Ka
The equilibrium constant expression is:
Acidsand
Bases
Calculating pH from Ka
Use the ICE table:
[C2H3O2], M [H3O+], M [C2H3O2−], M
Initial 0.30 0 0
Change –x +x +x
Equilibrium 0.30 – x x x
Acidsand
Bases
Calculating pH from Ka
Use the ICE table:
[C2H3O2], M [H3O+], M [C2H3O2−], M
Initial 0.30 0 0
Change –x +x +x
Equilibrium 0.30 – x x x
Simplify: how big is x relative to 0.30?
Acidsand
Bases
Calculating pH from Ka
Use the ICE table:
[C2H3O2], M [H3O+], M [C2H3O2−], M
Initial 0.30 0 0
Change –x +x +x
Equilibrium 0.30 – x ≈ 0.30 x x
Simplify: how big is x relative to 0.30?
Acidsand
Bases
Calculating pH from Ka
Now,
(1.8 10-5) (0.30) = x2
5.4 10-6 = x2
2.3 10-3 = x
Check: is approximation ok?
Acidsand
Bases
Calculating pH from Ka
pH = –log [H3O+]pH = – log (2.3 10−3)pH = 2.64
Acidsand
Bases
Polyprotic Acids
Have more than one acidic proton.
If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.
Dissociation of Strong Dissociation of Strong BasesBases
Strong bases are metallic hydroxidesGroup I hydroxides (NaOH, KOH) are
very solubleGroup II hydroxides (Ca, Ba, Mg, Sr)
are less solublepH of strong bases is calculated
directly from the concentration of the base in solution
MOH(s) M+(aq) + OH-(aq)
Reaction of Weak Bases with Reaction of Weak Bases with WaterWater
The generic reaction for a base The generic reaction for a base reacting with water, producing its reacting with water, producing its conjugate acid and hydroxide ion:conjugate acid and hydroxide ion:
B + H2O BH+ + OH-
[ ][ ]
[ ]b
BH OHK
B
(all weak bases do this)(all weak bases do this)
Weak Bases
Bases react with water to produce hydroxide ion.
Weak Bases
The equilibrium constant expression for this reaction is
where Kb is the base-dissociation constant.
Reaction of Weak Bases with Reaction of Weak Bases with WaterWater
The base reacts with water, producing The base reacts with water, producing its conjugate acid and hydroxide ion:its conjugate acid and hydroxide ion:
CH3NH2 + H2O CH3NH3+ + OH- Kb = 4.38 x 10-
4
4 3 3
3 2
[ ][ ]4.38 10
[ ]b
CH NH OHK x
CH NH
KKbb for Some Common Weak for Some Common Weak BasesBases
Base FormulaConjugat
e AcidKb
Ammonia NH3 NH4+ 1.8 x 10-5
Methylamine CH3NH2 CH3NH3+ 4.38 x 10-4
Ethylamine C2H5NH2 C2H5NH3+ 5.6 x 10-4
Diethylamine (C2H5)2NH (C2H5)2NH2+ 1.3 x 10-3
Triethylamine (C2H5)3N (C2H5)3NH+ 4.0 x 10-4
Hydroxylamine HONH2 HONH3+
1.1 x 10-8
Hydrazine H2NNH2 H2NNH3+
3.0 x 10-6
Aniline C6H5NH2 C6H5NH3+
3.8 x 10-10
Pyridine C5H5N C5H5NH+ 1.7 x 10-9
Many students struggle with identifying weak bases and their conjugate acids.What patterns do you see that may help you?
A Weak Base Equilibrium A Weak Base Equilibrium ProblemProblem
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
Step #1: Write the equation for the reaction
NH3 + H2O NH4+ + OH-
A Weak Base Equilibrium A Weak Base Equilibrium ProblemProblem
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
Step #2: ICE it!
II
CC
EE
0.50 0 0
- x +x +x
0.50 - x xx
NH3 + H2O NH4+ + OH-
A Weak Base Equilibrium A Weak Base Equilibrium ProblemProblem
Step #3: Set up the law of mass action
0.50 - x xxEE
)50.0()50.0(
))((108.1
25 x
x
xxx
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
NH3 + H2O NH4+ + OH-
A Weak Base Equilibrium A Weak Base Equilibrium ProblemProblem
Step #4: Solve for x, which is also [OH-]
0.50 - x xxEE
)50.0(108.1
25 x
x [OH-] = 3.0 x 10-3 M
NH3 + H2O NH4+ + OH-
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
A Weak Base Equilibrium A Weak Base Equilibrium ProblemProblem
52.4)100.3log( 5 xpOH
Step #5: Convert [OH-] to pH
0.50 - x xxEE
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
NH3 + H2O NH4+ + OH-
48.900.14 pOHpH
Acidsand
Bases
pH of Basic Solutions
What is the pH of a 0.15 M solution of NH3?
[NH4+] [OH−]
[NH3]Kb = = 1.8 10-5
Acidsand
Bases
pH of Basic Solutions
Tabulate the data.
[NH3], M [NH4+], M [OH−], M
Initial 0.15 0 0
Equilibrium 0.15 - x 0.15 x x
Simplify: how big is x relative to 0.15?
Acidsand
Bases
pH of Basic Solutions
(1.8 10-5) (0.15) = x2
2.7 10-6 = x2
1.6 10-3 = x2
(x)2
(0.15)1.8 10-5 =
Check: is approximation ok?
