Download Aqueous Equilibria: Acids and Bases and more Study notes Chemistry in PDF only on Docsity!
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
) donor
- Base: Proton (H+) acceptor
- Acid: Electron (e-^ ) acceptor
- Base: Electron (e
) donor
in water.
- Base: increases OH-^ in water
Ag+^ + 2 :NH 3 → [H 3 N:Ag:NH 3 ]+
Weak vs. Strong Acids/Bases
What does “strong” as in strong acids and bases mean?
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)
H 2
O
K’s large HCl is a strong acid
H
H
H +
HCl = 1.3x
6
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
H 2 O 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
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)
H 2 O
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 NH 3
, it itself is not
dissociated, however, it takes an H
from water to create OH
NH 3 in the liquid phase
NH
in the aqueous phase
H 2 O
Most NH 3 molecules are actually NOT converted
to NH 4 +
NH 3 (l) NH 4
(aq) K’s are very small NH 3 is a weak base
N
H O
H
N
NH 3 (l)+H 2 O(l) → NH 4
(aq) + OH
(aq)
NH 3 = 1.8x10-^5
Strong vs. Weak acids
If you know the strong acids/bases, assume the rest are weak
- HCl(aq)
- HBr(aq)
- HI(aq)
- HNO 3
- HClO 3
- HClO 4
- H 2 SO 4
**
Strong Acids Strong Bases
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
- CH 3 COOH (acetic)
- HCOOH (formic)
- NH 4
(ammonium)
- H 3 PO 4
- HClO 2
- HClO
- (small, highly-charged
metal ions)
bases)
- NH 2 R R = anything
- NH 3 (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 – K
w
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 K w
K w
= [H+][OH-]= 1x10-^14
Only 1.8x10-^6 % of H2O molecules within a container of water do this!!
The pH scale
A simple way of expressing concentration of H+^ ions in solution.
pH = - log[H
]
Notice the relationship:
[H+][OH-] = 1.0x
= K w
[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 10^3 , (or 1,000x) more
H+^ ions in solution than pH 7.
pH
Common solutions
[H
] = 10
pH and [H
]
Determine the pH of a solution with
1.55x
. Also, what will be the
concentration of OH
pH = - log(1.55x10-^6 M) = 5.
[H+][OH-] = 1.0x10-^14 = Kw
(1.55x10-^6 M)[OH-] = 1.0x10-^14
1.02x10M 0.977M
1.00x [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-^14 M) = 13.
6.45x10M 1.55x10M
1.00x [OH ]
9
14
Determine the H+ and OH- concentrations for a solution with a pH of 7.
pH = - log[H+]
[H+] = 10-pH [H ]^ 10 7.01^ 9.8x10^8 M 1.0x10M
9.8x10M
1.00x [OH ]
7
14
Notice that the closer to pH = 7, the closer the [H
] and [OH
]
pH and pOH
[H
][OH
] = 1.0x
= 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-^5 M)[OH-] = 1.00x10-^14
2.9x10M
1.0x
5
14
pOH = - log(3.4x10-^10 M) = 9.
pH = - log(2.9x
pH = 14.00 – pOH = 4.
pOH = 14 – 4.54 = 9.
[OH
- ] = 10 - 9. = 3.5x - 10 M
[OH-] = = 3.4x10-^10 M
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.
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.
pH = 14 - pOH
pH = 14.00 - 1.30 = 12.
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
From strong base …of a weak acid Basic salts
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
K a ≈ K b
K a > K b
K a < K b Basic salts
Acidic salts
Neutral salts
NaCl
NH 4 Cl
AlCl 3
NaClO
NH 4
CH 3
COO
NH 4 NO 2
NH 4 CN
“from” NaOH
“from” HCl
“from” HCl
“from” NH 3
“from” HCl
“from“NaOH
“from” HClO
“from” NH 3
“from” CH 3 COOH
“from” NH 3
“from” HNO 2
“from” NH 3
“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 H 3 O+^ while an OH-^ stays
bound to the complex. Thus, increasing H 3 O
concentration.
Al3+^ Al3+
H 2 O outside the
complex
H 3 O+^ outside the
complex
Acidity increases with higher charge and smaller size of ion
Determining the pH of a Weak Acid Solution
Determine the pH of a 0.123M aqueous solution of HClO. K a = 2.95x
I
E
C
nitial
hange
quilibirum
0.123M 0M
0.123M - x x
HClO(aq)+ H 2 O(l) ⇌ H 3 O+(aq) + ClO-(aq)
0M
+x
x
Write the balanced equation for the
acid in H 2 O…producing H 3 O+.
Just like any other equilibrium problem, plug in what we know.
