Chapter 14
• Arrhenius
– Acid – create H+ in water
– Base – create OH- in water
• Bronsted-Lowery
– Acid – donates proton (H+)
– Base – accepts proton (H+)
• Hydronium ion - H3O+
• Conjugate pairs
– HCN/CN- HCl/Cl- NH3/NH4+
• Acid dissociation constant
– Equilibrium expression for its dissociation
– Ka
– HF  H+ + F– Ka = [H+][F-] / [HF]
• Strong vs weak
• Oxyacids
– HNO3, HClO4, H2SO4
• Organic Acids
– COOH group HC2H3O2 (CH3COOH)
• Monoprotic, diprotic, triprotic
– HCl, H2SO4, H3PO4
• Water as an acid and base
– H2O + H2O  H3O+ + OH-
• Amphoteric
• Dissociation constant for water
– Kw = [H3O+][OH-] = 1.0 x 10-14
– [OH-] = 1.0 x 10-5 [H3O+] = ?
– [H3O+] = 1.0 x 10-14 / 1.0 x 10-5 = 1.0 x 10-9
• pH
– pH = - log[H+]
• pOH
– pOH = - log[OH-]
• pH + pOH = 14
• Calculate the pH and pOH of a 1.0 x10-9M
HCl solution.
– pH = - log(1.0x10-9) = 9.0 pOH = 14 – 9 = 5.0
– What is the [H+] if the pH = 4.4
– pH = -log[H+] [H+] = 10-pH
– [H+] = 10-4.4 = 3.98 x 10-5 = 4.0 x 10-5M
• pH of strong acids
– Completely dissociates so [acid] = [H+]
• pH of weak acids
– Have to do an equilibrium problem using Ka
– Follow the same process
– HF  H+ + F– Ka = [H+][F-] / [HF] = 7.2 x 10-4
– Calculate the pH of a 1.0 M HF solution
– Chem.
– [H+]
– [F-]
– [HF]
Initial
0
0
1.0
Equil.
+x
+x
1.0 – x = 1.0 (small x)
– Ka = [H+][F-] / [HF] = 7.2 x 10-4
– 7.2 x 10-4 = x2 / 1.0 x = 2.7 x 10-2 M
– pH = -log(2.7 x 10-2) = 1.57
• pH of a mixture of acids
– Focus on the strongest acid
– Our HF example also contained H2O but we
can ignore it since its constant is 1.0 x 10-14
and HF is much larger, 7.2 x 10-4
• Percent dissociation
– %diss. = [H+] / [HA]o x 100
• What is the pH of an aqueous solution of
1.00M HCN and 5.00M HNO2.
– HCN Ka = 6.2 x 10-10
– HNO2 Ka = 4.0 x 10-4
– H2O Kw = 1.0 x 10-14
– Strongest acid is HNO2 so we use it
• HNO2 Ka = 4.0 x 10-4
– HNO2  H+ + NO2-
•
•
•
•
Chem.
HNO2
H+
NO2-
[init.]
5.00
0
0
[equil]
5.00-x = 5.00
+x
+x
– 4.0 x 10-4 = x2 / 5.00
– x = 4.5 x 10-2
– pH = -log(4.5 x 10-2) = 1.35
(x is small)
• Bases
– We calculate [OH-] just like [H+] but use Kb
instead of Ka
• pH of strong bases
– Completely dissociates so the [Base] = #[OH-]
• pH of weak bases
– Focus on the strongest base present and do
an equilibrium problem
• Calculate the [OH-] for a 15.0 M NH3
solution, Kb = 1.8 x 10-5.
– NH3 + H2O  NH4+ + OH– Kb = [NH4+][OH-] / [NH3] = 1.8 x 10-5
– Chem.
– NH4+
– OH– NH3
[Init.]
0
0
15.0
[Equil]
+x
+x
15.0 – x = 15.0
– Chem.
– NH4+
– OH– NH3
[Init.]
0
0
15.0
[Equil]
+x
+x
15.0 – x = 15.0
– Kb = [NH4+][OH-] / [NH3] = 1.8 x 10-5
– 1.8 x 10-5 = x2 / 15.0
– x = 1.6 x 10-2
– pOH = -log(1.6 x 10-2) = 1.80
– pH = 14 – 1.80 = 12.20
• Strong bases make weak adics
• Strong acids make weak bases
• So the larger the Ka the smaller the Kb
• Ka x Kb = Kw
• Polyprotic acids
– Dissociate one proton at a time
– Follow the same steps as an equil. problem
– Calculate the pH as well as the concentration
of all other chemicals present in a 5.0M
H3PO4 aqueous solution.
