03/01/2017

Arrhenius Acid/Base
◦ Acid: Produces H+ in water
◦ Base: Produces OH- in water

Chem 112
Dr. Kevin Moore

Focus is generally in water
◦ Acid: Substance which donates H+ to another
◦ Base: Substance which accepts a donated H+

◦ In water, H+ has to be stabilized
H
H
O H+
H
H
◦ Represent as either [H+] or [H3O+]
O
+
Bronsted-Lowry Acid/Base
What happens when an acid dissolves in
water?
HCl ( g )  H 2 O(l )  H 3 O  (aq )  Cl  (aq )

H
What happens when base dissociates in
water?
2
NaOH ( s) 
 Na  (aq )  OH  (aq )
H O

What if the base does not show OH- in the
compound?
NH 3 ( g )  H 2 O(l )  NH 4  (aq )  OH  (aq )
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
Any acid and base pair which differ by H+
◦
◦
◦
◦

Some ions can be in more than one pair
◦
◦
◦
◦
◦

H2O/H3O+
NH3/NH4+
HClO2/ClO2HC2H3O2/C2H3O2-
HSO4-/SO4-2
HSO4-/H2SO4
H2O/H3O+
H2O/OHAmphoteric – substance can act as base or acid
Strong Acid – complete donation of protons
◦ Complete dissociation of acid
◦ HCl, HBr, HI, HNO3, H2SO4, HClO4

NH 3 ( g )  H2 O(l )  NH 4  (aq )  OH  (aq )
Base
Acid
Conj.
Acid
Conj.
Base
HC2 H 3 O2 (aq )  H 2 O(l )  H 3 O  (aq )  C2 H 3 O2  (aq )
Acid
Base
Strong Acid (HCl)
Conj.
Acid
Conj.
Base
Weak Acid (HF)
Weak Acid – partial donation of protons
◦ Incomplete dissociation of acid
 Typically very low (< 5%)

Strong Base – complete dissociation of OH◦ Group IA & IIA soluble Hydroxides (Sr, Ba)
◦ Solubility increases down the periodic table
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Strength
Acid
Base
Strength
Strong
HCl
Cl-
Negligible
H2SO4
HSO4-
H3O+
H2O
HNO3
NO3-
H3PO4
H2PO4-
HF
F-
HC2H3O2
C2H3O2-
H2S
S-2
H2CO3
HPO4-2
HPO4-2
PO4-3
NH4+
Stronger Bases have weaker conjugate acids
◦ Cation of a strong base is not considered acidic
◦ Na+, K+
NH3
H2O
OH-
OH-
O-2
H2
H-
CH4


HCO3-1
H2PO4-
Stronger Acids have weaker conjugate bases
◦ Anion of a strong acid is not considered basic
◦ Cl-, NO3-
SO4-2
HSO4-
Negligible

Strong
CH3-
Normal Water Dissociation
2 H2 O(l ) 
 H 3 O  (aq )  OH  (aq )
0.00
+x
x
0.00
+x
x
K w  [10
.  10  7 ][10
.  10  7 ]
K w  10
.  10 14
K w  [ H 3 O  ][OH  ]
At equilibrium: [H3O+]=1.0 x 10-7
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
[H3O+] = [1.0 x 10-7] is a neutral solution

◦ [H3O+] > [OH-] ===> Acidic
◦ [H3O+] < [OH-] ===> Basic
◦ [H3O+] = [OH-] ===> Neutral

pH – “Power of Hydrogen”
pH   log[ H 3 O  ]
Calculate the pH of 0.21 M HNO3.
pH   log[0.21]
pH  0.68
Blood has a pH of 7.40. Find the [H3O+] and
[OH-]
Kw
[OH  ] 

