I.
Acid Theory
A.
Classifications
1) Acids
a) Have a sour taste
b) Can dissolve metals
2)
Bases
a) Have a bitter taste
b) Feel slippery
3)
Arrhenius Definition
a) Acids produce H+ in water
b) Bases produce OH- in water
c) Applies only to aqueous solutions
d) Allows for only one kind of base (OH-)
4)
Bronsted-Lowery Definition
a) Acid is an H+ donor
b) Base is an H+ acceptor
c) HCl + H2O
acid
H O
+
base
H Cl
H
5)
“The differences between the various acid-base
concepts are not concerned with which is right,
but which is most convenient to use in a
particular situation.” James E. Huheey
H3O+ + Clhydronium ion
H O H
+
H
General Acid Equation
acid
base
HA(aq) + H2O(l)
a)
b)
c)
d)
conjugate conjugate
acid
base
+
H3O (aq) + A-(aq)
Conjugate base = what is left after H+ leaves acid
Conjugate acid = base + H+
Conjugate acid-base pair are related by loss/gain of H+
Competition for H+ by A- and H2O; strongest base wins
Cl
6)
Ka = acid dissociation constant
[H 3O ][A  ] [H  ][A  ]
Ka 

[HA]
[HA]
7) Example: Write simple Ionizations for:
HCl, HC2H3O2, NH4+, C6H5NH3+, Al(H2O)63+
8)
Bronsted-Lowery theory allows for non-aqueous solutions
conjugate conjugate
acid
base
H
acid
base
H
H N
+
H Cl
H
H N H
H
9)
More examples:
+
Cl
B.
Acid Strength
1) Acid strength describes the equilibrium position of the ionization reaction
HA + H2O
H3O+ + A2)
Strong Acid = equilibrium lies far to the right
a) Almost all HA has ionized to H+ and A- ([H+] = [HA]0)
b) A strong acid has a weak conjugate base
i. To ionize fully, the conjugate base must have low proton affinity
ii. The conjugate base must be weaker than water
3)
Weak Acid = equilibrium lies far to the left
a) Almost all HA remains unionized ([H+] << [HA]0)
b) A weak acid has a strong conjugate base
c) The conjugate base is much stronger than water
Strong acid
Weak Acid
4)
Common Strong Acids (Ka = very large)
a) Sulfuric Acid = H2SO4 is a diprotic acid (has 2 ionizable protons)
H2SO4
H+ + HSO4- Ka = ∞
HSO4H+ + SO42Ka = 1.2 x 10-2
b)
Hydrochloric Acid = HCl is a monoprotic acid (has 1 ionizable proton)
HCl
H+ + Cl-
c)
Nitric Acid = HNO3
d)
Perchloric Acid = HClO4
5)
Oxyacids = acids with the ionizable proton attached to oxygen
HClO4 HNO3 H2SO4 are all oxyacids
6)
Hydrohalic Acids = acids with the ionizable proton attached to a halide
HCl, HBr, HF, HI are all hydrohalic acids
7)
Common Weak Acids
a) Phosphoric Acid = H3PO4
H3PO4
H+ +
H2PO4H+ +
HPO42H+ +
8)
(triprotic acid)
H2PO4- Ka = 7.5 x 10-3
HPO42- Ka = 6.2 x 10-8
PO43Ka = 4.8 x 10-13
b)
Nitrous Acid = HNO2 Ka = 4.0 x 10-4
c)
Organic Acids have a carbon backbone and a carboxyl group
i. Acetic Acid = CH3COOH (HC2H3O2) Ka = 1.8 x10-5
O
ii. Benzoic Acid = C6H5COOH Ka = 6.4 x10-5
iii. Only the OH hydrogen is acidic
C
Example: Relative Basicity of H2O, F-, Cl-, NO2-, CN- (Table 15.5)
OH
C.
Water as an Acid and Base
1) An amphoteric substance can behave as an acid or a base (water)
2) Autoionization of water (reaction with itself)
H2O(l) + H2O(l)
H3O+(aq) + OH-(aq)
3)
Ionization constant for water = KW = [H3O+][OH-] = [H+][OH-]
a) For any water solution at 25 oC, [OH-] x [H+] = KW = 1 x 10-14
b) Neutral solutions (pure water) have [OH-] = [H+] = 1 x 10-7
c) Acidic solutions: [H+] > [OH-]
d) Basic solutions: [OH-] > [H+]
e)
Example: Calculate [OH-] or [H+] for the following:
i. [OH-] = 1 x 10-5 M
ii. [OH-] = 1 x 10-7 M
iii. [H+] = 10 M
4) Kw is temperature dependent. At 60 oC, KW = 1 x 10-13
Example: Is water autoionization exothermic or endothermic? [H+]??
D.
pH Scale
1) pH = -log[H+] (simplifies working with small numbers)
2) If [H+] = 1.0 x 10-7, pH = -log(1 x 10-7) = -(-7.00) = 7.00
3)
Number of decimal places in a log:
The number of sig. fig’s in the number is
how many decimal places you keep when
you take the log
4)
Other p Scales:
a) pOH = -log[OH-]
b) pKa = -logKa
5) pH changes by 1 unit for every power
of 10 change in [H+]
a) pH = 3 [H+] = 10 times the [H+] at pH = 4
b) pH decreases as [H+] increases
(pH = 2 more acidic than pH = 3)
6)
Example: Find pH and pOH for [OH-] = 1 x 10-3 and [H+] = 1.0 M
7)
Relationship of pH and pOH
a) KW = [H+][OH-]
b) -logKW = -log[H+] – log[OH-]
c) -log(1 x 10-14) = pH + pOH = 14
d)
II.
Example: Find [H+], [OH-], and pOH for sample of pH = 7.41
Acid-Base Problem Solving
A.
Calculating pH of Strong Acid Solutions
1) Solution labels tell what went into the solution, not what is present now
a) 1.0 M HCl
b) 1.0 M H+ and 1.0 M Cl2) Major Species = those present in large amount
a) For 1.0 M HCl = H+, Cl-, H2O (not OH-)
b) Identifying these is the key to solving acid-base problems
B.
3)
Find pH of 1.0 M HCl
a) Since HCl is a strong acid, we assume [H+] = [HA]0 = 1.0 M
b) Ionization of water only produces [H+] = 1 x 10-7 M (ignore this)
c) pH = - log(1) = 0
4)
Example: Find pH of 0.1 M HNO3 and 1 x 10-10 M HCl
pH of Weak Acid Solutions
1) pH of 1.00 M HF? Ka = 7.2 x 10-4
2) Write the major species: HF, H2O
3) Which can furnish H+?
a) Dominant = HF
H+ + F- Ka = 7.2 x 10-4
b) Small amount = H2O
H+ + OH- KW = 1 x 10-14
4)
Do equilibrium calculation based on HF only
2
HF
H+ + F[H  ][F  ]
x
4
K


