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2/23/2014
1
Aqueous Equilibria: Acids and Bases
Ch. 16
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.
Brønsted-Lowrey: Acid-Base need not be in aqueous solutions . Acid and bases are part of a related conjugate pair.
Lewis: No need for hydrogen in definitions
•Acid: Proton (H+) donor •Base: Proton (H+) acceptor
•Acid: Electron (e- ) acceptor •Base: Electron (e- ) donor
•Acid: increases H+ in water. •Base: increases OH- in water
Ag+ + 2 :NH3 → [H3N:Ag:NH3]+
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Weak vs. Strong Acids/Bases What does “strong” as in strong acids and bases mean?
•Highly Concentrated?
•Dangerous?
Strong just mean 100% dissociation That’s all.
H Cl
HCl in the gas phase
H +
-
-
-
-
HCl in the aqueous phase
H Cl
H Cl
H Cl
Cl
Cl
Cl Cl
HCl(g) HCl(aq)
H2O
K’s large HCl is a strong acid
H +
H +
H +
HCl = 1.3x106
Weak vs. Strong Acids/Bases
Weak just means much less than 100% dissociation.
HF in the gas phase
HF in the aqueous phase
H F
H2O H F
H F H F
H F
H F
H F
F -
Most HF molecules are actually NOT dissociated
HF(g) HF(aq) K’s are very small HF is a weak acid
H F
H F H +
HF = 6.6x10-4
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Weak vs. Strong Acids/Bases
Just like with acids, strong bases are those with100% dissociation.
NaOH in the solid phase
NaOH in the aqueous phase
NaOH(s) NaOH(aq)
H2O
K’s large NaOH is a strong base
H O Na
H O Na
H O Na
H O Na
H O Na
H O Na
Na
+
H O
-
Na
+
H O -
H O -
H O - Na
+
Na
+
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-
NH3 in the liquid phase
NH+ in the aqueous phase
H2O
Most NH3 molecules are actually NOT converted to NH4
+
NH3(l) NH4+(aq) K’s are very small NH3 is a weak base
N
H O -
H +
N
NH3(l)+H2O(l) → NH4+(aq) + OH-(aq)
NH3 = 1.8x10-5
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Strong vs. Weak acids
If you know the strong acids/bases, assume the rest are weak
• HCl(aq) • HBr(aq) • HI(aq) • HNO3 • HClO3 • HClO4 • H2SO4
**
Strong Acids Strong Bases
• (soluble metal hydroxides)
• LiOH • NaOH • KOH • Ca(OH)2
(kinda-ish)
• Sr(OH)2 • Ba(OH)2
100% ionization
Examples of weak acids
Examples of weak bases
• HF • CH3COOH (acetic) • HCOOH (formic) • NH4
+ (ammonium) • H3PO4 • HClO2 • HClO • (small, highly-charged
metal ions)
• Al3+
• (nitrogen-containing bases)
• NH2R R = anything • NH3 (ammonia) • (“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”
Auto ionization of water – Kw
At any time in a sample of pure water, some water molecules, very few, will dissociate by themselves
The amount (concentration) that dissociates is quantified by Kw
Kw = [H+][OH-]= 1x10-14
Only 1.8x10-6% of H2O molecules within a container of water do this!!
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The pH scale A simple way of expressing concentration of H+ ions in solution.
pH = -log[H+]
0.00000000000001
0.0000000000001
0.000000000001
0.00000000001
0.0000000001
0.000000001
0.00000001
0.0000001
0.000001
0.00001
0.0001
0.001
0.01
0.1
1 0.00000000000001
0.0000000000001
0.000000000001
0.00000000001
0.0000000001
0.000000001
0.00000001
0.0000001
0.000001
0.00001
0.0001
0.001
0.01
0.1
1
Notice the relationship:
[H+][OH-] = 1.0x10-14 = Kw
[H+] in M [OH-] in M
Also notice the logarithmic nature
of pH vs. [H+]
For example, what is the
difference in [H+] between the pH
of 7 and 4?
pH of 4 has 103, (or 1,000x) more
H+ ions in solution than pH 7.