 7 .40
[ H3O  ]
[ H 3 O ]  10

[ H 3 O  ]  3.98  10  8
K w  [ H 3 O  ][OH  ]

pH  7  Acidic
pH  7  Basic
.  10 14
10
3.98  10  8
[OH  ]  2.51  10  7
[OH  ] 
Acid-Base Indicator
◦ Chemical which changes color in certain pH ranges
◦ Many known indicators
◦ pH meter – device which detects [H+] using
electrical potential
pH  7  Neutral
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pOH   log[OH  ]

pK w   log[10
.  10 14 ]
pK w  14.
Calculate the pOH of 0.28g of CaO in a 1.00 L
solution
CaO( s)  H 2 O(l )  Ca  2 (aq )  2OH  (aq )
pH  pOH  pK w  14.
 1 mol 
  0.0050 mol CaO
0.28 g 
. g
 561
 2 mol OH  
  0.010 mol OH 
0.0050 mol CaO
 1 mol CaO 
pOH  2.00

Calculate the pH & pOH of a 0.15 M HCl
solution.
pH   log 015
.   0.82
pOH  14. 0.82  1318
.

H+ is not all dissociated
◦ Requires an equilibrium constant
◦ Ka
HA(aq )  H 2 O(l ) 
 H 3 O  (aq )  A  (aq )
[ H 3 O  ][ A  ]
Ka 
[ HA]
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
A 0.25 M HF solution has a pH=2.04. What is
Ka?
HF (aq )  H 2 O(l ) 
 H 3 O  (aq )  F  (aq )

HOCl (aq )  H 2 O(l ) 
 H 3 O  (aq )  ClO  (aq )
[ H 3 O  ]  10  4.23  589
.  10  5
[ H 3 O  ]  10  2.04  0.0092
[0.0092][0.0092]
Ka 
 3.4  10  4
[0.25  0.0092]

What is the [H+] for a 0.10 M HCN solution
which has a Ka=4.9 x 10-10?
HCN (aq )  H 2 O(l ) 
 H 3 O  (aq )  CN  (aq )
0.10
-x
0.10 – x
Ka 
[ H 3 O  ][CN  ]
[ HCN ]
4.9  10 10
[ x ][ x ]

010
. x
0.00
+x
x
0.00
+x
x
(010
. )(4.9  10 10 )  x 2
A 0.10 M HOCl solution has a pH of 4.23.
What is Ka?
[589
.  10  5 ]2
Ka 
 35
.  10  8
[010
.  589
.  10  5 ]

Dissociation has two sources of H+
◦ Acid itself
◦ Water

Relationship between Ka & Kw
◦ Ka>>Kw
 Acid normally dominates production of H+ (ignore H2O)
◦ Ka~Kw
 Depends on the concentration of the acid
x  7.0  10  6
[H3O+]=[CN-]=7.0x10-6
[HCN]=0.10
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
Water’s contribution is based on [OH-]
allowed by the weak acid
pH  515
.
pOH  8.85
[OH  ] H2O  14
.  10  9

[ H 3 O  ] H2O  14
.  10  9
Consider a 0.050 M HF solution. What is the
pH? (Ka = 3.4 x 10-4)
HF (aq )  H 2 O(l ) 
 H 3 O  (aq )  F  (aq )
0.050
-x
0.050 – x
Water contribution is
barely negligible
0.00
+x
x
0.00
+x
x
x2
 3.4  10  4
0.050  x
x  4.1  10  3 (ignoring x )
pH (ignoring x )  2.39
0.050  0.0041  0.046
pH ( with x )  2.40
Ka 
Actually: 0.0040

If the H+ does not change the original acid
concentration by more than 5%, then not
using quadratic formula is valid.
4.1  10  3
 100  8.2%
0.050

A Vitamin C tablet contains 250. mg of
C6H8O6; Ka= 8.0 x 10-5. What is the pH in a
250. mL solution?
 1 mol  141
.  10  3 mol
 
 5.64  10  3 M
0.250 g 
0.250 L
 177.0 g 
8.0  10  5 
x2
5.64  10  3  x x  7.2  10  4
6.7  10  4
[ H 3 O  ]  [ Ascorbate]  7.2  10  4

