pH Curves (SL IB Chemistry)

• 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⁺] = = 0.0333 mol dm^-3
    • so pH = 1.5
      
      
• After 49.00 cm³ of NaOH has been 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
  • 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^–] = = 0.00099 mol dm^-3
  • pOH = 3.0
  • so pH = 11.0