Since H 2 O is a pure liquid (and does not change appreciably over time), we don’t need to deal with it
[HClO]
[H 3 O][ClO] a
K
0.123-x
(x)(x) 2.95x
8
x 2.95x
2 8
9 2 3.63x10 x
x = 6.02x10-^5
[H 3 O
+ ] = x = 6.02x
- 5 M
pH = - log(6.02x10-^5 M)
pH = 4.
x1000.0487%
6.02x
Less than 5%, good assumption -^5
-5 2
0.123- 6.02x
(6.02x10)
Determining the pH of a Weak Base Solution
Determine the pH of a 0.555M aqueous solution of NH 3. p Kb = 4.75.
I
E
C
nitial
hange
quilibirum
0.555M 0M
0.555M - x x
NH 3 (aq)+ H 2 O(l) ⇌ OH-(aq) + NH 4 +(aq)
0M
+x
x
Write the balanced equation for the base in
H 2 O…producing OH-^ and the conjugate acid, NH 4 +^.
[NH]
[OH][NH]
3
4 b
K
0.555-x
(x)(x) 1.78x
x 1.78x
2 5
6 2 9.88x10 x
x = 3.14x10-^3
[OH-] = x = 3.14x10-^3 M
pOH = - log(3.14x
- 3 M)
pOH = 2.
pH = 14.00-2.503 = 11.
x1000.566%
3.14x
Less than 5%, good assumption -^4
-3 2
0.555- 3.14x
(3.14x10)
First, we need to deal with pKb.
4.75 5 b 10 1.78x
K
Calculating K
a
from pH
If 1.55M solution of an acid has a pH of 5.15, determine the K a for the acid
(assume it is monoprotic).
I
E
C
nitial
hange
quilibirum
1.55M 0M
1.55M - x x
HA(aq)+ H 2 O(l) ⇌ H 3 O
+ (aq) + A
- (aq)
0M
+x
x
[H+] = 10-5.15^ = 7.1x10-^6 M
Since we know pH, we also
know [H+] at equilibrium
7.1x
7.1x10-^6 M = x
1.55M – 7.1x
[HA]
[H 3 O][A]
a
K
(1.55-1.7x10)
(7.1x10 )(7.1x10) 6 -
-6 -
K a
a 3.3x
K
“Short-Cuts”
initial
2
c [weakacidorbase]
x K
This assumes that K c 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!!
Percent ionization
x
[HA]
[HO]
%ionizationofanacid
initial
3 equilibrium
x [B]
[OH]
%ionizationofabase
initial
equilibrium
Percent ionization is the amount of a weak acid or base that has become H 3 O
or OH
Determine the percent ionization
if 0.76M aqueous solution of an
unknown acid mixed with water
resulted in a an [H 3 O+] of
9.05x10-^4 M
x1000.12% 0.76M
9.05x10M %ionization
4
Determine the percent ionization
if 5.00M aqueous solution of an
unknown acid mixed with water
resulted in a pH of 3.
x1000.015% 5.00M
7.6x10M %ionization
4
[H 3 O
] = 10
= 7.6x
M
Determine the percent ionization of a 0.100M
aqueous solution of NH 3. p K b = 4.75.
initial
2
c [weakacidorbase]
x Let’s use the “short-cut” K
K b = 10-4.75^ = 1.8x10-^5
x 1.8x
2 5
2 6 x 1.8x
x = 1.3x10-^3 = [OH-]
x1001.3% 0.100M
1.3x10M %ionization
3
Percent Ionization and Concentration
initial
2
c [weakacidorbase]
x K
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.
x 1.0x
2 5
x 0.0036M [H 3 O^ ]
x 1.0x
2 5
x 0.00063M
x 1.0x
2 5
x 0.00032M
0
1
2
3
0 0.5 1 1.
% ionization
Initial acid concentration (M)
x1000.28% 1.28M
0.0036M
%ionization
x1001.6% 0.040M
0.00063M
%ionization
x1003.2% 0.010M
0.00032M
%ionization
The more concentrated a weak base solution, the lower
the % ionization
0
Hydronium concentration at 0 0.2 0.4 0.6 0.8 1 1.2 1.
equilibrium (M)
Initial weak acid concentration (M)
Binary Acid Strengths
H 2
O HF
HCl
HBr
HI
p K a =15.74 p K a = 3.
H 2 S
H 2
Te
p K a ≈ - 10
p K a ≈ - 8
p K a = 6.9 p K a ≈ - 6
p K a = 2.
Acid Strength
Acid Strength
Electronegativity
Bond Length
Moving down a group, the bond
length gets longer. Longer bond
length = weaker X-H bond
p K a = 3.
H 2 Se
0
5
10
15
20
80 100 120 140 160 180
pKa
Bond Length (pm)
H 2 O
H 2 S
H 2 Se
H 2 Te
HCl
HF
HBr HI
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.
Shift in
electron
density away
from O-H
makes the
bond weaker
(more acidic)
Less electron
density at
the H
Carboxylic Acid Strengths
Explain the following trend in acid strength:
Oxoacid Strengths – Inductive effects
Acid Strength
Electronegativity of central atom
Across each row, electron
density of being shifted
away from the O-H bond.
Stabilizes negative charge
after proton loss
