1 / 12

Acids and Bases intro

Acids and Bases intro. Acid/Base Definitions. Arrhenius Model Acids produce hydrogen ions in aqueous solutions Bases produce hydroxide ions in aqueous solutions Bronsted-Lowry Model Acids are proton donors Bases are proton acceptors. Acid Dissociation. HA  H + + A -

aginger
Download Presentation

Acids and Bases intro

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. Acids and Bases intro

  2. Acid/Base Definitions • Arrhenius Model • Acids produce hydrogen ions in aqueous solutions • Bases produce hydroxide ions in aqueous solutions • Bronsted-Lowry Model • Acids are proton donors • Bases are proton acceptors

  3. Acid Dissociation HA  H+ + A- AcidProtonConjugate base Alternately, H+ may be written in its hydrated form, H3O+ (hydronium ion)

  4. for example… HCl  H+ + Cl- AcidProtonConjugate base

  5. Dissociation of Strong Acids Strong acids are assumed to dissociate completely in solution. Large Ka or small Ka? Reactant favored or product favored?

  6. Dissociation Constants: Strong Acids

  7. Dissociation of Weak Acids Weak acids are assumed to dissociate only slightly (less than 5%) in solution. Large Ka or small Ka? Reactant favored or product favored?

  8. Dissociation Constants: Weak Acids

  9. Self-Ionization of Water H2O + H2O  H3O+ + OH- At 25, [H3O+] = [OH-] = 1 x 10-7 Kw is a constant at 25 C: Kw = [H3O+][OH-] Kw = (1 x 10-7)(1 x 10-7) = 1 x 10-14

  10. Calculating pH, pOH pH = -log10(H3O+) pOH = -log10(OH-) Relationship between pH and pOH pH + pOH = 14 Finding [H3O+], [OH-] from pH, pOH [H3O+] = 10-pH [OH-] = 10-pOH

  11. pH and pOH Calculations

  12. pH Scale

More Related