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AP Chemistry Unit 8.4 Acid-Base Reactions and Buffers

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Acid-Base Reactions and Buffers

  • Equivalence point: enough titrant has been added to react exactly with the solution being analyzed (analyte)
    • In acid-base reactions: moles acid = moles base
  • In acid-base titrations, are dealing with neutralization reactions (NmN → water is not reactant)
    • N: neutralization 
    • M: moles
    • N: numbers
      • In ice table, subtract number of mmols of limiting reactant
  • Millimoles:
    • Ex: 40 mL x 0.2 M = 8 mmol
  • Halfway point/halfway equivalence point: is halfway to eq point (half of analyte is neutralized)
    • At halfway point pH = pKa;
  • Before eq point → have left over H+; after eq point → have left over OH-

Strong Acid-Strong Base Titrations

  • They both dissociate completely
    • Strong acids: molarity of acid = M of H+
    • Strong Base: molarity of base = molarity of OH-
  • The net ionic reaction for a strong acid–strong base titration is:
  • Do the stoichiometry and there is no equilibrium
  • pH is equal to 7 at the eq point (only strong + strong)

Weak Acid-Strong Base Titration

  • The weaker the acid being titrated, the smaller the vertical area around the equivalence point
  • At halfway point, the molarity of weak acid and conjugate base is equal
  • At 0 mL base added: use WMX ICE table, use Ka to determine pH
  • pH at equivalence is above 7
  • Weak acid before eq point:
    • NmN ICE table first
    • Then Henderson-Hasselbalch
  • At eq point: Have to do double ICE tables only at the eq point!
    • NmN ICE table first
    • Then, WMS ICE table, use Kb to find [OH-]
  • pH after equivalence:
    • NmN ICE table first → then do mmol/mL to find moles of [OH-] → find pOH- → find pH

Weak Base-Strong Acid Titration

  • Process is practically the same with some differences
  • pH at equivalence is below 7
  • At 0 mL acid added:
    • WMX ICE table, use Kb to determine pOH, then pH
  • Weak base before equivalence point:
    • NmN ICE table first → then henderson-Hasselbalch
  • At equivalence point:
    • NmN ICE table first → then WMX ICE table → use Ka to find [H+]
  • pH after equivalence:
    • NmN ICE table first → leftover strong acid → do mmol/mL to find moles of [H+] → find pH

Titration Curves

Strong Acid Strong Base Titration

  • Some base has started to be added but before the equivalence point:
    • Have left over strong acid (H+) which is why the pH is below 7
  • Everywhere beyond equivalence point: have excess OH- which is why the pH is above 7
  • At eq point, for Strong + strong the pH at eq point is exact 7

Weak Acid-Strong Base Titration

  • After base has started to be added but before eq point → what was left was buffer solution → buffer zone 25mL is halfway point and is where the buffer that is created is the most effective
  • At eq point only the conjugate base is present which is why the eq point is above 8
    • This reaction takes place

Weak Base-Strong Acid Titration

  • In buffer zone 25mL is halfway point pH = pKa
  • Will have left over conjugate acid at eq point → pH will be below 7
    • Justify: because this reaction takes place
  • Beyond eq point:

Acid-Base Titration Problems

  • How to know whether to use Ka or Kb? Look at equation → is species accepting or donating H+

Sketch a Titration Curve

  • Need to plot 3 data point:
  1. Starting PH
  2. pH at equivalence
  3. pH at the halfway point
    • If given pkB, can solve for pKa to find pH

Solve for what volume of titrant (acid or base) needed to reach equivalence point

  • Write out major species → write net ionic equation → use stoichiometry
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