pH Common
solutions
[H+] = 10-pH
pH and [H+]
Determine the pH of a solution with 1.55x10-6 M H+. Also, what will be the
concentration of OH-?
pH = -log(1.55x10-6M) = 5.810
[H+][OH-] = 1.0x10-14 = Kw
(1.55x10-6M)[OH-] = 1.0x10-14
M1.02x100.977M
1.00x10][H 14
14
Determine the pH of a solution with 0.977 M OH-. Also, what will be the concentration of
H+?
pH = -log(1.02x10-14M) = 13.991
M6.452x10M1.55x10
1.00x10][OH 9
6-
14-
Determine the H+ and OH- concentrations for a solution with a pH of 7.02
pH = -log[H+]
[H+] = 10-pH M9.8x1010][H 87.01 M1.0x10M9.8x10
1.00x10][OH 7
8-
14-
Notice that the closer to pH = 7, the closer the [H+] and [OH-]
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pH and pOH
[H+][OH-] = 1.0x10-14 = Kw
Although most people know about pH, we can also calculate something called pOH
take the -log
pH + pOH = 14
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
M2.9x10
1.0x105
14
pOH = -log(3.4x10-10M) = 9.47
pH = -log(2.9x10-5M) = 4.54
pH = 14.00 – pOH = 4.53
pOH = 14 – 4.54 = 9.46
[OH-] = 10-4.54 = 3.5x10-10M
[OH-] = = 3.4x10-10M
Difference in last digit due to rounding (keep an extra sig fig in calculations)
Qualitatively: •The larger the [H+], to lower the [OH-] •The smaller the pH, the larger the pOH
or
pH of a strong acid/strong base 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+.
What is the pH of a 0.050M solution of HCl?
Since HCl is 100% dissociated, the [H+] is also 0.050M
pH = -log[H+]
pH = -log(0.050) = 1.30
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 NaOH?
Since NaOH is 100% dissociated, the [OH-] is also 0.050M
pOH = -log[OH-]
pOH = -log(0.050) = 1.30
pH = 14 - pOH
pH = 14.00 -1.30 = 12.70
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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+
acid base
Conjugate acid/base pair
acid base
Conjugate acid/base pair
The stronger the acid, the weaker
its conjugate base.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
stronger
acid
stronger
base
weaker
acid weaker
base
Conjugate Acid/Base pairs
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
stronger
acid
stronger
base
weaker
acid weaker
base
Reaction proceeds in direction that results in a weaker acid and weaker base produced
NO3- (aq) + H2O(l) → OH-(aq) + HNO3(aq)
stronger
acid
stronger
base
weaker
acid weaker
base
The reaction will not proceed in the forward direction
NH4+ (aq) + CH3COO-(aq) ⇌ NH3(aq) + CH3COOH(aq)
stronger
acid
stronger
base
weaker
acid
weaker
base
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The Acid and Base Equilibrium constants: Ka and Kb
[HA]
]][AO[H 3
a
K
The strength of a acid or base can be related to the an equilibrium constant
HA(aq) + H2O(l) → H3O+(aq) + A-(aq)
[B]
]][BH[OH
-
b
K
B(aq) + H2O(l) → OH-(aq) + BH+(aq)
pKa = -log(Ka) pKb = -log(Kb) Ka = 10-pKa Kb = 10-pKb
The larger the Ka, and thus smaller the pKa, the stronger the acid
The larger the Kb, and thus smaller the pKb, the stronger the base
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.
Based on pKa or Ka values, which acid is stronger?
CH3COOH Ka = 1.6x10-5 CH3COO- Kb = 6.2x10-10
How are conjugate acid/base pairs related?
Kb*Ka = Kw = 1x10-14
For conjugate acid/base pairs
pKb +pKa = pKw = 14
We can also take the negative log of
Ka, Kb, and Kw.
CH3COOH pKa CH3COO- pKb
Conjugate Acid/Base pair relationship – Ka and Kb
NH3 Kb = 1.9x10-5
NH3 pKb
NH4+
Ka = 5.4x10-10
H3PO4 Ka = 7.6x10-3
H3PO4 pKa
H2PO4- Kb = 1.3x10-12
H2PO4- pKb
-
2.12 11.88
4.79 9.21
9.27 4.73 NH4+ pKa
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Polyprotic Acids While binary acids only have one H+ to donate, some oxoacids have multiple protons. They
can be diprotic or triprotic.
