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Module 5 Chapter 20 (20.1 Bronsted-Lowry acids and bases (Ionic equations,…

- - - - 1 H+ ion can be replaced
      - e.g. HCl or CH3COOH
    - - 2 H+ ions can be replaced
      - e.g, H2SO4
    - - 3 H+ ions can be replaced
      - e.g. H3PO4
- - - - pH of solutions of strong bases
        
        pH can be calculated from conc. of base and Kw
        
        Have to be careful about number of moles of OH-, use ratio to work out moles of solution
      - Weak bases
        
        1) Calculate pOH from [OH- ]- pOH= -log[OH-]
        
        2) Use pH + pOH =14
- - - - At equilibrium concentrations of products are the same
      - [H+ (aq)] = [A- (aq)]
      - Limit- very small amount of H+ produced from water- ignored as its negligible compared to [HA (aq)], after pH 6 this becomes significant however
    - - So little HA dissociates you can assume [HA (aq)]start =[HA(aq)]end
      - Limit- small amounts of HA dissociate no matter how weak the acid is so not equal. Weak acids with small Ka fit second part but if Ka > x10-2 [H+] becomes more significant
      - Conc. of H+ very small compared to conc. HA