10/7/2014
What is an acid? What is a base?
There are actually multiple definitions
Arrhenius: Dealt with species in aqueous solutions. Most basic definition of acis-base.
•Acid: increases H+ in water.
•Base: increases OH- in water
Aqueous Equilibria: Acids and
Bases
Brønsted-Lowrey: Acid-Base need not be in aqueous solutions . Acid and bases
are part of a related conjugate pair.
•Acid: Proton (H+) donor
•Base: Proton (H+) acceptor
Ch. 16
Lewis: No need for hydrogen in definitions
•Acid: Electron (e- ) acceptor
•Base: Electron (e- ) donor
Weak vs. Strong Acids/Bases
Ag+ + 2 :NH3 → [H3N:Ag:NH3]+
Weak vs. Strong Acids/Bases
What does “strong” as in strong acids and bases mean?
•Highly Concentrated?
Weak just means much less than 100% dissociation.
•Dangerous?
Strong just mean 100% dissociation
That’s all.
+
H
H
Most HF molecules are actually NOT dissociated
H
H
Cl
Cl
F
H2O
F
F
H
+
H
H
Cl
Cl
-
H
F
H
F
F
H
Cl
H
HCl in the gas phase
-
+
Cl
H
H
H
-
Cl
H2O
H
F
-
Cl
H
F
H
F
H
HF in the gas phase
+
F
+
HF in the aqueous phase
HCl in the aqueous phase
HF(g)
HCl(g)
HCl(aq)
K’s large
HF(aq)
K’s are very small
HCl is a strong acid
Weak vs. Strong Acids/Bases
Weak vs. Strong Acids/Bases
Weak just means much less than 100% dissociation! For NH3, it itself is not
dissociated, however, it takes an H+ from water to create OH-
Just like with acids, strong bases are those with100% dissociation.
+
Na
H
O
Na
H
O
H
Na
O
Na
H
O
Na
H
H
O
Na
H
O
O
Most NH3 molecules are actually NOT converted
to NH4+
NH3(l)+H2O(l) → NH4+(aq) + OH-(aq)
-
N
N
H
+
H2O
H
O
Na
-
Na
H
+
Na
NaOH in the solid phase
H2O
+
O
NaOH(aq)
+
Na
-
H
O
-
H
O
NH3 in the liquid phase
NH+ in the aqueous phase
NaOH in the aqueous phase
NaOH(s)
HF is a weak acid
HF = 6.6x10-4
HCl = 1.3x106
K’s large
NaOH is a strong base
NH3(l)
NH 4
+(aq)
K’s are very small
NH3 is a weak base
NH3 = 1.8x10-5
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10/7/2014
Strong vs. Weak acids
Auto ionization of water – Kw
If you know the strong acids/bases, assume the rest are weak
At any time in a sample of pure water, some water molecules, very few, will
dissociate by themselves
Strong Acids
•
•
•
•
•
•
•
Strong Bases
HCl(aq)
HBr(aq)
HI(aq)
HNO3
HClO3
HClO4
H2SO4**
Examples of
weak bases
Examples of
weak acids
•
(soluble metal
hydroxides)
•
•
•
•
•
•
LiOH
NaOH
KOH
Ca(OH)2 (kinda-ish)
Sr(OH)2
Ba(OH)2
100%
ionization
•
•
•
•
•
•
•
HF
CH3COOH (acetic)
HCOOH (formic)
NH4+ (ammonium)
H3PO4
HClO2
HClO
•
(small, highly-charged
metal ions)
•
Al3+
•
•
•
•
Kw = [H+][OH-]= 1x10-14
(nitrogen-containing
bases)
NH2R R = anything
NH3 (ammonia)
Only 1.8x10-6% of H2O molecules within a
container of water do this!!
The amount (concentration) that dissociates is quantified by Kw
(“Insoluble” hydroxides)
<<100%
ionization
A common rule for oxo-acids: If
there are at least 2 more O’s than
H+’s in the molecule, it is strong
**Only the 1st H+ removed is “strong”
The pH scale
pH and [H+]
A simple way of expressing concentration of H + ions in solution.