Acidsand
Bases
pH of Basic Solutions
Therefore,
[OH–] = 1.6 10-3 M
pOH = –log (1.6 10-3)
pOH = 2.80
pH = 14.00 – 2.80
pH = 11.20
Acidsand
Bases
Ka and Kb are linked:
Combined reaction = ?
Acidsand
Bases
Ka and Kb are linked:
Combined reaction = ?
Acidsand
Bases
Ka and Kb
Ka and Kb are related in this way:
Ka Kb = Kw
Therefore, if you know one of them, you can calculate the other.
Acid-Base Properties of SaltsAcid-Base Properties of Salts
Type of Salt Examples
Comment pH of solution
Cation is from a strong base, anion from a strong acid
KCl, KNO3
NaClNaNO3
Both ions are neutral
Neutral
These salts simply dissociate in water:
KCl(s) K+(aq) + Cl-(aq)
Acid-Base Properties of SaltsAcid-Base Properties of Salts
Type of Salt Examples
Comment pH of solution
Cation is from a strong base, anion from a weak acid
NaC2H3O2
KCN, NaF
Cation is neutral,Anion is basic
Basic
C2H3O2- + H2O HC2H3O2 + OH-
base acid acid base
The basic anion can accept a proton from water:
Acid-Base Properties of SaltsAcid-Base Properties of Salts
Type of Salt Examples
Comment pH of solution
Cation is the conjugate acid of a weak base, anion is from a strong acid
NH4Cl,
NH4NO3
Cation is acidic,Anion is neutral
Acidic
NH4+(aq) NH3(aq) + H+(aq)
Acid Conjugate Proton base
The acidic cation can act as a proton donor:
Acid-Base Properties of SaltsAcid-Base Properties of Salts
Type of Salt Examples
Comment pH of solution
Cation is the conjugate acid of a weak base, anion is conjugate base of a weak acid
NH4C2H3O2
NH4CN
Cation is acidic,Anion is basic
See below
IF Ka for the acidic ion is greater than Kb for the basic ion, the solution is acidic
IF Kb for the basic ion is greater than Ka for the acidic ion, the solution is basic
IF Kb for the basic ion is equal to Ka for the acidic ion, the solution is neutral
Acid-Base Properties of SaltsAcid-Base Properties of SaltsType of Salt Exampl
esComment pH of
solutionCation is a highly charged metal ion; Anion is from strong acid
Al(NO3)2
FeCl3
Hydrated cation acts as an acid;Anion is neutral
Acidic
Step #1:AlCl3(s) + 6H2O Al(H2O)6
3+(aq) + Cl-(aq)Salt water Complex ion anion
Step #2:Al(H2O)6
3+(aq) Al(OH)(H2O)52+(aq) + H+(aq)
Acid Conjugate base Proton
Reactions of Anions with Water
• Anions are bases.
• As such, they can react with water in a hydrolysis reaction to form OH– and the conjugate acid:
X–(aq) + H2O(l) HX(aq) + OH–(aq)
Reactions of Cations with Water• Cations with acidic protons
(like NH4+) lower the pH of a
solution by releasing H+.
• Most metal cations (like Al3+) that are hydrated in solution also lower the pH of the solution; they act by associating with H2O and making it release H+.
Reactions of Cations with Water
• Attraction between nonbonding electrons on oxygen and the metal causes a shift of the electron density in water.
• This makes the O-H bond more polar and the water more acidic.
• Greater charge and smaller size make a cation more acidic.
Effect of Cations and Anions
1. An anion that is the conjugate base of a strong acid will not affect the pH.
2. An anion that is the conjugate base of a weak acid will increase the pH.
3. A cation that is the conjugate acid of a weak base will decrease the pH.
Effect of Cations and Anions
4. Cations of the strong Arrhenius bases will not affect the pH.
5. Other metal ions will cause a decrease in pH.
6. When a solution contains both the conjugate base of a weak acid and the conjugate acid of a weak base, the affect on pH depends on the Ka and Kb values.
What effect on pH? Why?An anion that is the conjugate base of a strong acid does not affect pH.
= very weak base
An anion that is the conjugate base of a weak acid increases pH. = strong base
A cation that is the conjugate acid of a weak base decreases pH. = strong acid
Cations of the strong Arrhenius bases (Na+, Ca2+) do not affect pH.
= very weak acid(not really acidic at all)
Other metal ions cause a decrease
in pH.= moderate bases(cations)
Weak acid + weak base Depends on Ka and Kb
Factors Affecting Acid Strength
• The more polar the H-X bond and/or the weaker the H-X bond, the more acidic the compound.
• Acidity increases from left to right across a row and from top to bottom down a group.
Factors Affecting Acid Strength
In oxyacids, in which an OH is bonded to another atom, Y,
the more electronegative Y is, the more acidic the acid.
Factors Affecting Acid Strength
For a series of oxyacids, acidity increases with the number of oxygens.
Factors Affecting Acid Strength
Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic.
Lewis Acids
• Lewis acids are defined as electron-pair acceptors.
• Atoms with an empty valence orbital can be Lewis acids.
• A compound with no H’s can be a Lewis acid.
Lewis Bases
• Lewis bases are defined as electron-pair donors.• Anything that is a Brønsted–Lowry base is also a
Lewis base. (B-L bases also have a lone pair.)• Lewis bases can interact with things other than
protons.