I. Acid TheoryA. 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 H3O+ + Cl-
5) General Acid Equation
HA(aq) + H2O(l) H3O+(aq) + A-
(aq)
a) Conjugate base = what is left after H+ leaves acid
b) Conjugate acid = base + H+
c) Conjugate acid-base pair are related by loss/gain of H+
d) Competition for H+ by A- and H2O; strongest base wins
acid base hydronium ion
H O
H
H Cl H O
H
H Cl+ +
acid baseconjugateacid
conjugatebase
“The differences between the various acid-baseconcepts are not concerned with which is right,but which is most convenient to use in a particular situation.” James E. Huheey
6) Ka = acid dissociation constant
7) Example: Write simple Ionizations for:
HCl, HC2H3O2, NH4+, C6H5NH3
+, Al(H2O)63+
8) Bronsted-Lowery theory allows for non-aqueous solutions
9) More examples:
[HA]
]][A[H
[HA]
]][AO[HK 3
a
H N
H
H
H Cl H N
H
H
H
Cl+ +
acidbase
conjugateacid
conjugatebase
B. Acid Strength
1) Acid strength describes the equilibrium position of the ionization reaction
HA + H2O H3O+ + A-
2) 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 = ∞
HSO4- H+ + SO4
2- Ka = 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 (triprotic acid)
H3PO4 H+ + H2PO4- Ka = 7.5 x 10-3
H2PO4- H+ + HPO4
2- Ka = 6.2 x 10-8
HPO42- H+ + PO4
3- Ka = 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
ii. Benzoic Acid = C6H5COOH Ka = 6.4 x10-5
iii. Only the OH hydrogen is acidic
8) Example: Relative Basicity of H2O, F-, Cl-, NO2-, CN- (Table 15.5)
C
O
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 + H2O H3O+ + OH-
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) Example: Find [H+], [OH-], and pOH for sample of pH = 7.41
II. Acid-Base Problem SolvingA. 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 Cl-
2) 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
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
B. 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
HF H+ + F-
Inititial 1 0 0
Change -x +x +x
Equil. 1-x x x
x
x
110 x 7.2
[HF]
]][F[HK
24
a
Assume x = small, then 1- x ≈ 1x2 = 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?
If we dilute by 10 times:
Now if we calculate Q, we can see which way the equilibrium will shift
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%
0
2
a [HA]
x
[HA]
]][A[HK
10
[HA][HA] and
10
x][A][H 0
a0
2
0
K10
1
[HA]
x
10
1
10[HA]
10x
10x
Q
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