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Titrations

titration - known concentration of base is added to acid (or acid to base)  

  • equivalence point - where amount of acid/base are stoichiometrically equivalent
  • pH titration curve - graph of pH as a function of volume of added titrant
  • strong acid-strong base titration - characterized by sudden, very steep change in pH
    • initial pH based on initial concentration of acid
    • pH increases slowly when base first added
    • pH increases dramatically near equivalence point
    • only the salt exists at the equivalence point, pH = 7
    • pH after equivalence point determined by amount of base in solution
  • weak acid-strong base titration - less change at equivalence point than strong acid-strong base
    • higher initial pH than strong acid-strong base
    • equivalence point always > 7 due to strong base

Find the pH of solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M acetic acid.  

  • Given:
    • Ka = 1.8 x 10-5
  • mol of NaOH = 0.045 L (0.100 mol / 1 L) = 4.5 x 10-3 mol
  • mol of acetic acid = 0.050 L (0.100 mol / 1 L) = 5 x 10-3 mol
  • HC2H3O2 + NaOH >> NaC2H3O2 + H2O
  • 5 x 10-3 - 4.5 x 10-3 = 5 x 10-4 mol of acetic acid not consumed by NaOH after reaction
  • 4.5 x 10-3 mol of acetate after reaction
  • 45.0 mL + 50.0 mL = 0.095 L solution
  • [acetic acid] = 5 x 10-4 / 0.095 = 0.0053
  • [acetate] = 4.5 x 10-3 / 0.095 = 0.0474
  • Ka = [H+] [acetate] / [acetic acid]
  • [H+] = (1.8 x 10-5) ( 0.0053) / (0.0474) = 2.0 x 10-6
  • pH = -log [H+] = 5.70

Find the pH at the equivalence point in the titration of 50.0 mL of 0.100 M acetic acid w/ 0.100 M NaOH  

  • Given:
    • same molarity >> same volume needed to neutralize >> 50.0 mL of NaOH used at equivalence point
    • no acetic acid will be left over at the equivalence point
    • Ka = 1.8 x 10-5
  • mol of acetic acid at beginning = 0.0500 L (0.100 mol / 1 L) = 5 x 10-3 = mol of acetate at point
  • [acetate] = 5 x 10-3 mol / 0.100 L = 0.05 M
  • Kb = Kw / Ka = 10-14 / 1.8 x 10-5 = 5.6 x 10-10
  • Kb = [OH-] [acetic acid] / [acetate]
  • [OH-] = [(5.6 x 10-10) (0.05)]1/2 = 5.3 x 10-6
  • pOH = -log [OH-] = 5.28
  • pH = 14 - 5.28 = 8.72

When titrating a 0.1 M acetic acid solution with 0.005 M NaOH, what is the pH 1/2 way through the titration?  

  • Given:
    • 1/2 of acetic acid will have reacted w/ NaOH
    • Ka = 1.8 x 10-5
  • for comparison's sake, assume there's 100 mL of acetic acid
  • mol of acetic acid = 0.100 L (0.1 mol / 1 L) = 0.01 mol
  • mol of NaOH used = 0.005 (1/2 of acetic acid)
  • volume of NaOH used = 0.005 mol / 0.005 M = 1 L
  • HC2H3O2 + NaOH >> NaC2H3O2 + H2O
  • 0.005 mol acetic acid left after reaction
  • 0.005 mol acetate left after reaction
  • [acetic acid] = 0.005 mol / 1.1 L = 0.0045
  • [acetate] = 0.005 mol / 1.1 L = 0.0045
  • Ka = [H+] [acetate] / [acetic acid]
  • [H+] = 1.8 x 10-5
  • pH = -log [H+] = 4.74
Subject: 
Chemistry
Subject X2: 
Chemistry

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