Strong Base–Strong Acid Titration

A strong “monohydroxic” base such as NaOH is titrated with a strong monoprotic acid such as HCl. Both dissociate completely, so until the equivalence point the pH is set by the excess OH−, and after it by the excess H+, each diluted by the growing total volume. At the equivalence point, the solution is just NaCl (or other salt) in water, at pH 7 (for Kw = 10−14, 25 °C; the calculation considers water self-dissociation, and you can change Kw in the dropdown at the bottom). The right-hand plot zooms in on the equivalence point, where a single drop of titrant in a typical experiment changes the pH by several units. Move the pointer over either plot to read off the pH and titrant volume.

Example
mL
M
M
× eq.

Titration curve

pH of titrant

Strong base titrated with strong acid

Zoom in on the equivalence point

Range, % of equivalence volume:

Near the equivalence point

Complete table of values
How this is calculated

The base contributes Na+ and the acid contributes Cl−; neither reacts further. After Va mL of acid has been added to Vb mL of base, their concentrations, and the difference between them, are:

[Na+] = Cb Vb/(Vb+Va); [Cl-] = Ca Va/(Vb+Va); Delta = [Na+] - [Cl-]

The solution must be electrically neutral, and water's self-dissociation links [OH−] and [H+]:

[H+] + [Na+] = [Cl-] + [OH-]; [H+] = Kw/[OH-]

Combining these gives a quadratic in [OH−], with one positive root:

[OH-] - Kw/[OH-] = Delta, so [OH-] = Delta/2 + sqrt(Delta^2/4 + Kw)

When Δ is much larger than √Kw (10−7 M at 25 °C) this is simply [OH−] = Δ (excess base) or [H+] = −Δ (excess acid). The exact form also holds at the equivalence point (Δ = 0, so [OH−] = √Kw) and at very low concentrations, where water self-dissociation to give H+ and OH− matters. Try the "Very dilute" example: with 10−7 M NaOH and Kw = 1.0 × 10−14 the starting pH is 7.21, not the 7.00 that [OH−] = Cb would give, because water's own OH− adds to the base's.

1.0 × 10−14 at 25 °C; about 2.4 × 10−14 at 37 °C and 5.5 × 10−14 at 50 °C. The pH at equivalence is ½ pKw.

Activities are taken equal to concentrations. A drop is taken to be 0.05 mL.

Jason D. Kahn, Dept. of Chemistry and Biochemistry, University of Maryland College Park, using Claude