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
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 cm3 were 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 =
M = = 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