100% dissociation
⇌
HSO4- H2SO4 SO4
2- pKa2 = 1.99 pKa1 ≈ -3
<100% dissociation
- 2-
H3PO4 H2PO4- HPO4
2- PO43-
⇌ ⇌ ⇌
pKa1 = 2.12 pKa2 = 7.20 pKa3 = 12.38
- 2- 3-
Only the first H+ is considered
strong.
Each subsequent H+
dissociates less.
Acidic, Basic, and Neutral Salts 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
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Acidic, Basic, and Neutral Salts Whether a salt is neutral, acidic, or basic will depend on the specific anion and
anion
Neutral salts
Cation Anion (conjugate base)
From strong base …of a strong acid
Acidic salts Conjugate acid of
weak base …of a strong acid
Basic salts From strong base …of a weak acid
Acidic salts Small/highly-charged metal on
…of a strong acid
?????? Conjugate acid weak base or small/highly-
charged metal ion
…of a weak acid
Ka ≈ Kb
Ka > Kb
Ka < Kb Basic salts
Acidic salts
Neutral salts
NaCl
NH4Cl
AlCl3
NaClO
NH4CH3COO
NH4NO2
NH4CN
“from” NaOH
“from” HCl
“from” HCl
“from” NH3
“from” HCl
“from“NaOH
“from” HClO
“from” NH3
“from” CH3COOH
“from” NH3
“from” HNO2
“from” NH3
“from” HCN
Salt
Metal ions as acids Certain metal ions act as acids, increasing the H+ concentration in water.
When these ions dissolve in water, water ions “coordinate” with the ions
creating a “complex”.
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.
Al3+ Al3+
H2O outside the complex
H3O+ outside the complex
Acidity increases with higher charge and smaller size of ion
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Determining the pH of a Weak Acid Solution
Determine the pH of a 0.123M aqueous solution of HClO. Ka = 2.95x10-8.
I
E C
nitial
hange
quilibirum
0.123M 0M
-x +x
0.123M - x x
HClO(aq)+ H2O(l)⇌ H3O+(aq) + ClO-(aq)
0M
+x
x
Write the balanced equation for the acid in H2O…producing H3O+.
Just like any other equilibrium problem, plug in what we know.
Since H2O is a pure liquid (and does not change appreciably over time), we don’t need
to deal with it
[HClO]
]][ClOO[H3a
K
x-0.123
(x)(x)2.95x10 8
0.123
x2.95x10
28
29 x3.63x10
x = 6.02x10-5
[H3O+] = x = 6.02x10-5M
pH = -log(6.02x10-5M)
pH = 4.220
0.0487% x1000.123
6.02x10-5
Less than 5%, good assumption 5-
2-5
6.02x10-0.123
)(6.02x10
Determining the pH of a Weak Base Solution
Determine the pH of a 0.555M aqueous solution of NH3. pKb = 4.75.
I
E C
nitial
hange
quilibirum
0.555M 0M
-x +x
0.555M - x x
NH3(aq)+ H2O(l)⇌ OH-(aq) + NH4
+(aq)
0M
+x
x
Write the balanced equation for the base in H2O…producing OH- and the conjugate acid, NH4
+ .
][NH
]][NH[OH
3
4b
K
x-0.555
(x)(x)1.78x10 5-
0.555
x1.78x10
25
26 x9.88x10
x = 3.14x10-3
[OH-] = x = 3.14x10-3M
pOH = -log(3.14x10-3M)
pOH = 2.503 pH = 14.00-2.503 = 11.497
0.566% x1000.555
3.14x10-3
Less than 5%, good assumption 4-
2-3
3.14x10-0.555
)(3.14x10
First, we need to deal with pKb.
54.75b 1.78x10 10 K
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Calculating Ka from pH If 1.55M solution of an acid has a pH of 5.15, determine the Ka for the acid (assume it is monoprotic).
