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. Initial Equil.– [H+] 0 +x– [F-] 0 +x– [HF] 1.0 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. [init.] [equil]• HNO2 5.00 5.00-x = 5.00 (x is
small)
• H+ 0 +x
• NO2- 0 +x– 4.0 x 10-4 = x2 / 5.00– x = 4.5 x 10-2
– pH = -log(4.5 x 10-2) = 1.35
• 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. [Init.] [Equil]– NH4+ 0 +x– OH- 0 +x– NH3 15.0 15.0 – x = 15.0
– Chem. [Init.] [Equil]– NH4+ 0 +x– OH- 0 +x– NH3 15.0 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. [init] [equil]– H+ 0 +x– H2PO4- 0 +x– H3PO4 5.0 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 [init] [equil]– H+ 0.19 0.19+x = 0.19– HPO42- 0 +x– H2PO4- 0.19 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 [init.] [equil.]– H+ 0.19 0.19 + x = 0.19– PO43- 0 +x– HPO42- 6.2 x 10-8 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
• 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. [init.] [equil.]– H+ 1.0 1.0 + x = 1.0– SO42- 0 + x– HSO4- 1.0 1.0 – x = 1.0
– 1.2 x 10-2 = (1.0)x / 1.0 x = 1.2 x 10-2
– [SO42-] = 1.2 x 10-2 M
• 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 0 + x– H3O+ 0 + 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