Titration Curves & Equivalence Points

During a titration, a pH meter can be used and a pH curve plotted
A pH curve is a graph showing how the pH of a solution changes as the acid (or base) is added

Graph to show the pH curve for the addition of a alkali to an acid

The features of a pH curve

All pH curves show an s-shape curve
There are four main points in a titration curve:
- The start where the solution only contains either acid or base
- The region where the titrant is added up to the equivalence point, and the solution now contains a mixture of reactant and products
- The equivalence point occurs when the number of moles of titrant (the solution that has been added) added is equal to the number of moles of analyte originally present
- The region after the equivalence point where the solution contains product and excess titrant
pH curves yield useful information about how the acid and alkali react together with stoichiometric information
The midpoint of the inflection is called the equivalence point
From the curves, you can:
- Determine the pH of the acid by looking at where the curve starts on the y-axis
- Find the pH at the equivalence point
- Find the volume of base at the equivalence point
- Obtain the range of pH at the vertical section of the curve

pH curve of a strong acid - strong base

For example odium hydroxide, NaOH (aq), is being added to hydrochloric acid, HCl (aq)

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

Strong acid - strong base pH curve

Strong acid strong base (1), downloadable IB Chemistry revision notes

The pH increases as the alkali is added to the base

Worked example

A 0.675 g sample of a solid acid, HA, was dissolved in distilled water and made up to 100.0 cm³ in a volumetric flask. 25.0 cm³ of this solution was titrated against 0.100 mol dm^-3 NaOH solution and 12.1 cm³were required for complete reaction. Determine the molar mass of the acid.

Answer:

Step 1: Write the equation for the reaction:

HA (aq) + NaOH (aq) → NaA (aq) + H₂O (l)

Step 2: Calculate the number of moles of the NaOH

n(NaOH)_sample = 0.0121 mL x 0.100 M = 1.21 x 10^-3 mol

Step 3: Deduce the number of moles of the acid

Since the acid is monoprotic the number of moles of HA is also 1.21 x 10^-3 mol
This is present in 25.0 cm³ of the solution

Step 4: Scale up to find the amount in the original 100 mL solution

n(HA) = 4.84 x 10^-3 mol

Step 5: Calculate the molar mass

moles = $\frac{mass}{M}$

M = $\frac{0.675}{4.84 \times 10^{- 3} mol}$ = 139 g mol^-1

Exam Tip

When performing titration calculations using monoprotic acids (meaning one H⁺) such as HCl, the number of moles of the acid and alkali will be the same
This allows you to use the relationship:

C₁V₁ =C₂V₂

- C₁ and V₁ are the concentration and volume of the acid
- C₂ and V₂ are the concentration and volume of the alkali

There is no need to convert the units of volume to dm³ as this is a ratio
- Simply re-arrange the formula to solve for the unknown quantity

Titration Curves & Equivalence Points (College Board AP Chemistry)

Revision Note