7.2
x
10

a
Inititial 1
0
0
[HF]
1 x
Change -x
+x
+x
Equil. 1-x
x
x
Assume x = small, then 1- x ≈ 1
x2 = 7.2 x 10–4
x = 2.7 x 10–2
pH = 1.57
5)
Is our approximation valid?
a) Most Ka values are only known to within 5% error
b) If x is really small, x/[HA]0 < 5% and our approximation is valid
c) If x/[HA]0 > 5%, we have to use the quadratic, no approximation
d) 2.7 x 10-2 / 1.00 x 100% = 2.7% so our approximation is ok
6)
Example: Find pH of 0.1 M HOCl Ka = 3.5 x 10-8
7) Example: Find pH and [CN-] of a solution of
1.0 M HCN (Ka = 6.2 x 10-10) and 5.0 M HNO2 (Ka = 4.0 x 10-4)
8)
Percent Dissociation = amount dissociated / initial conc. x 100%
a) For 1.00 M HF we found [H+] = 2.7 x 10-2
Percent Dissociated = (2.7 x 10-2 / 1.00) x 100% = 2.7%
b) Strong acids are 100% dissociated always
c) For weak acids, percent dissociation increases as acid is diluted
Example: %Dissociation of 1 M and 0.1 M Acetic acid Ka = 1.8 x 10-5
d)
Why does dilution increase percent dissociation of weak acids?
[H  ][A  ]
x2
Ka 

[HA]
[HA] 0
If we dilute by 10 times:
[HA] 0
x
[H ]  [A ] 
and [HA] 
10
10


Now if we calculate Q, we can see which way the equilibrium will shift
 x  x 
10  10  1  x 2  1
  K a
Q
 
 [HA] 0  10  [HA] 0  10


 10 
Since Q < Ka, the equilibrium will shift to the right; %Dissoc.Increases
e)
Example: Find Ka for 0.1 M Lactic Acid when %Dissoc. = 3.7%