[H+] in M
pH
Common
solutions
0.00000000000001
1
0.0000000000001
0.1
0.000000000001
0.01
0.00000000001
0.001
0.0000000001
0.0001
0.000000001
0.00001
0.000001
(1.55x10-6M)[OH-] = 1.0x10-14
[H+][OH-] = 1.0x10-14 = Kw
0.00000001
0.000000001
0.0001
0.0000000001
0.001
0.00000000001
0.01
0.1
0.0000000000001
1
0.00000000000001
For example, what is the
difference in [H+] between the pH
of 7 and 4?
0.000000000001
pH of 4 has 103, (or 1,000x) more
H+ ions in solution than pH 7.
Although most people know about pH, we can also calculate something called pOH
So…let’s say we have a solution with a [H+] of 2.9x10-5 M. Determine
the concentration of [OH], find pH and pOH.
(2.9x10-5M)[OH-] = 1.00x10-14
1.0x10 14
[OH-] =
= 3.4x10-10M
2.9x10 5 M
pOH = -log(3.4x10-10M) = 9.47
pH = 14.00 – pOH = 4.53
1.00x1014
 6.45x109 M
1.55x10-6 M
Determine the H+ and OH- concentrations for a solution with a pH of 7.02
[H+] = 10-pH
[H ]  107.01  9.8x10 8 M
[OH- ] 
1.00x10 14
 1.0x10 7 M
9.8x10 -8 M
Notice that the closer to pH = 7, the closer the [H +] and [OH-]
pH of a strong acid/strong base
pH and pOH
pH + pOH = 14
[OH- ] 
Also notice the logarithmic nature
of pH vs. [H+]
0.00001
1.00x10 14
 1.02x10 14 M
0.977M
pH = -log(1.02x10-14M) = 13.991
[H+][OH-] = 1.0x10-14 = Kw
Notice the relationship:
0.0000001
take the -log
[H ] 
pH = -log(1.55x10-6M) = 5.810
0.0000001
[H+][OH-] = 1.0x10-14 = Kw
Determine the pH of a solution with 0.977 M
OH-. Also, what will be the concentration of
H+?
pH = -log[H+]
[H+] = 10-pH
0.00000001
0.000001
Determine the pH of a solution with
1.55x10-6 M H+. Also, what will be the
concentration of OH-?
pH = -log[H+]
[OH-] in M
or
pH =
-log(2.9x10-5M)
= 4.54
pOH = 14 – 4.54 = 9.46
[OH-] = 10-9.46 = 3.5x10-10M
Difference in last digit due to rounding (keep an extra
sig fig in calculations)
Remember that when a strong acid or a strong base is placed in water, we assume
that it is 100% dissociated.
So, for HCl, all of the “H” becomes dissociated
as H+. If we know the concentration of HCl, we
also know the concentration of H+.
And for NaOH, all of the “OH” becomes dissociated
as OH-. If we know the concentration of NaOH, we
also know the concentration of OH-.
What is the pH of a 0.050M solution of HCl?
What is the pH of a 0.050M solution of NaOH?
Since HCl is 100% dissociated, the [H+] is also
0.050M
Since NaOH is 100% dissociated, the [OH-] is
also 0.050M
pH = -log[H+]
pOH = -log[OH-]
pH = -log(0.050) = 1.30
pOH = -log(0.050) = 1.30
pH = 14 - pOH
pH = 14.00 -1.30 = 12.70
Qualitatively:
•The larger the [H+], to lower the [OH-]
•The smaller the pH, the larger the pOH
2
10/7/2014
Conjugate Acid/Base pairs
Conjugate Acid/Base pairs
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
Conjugate acid/base pairs are acids and bases that differ by only an H +
stronger
acid
Conjugate acid/base pair
base
acid
base
weaker
acid
weaker
base
How are conjugate acid/base pairs related?
