Titration of a strong acid by a strong base

The applet below draws the graph of pH against volume of titrant added for the titration of strong acid by a strong base. Specifically:

  • a solution of 0.1 mol/L NaOH (aq) is the titrant added from a burette

  • 25 cm3 of a solution of 0.1 mol/L HCl (aq) is the analyte in a flask below the burette

  • titrant is added beyond the equivalence point when 25 cm3 of titrant has been added.

  • turn on the data table feature to see the pH value of the equivalence point.


Press the run button to draw the pH curve for the strong acid/strong base titration.

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Note that for this titration:

  • the equivalence point is at exactly pH 7.0

The shape of the graph is explained as follows:

Before the equivalence point:

Added OH-(aq) present in the NaOH(aq) titratant is neutralised by the H3O+(aq) in the analyte solution to produce water:

OH-(aq) + H3O+(aq) --> 2H2O(l)

Na+ (from the NaOH solution) and Cl- (from the HCl solution) are spectator ions in the process. The overall result is that the added base is neutralised, acid is consumed and a solution of the salt sodium chloride, NaCl(aq), is produced.

The resulting pH is simply given by [H3O+], the concentration of hydronium ions remaining:

pH = -log10[H+]

Where [H+] is equivalent to [H3O+].

At the equivalence point:

All the acid has just been neutralised by the base to give a solution of sodium chloride in water only. The only source of hydronium ions is through the dissociation of water:

2H2O(l)   H3O+(aq) + OH-(aq)

given that:

[H3O+] x [OH-] = 1 x10-14 mol2/L2

and that:
[H3O+] = [OH-]

= 1 x 10-7 mol/L

pH = -log10[H+]

= 7

After the equivalence point:

The pH of the solution is determined by the equilibrium for the dissociation of water with the concentration of H3O+(aq) being progressively decreased as the increasing concentration of OH-(aq) shifts the equilibrium to the left of the equation below:

2H2O(l)   H3O+(aq) + OH-(aq)