Strong Acid–Strong Base Titration

A strong monoprotic acid such as HCl is titrated with a strong “monohydroxic” base such as NaOH. Both dissociate completely, so until the equivalence point the pH is set by the excess H+, and after it by the excess OH−, 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 acid titrated with strong base

Zoom in on the equivalence point

Range, % of equivalence volume:

Near the equivalence point

Complete table of values
How this is calculated

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

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

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

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

Combining these gives a quadratic in [H+], with one positive root:

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

When Δ is much larger than √Kw (10−7 M at 25 °C) this is simply [H+] = Δ (excess acid) or [OH−] = −Δ (excess base). The exact form also holds at the equivalence point (Δ = 0, so [H+] = √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 HCl and Kw = 1.0 × 10−14 the starting pH is 6.79, not the 7.00 that [H+] = Ca would give, because water's own H+ adds to the acid'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