Titration Curves & Equivalence Points (College Board AP® Chemistry)
Study Guide
Written by: Philippa Platt
Reviewed by: Stewart Hird
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
Examiner Tips and Tricks
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
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