– H3PO4 Ka1 = 7.5 x 10-3
– H2PO4- Ka2 = 6.2 x 10-8
– HPO42- Ka3 = 4.8 x 10-13
• H3PO4  H+ + H2PO4• Ka1 = 7.5 x 10-3 = [H+][H2PO4-] / [H3PO4]
– Chem.
– H+
– H2PO4– H3PO4
[init]
0
0
5.0
[equil]
+x
+x
5.0 – x = 5.0
– 7.5 x 10-3 = x2 / 5.0
– x = 0.19 = [H+] = [H2PO4-]
• H2PO4-  H+ + HPO42• Ka2 = 6.2 x10-8 = [H+][HPO42-]/[H2PO4-]
– Chem
– H+
– HPO42– H2PO4-
[init]
0.19
0
0.19
[equil]
0.19+x = 0.19
+x
0.19 – x = 0.19
– 6.2 x10-8 = (0.19)x / 0.19 x = 6.2 x10-8
– [HPO42-] = 6.2 x 10-8
• HPO42-  H+ + PO43• Ka3 = 4.8 x 10-13 = [H+][PO43-]/[HPO42-]
– Chem
– H+
– PO43– HPO42-
[init.]
0.19
0
6.2 x 10-8
[equil.]
0.19 + x = 0.19
+x
6.2 x10-8 -x = 6.2x10-8
– 4.8 x 10-13 = (0.19)x / 6.2 x 10-8
– x = 1.6x10-19
– [PO43-] = 1.6 x 10-19
• [OH-]
– Kw = 1.0 x 10-14 = [H+][OH-]
– 1.0 x 10-14 = (0.19)[OH-]
– [OH-] = 5.3 x 10-14
• Sulfuric acid
– Similar to phosphoric acid except the first
dissociation is complete and we can use that
info for the second dissociation problem.
– Calculate the [SO42-] in a 1.0M aqueous
H2SO4 solution.
– H2SO4  H+ + HSO4– [H2SO4] = [H+] = [HSO4-] = 1.0
• HSO4-  H+ + SO42• Ka2 = 1.2 x 10-2 = [H+][SO42-] / [HSO4-]
– Chem.
– H+
– SO42– HSO4-
[init.]
1.0
0
1.0
– 1.2 x 10-2 = (1.0)x / 1.0
– [SO42-] = 1.2 x 10-2 M
[equil.]
1.0 + x = 1.0
+x
1.0 – x = 1.0
x = 1.2 x 10-2
• Acid/Base properties of salts
– Salts from strong acids/bases produce neutral
solution, NaCl, KNO3 etc
– Salts from weak acid, strong base produce
basic solutions, NaF, KC2H3O2
– F- + H20  HF + OH– Salts from strong acid, weak base produce
acidic solutions, NH4Cl
– NH4+ + H2O  NH3 + H3O+
– Highly charged metal ions like Al3+ produce
acidic solutions when hydrated, because of
the large charge it is easier to release H+
• Ka x Kb = Kw
• So if you know a chemicals Ka or Kb you
can determine the corresponding Ka / Kb
as needed
• Calculate the pH for a 0.1 M NH4Cl
aqueous solution. Kb = 1.8 x 10-5 for NH3
• NH4+ reacts with water like an acid so we
need it’s Ka. We get it from the Kb
–
–
–
–
Ka = Kw / Kb = 1.0 x 10-14 / 1.8 x 10-5 = 5.6 x 10-10
NH4+ + H2O  NH3 + H3O+
Chem
[init.]
[equil.]
NH4+
0.1
0.1 – x = 0.1
– NH3
– H3O+
0
0
+x
+x
– Ka = 5.6 x 10-10 = [NH3][H3O+] / [NH4+]
– 5.6 x 10-10 = x2 / 0.1
– x = 7.5 x 10-6 = [H3O+]
– pH = -log (7.5 x 10-6) = 5.13
• Structure effect HClO vs HClO4
– The extra oxygens draw the electrons away
from the hydrogen allowing it to be released
more easily. Thus HClO4 is stronger
– How do HNO2 and HNO3 compare?
• Acid/Base properties of oxides
– Non-metal oxides produce acids when mixed
with water
– SO3 + H2O  H2SO4
– CO2 + H2O  H2CO3
– Metal oxides produce bases when mixed with
water
– CaO + H2O  Ca(OH)2
– Na2O + H2O  NaOH
• Lewis Acid/Bases
– Acids accept electron pairs
– Bases donate electron pairs
– BH3 + NH3  BH3NH3
– BH3 is the acid
– NH3 is the base