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pH Curves (HL IB Chemistry)

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pH Curves

Strong acid - strong base pH curve

  • During a titration, a pH meter can be used and a pH curve plotted
  • A pH curve is a graph showing how the pH of a solution changes as the acid (or base) is added in a strong acid - strong base titration, e.g.

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

Diagram to show the general characteristics of a strong acid-strong base pH curve

Diagram to show the general characteristics of a strong acid-strong base pH curve

The characteristics of a pH curve

  • All pH curves show an s-shape curve
  • pH curves yield useful information about how the acid and alkali react together with stoichiometric information
  • The midpoint of the inflection is called the equivalence or stoichiometric point
  • From the curves you can:
    • Determine the pH of the acid by looking where the curve starts on the y-axis
    • Find the pH at the equivalence point
    • Find volume of base at the equivalence point
    • Obtain the range of pH at the vertical section of the curve

How to calculate the pH depending on the volume of base added

  • If base is added to the conical flask then the pH of the solution will rise during the titration
  • Let's look at the reaction between 50 cm3 of 0.10 mol dm-3 HCl (aq) and 50 cm3 of 0.10 mol dm-3 of NaOH (aq)

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

  • At the start:
    • At the start of the titration, the conical flask will only contain a strong acid so the pH can be calculated by
      • pH = -log10[H+]
      • pH = -log10[0.10] = 1.0

  • After 25.00 cm3 of NaOH has been added
    • Now, we must consider what is in excess 
    • There is more acid in the flask than base in terms of volume, some of the acid has been neutralised, so we must calculate the excess moles of one of the reactants using n = c (mol dm-3) x v (dm3)
      • n(HCl) = 0.10 x 0.050 = 0.0050 mol
      • n(NaOH) = 0.10 x 0.025 = 0.0025 mol
      • n(Excess HCl) = 0.0050 - 0.0025 = 0.00250 mol
        • New volume = 0.0750 dm3
      • [H+] = fraction numerator 0.0025 over denominator 0.0750 end fraction= 0.0333 mol dm-3
      • so pH = 1.5

  • After 49.00 cm3 of NaOH has been added 
    • n(HCl) = 0.10 x 0.045 = 0.0050 mol
    • n(NaOH) = 0.10 x 0.049 = 0.0049 mol
    • n(Excess HCl) = 0.0050 - 0.0049 = 0.0001 mol
      • New volume = 0.0990 dm3
    • [H+] = fraction numerator 0.0001 over denominator 0.0990 end fraction= 0.000101 mol dm-3
    • so pH = 3.0

  • After 50.00 cm3 of NaOH has been added the acid has been completely neutralised by the base, so the solution only contains NaCl and H2O, therefore the pH = 7.0
  • After 51.00 cm3 of NaOH has been added 
    • n(Added NaOH) = 0.10 x 0.051 = 0.0051 mol
    • n(Excess NaOH) = 0.0051 - 0.0050 = 0.0001 mol
      • New volume = 0.101 dm3
    • [OH–] = fraction numerator 0.0001 over denominator 0.101 end fraction= 0.00099 mol dm-3
    • pOH = 3.0
    • so pH = 11.0
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Philippa

Author: Philippa

Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener.