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

• [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. [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

• Lewis Acid/Bases– Acids accept electron pairs– Bases donate electron pairs

– BH3 + NH3 BH3NH3

– BH3 is the acid– NH3 is the base