B(aq) + H2O(l) → OH-(aq) + BH+(aq)
HA(aq) + H2O(l) → H3O+(aq) + A-(aq)
[H3O ][A ]
[HA]
Kb 
Ka = 10-pKa
H3PO4 Ka =
[OH- ][BH ]
[B]
H2PO4- Kb = 1.3x10-12
7.6x10-3
CH3COOH Ka = 1.6x10-5
CH3COO- Kb = 6.2x10-10
NH4+ Ka = 5.4x10-10
Kb = 10-pKb
pKb = -log(Kb)
NH3 Kb = 1.9x10-5
The larger the Kb, and thus smaller the
pKb, the stronger the base
The larger the Ka, and thus smaller the
pKa, the stronger the acid
A certain acid has a Ka of 1.2x10-7.
Calculate it’s pKa.
A certain acid has a pKa of 9.23.
Calculate it’s Ka.
H3PO4 pKa
2.12
11.88
CH3COOH pKa
4.79
9.21
NH4+ pKa
9.27
4.73
For conjugate acid/base pairs
Based on pKa or Ka values, which acid is stronger?
Kb*Ka = Kw = 1x10-14
-
pKa1 ≈ -3
HSO4-
100% dissociation
2-
⇌
⇌
pKa1 = 2.12
SO42-
Each
subsequent H+
dissociates less.
<100% dissociation
-
H3PO4
Only the first H+
is considered
strong.
pKa2 = 1.99
2-
⇌
H2PO4-
pKa2 = 7.20
H2PO4- pKb
CH3COO- pKb
NH3 pKb
We can also take the negative log of
Ka, Kb, and Kw.
pKb +pKa = pKw = 14
Acidic, Basic, and Neutral Salts
Polyprotic Acids
While binary acids only have one H+ to donate, some oxoacids have multiple protons. They
can be diprotic or triprotic.
H2SO4
stronger
acid
Conjugate Acid/Base pair relationship – Ka and Kb
The strength of a acid or base can be related to the an equilibrium constant
pKa = -log(Ka)
stronger
base
weaker
base
The Acid and Base Equilibrium constants: Ka and Kb
Ka 
stronger
acid
NH4+ (aq) + CH3COO-(aq) ⇌ NH3(aq) + CH3COOH(aq)
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
weaker
acid
Reaction proceeds in direction that results in a weaker
acid and weaker base produced
The reaction will not proceed in the forward direction
The stronger the acid, the weaker
its conjugate base.
stronger
base
weaker
base
stronger
base
weaker
acid
weaker
base
Conjugate acid/base pair
stronger
acid
weaker
acid
NO3- (aq) + H2O(l) → OH-(aq) + HNO3(aq)
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
acid
stronger
base
Remember: in chemistry, a salt is just an ionic compound containing an anion and
cation, but is neutral overall.
The anion and cation in this case can be viewed as conjugate acid/bases of
something else…
Rules to remember:
•The child of a strong acid or base is negligible in strength.
•The child of weak acid or base is also weak
3-
⇌
HPO42-
pKa3 = 12.38
PO43-
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10/7/2014
Acidic, Basic, and Neutral Salts
Cation
Anion
Whether a salt is neutral, acidic, or basic
will depend on the specific anion and
anion
Salt
(conjugate base)
“from” HCl
NaCl
From strong base
…of a strong acid
“from” NaOH
Neutral salts
Conjugate acid of
weak base
…of a strong acid
Acidic salts
Small/highly-charged
metal on
…of a strong acid
Acidic salts
From strong base
…of a weak acid
Basic salts
Conjugate acid weak
base or small/highlycharged metal ion
…of a weak acid
??????
Metal ions as acids
Certain metal ions act as acids, increasing the H + concentration in water.
“from” HCl
NH4Cl
When these ions dissolve in water,
water ions “coordinate” with the ions
creating a “complex”.
