pH Curves (DP IB Chemistry): Revision Note

pH curves

Strong acid - strong base pH curve

• During a titration, a pH meter can be used to record changes in pH and create a pH curve

• A pH curve shows how the pH of a solution changes as acid or base is gradually added

  • For example, in a strong acid–strong base titration such as:

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

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

• All pH curves show an S-shaped curve

• They provide information on how an acid and base react, including stoichiometric data

• The midpoint of the vertical section is called the equivalence point

• From a pH curve, you can:

  • Identify the initial pH of the acid (where the curve starts on the y-axis)

  • Determine the pH at the equivalence point

  • Find the volume of base added at the equivalence point

  • Estimate the pH range over which the sharp change (vertical section) occurs

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

• If a base is added to the conical flask, the pH will increase during the titration.

• Example:

  • Titrating 50 cm³ of 0.10 mol dm^-3 HCl with 50 cm³ of 0.10 mol dm^-3 NaOH

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

At the start (0.00 cm³ NaOH)

• Only strong acid (HCl) is present

  • [H⁺] = 0.10 dm³

  • pH = -log₁₀[H⁺]

  • pH = 1.0

After 25.00 cm³ of NaOH added

• Some acid us neutralised, HCl in excess

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

• Total volume = 50.00 cm³ + 25.00 cm³ = 75.00 cm³ = 0.0750 dm³

• [H⁺] = = 0.0333 mol dm^-3

• pH = –log₁₀(0.0333) = 1.5

After 49.00 cm³ of NaOH added 

• n(HCl) = 0.10 x 0.050 = 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⁺] = = 0.00101 mol dm^-3

• pH = –log₁₀(0.00101 ) = 3.0

After 50.00 cm3 of NaOH added

• 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(HCl) = 0.10 x 0.050 = 0.0050 mol

• n(Added NaOH) = 0.10 x 0.051 = 0.0051 mol

• n(Excess NaOH) = 0.0051 - 0.0050 = 0.0001 mol

• Total volume = 50.00 cm³ + 51.00 cm³ = 101.00 cm³ = 0.101 dm³

• [OH^–] = = 0.00099 mol dm^-3

• pOH = 3.0

• so pH = 14.0 - 3.0 = 11.0