100

119%
.
5.64  10  3
pH  315
.
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
[ H3O  ]
 100  % diss
[ HA]0

Dissociation of Acids containing multiple
acidic hydrogens is determined sequentially
◦ H2SO3
◦ H2CO3
◦ H3PO4

Ka values are always weaker for each
successive hydrogen
Calculate the % dissociation in 0.050 M HF.
0.0040
 100  8.0 %
0.050

General Form
H 2 A(aq )  H 2 O(l ) 
 H 3 O  (aq )  HA  (aq )
K a1
HA  (aq )  H 2 O(l ) 
 H 3 O  (aq )  A  2 (aq )
Ka 2
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
Calculate the pH and the concentration of each
ion found in a solution of 0.10 M H2SO3
Ka1=0.015
Ka2=6.3x10-8
2
Ka1 
x
010
. x
x=0.032
[H3O+]=0.032
[HSO3-]=0.032
[H2SO3]=0.068


◦ NH3
◦ CH3NH2
◦ (CH3)2NH
H 2 SO3 (aq )  H 2 O(l ) 
 H 3 O  (aq )  HSO3 (aq )
HSO3 (aq )  H 2 O(l ) 
 H 3 O  (aq )  SO3 2 (aq )

Ka 2 
0.032
+x
[0.032  x ]x
[0.032  x ]
Conjugate Base of weak acid
◦
◦
◦
◦
HSO3 (aq )  H 2 O(l ) 
 H 3 O  (aq )  SO3 2 (aq )
0.032
-x
Nitrogen based Organic compounds
0.00
+x
pH  149
.
NO2SO3-2
FS-2
Treat x as negligible: x = 6.3 x 10-8
[SO3-2]=6.3 x 10-8
Base Dissociation
B(aq )  H 2 O(l ) 
 BH  (aq )  OH  (aq )


Calculation involves OH- rather than H3O+
Calculate the pH of 0.15 M NH3 (Kb=1.8 x 10-5)
NH 3 (aq )  H 2 O(l ) 
 NH 4 (aq )  OH  (aq )

NH3
NH 3 (aq )  H 2 O(l ) 
 NH 4 (aq )  OH  (aq )
0.15
-x
x2
Kb 
015
. x
0.00
+x
0.00
+x
x  0.0016  [OH  ]
pOH  2.80
0.0016
.
 100  11%
.
015
pH  1120
.
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
Calculate the pH of a 0.045 M solution of
C5H11N (Kb = 1.3 x 10-3).
C5 H11 N (aq )  H 2 O(l ) 
 C5 H11 NH  (aq )  OH  (aq )
0.045
-x
Kb 
0.00
+x
x2
0.045  x
x  0.0076
x 2  13
.  10  3 x  585
.  10  5  0
x
 13
.  10
0.0076
 100  17% x  7.0  10  3
0.045

0.00
+x
3
 2.36  10
2
  OH  
4
pOH  2.15
pH  1185
.
NH 3 (aq )  H 2 O(l ) 
 NH 4 (aq )  OH  (aq ) Kb
NH 4 (aq )  H 2 O(l ) 
 H3 O  (aq )  NH 3 (aq ) Ka/
2 H 2 O(l ) 
 H 3 O  (aq )  OH  (aq )
K w  Ka/ Kb
Kw
Ka/ 
Kb
Ka/ 
Kw
10
.  10 14
 5.6  10 10
18
.  10  5
C5H11N has a Kb=1.3 x 10-3, what is the
/
conjugate acid and its K a
C5 H11 N (aq )  H 2 O(l ) 
 C5 H11 NH  (aq )  OH  (aq )
Ka/ 