I
E C
nitial
hange
quilibirum
1.55M 0M
-x +x
1.55M - x x
HA(aq)+ H2O(l) ⇌ H3O+(aq) + A-(aq)
0M
+x
x
[H+] = 10-5.15 = 7.1x10-6M
Since we know pH, we also know [H+] at equilibrium
7.1x10-6M 7.1x10-6M
7.1x10-6 M = x
1.55M – 7.1x10-6M
[HA]
]][AO[H3a
K
)1.7x10-(1.55
))(7.1x10(7.1x106-
-6-6
a K
11a 3.3x10K
“Short-Cuts”
initial
2
cbase] or acid [weak
xK
This assumes that Kc is small enough where x will also be small. Therefore we
leave it out.
If you complete enough ice tables, you will start noticing trends.
For example, placing a weak acid or base in water will give you the following expression using an ice table:
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!!
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Percent ionization
x100[HA]
]O[Hacid anof ionization %
initial
mequilibriu3
x100[B]
][OHbase aof ionization %
initial
mequilibriu-
Percent ionization is the amount of a weak acid or base that has become H3O+ or OH-.
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
0.12%x1000.76M
M9.05x10 ionization %
4
Determine the percent ionization if 5.00M aqueous solution of an unknown acid mixed with water
resulted in a pH of 3.12
0.015%x1005.00M
M7.6x10 ionization %
4
[H3O+] = 10-pH = 10-3.12 = 7.6x10-4 M
Determine the percent ionization of a 0.100M aqueous solution of NH3. pKb = 4.75.
initial
2
cbase] or acid [weak
xKLet’s use the “short-cut”
Kb = 10-4.75 = 1.8x10-5
0.100
x1.8x10
25
62 1.8x10x
x = 1.3x10-3 = [OH-]
1.3%x1000.100M
M1.3x10 ionization %
3
Percent Ionization and Concentration
initial
2
cbase] or acid [weak
xK
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.
Let’s use the “short-cut” 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.
1.28
x1.0x10
25
]O[H 0.0036Mx 3
0.040
x1.0x10
25
0.00063Mx
0.010
x1.0x10
25
0.00032Mx
0
0.5
1
1.5
2
2.5
3
3.5
0 0.5 1 1.5
% io
niz
atio
n
Initial acid concentration (M)
0.28%x1001.28M
0.0036Mionization %
1.6%x1000.040M
0.00063Mionization %
3.2%x1000.010M
0.00032Mionization %
The more concentrated a weak base solution, the lower
the % ionization
0
0.0005
0.001
0.0015
0.002
0.0025
0.003
0.0035
0.004
0 0.2 0.4 0.6 0.8 1 1.2 1.4Hyd
ron
ium
co
nce
ntr
ati
on
at
equ
ilib
riu
m (M
)
Initial weak acid concentration (M)
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14
Binary Acid Strengths
H2O HF
HCl
HBr
HI
pKa =15.74 pKa = 3.17
H2S
H2Te
pKa ≈ -10
pKa ≈ -8
pKa ≈ -6 pKa = 6.9
pKa = 2.6
Acid Strength
Acid
Strength
Electronegativity
Bo
nd
Len
gth
Moving down a group, the bond
length gets longer. Longer bond
length = weaker X-H bond
pKa = 3.9
H2Se
-15
-10
-5
0
5
10
15
20
80 100 120 140 160 180
pK
a
Bond Length (pm)
H2O
H2S
H2Se H2Te
HCl
HF
HBr HI
Carboxylic Acid Strengths
4.79 2.86 1.35 0.77
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Oxoacid Strengths – Inductive effects
In oxoacids, The weaker the O-H bond, the stronger the acid
What can make the O-H bond weaker? Highly electronegative atoms withdraw
electron density from the O-H bond, making it weaker.
≈ -1 ≈ -8 7.53
Shift in electron
density away from O-H makes the
bond weaker (more acidic)
Less electron density at
the H
1.96
Carboxylic Acid Strengths
4.74 2.86 1.35 0.77
Explain the following trend in acid strength:
4.74
NaOH
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