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        Buffers
       
 
       
        Buffers Titrations Chapter 18 Buffers A soln that resists change in pH when strong acid or strong base is added Made from weak acid/conjugate-base Or Made from weak base/conjugate-acid Strong acids and strong bases don’t make buffers! So how does it work? How buffers work Acetic acid/acetate buffer system: C2H3O2H(aq) + H2O(l) ? C2H3O2-(aq) + H3O+(l) Add base: C2H3O2H(aq) + OH-(aq) ? C2H3O2-(aq) + H2O(l) Add acid: H3O+(aq) + C2H3O2-(aq) ? H2O(l) +C2H3O2H(aq) Common Ion Effect: CIE The ionization of an acid or a base is limited by the presence of its conjugate base or acid. HAc(aq) + H2O(l) ? Ac-(aq) + H3O+(aq) Acetate ion is added in form of NaAc Which way will this shift the rxn? Would you expect a greater or lesser acidity if the CIE was lacking? Let’s look at the next problem Calculating pH of a buffer soln You have an acetic acid/acetate buffer with a 0.700 M conc of acetic acid and a 0.600 M conc of acetate ion. What’s the pH of the buffer? (Ka = 1.8 x 10-5) Solution Let’s work on this Consider 100.0 mL of a buffer solution that is 1.00M in HAc and 1.00M in NaAc. What is the pH after addition of 25.0 mL of 1.00M NaOH? Solution Henderson-Hasselbalch equation Useful for previous problem Let’s take a look Titrations Used to determine quantity of acid or base in unknown (analyte) When sample is neutralized (H3O+ = OH-) Equivalence point Determined by use of pH indicator Different titration types Strong acid-strong base: equivalence pt = 7.0 (contains neutral salt) Different titration types Weak acid-strong base: equivalence pt is greater than 7 (basic salt) pH @ half-equivalence pt (halfway pt) = pKa Problem We start with 50.0 mL of 0.100 M HAc. 25.0 mL of 0.100 M NaOH is then added. What is the pH of the resulting solution? Solution * * * * * 
       
 
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 有哪些信誉好的足球投注网站
有哪些信誉好的足球投注网站 
  
       
      
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