pH Curves

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)

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

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⁺] = $\frac{0.0025}{0.0750}$ = 0.0333 mol dm^-3
  - so pH = 1.5
After 49.00 cm³of 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⁺] = $\frac{0.0001}{0.0990}$ = 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

pH Curves (SL IB Chemistry)