10
.  10 14
 7.7  10 12
13
.  10  3
Calculate the pH of a solution of 0.21 M
C5H11NH+.
x2
0.21  x
x  13
.  10  6
Ka/ 
[ H 3 O  ]  13
.  10  6
pH  5.90
[OH  ]  7.7  10  9
[ H3 O  ] H2O  7.7  10  9
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
Ions of Strong Acids & Bases are Neutral
Strong Acid + Strong Base ---> Neutral Salt

Strong Acid + Weak Base ---> Acidic Salt



◦ Neutral Ion + Conjugate Acid

Weak Acid + Strong Base ---> Basic Salt
Weak Acid + Weak Base ---> It depends
◦ Conjugate Base + Conjugate Acid

Conjugate Acid from a Weak Base



Neutral Anion
NH4Cl
CH3NH3NO3
(CH3)3NHBr
Anions are neutral
◦ Ions from strong acid




◦ Nitrogen containing…positive charged

Cations are neutral
◦ Group IA & IIA cations
◦ Neutral Ions
◦ Conjugate Base + Neutral Ion


NaNO3
KBr
BaCl2
Sr(ClO4)2
Conjugate base from a Weak Acid
◦ Anion not from strong acid





Neutral Cation
KNO2
SrSO3
Na2CO3
Li2HPO4
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
Conjugate Acid from Weak Base
/
◦ K





/
◦ Kb
/
/
K a>K b---> Acidic
/
/
K a<K b---> Basic
/
/
K a=K b---> Neutral
NH4F (Kb = 1.8 x 10-5; Ka = 6.7 x 10-4)


Higher charge

Smaller ion
◦ More acidic
◦ More acidic

Na+<Mg+2<Al+3
Cr+2<Cr+3<Cr+6

Size of Atom (HX)

CH3NH3NO2 (Kb = 4.4 x 10-4; Ka = 6.0 x 10-4)


Acidic
Metals hydrate water
◦ Al(H2O)6+3
a
Conjugate Base from Weak Acid


Acidic
Bond Polarity
◦ For similar sized atoms, the polarity of bond
determines acid strength
 Difference of Electronegativity
Compound
HF
H2O
NH3
CH4
Ka
◦ polarity increases as the atom becomes larger
 Energy to dissociate decreases
Compound
ΔEN
7.2x10-4
1.8
1.0x10-14
1.4
1.0x10-33
0.9
1.0x10-49
0.4
Ka
BDE (kJ/mol)
7.2x10-4
569
HCl
1.0x106
430
HBr
1.0x109
370
HI
3.0x109
300
HF
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
Hydrogen Attached to Oxygen

◦ EN of central atom determines pull of electrons
Compound


Ka
EN
3.0x10-8
3.0
HBrO
2.8x10-9
2.8
HIO
3.2x10-11
2.5
HClO
Often acid strength is represented as pKa
Increases exactly like pH
◦ Strongest is lowest value
◦ 2.5 > 3.5

What is the pKa of Vinegar (Ka = 1.8 x 10-5)
pKa   log[18
.  10  5 ]  4.74
Oxidation State of Central Atom
Compound
Ka
Charge on Cl
~1.0 x 1010
+7
HClO3
1.0x101
+5
1.0x10-2
+3
HClO
3.0x10-8
+1
HClO4
HClO2

Use pKa to determine relative acid strength
Compound
Ka
pKa
HC2H3O2
1.8 x 10-5
4.74
HIO3
1.7 x 10-1
0.77
H2SO3
1.5 x 10-2
1.82
H3BO3
5.8 x 10-10
9.24
HNO2
4.5 x 10-4
3.35
HC6H7O6
8.0 x 10-5
4.10
HCO2H
1.8 x 10-4
3.74
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03/01/2017
Lewis Acid
◦ Substance which accepts an electron pair
◦ More than just H+
◦ Metal ions (Al+3, Cr+6)

Lewis Base
◦ Substance which donates an electron pair
◦ Organic Nitrogen compounds are pair donors
F
F
B
F
H
+
N H
••

F
H
acid
base
F
H
B
N H
F
H
No protons donated or accepted!
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