1 / 28

Chapter 14

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 - H 3 O +. Conjugate pairs HCN/CN - HCl/Cl - NH 3 /NH 4 + Acid dissociation constant

enya
Download Presentation

Chapter 14

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 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+

  2. 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

  3. Oxyacids • HNO3, HClO4, H2SO4 • Organic Acids • COOH group HC2H3O2 (CH3COOH) • Monoprotic, diprotic, triprotic • HCl, H2SO4, H3PO4

  4. 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

  5. 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

  6. 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

  7. 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

  8. 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

  9. 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

  10. 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

  11. 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

  12. 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

  13. 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

  14. Strong bases make weak adics • Strong acids make weak bases • So the larger the Ka the smaller the Kb • Ka x Kb = Kw

  15. 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

  16. 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-]

  17. 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

  18. 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

  19. [OH-] • Kw = 1.0 x 10-14 = [H+][OH-] • 1.0 x 10-14 = (0.19)[OH-] • [OH-] = 5.3 x 10-14

  20. 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

  21. 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

  22. 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+

  23. 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

  24. 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

  25. 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

  26. 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?

  27. 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

  28. Lewis Acid/Bases • Acids accept electron pairs • Bases donate electron pairs • BH3 + NH3 BH3NH3 • BH3 is the acid • NH3 is the base

More Related