“from” NH3
“from” HCl
AlCl3
“from” HClO
NaClO
The metal ions pulls enough electron density from one water molecule
outside the complex to allow it to dissociate into H3O+ while an OH- stays
bound to the complex. Thus, increasing H3O+ concentration.
“from“NaOH
Ka
Ka
≈
Kb
>
Ka
Kb
<
NH4NO2
Acidic salts
Kb
“from” CH3COOH
NH4CH3COO
Neutral salts
Basic salts
“from” NH3
Determine the pH of a 0.123M aqueous solution of HClO. Ka = 2.95x10-8.
Ka 

HClO(aq)+ H2O(l)⇌ H3O+(aq) + ClO-(aq)
I
C
E

[H3O ][ClO ]
[HClO]
nitial
(x)(x)
2.95x10 8 
0.123 - x
0.123M
0M
-x
hange
quilibirum
Determining the pH of a Weak Base Solution
First, we need to deal with pKb.
+x
x
Write the balanced equation for the base in
H2O…producing OH- and the conjugate acid, NH4+ .
[OH ][NH4  ]
Kb 
[NH3 ]
x
1.78x10 5 
2.95x10 8 
3.63x10
9
x
0.123
 x
6.02x10
x100  0.0487%
0.123
x2
0.555
x = 6.02x10-5
[H3O+] = x = 6.02x10-5M
(6.02x10-5 )2
0.123 - 6.02x10 -5
Less than 5%, good assumption
9.88x10 6  x2
-x
hange
0.555M - x
0M
0M
+x
+x
x
x
pOH = 2.503
pH = 14.00-2.503 = 11.497
3.14x10 -3
x100  0.566%
0.555
Less than 5%, good assumption
(3.14x10-3 )2
0.555 - 3.14x10 -4
pH = 4.220
“Short-Cuts”
If you complete enough ice tables, you will start noticing trends.
If 1.55M solution of an acid has a pH of 5.15, determine the Ka for the acid
(assume it is monoprotic).
Since we know pH, we also
know [H+] at equilibrium
[H+] = 10-5.15 = 7.1x10-6M
Ka 
0.555M
x = 3.14x10-3
Calculating Ka from pH
Ka 
nitial
[OH-] = x = 3.14x10-3M
pOH = -log(3.14x10-3M)
pH = -log(6.02x10-5M)
2
I
C
E
quilibirum
(x)(x)
1.78x10 
0.555 - x
-5
-5
2
NH3(aq)+ H2O(l)⇌ OH-(aq) + NH4+(aq)
Kb  104.75  1.78x10 5
0M
+x
0.123M - x
Acidity increases with higher charge and smaller size of ion
Determine the pH of a 0.555M aqueous solution of NH 3. pKb = 4.75.
Since H2O is a pure liquid (and does not
change appreciably over time), we don’t need
to deal with it
Just like any other equilibrium problem, plug
in what we know.
H3O+ outside the
complex
H2O outside the
complex
“from” NH3
Determining the pH of a Weak Acid Solution
Write the balanced equation for the
acid in H2O…producing H3O+.
Al3+
“from” NH3
“from” HCN
NH4CN
Al3+
“from” HNO2
[H3O ][A ]
[HA]
(7.1x10-6 )(7.1x10-6 )
(1.55 - 1.7x10 -6 )
For example, placing a weak acid or base in water will give you the following expression
using an ice table:
HA(aq)+ H2O(l) ⇌ H3O+(aq) + A-(aq)
I
C
E
nitial
hange
quilibirum
1.55M
-x
1.55M - x
1.55M – 7.1x10-6M
0M
0M
+x
+x
x
x
7.1x10-6M
7.1x10-6M
Kc 
x2
[weakacidor base]initial
This assumes that Kc is small enough
where x will also be small. Therefore we
leave it out.
This also assumes that there are no
products of ionization in the water
before the weak base or acid is added.
If you are ever unsure of a short cut…always do it the
long way!!
7.1x10-6 M = x
Ka  3.3x10 11
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10/7/2014
Percent Ionization and Concentration
Percent ionization
Percent ionization is the amount of a weak acid or base that has become H 3O+ or OH-.
[HA]initial
Determine the percent ionization
if 0.76M aqueous solution of an
unknown acid mixed with water
resulted in a an [H3O+] of
9.05x10-4M
% ionization
% ionizationof a base
x100
[B]initial
Weak acids and bases do not ionize to the same extent in every solution. The amount to
which it ionizes depends on, among other things, the initial concentration of the weak acid or
base.
x2
Let’s use the “short-cut” K c  [weakacidor base]
initial
x100
Determine the % ionization of a weak acid with
a Ka of 1.0x10-5 with the initial concentrations of
1.28M, 0.040M, and 0.010M.
Determine the percent ionization of a 0.100M
aqueous solution of NH3. pKb = 4.75.
Kb =
10-4.75
1.0x10 5 
1.8x10-5
=
Kc 
Let’s use the “short-cut”
9.05x10 4 M
x100  0.12%
0.76M
1.8x10 5 
Determine the percent ionization
if 5.00M aqueous solution of an
unknown acid mixed with water
resulted in a pH of 3.12
1.3x10-3
1.0x10 5 
x2
0.100
[H3O+] = 10-pH = 10-3.12 = 7.6x10-4 M
=
% ionization
% ionization
0.00032M
x100  3.2%
0.010M
x2
0.040
% ionization
0.00063M
x100  1.6%
0.040M
0.004
0.0035
0.003
0.0025
0.002
0.0015
0.001
0.0005
0
0
x  0.00063M
1.0x10 5 
[OH-]
7.6x10 4 M
x100  0.015%
5.00M
x2
0.010
x  0.00032M
x2
[weakacidor base]initial
x2  1.8x10 6
x=
% ionization
[OH- ]equilibrium
Hydronium concentration at
equilibrium (M)
[H3O ]equilibrium
x2
1.28
% ionization
x  0.0036M  [H3O ]
1.3x10 3 M
x100  1.3%
0.100M
% ionization
% ionizationof anacid
0.0036M
x100  0.28%
1.28M
The more concentrated a weak base solution, the lower
the % ionization
Binary Acid Strengths
0.2
0.4
0.6
0.8
1
1.2
Initial weak acid concentration (M)
1.4
3.5
3
2.5
2
1.5
1
0.5
0
0
0.5
1
1.5
Initial acid concentration (M)
Oxoacid Strengths – Inductive effects
In oxoacids, The weaker the O-H bond, the stronger the acid
H2O
HF
H2O
pKa = 6.9
pKa ≈ -6
H2Se
HBr
pKa = 3.9
pKa ≈ -8
15
H2S
HF
5
H2Te
H2Se
0
80
100
120
140
160
HBr
180
HI
-10
pKa ≈ -10
What can make the O-H bond weaker? Highly electronegative atoms withdraw
electron density from the O-H bond, making it weaker.
-5
HCl
HI
H2Te
10
pKa
HCl
Acid Strength
H2S
pKa = 2.6
20
pKa = 3.17
Bond Length
pKa =15.74
7.53
-15
≈ -1
1.96
≈ -8
Shift in
electron
density away
from O-H
makes the
bond weaker
(more acidic)
Bond Length (pm)
Acid Strength
Less electron
density at
the H
Moving down a group, the bond
length gets longer. Longer bond
length = weaker X-H bond
Electronegativity
Carboxylic Acid Strengths
Oxoacid Strengths – Inductive effects
Electronegativity of central atom
Across each row, electron
density of being shifted
away from the O-H bond.
Acid Strength
15.7
Explain the following trend in acid strength:
4.74
2.86
Stabilizes negative charge
after proton loss
4.74
1.35
0.77
9.24
6.37
-1.4
9.84
2.15
≈ -3
≈ -8
5